Search: a000027 -id:a000027
|
| |
|
|
A000007
|
|
The characteristic function of {0}: a(n) = 0^n.
(Formerly M0002)
|
|
+0
1026
|
|
|
|
1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
(list;
constant;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,1
|
|
|
COMMENTS
|
Changing the offset to 1 gives the arithmetical function a(1) = 1, a(n) = 0 for n > 1, the identity function for Dirichlet multiplication (see Apostol). - N. J. A. Sloane
Changing the offset to 1 makes this the decimal expansion of 1. - N. J. A. Sloane, Nov 13 2014
Hankel transform (see A001906 for definition) of A000007 (powers of 0), A000012 (powers of 1), A000079 (powers of 2), A000244 (powers of 3), A000302 (powers of 4), A000351 (powers of 5), A000400 (powers of 6), A000420 (powers of 7), A001018 (powers of 8), A001019 (powers of 9), A011557 (powers of 10), A001020 (powers of 11), etc. - Philippe Deléham, Jul 07 2005
This is the identity sequence with respect to convolution. - David W. Wilson, Oct 30 2006
The alternating sum of the n-th row of Pascal's triangle gives the characteristic function of 0, a(n) = 0^n. - Daniel Forgues, May 25 2010
The number of maximal self-avoiding walks from the NW to SW corners of a 1 X n grid. - Sean A. Irvine, Nov 19 2010
Historically there has been some disagreement as to whether 0^0 = 1. Graphing x^0 seems to support that conclusion, but graphing 0^x instead suggests that 0^0 = 0. Euler and Knuth have argued in favor of 0^0 = 1. For some calculators, 0^0 triggers an error, while in Mathematica, 0^0 is Indeterminate. - Alonso del Arte, Nov 15 2011
Another consequence of changing the offset to 1 is that then this sequence can be described as the sum of Moebius mu(d) for the divisors d of n. - Alonso del Arte, Nov 28 2011
With the convention 0^0 = 1, 0^n = 0 for n > 0, the sequence a(n) = 0^|n-k|, which equals 1 when n = k and is 0 for n >= 0, has g.f. x^k. A000007 is the case k = 0. - George F. Johnson, Mar 08 2013
A fixed point of the run length transform. - Chai Wah Wu, Oct 21 2016
|
|
|
REFERENCES
|
T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 30.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 55.
|
|
|
LINKS
|
|
|
|
FORMULA
|
As a function of Bernoulli numbers (cf. A027641: (1, -1/2, 1/6, 0, -1/30, ...)), triangle A074909 (the beheaded Pascal's triangle) * B_n as a vector = [1, 0, 0, 0, 0, ...]. - Gary W. Adamson, Mar 05 2012
a(n) = Sum_{k = 0..n} exp(2*Pi*i*k/(n+1)) is the sum of the (n+1)th roots of unity. - Franz Vrabec, Nov 09 2012
Inverse binomial transform of A000012. (End)
|
|
|
MAPLE
|
A000007 := proc(n) if n = 0 then 1 else 0 fi end: seq(A000007(n), n=0..20);
spec := [A, {A=Z} ]: seq(combstruct[count](spec, size=n+1), n=0..20);
|
|
|
MATHEMATICA
|
Table[If[n == 0, 1, 0], {n, 0, 99}]
Table[Boole[n == 0], {n, 0, 99}] (* Michael Somos, Aug 25 2012 *)
Join[{1}, LinearRecurrence[{1}, {0}, 102]] (* Ray Chandler, Jul 30 2015 *)
|
|
|
PROG
|
(PARI) {a(n) = !n};
(Magma) [1] cat [0:n in [1..100]]; // Sergei Haller, Dec 21 2006
(Haskell)
a000007 = (0 ^)
a000007_list = 1 : repeat 0
(Python)
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
| |
|
|
A000012
|
|
The simplest sequence of positive numbers: the all 1's sequence.
(Formerly M0003)
|
|
+0
2496
|
|
|
|
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
(list;
table;
constant;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,1
|
|
|
COMMENTS
|
Number of ways of writing n as a product of primes.
Number of ways of writing n as a sum of distinct powers of 2.
Continued fraction for golden ratio A001622.
An example of an infinite sequence of positive integers whose distinct pairwise concatenations are all primes! - Don Reble, Apr 17 2005
For n >= 0, let M(n) be the matrix with first row = (n n+1) and 2nd row = (n+1 n+2). Then a(n) = absolute value of det(M(n)). - K.V.Iyer, Apr 11 2009
a(n) is also tau_1(n) where tau_2(n) is A000005.
a(n) is a completely multiplicative arithmetical function.
a(n) is both squarefree and a perfect square. See A005117 and A000290. (End)
a(n) is also the number of complete graphs on n nodes. - Pablo Chavez (pchavez(AT)cmu.edu), Sep 15 2009
Totally multiplicative sequence with a(p) = 1 for prime p. Totally multiplicative sequence with a(p) = a(p-1) for prime p. - Jaroslav Krizek, Oct 18 2009
n-th prime minus phi(prime(n)); number of divisors of n-th prime minus number of perfect partitions of n-th prime; the number of perfect partitions of n-th prime number; the number of perfect partitions of n-th noncomposite number. - Juri-Stepan Gerasimov, Oct 26 2009
For all n>0, the sequence of limit values for a(n) = n!*Sum_{k>=n} k/(k+1)!. Also, a(n) = n^0. - Harlan J. Brothers, Nov 01 2009
a(n) is also the number of 0-regular graphs on n vertices. - Jason Kimberley, Nov 07 2009
1) When sequence is read as a regular triangular array, T(n,k) is the coefficient of the k-th power in the expansion of (x^(n+1)-1)/(x-1).
2) Sequence can also be read as a uninomial array with rows of length 1, analogous to arrays of binomial, trinomial, etc., coefficients. In a q-nomial array, T(n,k) is the coefficient of the k-th power in the expansion of ((x^q -1)/(x-1))^n, and row n has a sum of q^n and a length of (q-1)*n + 1. (End)
The number of maximal self-avoiding walks from the NW to SW corners of a 2 X n grid.
a(n) = A007310(n+1) (Modd 3) := A193680(A007310(n+1)), n>=0. For general Modd n (not to be confused with mod n) see a comment on A203571. The nonnegative members of the three residue classes Modd 3, called [0], [1], and [2], are shown in the array A088520, if there the third row is taken as class [0] after inclusion of 0. - Wolfdieter Lang, Feb 09 2012
Let M = Pascal's triangle without 1's (A014410) and V = a variant of the Bernoulli numbers A027641 but starting [1/2, 1/6, 0, -1/30, ...]. Then M*V = [1, 1, 1, 1, ...]. - Gary W. Adamson, Mar 05 2012
As a lower triangular array, T is an example of the fundamental generalized factorial matrices of A133314. Multiplying each n-th diagonal by t^n gives M(t) = I/(I-t*S) = I + t*S + (t*S)^2 + ... where S is the shift operator A129184, and T = M(1). The inverse of M(t) is obtained by multiplying the first subdiagonal of T by -t and the other subdiagonals by zero, so A167374 is the inverse of T. Multiplying by t^n/n! gives exp(t*S) with inverse exp(-t*S). - Tom Copeland, Nov 10 2012
The original definition of the meter was one ten-millionth of the distance from the Earth's equator to the North Pole. According to that historical definition, the length of one degree of latitude, that is, 60 nautical miles, would be exactly 111111.111... meters. - Jean-François Alcover, Jun 02 2013
Consider n >= 1 nonintersecting spheres each with surface area S. Define point p on sphere S_i to be a "public point" if and only if there exists a point q on sphere S_j, j != i, such that line segment pq INTERSECT S_i = {p} and pq INTERSECT S_j = {q}; otherwise, p is a "private point". The total surface area composed of exactly all private points on all n spheres is a(n)*S = S. ("The Private Planets Problem" in Zeitz.) - Rick L. Shepherd, May 29 2014
A fixed point of the run length transform. - Chai Wah Wu, Oct 21 2016
a(n) is also the determinant of the (n+1) X (n+1) matrix M defined by M(i,j) = binomial(i,j) for 0 <= i,j <= n, since M is a lower triangular matrix with main diagonal all 1's. - Jianing Song, Jul 17 2018
a(n) is also the determinant of the symmetric n X n matrix M defined by M(i,j) = min(i,j) for 1 <= i,j <= n (see Xavier Merlin reference). - Bernard Schott, Dec 05 2018
a(n) is also the determinant of the symmetric n X n matrix M defined by M(i,j) = tau(gcd(i,j)) for 1 <= i,j <= n (see De Koninck & Mercier reference). - Bernard Schott, Dec 08 2020
|
|
|
REFERENCES
|
J.-M. De Koninck & A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Problème 692 pp. 90 and 297, Ellipses, Paris, 2004.
Xavier Merlin, Méthodix Algèbre, Exercice 1-a), page 153, Ellipses, Paris, 1995.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 55.
Paul Zeitz, The Art and Craft of Mathematical Problem Solving, The Great Courses, The Teaching Company, 2010 (DVDs and Course Guidebook, Lecture 6: "Pictures, Recasting, and Points of View", pp. 32-34).
|
|
|
LINKS
|
|
|
|
FORMULA
|
a(n) = 1.
G.f.: 1/(1-x).
E.g.f.: exp(x).
G.f.: Product_{k>=0} (1 + x^(2^k)). - Zak Seidov, Apr 06 2007
Completely multiplicative with a(p^e) = 1.
Regarded as a square array by antidiagonals, g.f. 1/((1-x)(1-y)), e.g.f. Sum T(n,m) x^n/n! y^m/m! = e^{x+y}, e.g.f. Sum T(n,m) x^n y^m/m! = e^y/(1-x). Regarded as a triangular array, g.f. 1/((1-x)(1-xy)), e.g.f. Sum T(n,m) x^n y^m/m! = e^{xy}/(1-x). - Franklin T. Adams-Watters, Feb 06 2006
a(n) = Sum_{l=1..n} (-1)^(l+1)*2*cos(Pi*l/(2*n+1)) = 1 identically in n >= 1 (for n=0 one has 0 from the undefined sum). From the Jolley reference, (429) p. 80. Interpretation: consider the n segments between x=0 and the n positive zeros of the Chebyshev polynomials S(2*n, x) (see A049310). Then the sum of the lengths of every other segment starting with the one ending in the largest zero (going from the right to the left) is 1. - Wolfdieter Lang, Sep 01 2016
|
|
|
EXAMPLE
|
1 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + ...)))) = A001622.
1/9 = 0.11111111111111...
Modd 7 for nonnegative odd numbers not divisible by 3:
A007310: 1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, ...
Modd 3: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
(End)
|
|
|
MAPLE
|
seq(1, i=0..150);
|
|
|
MATHEMATICA
|
Array[1 &, 50] (* Joseph Biberstine (jrbibers(AT)indiana.edu), Dec 26 2006 *)
|
|
|
PROG
|
(Magma) [1 : n in [0..100]];
(PARI) {a(n) = 1};
(Haskell)
a000012 = const 1
(Maxima) makelist(1, n, 1, 30); /* Martin Ettl, Nov 07 2012 */
|
|
|
CROSSREFS
|
Cf. A000004, A007395, A010701, A000027, A027641, A014410, A211216, A212393, A060544, A051801, A104684.
|
|
|
KEYWORD
|
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
| |
|
|
A000041
|
|
a(n) is the number of partitions of n (the partition numbers).
(Formerly M0663 N0244)
|
|
+0
3685
|
|
|
|
1, 1, 2, 3, 5, 7, 11, 15, 22, 30, 42, 56, 77, 101, 135, 176, 231, 297, 385, 490, 627, 792, 1002, 1255, 1575, 1958, 2436, 3010, 3718, 4565, 5604, 6842, 8349, 10143, 12310, 14883, 17977, 21637, 26015, 31185, 37338, 44583, 53174, 63261, 75175, 89134, 105558, 124754, 147273, 173525
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,3
|
|
|
COMMENTS
|
Also number of nonnegative solutions to b + 2c + 3d + 4e + ... = n and the number of nonnegative solutions to 2c + 3d + 4e + ... <= n. - Henry Bottomley, Apr 17 2001
a(n) is also the number of conjugacy classes in the symmetric group S_n (and the number of irreducible representations of S_n).
Also the number of rooted trees with n+1 nodes and height at most 2.
Coincides with the sequence of numbers of nilpotent conjugacy classes in the Lie algebras gl(n). A006950, A015128 and this sequence together cover the nilpotent conjugacy classes in the classical A,B,C,D series of Lie algebras. - Alexander Elashvili, Sep 08 2003
Number of distinct Abelian groups of order p^n, where p is prime (the number is independent of p). - Lekraj Beedassy, Oct 16 2004
Number of graphs on n vertices that do not contain P3 as an induced subgraph. - Washington Bomfim, May 10 2005
Numbers of terms to be added when expanding the n-th derivative of 1/f(x). - Thomas Baruchel, Nov 07 2005
Sequence agrees with expansion of Molien series for symmetric group S_n up to the term in x^n. - Maurice D. Craig (towenaar(AT)optusnet.com.au), Oct 30 2006
Also the number of nonnegative integer solutions to x_1 + x_2 + x_3 + ... + x_n = n such that n >= x_1 >= x_2 >= x_3 >= ... >= x_n >= 0, because by letting y_k = x_k - x_(k+1) >= 0 (where 0 < k < n) we get y_1 + 2y_2 + 3y_3 + ... + (n-1)y_(n-1) + nx_n = n. - Werner Grundlingh (wgrundlingh(AT)gmail.com), Mar 14 2007
Let P(z) := Sum_{j>=0} b_j z^j, b_0 != 0. Then 1/P(z) = Sum_{j>=0} c_j z^j, where the c_j must be computed from the infinite triangular system b_0 c_0 = 1, b_0 c_1 + b_1 c_0 = 0 and so on (Cauchy products of the coefficients set to zero). The n-th partition number arises as the number of terms in the numerator of the expression for c_n: The coefficient c_n of the inverted power series is a fraction with b_0^(n+1) in the denominator and in its numerator having a(n) products of n coefficients b_i each. The partitions may be read off from the indices of the b_i. - Peter C. Heinig (algorithms(AT)gmx.de), Apr 09 2007
a(n) is the number of different ways to run up a staircase with n steps, taking steps of sizes 1, 2, 3, ... and r (r <= n), where the order is not important and there is no restriction on the number or the size of each step taken. - Mohammad K. Azarian, May 21 2008
A sequence of positive integers p = p_1 ... p_k is a descending partition of the positive integer n if p_1 + ... + p_k = n and p_1 >= ... >= p_k. If formally needed p_j = 0 is appended to p for j > k. Let P_n denote the set of these partition for some n >= 1. Then a(n) = 1 + Sum_{p in P_n} floor((p_1-1)/(p_2+1)). (Cf. A000065, where the formula reduces to the sum.) Proof in Kelleher and O'Sullivan (2009). For example a(6) = 1 + 0 + 0 + 0 + 0 + 1 + 0 + 0 + 1 + 1 + 2 + 5 = 11. - Peter Luschny, Oct 24 2010
Let n = Sum( k_(p_m) p_m ) = k_1 + 2k_2 + 5k_5 + 7k_7 + ..., where p_m is the m-th generalized pentagonal number (A001318). Then a(n) is the sum over all such pentagonal partitions of n of (-1)^(k_5+k_7 + k_22 + ...) ( k_1 + k_2 + k_5 + ...)! /( k_1! k_2! k_5! ...), where the exponent of (-1) is the sum of all the k's corresponding to even-indexed GPN's. - Jerome Malenfant, Feb 14 2011
The matrix of a(n) values
a(0)
a(1) a(0)
a(2) a(1) a(0)
a(3) a(2) a(1) a(0)
....
a(n) a(n-1) a(n-2) ... a(0)
is the inverse of the matrix
1
-1 1
-1 -1 1
0 -1 -1 1
....
-d_n -d_(n-1) -d_(n-2) ... -d_1 1
where d_q = (-1)^(m+1) if q = m(3m-1)/2 = the m-th generalized pentagonal number (A001318), = 0 otherwise. (End)
Let k > 0 be an integer, and let i_1, i_2, ..., i_k be distinct integers such that 1 <= i_1 < i_2 < ... < i_k. Then, equivalently, a(n) equals the number of partitions of N = n + i_1 + i_2 + ... + i_k in which each i_j (1 <= j <= k) appears as a part at least once. To see this, note that the partitions of N of this class must be in 1-to-1 correspondence with the partitions of n, since N - i_1 - i_2 - ... - i_k = n. - L. Edson Jeffery, Apr 16 2011
a(n) is the number of distinct degree sequences over all free trees having n + 2 nodes. Take a partition of the integer n, add 1 to each part and append as many 1's as needed so that the total is 2n + 2. Now we have a degree sequence of a tree with n + 2 nodes. Example: The partition 3 + 2 + 1 = 6 corresponds to the degree sequence {4, 3, 2, 1, 1, 1, 1, 1} of a tree with 8 vertices. - Geoffrey Critzer, Apr 16 2011
a(n) is number of distinct characteristic polynomials among n! of permutations matrices size n X n. - Artur Jasinski, Oct 24 2011
Conjecture: starting with offset 1 represents the numbers of ordered compositions of n using the signed (++--++...) terms of A001318 starting (1, 2, -5, -7, 12, 15, ...). - Gary W. Adamson, Apr 04 2013 (this is true by the pentagonal number theorem, Joerg Arndt, Apr 08 2013)
a(n) is also number of terms in expansion of the n-th derivative of log(f(x)). In Mathematica notation: Table[Length[Together[f[x]^n * D[Log[f[x]], {x, n}]]], {n, 1, 20}]. - Vaclav Kotesovec, Jun 21 2013
Conjecture: No a(n) has the form x^m with m > 1 and x > 1. - Zhi-Wei Sun, Dec 02 2013
Partitions of n that contain a part p are the partitions of n - p. Thus, number of partitions of m*n - r that include k*n as a part is A000041(h*n-r), where h = m - k >= 0, n >= 2, 0 <= r < n; see A111295 as an example. - Clark Kimberling, Mar 03 2014
a(n) is the number of compositions of n into positive parts avoiding the pattern [1, 2]. - Bob Selcoe, Jul 08 2014
Conjecture: For any j there exists k such that all primes p <= A000040(j) are factors of one or more a(n) <= a(k). Growth of this coverage is slow and irregular. k = 1067 covers the first 102 primes, thus slower than A000027. - Richard R. Forberg, Dec 08 2014
a(n) is the number of nilpotent conjugacy classes in the order-preserving, order-decreasing and (order-preserving and order-decreasing) injective transformation semigroups. - Ugbene Ifeanyichukwu, Jun 03 2015
Define a segmented partition a(n,k, <s(1)..s(j)>) to be a partition of n with exactly k parts, with s(j) parts t(j) identical to each other and distinct from all the other parts. Note that n >= k, j <= k, 0 <= s(j) <= k, s(1)t(1) + ... + s(j)t(j) = n and s(1) + ... + s(j) = k. Then there are up to a(k) segmented partitions of n with exactly k parts. - Gregory L. Simay, Nov 08 2015
(End)
The polynomials for a(n, k, <s(1), ..., s(j)>) have degree j-1.
a(n, k, <k>) = 1 if n = 0 mod k, = 0 otherwise
a(rn, rk, <r*s(1), ..., r*s(j)>) = a(n, k, <s(1), ..., s(j)>)
a(n odd, k, <all s(j) even>) = 0
Established results can be recast in terms of segmented partitions:
For j(j+1)/2 <= n < (j+1)(j+2)/2, A000009(n) = a(n, 1, <1>) + ... + a(n, j, <j 1's>), j < n
a(n, k, <j 1's>) = a(n - j(j-1)/2, k)
(End)
a(10^20) was computed using the NIST Arb package. It has 11140086260 digits and its head and tail sections are 18381765...88091448. See the Johansson 2015 link. - Stanislav Sykora, Feb 01 2016
Satisfies Benford's law [Anderson-Rolen-Stoehr, 2011]. - N. J. A. Sloane, Feb 08 2017
The partition function p(n) is log-concave for all n>25 [DeSalvo-Pak, 2014]. - Michel Marcus, Apr 30 2019
a(n) is also the dimension of the n-th cohomology of the infinite real Grassmannian with coefficients in Z/2. - Luuk Stehouwer, Jun 06 2021
Equivalently, number of idempotent mappings f from a set X of n elements into itself (i.e., satisfying f o f = f) up to permutation (i.e., f~f' :<=> There is a permutation sigma in Sym(X) such that f' o sigma = sigma o f). - Philip Turecek, Apr 17 2023
Conjecture: Each integer n > 2 different from 6 can be written as a sum of finitely many numbers of the form a(k) + 2 (k > 0) with no summand dividing another. This has been verified for n <= 7140. - Zhi-Wei Sun, May 16 2023
a(n) is also the number of partitions of n*(n+3)/2 into n distinct parts. - David García Herrero, Aug 20 2024
|
|
|
REFERENCES
|
George E. Andrews, The Theory of Partitions, Addison-Wesley, Reading, Mass., 1976.
George E. Andrews and K. Ericksson, Integer Partitions, Cambridge University Press 2004.
T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 307.
R. Ayoub, An Introduction to the Analytic Theory of Numbers, Amer. Math. Soc., 1963; Chapter III.
Mohammad K. Azarian, A Generalization of the Climbing Stairs Problem, Mathematics and Computer Education Journal, Vol. 31, No. 1, pp. 24-28, Winter 1997.
Mohammad K. Azarian, A Generalization of the Climbing Stairs Problem II, Missouri Journal of Mathematical Sciences, Vol. 16, No. 1, Winter 2004, pp. 12-17. Zentralblatt MATH, Zbl 1071.05501.
Bruce C. Berndt, Ramanujan's Notebooks Part V, Springer-Verlag.
B. C. Berndt, Number Theory in the Spirit of Ramanujan, Chap. I Amer. Math. Soc. Providence RI 2006.
Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 999.
J. M. Borwein, D. H. Bailey and R. Girgensohn, Experimentation in Mathematics, A K Peters, Ltd., Natick, MA, 2004. x+357 pp. See p. 183.
L. E. Dickson, History of the Theory of Numbers, Vol.II Chapter III pp. 101-164, Chelsea NY 1992.
N. J. Fine, Basic Hypergeometric Series and Applications, Amer. Math. Soc., 1988; p. 37, Eq. (22.13).
H. Gupta et al., Tables of Partitions. Royal Society Mathematical Tables, Vol. 4, Cambridge Univ. Press, 1958, p. 90.
G. H. Hardy and S. Ramanujan, Asymptotic formulas in combinatorial analysis, Proc. London Math. Soc., 17 (1918), 75-.
G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers (Fifth edition), Oxford Univ. Press (Clarendon), 1979, 273-296.
D. E. Knuth, The Art of Computer Programming, vol. 4A, Combinatorial Algorithms, Section 7.2.1.4, p. 396.
D. S. Mitrinovic et al., Handbook of Number Theory, Kluwer, Section XIV.1, p. 491.
S. Ramanujan, Collected Papers, Chap. 25, Cambridge Univ. Press 1927 (Proceedings of the Camb. Phil. Soc., 19 (1919), pp. 207-213).
S. Ramanujan, Collected Papers, Chap. 28, Cambridge Univ. Press 1927 (Proceedings of the London Math. Soc., 2, 18(1920)).
S. Ramanujan, Collected Papers, Chap. 30, Cambridge Univ. Press 1927 (Mathematische Zeitschrift, 9 (1921), pp. 147-163).
S. Ramanujan, Collected Papers, Ed. G. H. Hardy et al., Cambridge 1927; Chelsea, NY, 1962. See Table IV on page 308.
J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 122.
J. E. Roberts, Lure of the Integers, pp. 168-9 MAA 1992.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
R. E. Tapscott and D. Marcovich, "Enumeration of Permutational Isomers: The Porphyrins", Journal of Chemical Education, 55 (1978), 446-447.
Robert M. Young, "Excursions in Calculus", Mathematical Association of America, p. 367.
|
|
|
LINKS
|
Milton Abramowitz and Irene A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972, p. 836. [scanned copy]
James Grime and Brady Haran, Partitions, Numberphile video (2016).
Johannes W. Meijer, Euler's ship on the Pentagonal Sea, pdf and jpg.
|
|
|
FORMULA
|
G.f.: Product_{k>0} 1/(1-x^k) = Sum_{k>= 0} x^k Product_{i = 1..k} 1/(1-x^i) = 1 + Sum_{k>0} x^(k^2)/(Product_{i = 1..k} (1-x^i))^2.
G.f.: 1 + Sum_{n>=1} x^n/(Product_{k>=n} 1-x^k). - Joerg Arndt, Jan 29 2011
a(n) - a(n-1) - a(n-2) + a(n-5) + a(n-7) - a(n-12) - a(n-15) + ... = 0, where the sum is over n-k and k is a generalized pentagonal number (A001318) <= n and the sign of the k-th term is (-1)^([(k+1)/2]). See A001318 for a good way to remember this!
a(n) = (1/n) * Sum_{k=0..n-1} sigma(n-k)*a(k), where sigma(k) is the sum of divisors of k (A000203).
a(n) ~ 1/(4*n*sqrt(3)) * e^(Pi * sqrt(2n/3)) as n -> infinity (Hardy and Ramanujan). See A050811.
a(n) = a(0)*b(n) + a(1)*b(n-2) + a(2)*b(n-4) + ... where b = A000009.
It appears that the above approximation from Hardy and Ramanujan can be refined as
a(n) ~ 1/(4*n*sqrt(3)) * e^(Pi * sqrt(2n/3 + c0 + c1/n^(1/2) + c2/n + c3/n^(3/2) + c4/n^2 + ...)), where the coefficients c0 through c4 are approximately
c0 = -0.230420145062453320665537
c1 = -0.0178416569128570889793
c2 = 0.0051329911273
c3 = -0.0011129404
c4 = 0.0009573,
as n -> infinity. (End)
c0 = -0.230420145062453320665536704197233... = -1/36 - 2/Pi^2
c1 = -0.017841656912857088979502135349949... = 1/(6*sqrt(6)*Pi) - sqrt(3/2)/Pi^3
c2 = 0.005132991127342167594576391633559... = 1/(2*Pi^4)
c3 = -0.001112940489559760908236602843497... = 3*sqrt(3/2)/(4*Pi^5) - 5/(16*sqrt(6)*Pi^3)
c4 = 0.000957343284806972958968694349196... = 1/(576*Pi^2) - 1/(24*Pi^4) + 93/(80*Pi^6)
a(n) ~ exp(Pi*sqrt(2*n/3))/(4*sqrt(3)*n) * (1 - (sqrt(3/2)/Pi + Pi/(24*sqrt(6)))/sqrt(n) + (1/16 + Pi^2/6912)/n).
a(n) ~ exp(Pi*sqrt(2*n/3) - (sqrt(3/2)/Pi + Pi/(24*sqrt(6)))/sqrt(n) + (1/24 - 3/(4*Pi^2))/n) / (4*sqrt(3)*n).
(End)
a(n) < exp( (2/3)^(1/2) Pi sqrt(n) ) (Ayoub, p. 197).
a(n) = Sum_{i=0..n-1} P(i, n-i), where P(x, y) is the number of partitions of x into at most y parts and P(0, y)=1. - Jon Perry, Jun 16 2003
G.f.: Product_{i>=1} Product_{j>=0} (1+x^((2i-1)*2^j))^(j+1). - Jon Perry, Jun 06 2004
a(n) = determinant of the n X n Toeplitz matrix:
1 -1
1 1 -1
0 1 1 -1
0 0 1 1 -1
-1 0 0 1 1 -1
. . .
d_n d_(n-1) d_(n-2)...1
where d_q = (-1)^(m+1) if q = m(3m-1)/2 = p_m, the m-th generalized pentagonal number (A001318), otherwise d_q = 0. Note that the 1's run along the diagonal and the -1's are on the superdiagonal. The (n-1) row (not written) would end with ... 1 -1. (End)
Empirical: let F*(x) = Sum_{n=0..infinity} p(n)*exp(-Pi*x*(n+1)), then F*(2/5) = 1/sqrt(5) to a precision of 13 digits.
F*(4/5) = 1/2+3/2/sqrt(5)-sqrt(1/2*(1+3/sqrt(5))) to a precision of 28 digits. These are the only values found for a/b when a/b is from F60, Farey fractions up to 60. The number for F*(4/5) is one of the real roots of 25*x^4 - 50*x^3 - 10*x^2 - 10*x + 1. Note here the exponent (n+1) compared to the standard notation with n starting at 0. - Simon Plouffe, Feb 23 2011
The constant (2^(7/8)*GAMMA(3/4))/(exp(Pi/6)*Pi^(1/4)) = 1.0000034873... when expanded in base exp(4*Pi) will give the first 52 terms of a(n), n>0, the precision needed is 300 decimal digits. - Simon Plouffe, Mar 02 2011
G.f.: A(x)=1+x/(G(0)-x); G(k) = 1 + x - x^(k+1) - x*(1-x^(k+1))/G(k+1); (continued fraction Euler's kind, 1-step ). - Sergei N. Gladkovskii, Jan 25 2012
G.f.: 1 + x*(1 - G(0))/(1-x) where G(k) = 1 - 1/(1-x^(k+1))/(1-x/(x-1/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jan 22 2013
G.f.: Q(0) where Q(k) = 1 + x^(4*k+1)/( (x^(2*k+1)-1)^2 - x^(4*k+3)*(x^(2*k+1)-1)^2/( x^(4*k+3) + (x^(2*k+2)-1)^2/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Feb 16 2013
a(n-1) = Sum_{parts k in all partitions of n} mu(k), where mu(k) is the arithmetical Möbius function (see A008683).
Let P(2,n) denote the set of partitions of n into parts k >= 2. Then a(n-2) = -Sum_{parts k in all partitions in P(2,n)} mu(k).
n*( a(n) - a(n-1) ) = Sum_{parts k in all partitions in P(2,n)} k (see A138880).
Let P(3,n) denote the set of partitions of n into parts k >= 3. Then
a(n-3) = (1/2)*Sum_{parts k in all partitions in P(3,n)} phi(k), where phi(k) is the Euler totient function (see A000010). Using this result and Mertens's theorem on the average order of the phi function, we can find an approximate 3-term recurrence for the partition function: a(n) ~ a(n-1) + a(n-2) + (Pi^2/(3*n) - 1)*a(n-3). For example, substituting the values a(47) = 124754, a(48) = 147273 and a(49) = 173525 into the recurrence gives the approximation a(50) ~ 204252.48... compared with the true value a(50) = 204226. (End)
A production matrix for the sequence with offset 1 is M, an infinite n x n matrix of the following form:
a, 1, 0, 0, 0, 0, ...
b, 0, 1, 0, 0, 0, ...
c, 0, 0, 1, 0, 0, ...
d, 0, 0, 0, 1, 0, ...
.
.
... such that (a, b, c, d, ...) is the signed version of A080995 with offset 1: (1,1,0,0,-1,0,-1,...)
and a(n) is the upper left term of M^n.
This operation is equivalent to the g.f. (1 + x + 2x^2 + 3x^3 + 5x^4 + ...) = 1/(1 - x - x^2 + x^5 + x^7 - x^12 - x^15 + x^22 + ...). (End)
a(n) = Sum_{k=-inf..+inf} (-1)^k a(n-k(3k-1)/2) with a(0)=1 and a(negative)=0. The sum can be restricted to the (finite) range from k = (1-sqrt(1-24n))/6 to (1+sqrt(1-24n))/6, since all terms outside this range are zero. - Jos Koot, Jun 01 2016
G.f.: (conjecture) (r(x) * r(x^2) * r(x^4) * r(x^8) * ...) where r(x) is A000009: (1, 1, 1, 2, 2, 3, 4, ...). - Gary W. Adamson, Sep 18 2016; Doron Zeilberger observed today that "This follows immediately from Euler's formula 1/(1-z) = (1+z)*(1+z^2)*(1+z^4)*(1+z^8)*..." Gary W. Adamson, Sep 20 2016
a(n) ~ 2*Pi * BesselI(3/2, sqrt(24*n-1)*Pi/6) / (24*n-1)^(3/4). - Vaclav Kotesovec, Jan 11 2017
G.f.: Product_{k>=1} (1 + x^k)/(1 - x^(2*k)). - Ilya Gutkovskiy, Jan 23 2018
a(n) = p(1, n) where p(k, n) = p(k+1, n) + p(k, n-k) if k < n, 1 if k = n, and 0 if k > n. p(k, n) is the number of partitions of n into parts >= k. - Lorraine Lee, Jan 28 2020
Sum_{n>=0} a(n)/exp(Pi*n) = 2^(3/8)*Gamma(3/4)/(Pi^(1/4)*exp(Pi/24)).
Sum_{n>=0} a(n)/exp(2*Pi*n) = 2^(1/2)*Gamma(3/4)/(Pi^(1/4)*exp(Pi/12)).
[These are the reciprocals of phi(exp(-Pi)) (A259148) and phi(exp(-2*Pi)) (A259149), where phi(q) is the Euler modular function. See B. C. Berndt (RLN, Vol. V, p. 326), and formulas (13) and (14) in I. Mező, 2013. - Peter Luschny, Mar 13 2021]
a(n) = A008284(2*n,n) is also the number of partitions of 2n into n parts. - Ryan Brooks, Jun 11 2022
a(n) ~ exp(Pi*sqrt(2*n/3)) / (4*n*sqrt(3)) * (1 + Sum_{r>=1} w(r)/n^(r/2)), where w(r) = 1/(-4*sqrt(6))^r * Sum_{k=0..(r+1)/2} binomial(r+1,k) * (r+1-k) / (r+1-2*k)! * (Pi/6)^(r-2*k) [Cormac O'Sullivan, 2023, pp. 2-3]. - Vaclav Kotesovec, Mar 15 2023
|
|
|
EXAMPLE
|
a(5) = 7 because there are seven partitions of 5, namely: {1, 1, 1, 1, 1}, {2, 1, 1, 1}, {2, 2, 1}, {3, 1, 1}, {3, 2}, {4, 1}, {5}. - Bob Selcoe, Jul 08 2014
G.f. = 1 + x + 2*x^2 + 3*x^3 + 5*x^4 + 7*x^5 + 11*x^6 + 15*x^7 + 22*x^8 + ...
G.f. = 1/q + q^23 + 2*q^47 + 3*q^71 + 5*q^95 + 7*q^119 + 11*q^143 + 15*q^167 + ...
There are up to a(4)=5 segmented partitions of the partitions of n with exactly 4 parts. They are a(n,4, <4>), a(n,4,<3,1>), a(n,4,<2,2>), a(n,4,<2,1,1>), a(n,4,<1,1,1,1>).
The partition 8,8,8,8 is counted in a(32,4,<4>).
The partition 9,9,9,5 is counted in a(32,4,<3,1>).
The partition 11,11,5,5 is counted in a(32,4,<2,2>).
The partition 13,13,5,1 is counted in a(32,4,<2,1,1>).
The partition 14,9,6,3 is counted in a(32,4,<1,1,1,1>).
a(n odd,4,<2,2>) = 0.
a(12, 6, <2,2,2>) = a(6,3,<1,1,1>) = a(6-3,3) = a(3,3) = 1. The lone partition is 3,3,2,2,1,1.
(End)
|
|
|
MAPLE
|
A000041 := n -> combinat:-numbpart(n): [seq(A000041(n), n=0..50)]; # Warning: Maple 10 and 11 give incorrect answers in some cases: A110375.
spec := [B, {B=Set(Set(Z, card>=1))}, unlabeled ];
[seq(combstruct[count](spec, size=n), n=0..50)];
with(combstruct):ZL0:=[S, {S=Set(Cycle(Z, card>0))}, unlabeled]: seq(count(ZL0, size=n), n=0..45); # Zerinvary Lajos, Sep 24 2007
G:={P=Set(Set(Atom, card>0))}: combstruct[gfsolve](G, labeled, x); seq(combstruct[count]([P, G, unlabeled], size=i), i=0..45); # Zerinvary Lajos, Dec 16 2007
# Using the function EULER from Transforms (see link at the bottom of the page).
|
|
|
MATHEMATICA
|
Table[ PartitionsP[n], {n, 0, 45}]
a[ n_] := SeriesCoefficient[ q^(1/24) / DedekindEta[ Log[q] / (2 Pi I)], {q, 0, n}]; (* Michael Somos, Jul 11 2011 *)
a[ n_] := SeriesCoefficient[ 1 / Product[ 1 - x^k, {k, n}], {x, 0, n}]; (* Michael Somos, Jul 11 2011 *)
a[0] := 1; a[n_] := a[n] = Block[{k=1, s=0, i=n-1}, While[i >= 0, s=s-(-1)^k (a[i]+a[i-k]); k=k+1; i=i-(3 k-2)]; s]; Map[a, Range[0, 49]] (* Oliver Seipel, Jun 01 2024 after Euler *)
|
|
|
PROG
|
(Magma) a:= func< n | NumberOfPartitions(n) >; [ a(n) : n in [0..10]];
(PARI) {a(n) = if( n<0, 0, polcoeff( 1 / eta(x + x * O(x^n)), n))};
(PARI) /* The Hardy-Ramanujan-Rademacher exact formula in PARI is as follows (this is no longer necessary since it is now built in to the numbpart command): */
Psi(n, q) = local(a, b, c); a=sqrt(2/3)*Pi/q; b=n-1/24; c=sqrt(b); (sqrt(q)/(2*sqrt(2)*b*Pi))*(a*cosh(a*c)-(sinh(a*c)/c))
L(n, q) = if(q==1, 1, sum(h=1, q-1, if(gcd(h, q)>1, 0, cos((g(h, q)-2*h*n)*Pi/q))))
g(h, q) = if(q<3, 0, sum(k=1, q-1, k*(frac(h*k/q)-1/2)))
part(n) = round(sum(q=1, max(5, 0.5*sqrt(n)), L(n, q)*Psi(n, q)))
(PARI) {a(n) = numbpart(n)};
(PARI) {a(n) = if( n<0, 0, polcoeff( sum( k=1, sqrtint(n), x^k^2 / prod( i=1, k, 1 - x^i, 1 + x * O(x^n))^2, 1), n))};
(PARI) f(n)= my(v, i, k, s, t); v=vector(n, k, 0); v[n]=2; t=0; while(v[1]<n, i=2; while(v[i]==0, i++); v[i]--; s=sum(k=i, n, k*v[k]); while(i>1, i--; s+=i*(v[i]=(n-s)\i)); t++); t \\ Thomas Baruchel, Nov 07 2005
(PARI) a(n)=if(n<0, 0, polcoeff(exp(sum(k=1, n, x^k/(1-x^k)/k, x*O(x^n))), n)) \\ Joerg Arndt, Apr 16 2010
(MuPAD) combinat::partitions::count(i) $i=0..54 // Zerinvary Lajos, Apr 16 2007
(Sage) [number_of_partitions(n) for n in range(46)] # Zerinvary Lajos, May 24 2009
(Sage)
@CachedFunction
if n == 0: return 1
S = 0; J = n-1; k = 2
while 0 <= J:
S = S+T if is_odd(k//2) else S-T
J -= k if is_odd(k) else k//2
k += 1
return S
(Sage) # uses[EulerTransform from A166861]
a = BinaryRecurrenceSequence(1, 0)
b = EulerTransform(a)
(Haskell)
import Data.MemoCombinators (memo2, integral)
a000041 n = a000041_list !! n
a000041_list = map (p' 1) [0..] where
p' = memo2 integral integral p
p _ 0 = 1
p k m = if m < k then 0 else p' k (m - k) + p' (k + 1) m
(GAP) List([1..10], n->Size(OrbitsDomain(SymmetricGroup(IsPermGroup, n), SymmetricGroup(IsPermGroup, n), \^))); # Attila Egri-Nagy, Aug 15 2014
(Perl) use ntheory ":all"; my @p = map { partitions($_) } 0..100; say "[@p]"; # Dana Jacobsen, Sep 06 2015
(Racket)
#lang racket
; SUM(k, -inf, +inf) (-1)^k p(n-k(3k-1)/2)
; For k outside the range (1-(sqrt(1-24n))/6 to (1+sqrt(1-24n))/6) argument n-k(3k-1)/2 < 0.
; Therefore the loops below are finite. The hash avoids repeated identical computations.
(define (p n) ; Nr of partitions of n.
(hash-ref h n
(λ ()
(define r
(+
(let loop ((k 1) (n (sub1 n)) (s 0))
(if (< n 0) s
(loop (add1 k) (- n (* 3 k) 1) (if (odd? k) (+ s (p n)) (- s (p n))))))
(let loop ((k -1) (n (- n 2)) (s 0))
(if (< n 0) s
(loop (sub1 k) (+ n (* 3 k) -2) (if (odd? k) (+ s (p n)) (- s (p n))))))))
(hash-set! h n r)
r)))
(define h (make-hash '((0 . 1))))
; (for ((k (in-range 0 50))) (printf "~s, " (p k))) runs in a moment.
(Python)
from sympy.ntheory import npartitions
print([npartitions(i) for i in range(101)]) # Indranil Ghosh, Mar 17 2017
(Julia) # DedekindEta is defined in A000594
A000041List(len) = DedekindEta(len, -1)
|
|
|
CROSSREFS
|
Cf. A000009, A000079, A000203, A001318, A008284, A026820, A065446, A078506, A113685, A132311, A000248.
|
|
|
KEYWORD
|
core,easy,nonn,nice,changed
|
|
|
AUTHOR
|
|
|
|
EXTENSIONS
|
Additional comments from Ola Veshta (olaveshta(AT)my-deja.com), Feb 28 2001
Additional comments from Dan Fux (dan.fux(AT)OpenGaia.com or danfux(AT)OpenGaia.com), Apr 07 2001
|
|
|
STATUS
|
approved
|
| |
|
|
| |
|
|
A000123
|
|
Number of binary partitions: number of partitions of 2n into powers of 2.
(Formerly M1011 N0378)
|
|
+0
108
|
|
|
|
1, 2, 4, 6, 10, 14, 20, 26, 36, 46, 60, 74, 94, 114, 140, 166, 202, 238, 284, 330, 390, 450, 524, 598, 692, 786, 900, 1014, 1154, 1294, 1460, 1626, 1828, 2030, 2268, 2506, 2790, 3074, 3404, 3734, 4124, 4514, 4964, 5414, 5938, 6462, 7060, 7658, 8350, 9042, 9828
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,2
|
|
|
COMMENTS
|
Also, a(n) = number of "non-squashing" partitions of 2n (or 2n+1), that is, partitions 2n = p_1 + p_2 + ... + p_k with 1 <= p_1 <= p_2 <= ... <= p_k and p_1 + p_2 + ... + p_i <= p_{i+1} for all 1 <= i < k [Hirschhorn and Sellers].
Equals infinite convolution product of [1,2,2,2,2,2,2,2,2] aerated A000079 - 1 times, i.e., [1,2,2,2,2,2,2,2,2] * [1,0,2,0,2,0,2,0,2] * [1,0,0,0,2,0,0,0,2]. - Mats Granvik and Gary W. Adamson, Aug 04 2009
Which can be further decomposed to the infinite convolution product of finally supported sequences, namely [1,1] aerated A000079 - 1 times with multiplicity A000027 + 1 times, i.e., [1,1] * [1,1] * [1,0,1] * [1,0,1] * [1,0,1] * ... (next terms are [1,0,0,0,1] 4 times, etc.). - Eitan Y. Levine, Jun 18 2023
Given A018819 = A000123 with repeats, polcoeff (1, 1, 2, 2, 4, 4, ...) * (1, 1, 1, ...) = (1, 2, 4, 6, 10, ...) = (1, 0, 2, 0, 4, 0, 6, ...) * (1, 2, 2, 2, ...). - Gary W. Adamson, Dec 16 2009
Let M = an infinite lower triangular matrix with (1, 2, 2, 2, ...) in every column shifted down twice. A000123 = lim_{n->infinity} M^n, the left-shifted vector considered as a sequence. Replacing (1, 2, 2, 2, ...) with (1, 3, 3, 3, ...) and following the same procedure, we obtain A171370: (1, 3, 6, 12, 18, 30, 42, 66, 84, 120, ...). - Gary W. Adamson, Dec 06 2009
First differences of the sequence are (1, 2, 2, 4, 4, 6, 6, 10, ...), A018819, i.e., the sequence itself with each term duplicated except for the first one (unless a 0 is prefixed before taking the first differences), as shown by the formula a(n) - a(n-1) = a(floor(n/2)), valid for all n including n = 0 if we let a(-1) = 0. - M. F. Hasler, Feb 19 2019
Sum over k <= n of number of partitions of k into powers of 2, A018819. - Peter Munn, Feb 21 2020
|
|
|
REFERENCES
|
G. E. Andrews, The Theory of Partitions, Addison-Wesley, 1976.
R. F. Churchhouse, Binary partitions, pp. 397-400 of A. O. L. Atkin and B. J. Birch, editors, Computers in Number Theory. Academic Press, NY, 1971.
N. G. de Bruijn, On Mahler's partition problem, Indagationes Mathematicae, vol. X (1948), 210-220.
G. Everest, A. van der Poorten, I. Shparlinski and T. Ward, Recurrence Sequences, Amer. Math. Soc., 2003; see esp. p. 255.
H. Gupta, A simple proof of the Churchhouse conjecture concerning binary partitions, Indian J. Pure Appl. Math. 3 (1972), 791-794.
H. Gupta, A direct proof of the Churchhouse conjecture concerning binary partitions, Indian J. Math. 18 (1976), 1-5.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
|
|
|
LINKS
|
M. D. Hirschhorn and J. A. Sellers, A different view of m-ary partitions, Australasian J. Combin., 30 (2004), 193-196.
M. Konvalinka and I. Pak, Cayley compositions, partitions, polytopes, and geometric bijections, Journal of Combinatorial Theory, Series A, Volume 123, Issue 1, April 2014, Pages 86-91.
B. Reznick, Some binary partition functions, in "Analytic number theory" (Conf. in honor P. T. Bateman, Allerton Park, IL, 1989), 451-477, Progr. Math., 85, Birkhäuser Boston, Boston, MA, 1990.
|
|
|
FORMULA
|
a(n) = a(n-1) + a(floor(n/2)). For proof see A018819.
2 * a(n) = a(n+1) + a(n-1) if n is even. - Michael Somos, Jan 07 2011
G.f.: (1-x)^(-1) Product_{n>=0} (1 - x^(2^n))^(-1).
a(n) = Sum_{i=0..n} a(floor(i/2)) [O'Shea].
Conjecture: Limit_{n ->infinity} (log(n)*a(2n))/(n*a(n)) = c = 1.63... - Benoit Cloitre, Jan 26 2003 [The constant c is equal to 2*log(2) = 1.38629436... =A016627. - Vaclav Kotesovec, Aug 07 2019]
G.f. A(x) satisfies A(x^2) = ((1-x)/(1+x)) * A(x). - Michael Somos, Aug 25 2003
G.f.: Product_{k>=0} (1+x^(2^k))/(1-x^(2^k)) = (Product_{k>=0} (1+x^(2^k))^(k+1) )/(1-x) = Product_{k>=0} (1+x^(2^k))^(k+2). - Joerg Arndt, Apr 24 2005
The asymptotic rate of growth is known precisely - see De Bruijn's paper. With p(n) the number of partitions of n into powers of two, the asymptotic formula of de Bruijn is: log(p(2*n)) = 1/(2*L2)*(log(n/log(n)))^2 + (1/2 + 1/L2 + LL2/L2)*log(n) - (1 + LL2/L2)*log(log(n)) + Phi(log(n/log(n))/L2), where L2=log(2), LL2=log(log(2)) and Phi(x) is a certain periodic function with period 1 and a tiny amplitude.
Numerically, Phi(x) appears to have a mean value around 0.66. An expansion up to O(1) term had been obtained earlier by Kurt Mahler. (End)
G.f.: exp( Sum_{n>=1} 2^A001511(n) * x^n/n ), where 2^A001511(n) is the highest power of 2 that divides 2*n. - Paul D. Hanna, Oct 30 2012
|
|
|
EXAMPLE
|
For non-squashing partitions and binary partitions see the example in A018819.
For n=3, the a(3)=6 admitted partitions of 2n=6 are 1+1+1+1+1+1, 1+1+1+1+2, 1+1+2+2, 2+2+2, 1+1+4 and 2+4. - R. J. Mathar, Aug 11 2021
|
|
|
MAPLE
|
# second Maple program: more efficient for large n; try: a( 10^25 );
g:= proc(b, n) option remember; `if`(b<0, 0, `if`(b=0 or
n=0, 1, `if`(b>=n, add((-1)^(t+1)*binomial(n+1, t)
*g(b-t, n), t=1..n+1), g(b-1, n)+g(2*b, n-1))))
end:
a:= n-> (t-> g(n/2^(t-1), t))(max(ilog2(2*n), 1)):
seq(a(n), n=0..60); # Alois P. Heinz, Apr 16 2009, revised Apr 14 2016
|
|
|
MATHEMATICA
|
Fold[Append[#1, Total[Take[Flatten[Transpose[{#1, #1}]], #2]]] &, {1}, Range[2, 49]] (* Birkas Gyorgy, Apr 18 2011 *)
|
|
|
PROG
|
(PARI) {a(n) = my(A, m); if( n<1, n==0, m=1; A = 1 + O(x); while(m<=n, m*=2; A = subst(A, x, x^2) * (1+x) / (1-x)); polcoeff(A, n))}; /* Michael Somos, Aug 25 2003 */
(PARI) {a(n) = if( n<1, n==0, a(n\2) + a(n-1))}; /* Michael Somos, Aug 25 2003 */
(PARI) A123=[]; A000123(n)={ n<3 && return(2^n); if( n<=#A123, A123[n] && return(A123[n]); A123[n-1] && return( A123[n] = A123[n-1]+A000123(n\2) ), n>2*#A123 && A123=concat(A123, vector((n-#A123)\2))); A123[if(n>#A123, 1, n)]=2*sum(k=1, n\2-1, A000123(k), 1)+(n%2+1)*A000123(n\2)} \\ Stores results in global vector A123 dynamically resized to at most 3n/4 when size is less than n/2. Gives a(n*10^6) in ~ n sec. - M. F. Hasler, Apr 30 2009
(PARI) {a(n)=polcoeff(exp(sum(m=1, n, 2^valuation(2*m, 2)*x^m/m)+x*O(x^n)), n)} \\ Paul D. Hanna, Oct 30 2012
(Haskell)
import Data.List (transpose)
a000123 n = a000123_list !! n
a000123_list = 1 : zipWith (+)
a000123_list (tail $ concat $ transpose [a000123_list, a000123_list])
(Magma) [1] cat [n eq 1 select n+1 else Self(n-1) + Self(n div 2): n in [1..70]]; // Vincenzo Librandi, Dec 17 2016
(Python)
from functools import lru_cache
@lru_cache(maxsize=None)
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn,easy,core,nice
|
|
|
AUTHOR
|
|
|
|
EXTENSIONS
|
More terms from Robin Trew (trew(AT)hcs.harvard.edu)
Values up to a(10^4) checked with given PARI code by M. F. Hasler, Apr 30 2009
|
|
|
STATUS
|
approved
|
| |
|
|
| |
|
|
A000124
|
|
Central polygonal numbers (the Lazy Caterer's sequence): n(n+1)/2 + 1; or, maximal number of pieces formed when slicing a pancake with n cuts.
(Formerly M1041 N0391)
|
|
+0
426
|
|
|
|
1, 2, 4, 7, 11, 16, 22, 29, 37, 46, 56, 67, 79, 92, 106, 121, 137, 154, 172, 191, 211, 232, 254, 277, 301, 326, 352, 379, 407, 436, 466, 497, 529, 562, 596, 631, 667, 704, 742, 781, 821, 862, 904, 947, 991, 1036, 1082, 1129, 1177, 1226, 1276, 1327, 1379
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,2
|
|
|
COMMENTS
|
These are Hogben's central polygonal numbers with the (two-dimensional) symbol
2
.P
1 n
The first line cuts the pancake into 2 pieces. For n > 1, the n-th line crosses every earlier line (avoids parallelism) and also avoids every previous line intersection, thus increasing the number of pieces by n. For 16 lines, for example, the number of pieces is 2 + 2 + 3 + 4 + 5 + ... + 16 = 137. These are the triangular numbers plus 1 (cf. A000217).
m = (n-1)(n-2)/2 + 1 is also the smallest number of edges such that all graphs with n nodes and m edges are connected. - Keith Briggs, May 14 2004
Also maximal number of grandchildren of a binary vector of length n+2. E.g., a binary vector of length 6 can produce at most 11 different vectors when 2 bits are deleted.
This is also the order dimension of the (strong) Bruhat order on the finite Coxeter group B_{n+1}. - Nathan Reading (reading(AT)math.umn.edu), Mar 07 2002
Number of 132- and 321-avoiding permutations of {1,2,...,n+1}. - Emeric Deutsch, Mar 14 2002
For n >= 1 a(n) is the number of terms in the expansion of (x+y)*(x^2+y^2)*(x^3+y^3)*...*(x^n+y^n). - Yuval Dekel (dekelyuval(AT)hotmail.com), Jul 28 2003
Also the number of terms in (1)(x+1)(x^2+x+1)...(x^n+...+x+1); see A000140.
Narayana transform (analog of the binomial transform) of vector [1, 1, 0, 0, 0, ...] = A000124; using the infinite lower Narayana triangle of A001263 (as a matrix), N; then N * [1, 1, 0, 0, 0, ...] = A000124. - Gary W. Adamson, Apr 28 2005
Number of interval subsets of {1, 2, 3, ..., n} (cf. A002662). - Jose Luis Arregui (arregui(AT)unizar.es), Jun 27 2006
Define a number of straight lines in the plane to be in general arrangement when (1) no two lines are parallel, (2) there is no point common to three lines. Then these are the maximal numbers of regions defined by n straight lines in general arrangement in the plane. - Peter C. Heinig (algorithms(AT)gmx.de), Oct 19 2006
Note that a(n) = a(n-1) + A000027(n-1). This has the following geometrical interpretation: Suppose there are already n-1 lines in general arrangement, thus defining the maximal number of regions in the plane obtainable by n-1 lines and now one more line is added in general arrangement. Then it will cut each of the n-1 lines and acquire intersection points which are in general arrangement. (See the comments on A000027 for general arrangement with points.) These points on the new line define the maximal number of regions in 1-space definable by n-1 points, hence this is A000027(n-1), where for A000027 an offset of 0 is assumed, that is, A000027(n-1) = (n+1)-1 = n. Each of these regions acts as a dividing wall, thereby creating as many new regions in addition to the a(n-1) regions already there, hence a(n) = a(n-1) + A000027(n-1). Cf. the comments on A000125 for an analogous interpretation. - Peter C. Heinig (algorithms(AT)gmx.de), Oct 19 2006
When constructing a zonohedron, one zone at a time, out of (up to) 3-d non-intersecting parallelepipeds, the n-th element of this sequence is the number of edges in the n-th zone added with the n-th "layer" of parallelepipeds. (Verified up to 10-zone zonohedron, the enneacontahedron.) E.g., adding the 10th zone to the enneacontahedron requires 46 parallel edges (edges in the 10th zone) by looking directly at a 5-valence vertex and counting visible vertices. - Shel Kaphan, Feb 16 2006
Binomial transform of (1, 1, 1, 0, 0, 0, ...) and inverse binomial transform of A072863: (1, 3, 9, 26, 72, 192, ...). - Gary W. Adamson, Oct 15 2007
If Y is a 2-subset of an n-set X then, for n >= 3, a(n-3) is the number of (n-2)-subsets of X which do not have exactly one element in common with Y. - Milan Janjic, Dec 28 2007
It appears that a(n) is the number of distinct values among the fractions F(i+1)/F(j+1) as j ranges from 1 to n and, for each fixed j, i ranges from 1 to j, where F(i) denotes the i-th Fibonacci number. - John W. Layman, Dec 02 2008
a(n) is the number of subsets of {1,2,...,n} that contain at most two elements. - Geoffrey Critzer, Mar 10 2009
For n >= 2, a(n) gives the number of sets of subsets A_1, A_2, ..., A_n of n = {1, 2, ..., n} such that Meet_{i = 1..n} A_i is empty and Sum_{j in [n]} (|Meet{i = 1..n, i != j} A_i|) is a maximum. - Srikanth K S, Oct 22 2009
The numbers along the left edge of Floyd's triangle. - Paul Muljadi, Jan 25 2010
Let A be the Hessenberg matrix of order n, defined by: A[1,j] = A[i,i]:=1, A[i,i-1] = -1, and A[i,j] = 0 otherwise. Then, for n >= 1, a(n-1) = (-1)^(n-1)*coeff(charpoly(A,x),x). - Milan Janjic, Jan 24 2010
Also the number of deck entries of Euler's ship. See the Meijer-Nepveu link. - Johannes W. Meijer, Jun 21 2010
(1 + x^2 + x^3 + x^4 + x^5 + ...)*(1 + 2x + 3x^2 + 4x^3 + 5x^4 + ...) = (1 + 2x + 4x^2 + 7x^3 + 11x^4 + ...). - Gary W. Adamson, Jul 27 2010
The number of length n binary words that have no 0-digits between any pair of consecutive 1-digits. - Jeffrey Liese, Dec 23 2010
Let b(0) = b(1) = 1; b(n) = max(b(n-1)+n-1, b(n-2)+n-2) then a(n) = b(n+1). - Yalcin Aktar, Jul 28 2011
Also number of distinct sums of 1 through n where each of those can be + or -. E.g., {1+2,1-2,-1+2,-1-2} = {3,-1,1,-3} and a(2) = 4. - Toby Gottfried, Nov 17 2011
This sequence is complete because the sum of the first n terms is always greater than or equal to a(n+1)-1. Consequently, any nonnegative number can be written as a sum of distinct terms of this sequence. See A204009, A072638. - Frank M Jackson, Jan 09 2012
The sequence is the number of distinct sums of subsets of the nonnegative integers, and its first differences are the positive integers. See A208531 for similar results for the squares. - John W. Layman, Feb 28 2012
Apparently the number of Dyck paths of semilength n+1 in which the sum of the first and second ascents add to n+1. - David Scambler, Apr 22 2013
Without 1 and 2, a(n) equals the terminus of the n-th partial sum of sequence 1, 1, 2. Explanation: 1st partial sums of 1, 1, 2 are 1, 2, 4; 2nd partial sums are 1, 3, 7; 3rd partial sums are 1, 4, 11; 4th partial sums are 1, 5, 16, etc. - Bob Selcoe, Jul 04 2013
Equivalently, numbers of the form 2*m^2+m+1, where m = 0, -1, 1, -2, 2, -3, 3, ... . - Bruno Berselli, Apr 08 2014
For n >= 2: quasi-triangular numbers; the almost-triangular numbers being A000096(n), n >= 2. Note that 2 is simultaneously almost-triangular and quasi-triangular. - Daniel Forgues, Apr 21 2015
n points in general position determine "n choose 2" lines, so A055503(n) <= a(n(n-1)/2). If n > 3, the lines are not in general position and so A055503(n) < a(n(n-1)/2). - Jonathan Sondow, Dec 01 2015
The digital root is period 9 (1, 2, 4, 7, 2, 7, 4, 2, 1), also the digital roots of centered 10-gonal numbers (A062786), for n > 0, A133292. - Peter M. Chema, Sep 15 2016
For n >= 0, a(n) is the number of weakly unimodal sequences of length n over the alphabet {1, 2}. - Armend Shabani, Mar 10 2017
Number of sequences (e(1), ..., e(n+1)), 0 <= e(i) < i, such that there is no triple i < j < k with e(i) < e(j) != e(k). [Martinez and Savage, 2.4]
Number of sequences (e(1), ..., e(n+1)), 0 <= e(i) < i, such that there is no triple i < j < k with e(i) < e(j) and e(i) < e(k). [Martinez and Savage, 2.4]
Number of sequences (e(1), ..., e(n+1)), 0 <= e(i) < i, such that there is no triple i < j < k with e(i) >= e(j) != e(k). [Martinez and Savage, 2.4]
(End)
The odd prime factors != 7 occur in an interval of p successive terms either never or exactly twice, while 7 always occurs only once. If a prime factor p appears in a(n) and a(m) within such an interval, then n + m == -1 (mod p). When 7 divides a(n), then 2*n == -1 (mod 7). a(n) is never divisible by the prime numbers given in A003625.
While all prime factors p != 7 can occur to any power, a(n) is never divisible by 7^2. The prime factors are given in A045373. The prime terms of this sequence are given in A055469.
(End)
a(n-1) is the greatest sum of arch lengths for the top arches of a semi-meander with n arches. An arch length is the number of arches covered + 1.
/\ The top arch has a length of 3. /\ The top arch has a length of 3.
/ \ Both bottom arches have a //\\ The middle arch has a length of 2.
//\/\\ length of 1. ///\\\ The bottom arch has a length of 1.
Example: for n = 4, a(4-1) = a(3) = 7 /\
//\\
/\ ///\\\ 1 + 3 + 2 + 1 = 7. (End)
a(n+1) is the a(n)-th smallest positive integer that has not yet appeared in the sequence. - Matthew Malone, Aug 26 2021
For n> 0, let the n-dimensional cube {0,1}^n be, provided with the Hamming distance, d. Given an element x in {0,1}^n, a(n) is the number of elements y in {0,1}^n such that d(x, y) <= 2. Example: n = 4. (0,0,0,0), (1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1), (0,0,1,1), (0,1,0,1), (0,1,1,0), (1,0,0,1), (1,0,1,0), (1,1,0,0) are at distance <= 2 from (0,0,0,0), so a(4) = 11. - Yosu Yurramendi, Dec 10 2021
a(n) is the sum of the first three entries of row n of Pascal's triangle. - Daniel T. Martin, Apr 13 2022
a(n-1) is the number of Grassmannian permutations that avoid a pattern, sigma, where sigma is a pattern of size 3 with exactly one descent. For example, sigma is one of the patterns, {132, 213, 231, 312}. - Jessica A. Tomasko, Sep 14 2022
a(n+4) is the number of ways to tile an equilateral triangle of side length 2*n with smaller equilateral triangles of side length n and side length 1. For example, with n=2, there are 22 ways to tile an equilateral triangle of side length 4 with smaller ones of sides 2 and 1, including the one tiling with sixteen triangles of sides 1 and the one tiling with four triangles of sides 2. - Ahmed ElKhatib and Greg Dresden, Aug 19 2024
|
|
|
REFERENCES
|
Robert B. Banks, Slicing Pizzas, Racing Turtles and Further Adventures in Applied Mathematics, Princeton Univ. Press, 1999. See p. 24.
Louis Comtet, Advanced Combinatorics, Reidel, 1974, p. 72, Problem 2.
Henry Ernest Dudeney, Amusements in Mathematics, Nelson, London, 1917, page 177.
Derrick Niederman, Number Freak, From 1 to 200 The Hidden Language of Numbers Revealed, A Perigee Book, NY, 2009, p. 83.
Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence - see "List of Sequences" in Vol. 2.
Alain M. Robert, A Course in p-adic Analysis, Springer-Verlag, 2000; p. 213.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane, On single-deletion-correcting codes, in Codes and Designs (Columbus, OH, 2000), 273-291, Ohio State Univ. Math. Res. Inst. Publ., 10, de Gruyter, Berlin, 2002.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
William Allen Whitworth, DCC Exercises in Choice and Chance, Stechert, NY, 1945, p. 30.
Akiva M. Yaglom and Isaak M. Yaglom, Challenging Mathematical Problems with Elementary Solutions. Vol. I. Combinatorial Analysis and Probability Theory. New York: Dover Publications, Inc., 1987, p. 13, #44 (First published: San Francisco: Holden-Day, Inc., 1964).
|
|
|
LINKS
|
M. F. Hasler, Interactive illustration of A000124. [Sep 06 2017: The user can choose the slices to make, but the program can suggest a set of n slices which should yield the maximum number of pieces. For n slices this obviously requires 2n endpoints, or 2n+1 if they are equally spaced, so if there are not enough "blobs", their number is accordingly increased. This is the distinction between "draw" (done when you change the slices or number of blobs by hand) and "suggest" (propose a new set of slices).]
Rodica Simion and Frank W. Schmidt, Restricted permutations, European J. Combin., 6, 383-406, 1985; see Example 3.5.
|
|
|
FORMULA
|
G.f.: (1 - x + x^2)/(1 - x)^3. Simon Plouffe in his 1992 dissertation
G.f.: (1 - x^6)/((1 - x)^2*(1 - x^2)*(1 - x^3)). a(n) = a(-1 - n) for all n in Z. - Michael Somos, Sep 04 2006
Euler transform of length 6 sequence [ 2, 1, 1, 0, 0, -1]. - Michael Somos, Sep 04 2006
a(n+3) = 3*a(n+2) - 3*a(n+1) + a(n) and a(1) = 1, a(2) = 2, a(3) = 4. - Artur Jasinski, Oct 21 2008
a(n) = a(n-1) + n. E.g.f.:(1 + x + x^2/2)*exp(x). - Geoffrey Critzer, Mar 10 2009
a(n) = Sum_{k = 0..n + 1} binomial(n+1, 2(k - n)). - Paul Barry, Aug 29 2004
a(n) = binomial(n+2, 1) - 2*binomial(n+1, 1) + binomial(n+2, 2). - Zerinvary Lajos, May 12 2006
a(n) = Sum_{l_1 = 0..n + 1} Sum_{l_2 = 0..n}...Sum_{l_i = 0..n - i}...Sum_{l_n = 0..1} delta(l_1, l_2, ..., l_i, ..., l_n) where delta(l_1, l_2, ..., l_i, ..., l_n) = 0 if any l_i != l_(i+1) and l_(i+1) != 0 and delta(l_1, l_2, ..., l_i, ..., l_n) = 1 otherwise. (End)
a(n) = 2*a(n-1) - a(n-2) + 1. - Eric Werley, Jun 27 2011
E.g.f.: exp(x)*(1+x+(x^2)/2) = Q(0); Q(k) = 1+x/(1-x/(2+x-4/(2+x*(k+1)/Q(k+1)))); (continued fraction). - Sergei N. Gladkovskii, Nov 21 2011
Dirichlet g.f.: (zeta(s-2) + zeta(s-1) + 2*zeta(s))/2.
Sum_{n >= 0} 1/a(n) = 2*Pi*tanh(sqrt(7)*Pi/2)/sqrt(7) = A226985. (End)
a(n) = 2*a(n-1) - binomial(n-1, 2) and a(0) = 1. - Armend Shabani, Mar 10 2017
a(n) = (Sum_{i=n-2..n+2} A000217(i))/5.
a(n) = (Sum_{i=n-2..n+2} A002378(i))/10.
a(n) = (Sum_{i=n..n+2} A002061(i)+1)/6.
a(n) = (Sum_{i=n-1..n+2} A000290(i)+2)/8.
(End)
Product_{n>=0} (1 + 1/a(n)) = cosh(sqrt(15)*Pi/2)*sech(sqrt(7)*Pi/2).
Product_{n>=1} (1 - 1/a(n)) = 2*Pi*sech(sqrt(7)*Pi/2). (End)
a((n^2-3n+6)/2) + a((n^2-n+4)/2) = a(n^2-2n+6)/2. - Charlie Marion, Feb 14 2023
|
|
|
EXAMPLE
|
a(3) = 7 because the 132- and 321-avoiding permutations of {1, 2, 3, 4} are 1234, 2134, 3124, 2314, 4123, 3412, 2341.
G.f. = 1 + 2*x + 4*x^2 + 7*x^3 + 11*x^4 + 16*x^5 + 22*x^6 + 29*x^7 + ...
|
|
|
MAPLE
|
|
|
|
MATHEMATICA
|
Table[PolygonalNumber[n] + 1, {n, 0, 52}] (* Michael De Vlieger, Jun 30 2016, Version 10.4 *)
|
|
|
PROG
|
(PARI) {a(n) = (n^2 + n) / 2 + 1}; /* Michael Somos, Sep 04 2006 */
(Haskell)
a000124 = (+ 1) . a000217
(Python)
def a(n): return n*(n+1)//2 + 1
|
|
|
CROSSREFS
|
Cf. A000096 (Maximal number of pieces that can be obtained by cutting an annulus with n cuts, for n >= 1).
Cf. A002061, A002522, A016028, A055503, A072863, A144328, A177862, A263883, A000127, A005408, A006261, A016813, A058331, A080856, A086514, A161701, A161702, A161703, A161704, A161706, A161707, A161708, A161710, A161711, A161712, A161713, A161715, A051601, A228918.
Cf. A055469 Quasi-triangular primes.
|
|
|
KEYWORD
|
nonn,core,easy,nice,changed
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
| |
|
|
A000125
|
|
Cake numbers: maximal number of pieces resulting from n planar cuts through a cube (or cake): C(n+1,3) + n + 1.
(Formerly M1100 N0419)
|
|
+0
79
|
|
|
|
1, 2, 4, 8, 15, 26, 42, 64, 93, 130, 176, 232, 299, 378, 470, 576, 697, 834, 988, 1160, 1351, 1562, 1794, 2048, 2325, 2626, 2952, 3304, 3683, 4090, 4526, 4992, 5489, 6018, 6580, 7176, 7807, 8474, 9178, 9920, 10701, 11522, 12384, 13288, 14235, 15226
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,2
|
|
|
COMMENTS
|
Note that a(n) = a(n-1) + A000124(n-1). This has the following geometrical interpretation: Define a number of planes in space to be in general arrangement when
(1) no two planes are parallel,
(2) there are no two parallel intersection lines,
(3) there is no point common to four or more planes.
Suppose there are already n-1 planes in general arrangement, thus defining the maximal number of regions in space obtainable by n-1 planes and now one more plane is added in general arrangement. Then it will cut each of the n-1 planes and acquire intersection lines which are in general arrangement. (See the comments on A000124 for general arrangement with lines.) These lines on the new plane define the maximal number of regions in 2-space definable by n-1 straight lines, hence this is A000124(n-1). Each of this regions acts as a dividing wall, thereby creating as many new regions in addition to the a(n-1) regions already there, hence a(n) = a(n-1) + A000124(n-1). - Peter C. Heinig (algorithms(AT)gmx.de), Oct 19 2006
More generally, we have: A000027(n) = binomial(n,0) + binomial(n,1) (the natural numbers), A000124(n) = binomial(n,0) + binomial(n,1) + binomial(n,2) (the Lazy Caterer's sequence), a(n) = binomial(n,0) + binomial(n,1) + binomial(n,2) + binomial(n,3) (Cake Numbers). - Peter C. Heinig (algorithms(AT)gmx.de), Oct 19 2006
If Y is a 2-subset of an n-set X then, for n>=3, a(n-3) is the number of 3-subsets of X which do not have exactly one element in common with Y. - Milan Janjic, Dec 28 2007
a(n) is the number of compositions (ordered partitions) of n+1 into four or fewer parts or equivalently the sum of the first four terms in the n-th row of Pascal's triangle. - Geoffrey Critzer, Jan 23 2009
a(n) is also the maximum number of different values obtained by summing n consecutive positive integers with all possible 2^n sign combinations. This maximum is first reached when summing the interval [n, 2n-1]. - Olivier Gérard, Mar 22 2010
a(n) contains only 5 perfect squares > 1: 4, 64, 576, 67600, and 75203584. The incidences of > 0 are given by A047694. - Frank M Jackson, Mar 15 2013
Given n tiles with two values - an A value and a B value - a player may pick either the A value or the B value. The particular tiles are [n, 0], [n-1, 1], ..., [2, n-2] and [1, n-1]. The sequence is the number of different final A:B counts. For example, with n=4, we can have final total [5, 3] = [4, _] + [_, 1] + [_, 2] + [1, _] = [_, 0] + [3, _] + [2, _] + [_, 3], so a(4) = 2^4 - 1 = 15. The largest and smallest final A+B counts are given by A077043 and A002620 respectively. - Jon Perry, Oct 24 2014
For n>=3, a(n) is also the number of maximal cliques in the (n+1)-triangular graph (the 4-triangular graph has a(3)=8 maximal cliques). - Andrew Howroyd, Jul 19 2017
a(n) is the number of binary words of length n matching the regular expression 1*0*1*0*. Coincidentally, A000124 counts binary words of the form 0*1*0*. See Alexandersson and Nabawanda for proof. - Per W. Alexandersson, May 15 2021
For n > 0, let the n-dimensional cube, {0,1}^n be provided with the Hamming distance, d. Given an element x in {0,1}^n, a(n) is the number of elements y in {0,1}^n such that d(x, y) <= 3. Example: n = 4. Let x = (0,0,0,0) be in {0,1}^4.
d(x,y) = 0: y in {(0,0,0,0)}.
d(x,y) = 1: y in {(1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1)}.
d(x,y) = 2: y in {(1,1,0,0), (1,0,1,0), (1,0,0,1), (0,1,1,0), (0,1,0,1), (0,0,1,1)}.
d(x,y) = 3: y in {(1,1,1,0), (1,1,0,1), (1,0,1,1), (0,1,1,1)}.
All these y are at a distance <= 3 from (0,0,0,0), so a(4) = 15. (See Peter C. Heinig's formula). - Yosu Yurramendi, Dec 14 2021
For n >= 2, a(n) is the number of distinct least squares regression lines fitted to n points (j,y_j), 1 <= j <= n, where each y_j is 0 or 1. The number of distinct lines with exactly k 1's among y_1, ..., y_n is A077028(n,k). The number of distinct slopes is A123596(n). - Pontus von Brömssen, Mar 16 2024
|
|
|
REFERENCES
|
V. I. Arnold (ed.), Arnold's Problems, Springer, 2004, comments on Problem 1990-11 (p. 75), pp. 503-510. Numbers N_3.
R. B. Banks, Slicing Pizzas, Racing Turtles and Further Adventures in Applied Mathematics, Princeton Univ. Press, 1999. See p. 27.
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 72, Problem 2.
H. E. Dudeney, Amusements in Mathematics, Nelson, London, 1917, page 177.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
T. H. Stickels, Mindstretching Puzzles. Sterling, NY, 1994 p. 85.
W. A. Whitworth, DCC Exercises in Choice and Chance, Stechert, NY, 1945, p. 30.
A. M. Yaglom and I. M. Yaglom: Challenging Mathematical Problems with Elementary Solutions. Vol. I. Combinatorial Analysis and Probability Theory. New York: Dover Publications, Inc., 1987, p. 13, #45 (First published: San Francisco: Holden-Day, Inc., 1964)
|
|
|
LINKS
|
Sebastian Mizera and Sabrina Pasterski, Celestial Geometry, arXiv:2204.02505 [hep-th], 2022.
|
|
|
FORMULA
|
a(n) = (n+1)*(n^2-n+6)/6 = (n^3 + 5*n + 6) / 6.
G.f.: (1 - 2*x + 2x^2)/(1-x)^4. - [Simon Plouffe in his 1992 dissertation.]
E.g.f.: (1 + x + x^2/2 + x^3/6)*exp(x).
a(n) = binomial(n,3) + binomial(n,2) + binomial(n,1) + binomial(n,0). - Peter C. Heinig (algorithms(AT)gmx.de), Oct 19 2006
Paraphrasing the previous comment: the sequence is the binomial transform of [1,1,1,1,0,0,0,...]. - Gary W. Adamson, Oct 23 2007
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
Inverse binomial transform of A134396.
Sum_{n>=0} a(n)/n! = 8*exp(1)/3. (End)
|
|
|
EXAMPLE
|
a(4)=15 because there are 15 compositions of 5 into four or fewer parts. a(6)=42 because the sum of the first four terms in the 6th row of Pascal's triangle is 1+6+15+20=42. - Geoffrey Critzer, Jan 23 2009
For n=5, (1, 3, 5, 7, 9, 11, 13, 17, 19, 21, 23, 25, 35) and their opposite are the 26 different sums obtained by summing 5,6,7,8,9 with any sign combination. - Olivier Gérard, Mar 22 2010
G.f. = 1 + 2*x + 4*x^2 + 8*x^3 + 15*x^4 + 26*x^5 + 42*x^6 + 64*x^7 + ... - Michael Somos, Jul 07 2022
|
|
|
MAPLE
|
|
|
|
MATHEMATICA
|
Table[(n^3 + 5 n + 6)/6, {n, 0, 50}] (* Harvey P. Dale, Jan 19 2013 *)
LinearRecurrence[{4, -6, 4, -1}, {1, 2, 4, 8}, 50] (* Harvey P. Dale, Jan 19 2013 *)
|
|
|
PROG
|
(PARI) Vec((1-2*x+2*x^2)/((1-x)^4) + O(x^100)) \\ Altug Alkan, Oct 16 2015
(Python)
def A000125_gen(): # generator of terms
a, b, c = 1, 1, 1
while True:
yield a
a, b, c = a+b, b+c, c+1
|
|
|
CROSSREFS
|
Cf. A000124, A003600, A005408, A016813, A086514, A058331, A002522, A161701-A161705, A000127, A161706-A161708, A080856, A161710-A161713, A161715, A006261, A063865, A051601, A077043, A002620, A123596.
|
|
|
KEYWORD
|
nonn,easy,nice
|
|
|
AUTHOR
|
|
|
|
EXTENSIONS
|
|
|
|
STATUS
|
approved
|
| |
|
|
| |
|
|
A000127
|
|
Maximal number of regions obtained by joining n points around a circle by straight lines. Also number of regions in 4-space formed by n-1 hyperplanes.
(Formerly M1119 N0427)
|
|
+0
50
|
|
|
|
1, 2, 4, 8, 16, 31, 57, 99, 163, 256, 386, 562, 794, 1093, 1471, 1941, 2517, 3214, 4048, 5036, 6196, 7547, 9109, 10903, 12951, 15276, 17902, 20854, 24158, 27841, 31931, 36457, 41449, 46938, 52956, 59536, 66712, 74519, 82993, 92171, 102091, 112792, 124314, 136698
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,2
|
|
|
COMMENTS
|
a(n) is the sum of the first five terms in the n-th row of Pascal's triangle. - Geoffrey Critzer, Jan 18 2009
Equals binomial transform of [1, 1, 1, 1, 1, 0, 0, 0, ...]. - Gary W. Adamson, Mar 02 2010
As a(n) = 2^(n-1) for n = 1..5, it is misleading to believe that a(n) = 2^(n-1) for n > 5 (see Patrick Popescu-Pampu link); other curiosities: a(6) = 2^5 - 1 and a(10) = 2^8.
The sequence of the first differences is A000125, the sequence of the second differences is A000124, the sequence of the third differences is A000027 and the sequence of the fourth differences is the all 1's sequence A000012 (see J. H. Conway and R. K. Guy reference, p. 80). (End)
a(n) is the number of binary words of length n matching the regular expression 0*1*0*1*0*. A000124 and A000125 count binary words of the form 0*1*0* and 1*0*1*0*, respectively. - Manfred Scheucher, Jun 22 2023
|
|
|
REFERENCES
|
R. B. Banks, Slicing Pizzas, Racing Turtles and Further Adventures in Applied Mathematics, Princeton Univ. Press, 1999. See p. 28.
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 72, Problem 2.
J. H. Conway and R. K. Guy, The Book of Numbers, Copernicus Press, NY, 1996, Chap. 3.
J. H. Conway and R. K. Guy, Le Livre des Nombres, Eyrolles, 1998, p. 80.
J.-M. De Koninck & A. Mercier, 1001 Problèmes en Théorie Classique Des Nombres, Problem 33 pp. 18; 128 Ellipses Paris 2004.
A. Deledicq and D. Missenard, A La Recherche des Régions Perdues, Math. & Malices, No. 22 Summer 1995 issue pp. 22-3 ACL-Editions Paris.
M. Gardner, Mathematical Circus, pp. 177; 180-1 Alfred A. Knopf NY 1979.
M. Gardner, The Colossal Book of Mathematics, 2001, p. 561.
James Gleick, Faster, Vintage Books, NY, 2000 (see pp. 259-261).
M. de Guzman, Aventures Mathématiques, Prob. B pp. 115-120 PPUR Lausanne 1990.
Ross Honsberger; Mathematical Gems I, Chap. 9.
Ross Honsberger; Mathematical Morsels, Chap. 3.
Jeux Mathématiques et Logiques, Vol. 3 pp. 12; 51 Prob. 14 FFJM-SERMAP Paris 1988.
J. N. Kapur, Reflections of a Mathematician, Chap.36, pp. 337-343, Arya Book Depot, New Delhi 1996.
C. D. Miller, V. E. Heeren, J. Hornsby, M. L. Morrow and J. Van Newenhizen, Mathematical Ideas, Tenth Edition, Pearson, Addison-Wesley, Boston, 2003, Cptr 1, 'The Art of Problem Solving, page 6.
I. Niven, Mathematics of Choice, pp. 158; 195 Prob. 40 NML 15 MAA 1965.
C. S. Ogilvy, Tomorrow's Math, pp. 144-6 OUP 1972.
Alfred S. Posamentier & Ingmar Lehmann, The (Fabulous) Fibonacci Numbers, Prometheus Books, NY, 2007, page 81-87.
A. M. Robert, A Course in p-adic Analysis, Springer-Verlag, 2000; p. 213.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
|
|
|
LINKS
|
Patrick Popescu-Pampu, Démarrage trompeur, Images des Mathématiques, CNRS, 2017, rediffusion 2021 (in French).
|
|
|
FORMULA
|
a(n) = C(n-1, 4) + C(n-1, 3) + ... + C(n-1, 0) = A055795(n) + 1 = C(n, 4) + C(n-1, 2) + n.
a(n) = Sum_{k=0..2} C(n, 2k). - Joel Sanderi (sanderi(AT)itstud.chalmers.se), Sep 08 2004
a(n) = (n^4 - 6*n^3 + 23*n^2 - 18*n + 24)/24.
G.f.: (1 - 3*x + 4*x^2 - 2*x^3 + x^4)/(1-x)^5. (for offset 0) - Simon Plouffe in his 1992 dissertation
E.g.f.: (1 + x + x^2/2 + x^3/6 + x^4/24)*exp(x) (for offset 0). [Typos corrected by Juan M. Marquez, Jan 24 2011]
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5), n > 4. - Harvey P. Dale, Aug 24 2011
For n > 2, a(n) = n + 1 + sum_{i=2..(n-2)}sum_{j=1..(n-i)}(1+(i-1)(j-1)). - Alec Jones, Nov 17 2019
|
|
|
EXAMPLE
|
a(7)=99 because the first five terms in the 7th row of Pascal's triangle are 1 + 7 + 21 + 35 + 35 = 99. - Geoffrey Critzer, Jan 18 2009
G.f. = x + 2*x^2 + 4*x^3 + 8*x^4 + 16*x^5 + 31*x^6 + 57*x^7 + 99*x^8 + 163*x^9 + ...
|
|
|
MAPLE
|
A000127 := n->(n^4 - 6*n^3 + 23*n^2 - 18*n + 24)/24;
with (combstruct):ZL:=[S, {S=Sequence(U, card<r), U=Set(Z, card>=1)}, unlabeled]: seq(count(subs(r=6, ZL), size=m), m=0..41); # Zerinvary Lajos, Mar 08 2008
|
|
|
MATHEMATICA
|
f[n_] := Sum[Binomial[n, i], {i, 0, 4}]; Table[f@n, {n, 0, 40}] (* Robert G. Wilson v, Jun 29 2007 *)
Total/@Table[Binomial[n-1, k], {n, 50}, {k, 0, 4}] (* or *) LinearRecurrence[ {5, -10, 10, -5, 1}, {1, 2, 4, 8, 16}, 50] (* Harvey P. Dale, Aug 24 2011 *)
Table[(n^4 - 6 n^3 + 23 n^2 - 18 n + 24) / 24, {n, 100}] (* Vincenzo Librandi, Feb 16 2015 *)
a[ n_] := Binomial[n, 4] + Binomial[n, 2] + 1; (* Michael Somos, Dec 23 2017 *)
|
|
|
PROG
|
(Haskell)
(Magma) [(n^4-6*n^3+23*n^2-18*n+24)/24: n in [1..50]]; // Vincenzo Librandi, Feb 16 2015
(PARI) {a(n) = binomial(n, 4) + binomial(n, 2) + 1}; /* Michael Somos, Dec 23 2017 */
(Python)
|
|
|
CROSSREFS
|
Cf. A000012, A000027, A000124, A000125, A002522, A005408, A016813, A086514, A058331, A161701, A161702, A161703, A161704, A161706, A161707, A161708, A161710, A080856, A161711, A161712, A161713, A161715, A006261, A007318, A008859-A008863, A219531, A223718.
|
|
|
KEYWORD
|
nonn,easy,nice
|
|
|
AUTHOR
|
|
|
|
EXTENSIONS
|
Formula corrected and additional references from torsten.sillke(AT)lhsystems.com
Additional correction from Jonas Paulson (jonasso(AT)sdf.lonestar.org), Oct 30 2003
|
|
|
STATUS
|
approved
|
| |
|
|
| |
|
|
A000384
|
|
Hexagonal numbers: a(n) = n*(2*n-1).
(Formerly M4108 N1705)
|
|
+0
441
|
|
|
|
0, 1, 6, 15, 28, 45, 66, 91, 120, 153, 190, 231, 276, 325, 378, 435, 496, 561, 630, 703, 780, 861, 946, 1035, 1128, 1225, 1326, 1431, 1540, 1653, 1770, 1891, 2016, 2145, 2278, 2415, 2556, 2701, 2850, 3003, 3160, 3321, 3486, 3655, 3828, 4005, 4186, 4371, 4560
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,3
|
|
|
COMMENTS
|
Number of edges in the join of two complete graphs, each of order n, K_n * K_n. - Roberto E. Martinez II, Jan 07 2002
The power series expansion of the entropy function H(x) = (1+x)log(1+x) + (1-x)log(1-x) has 1/a_i as the coefficient of x^(2i) (the odd terms being zero). - Tommaso Toffoli (tt(AT)bu.edu), May 06 2002
Sequence also gives the greatest semiperimeter of primitive Pythagorean triangles having inradius n-1. Such a triangle has consecutive longer sides, with short leg 2n-1, hypotenuse a(n) - (n-1) = A001844(n), and area (n-1)*a(n) = 6*A000330(n-1). - Lekraj Beedassy, Apr 23 2003
More generally, if p1 and p2 are two arbitrarily chosen distinct primes then a(n) is the number of divisors of (p1^2*p2)^(n-1) or equivalently of any member of A054753^(n-1). - Ant King, Aug 29 2011
Number of standard tableaux of shape (2n-1,1,1) (n>=1). - Emeric Deutsch, May 30 2004
It is well known that for n>0, A014105(n) [0,3,10,21,...] is the first of 2n+1 consecutive integers such that the sum of the squares of the first n+1 such integers is equal to the sum of the squares of the last n; e.g., 10^2 + 11^2 + 12^2 = 13^2 + 14^2.
Less well known is that for n>1, a(n) [0,1,6,15,28,...] is the first of 2n consecutive integers such that sum of the squares of the first n such integers is equal to the sum of the squares of the last n-1 plus n^2; e.g., 15^2 + 16^2 + 17^2 = 19^2 + 20^2 + 3^2. - Charlie Marion, Dec 16 2006
Sequence found by reading the line from 0, in the direction 0, 6, ... and the line from 1, in the direction 1, 15, ..., in the square spiral whose vertices are the generalized hexagonal numbers A000217. - Omar E. Pol, Jan 09 2009
Let Hex(n)=hexagonal number, T(n)=triangular number, then Hex(n)=T(n)+3*T(n-1). - Vincenzo Librandi, Nov 10 2010
For n>=1, 1/a(n) = Sum_{k=0..2*n-1} ((-1)^(k+1)*binomial(2*n-1,k)*binomial(2*n-1+k,k)*H(k)/(k+1)) with H(k) harmonic number of order k.
The number of possible distinct colorings of any 2 colors chosen from n colors of a square divided into quadrants. - Paul Cleary, Dec 21 2010
For n>0, a(n-1) is the number of triples (w,x,y) with all terms in {0,...,n} and max(|w-x|,|x-y|) = |w-y|. - Clark Kimberling, Jun 12 2012
a(n) is the number of positions of one domino in an even pyramidal board with base 2n. - César Eliud Lozada, Sep 26 2012
Let a triangle have T(0,0) = 0 and T(r,c) = |r^2 - c^2|. The sum of the differences of the terms in row(n) and row(n-1) is a(n). - J. M. Bergot, Jun 17 2013
With T_(i+1,i)=a(i+1) and all other elements of the lower triangular matrix T zero, T is the infinitesimal generator for A176230, analogous to A132440 for the Pascal matrix. - Tom Copeland, Dec 11 2013
a(n) is the number of length 2n binary sequences that have exactly two 1's. a(2) = 6 because we have: {0,0,1,1}, {0,1,0,1}, {0,1,1,0}, {1,0,0,1}, {1,0,1,0}, {1,1,0,0}. The ordinary generating function with interpolated zeros is: (x^2 + 3*x^4)/(1-x^2)^3. - Geoffrey Critzer, Jan 02 2014
For n > 0, a(n) is the largest integer k such that k^2 + n^2 is a multiple of k + n. More generally, for m > 0 and n > 0, the largest integer k such that k^(2*m) + n^(2*m) is a multiple of k + n is given by k = 2*n^(2*m) - n. - Derek Orr, Sep 04 2014
Binomial transform of (0, 1, 4, 0, 0, 0, ...) and second partial sum of (0, 1, 4, 4, 4, ...). - Gary W. Adamson, Oct 05 2015
a(n) also gives the dimension of the simple Lie algebras D_n, for n >= 4. - Wolfdieter Lang, Oct 21 2015
For n > 0, a(n) equals the number of compositions of n+11 into n parts avoiding parts 2, 3, 4. - Milan Janjic, Jan 07 2016
Also the number of minimum dominating sets and maximal irredundant sets in the n-cocktail party graph. - Eric W. Weisstein, Jun 29 and Aug 17 2017
As Beedassy's formula shows, this Hexagonal number sequence is the odd bisection of the Triangle number sequence. Both of these sequences are figurative number sequences. For A000384, a(n) can be found by multiplying its triangle number by its hexagonal number. For example let's use the number 153. 153 is said to be the 17th triangle number but is also said to be the 9th hexagonal number. Triangle(17) Hexagonal(9). 17*9=153. Because the Hexagonal number sequence is a subset of the Triangle number sequence, the Hexagonal number sequence will always have both a triangle number and a hexagonal number. n* (2*n-1) because (2*n-1) renders the triangle number. - Bruce J. Nicholson, Nov 05 2017
Also numbers k with the property that in the symmetric representation of sigma(k) the smallest Dyck path has a central valley and the largest Dyck path has a central peak, n >= 1. Thus all hexagonal numbers > 0 have middle divisors. (Cf. A237593.) - Omar E. Pol, Aug 28 2018
Consider all Pythagorean triples (X, Y, Z=Y+1) ordered by increasing Z: a(n+1) gives the semiperimeter of related triangles; A005408, A046092 and A001844 give the X, Y and Z values. - Ralf Steiner, Feb 25 2020
It appears that these are the numbers k with the property that the smallest subpart in the symmetric representation of sigma(k) is 1. - Omar E. Pol, Aug 28 2021
The n-th hexagonal number equals the sum of the n consecutive integers with the same parity starting at 2*n-1; for example, 1, 2+4, 3+5+7, 4+6+8+10, etc. In general, the n-th 2k-gonal number is the sum of the n consecutive integers with the same parity starting at (k-2)*n - (k-3). When k = 1 and 2, this result generates the positive integers, A000027, and the squares, A000290, respectively. - Charlie Marion, Mar 02 2022
Conjecture: For n>0, min{k such that there exist subsets A,B of {0,1,2,...,a(n)} such that |A|=|B|=k and A+B={0,1,2,...,2*a(n)}} = 2*n. - Michael Chu, Mar 09 2022
|
|
|
REFERENCES
|
Albert H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 189.
Louis Comtet, Advanced Combinatorics, Reidel, 1974, pp. 77-78. (In the integral formula on p. 77 a left bracket is missing for the cosine argument.)
E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 6.
L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 2, p. 2.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
|
|
|
LINKS
|
Jonathan M. Borwein, Dirk Nuyens, Armin Straub and James Wan, Random Walk Integrals, The Ramanujan Journal, October 2011, 26:109. DOI: 10.1007/s11139-011-9325-y.
Michel Waldschmidt, Continued fractions, Ecole de recherche CIMPA-Oujda, Théorie des Nombres et ses Applications, 18 - 29 mai 2015: Oujda (Maroc).
|
|
|
FORMULA
|
G.f.: x*(1+3*x)/(1-x)^3. - Simon Plouffe in his 1992 dissertation, dropping the initial zero
a(n) = right term of M^n * [1,0,0], where M = the 3 X 3 matrix [1,0,0; 1,1,0; 1,4,1]. Example: a(5) = 45 since M^5 *[1,0,0] = [1,5,45]. - Gary W. Adamson, Dec 24 2006
Starting with offset 1, = binomial transform of [1, 5, 4, 0, 0, 0, ...]. Also, A004736 * [1, 4, 4, 4, ...]. - Gary W. Adamson, Oct 25 2007
a(n)^2 + (a(n)+1)^2 + ... + (a(n)+n-1)^2 = (a(n)+n+1)^2 + ... + (a(n)+2n-1)^2 + n^2; e.g., 6^2 + 7^2 = 9^2 + 2^2; 28^2 + 29^2 + 30^2 + 31^2 = 33^2 + 34^2 + 35^2 + 4^2. - Charlie Marion, Nov 10 2007
a(n) = binomial(n+1,2) + 3*binomial(n,2).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), a(0)=0, a(1)=1, a(2)=6. - Jaume Oliver Lafont, Dec 02 2008
a(n) = 2*a(n-1) - a(n-2) + 4. - Ant King, Aug 26 2011
a(n+1) = trinomial(2*n+1, 2) = trinomial(2*n+1, 4*n), for n >= 0, with the trinomial irregular triangle A027907. a(n+1) = (n+1)*(2*n+1) = (1/Pi)*Integral_{x=0..2} (1/sqrt(4 - x^2))*(x^2 - 1)^(2*n+1)*R(4*n-2, x) with the R polynomial coefficients given in A127672. [Comtet, p. 77, the integral formula for q=3, n -> 2*n+1, k = 2, rewritten with x = 2*cos(phi)]. - Wolfdieter Lang, Apr 19 2018
Product_{n>=2} (1 - 1/a(n)) = 2/3. - Amiram Eldar, Jan 21 2021
a(n) = floor(Sum_{k=(n-1)^2..n^2} sqrt(k)), for n >= 1. - Amrit Awasthi, Jun 13 2021
|
|
|
MAPLE
|
|
|
|
MATHEMATICA
|
LinearRecurrence[{3, -3, 1}, {0, 1, 6}, 50] (* Harvey P. Dale, Sep 10 2015 *)
Join[{0}, Accumulate[Range[1, 312, 4]]] (* Harvey P. Dale, Mar 26 2016 *)
(* For Mathematica 10.4+ *) Table[PolygonalNumber[RegularPolygon[6], n], {n, 0, 48}] (* Arkadiusz Wesolowski, Aug 27 2016 *)
CoefficientList[Series[x*(1 + 3*x)/(1 - x)^3 , {x, 0, 100}], x] (* Stefano Spezia, Sep 02 2018 *)
|
|
|
PROG
|
(PARI) a(n)=n*(2*n-1)
(PARI) a(n) = binomial(2*n, 2) \\ Altug Alkan, Oct 06 2015
(Haskell)
a000384 n = n * (2 * n - 1)
a000384_list = scanl (+) 0 a016813_list
(Python) # Intended to compute the initial segment of the sequence, not isolated terms.
def aList():
x, y = 1, 1
yield 0
while True:
yield x
x, y = x + y + 4, y + 4
|
|
|
CROSSREFS
|
a(n)= A093561(n+1, 2), (4, 1)-Pascal column.
Cf. A002939 (twice a(n): sums of Pythagorean triples (X, Y, Z=Y+1)).
|
|
|
KEYWORD
|
nonn,easy,nice
|
|
|
AUTHOR
|
|
|
|
EXTENSIONS
|
|
|
|
STATUS
|
approved
|
| |
|
|
| |
|
|
A000435
|
|
Normalized total height of all nodes in all rooted trees with n labeled nodes.
(Formerly M4558 N1940)
|
|
+0
21
|
|
|
|
0, 1, 8, 78, 944, 13800, 237432, 4708144, 105822432, 2660215680, 73983185000, 2255828154624, 74841555118992, 2684366717713408, 103512489775594200, 4270718991667353600, 187728592242564421568, 8759085548690928992256, 432357188322752488126152, 22510748754252398927872000
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,3
|
|
|
COMMENTS
|
This is the sequence that started it all: the first sequence in the database!
The height h(V) of a node V in a rooted tree is its distance from the root. a(n) = Sum_{all nodes V in all n^(n-1) rooted trees on n nodes} h(V)/n.
In the trees which have [0, n-1] = (0, 1, ..., n-1) as their ordered set of nodes, the number of nodes at distance i from node 0 is f(n,i) = (n-1)...(n-i)(i+1)n^(j-1), 0 <= i < n-1, i+j = n-1 (and f(n,n-1) = (n-1)!): (n-1)...(n-i) counts the words coding the paths of length i from any node to 0, n^(j-1) counts the Pruefer codes of the rest, words build by iterated deletion of the greater node of degree 1 ... except the last one, (i+1), necessary pointing at the path. If g(n,i) = (n-1)...(n-i)n^j, i+j = n-1, f(n,i) = g(n,i) - g(n,i+1), g(n,i) = Sum_{k>=i} f(n,k), the sequence is Sum_{i=1..n-1} g(n,i). - Claude Lenormand (claude.lenormand(AT)free.fr), Jan 26 2001
If one randomly selects one ball from an urn containing n different balls, with replacement, until exactly one ball has been selected twice, the probability that this ball was also the second ball to be selected once is a(n)/n^n. See also A001865. - Matthew Vandermast, Jun 15 2004
a(n) is the number of connected endofunctions with no fixed points. - Geoffrey Critzer, Dec 13 2011
a(n) is the number of weakly connected simple digraphs on n labeled nodes where every node has out-degree 1. A digraph where all out-degrees are 1 can be called a functional digraph due to the correspondence with endofunctions. - Andrew Howroyd, Feb 06 2024
|
|
|
REFERENCES
|
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
|
|
|
LINKS
|
A. Rényi and G. Szekeres, On the height of trees, Journal of the Australian Mathematical Society 7.04 (1967): 497-507. See (4.7).
N. J. A. Sloane, Page from 1964 notebook showing start of OEIS [includes A000027, A000217, A000292, A000332, A000389, A000579, A000110, A007318, A000058, A000215, A000289, A000324, A234953 (= A001854(n)/n), A000435, A000169, A000142, A000272, A000312, A000111]
|
|
|
FORMULA
|
a(n) = (n-1)! * Sum_{k=0..n-2} n^k/k!.
|
|
|
EXAMPLE
|
For n = 3 there are 3^2 = 9 rooted labeled trees on 3 nodes, namely (with o denoting a node, O the root node):
o
|
o o o
| \ /
O O
The first can be labeled in 6 ways and contains nodes at heights 1 and 2 above the root, so contributes 6*(1+2) = 18 to the total; the second can be labeled in 3 ways and contains 2 nodes at height 1 above the root, so contributes 3*2=6 to the total, giving 24 in all. Dividing by 3 we get a(3) = 24/3 = 8.
For n = 4 there are 4^3 = 64 rooted labeled trees on 4 nodes, namely (with o denoting a node, O the root node):
o
|
o o o o
| | \ /
o o o o o o o
| \ / | \|/
O O O O
(1) (2) (3) (4)
Tree (1) can be labeled in 24 ways and contains nodes at heights 1, 2, 3 above the root, so contributes 24*(1+2+3) = 144 to the total;
tree (2) can be labeled in 24 ways and contains nodes at heights 1, 1, 2 above the root, so contributes 24*(1+1+2) = 96 to the total;
tree (3) can be labeled in 12 ways and contains nodes at heights 1, 2, 2 above the root, so contributes 12*(1+2+2) = 60 to the total;
tree (4) can be labeled in 4 ways and contains nodes at heights 1, 1, 1 above the root, so contributes 4*(1+1+1) = 12 to the total;
giving 312 in all. Dividing by 4 we get a(4) = 312/4 = 78.
|
|
|
MAPLE
|
A000435 := n-> (n-1)!*add (n^k/k!, k=0..n-2);
seq(simplify((n-1)*GAMMA(n-1, n)*exp(n)), n=1..20); # Vladeta Jovovic, Jul 21 2005
|
|
|
MATHEMATICA
|
f[n_] := (n - 1)! Sum [n^k/k!, {k, 0, n - 2}]; Array[f, 18] (* Robert G. Wilson v, Aug 10 2010 *)
nx = 18; Rest[ Range[0, nx]! CoefficientList[ Series[ LambertW[-x] - Log[1 + LambertW[-x]], {x, 0, nx}], x]] (* Robert G. Wilson v, Apr 13 2013 *)
|
|
|
PROG
|
(PARI) x='x+O('x^30); concat(0, Vec(serlaplace(lambertw(-x)-log(1+lambertw(-x))))) \\ Altug Alkan, Sep 05 2018
(Python)
from math import comb
def A000435(n): return ((sum(comb(n, k)*(n-k)**(n-k)*k**k for k in range(1, (n+1>>1)))<<1) + (0 if n&1 else comb(n, m:=n>>1)*m**n))//n # Chai Wah Wu, Apr 25-26 2023
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn,easy,nice
|
|
|
AUTHOR
|
|
|
|
EXTENSIONS
|
|
|
|
STATUS
|
approved
|
| |
|
|
| |
|
|
A000578
|
|
The cubes: a(n) = n^3.
(Formerly M4499 N1905)
|
|
+0
1018
|
|
|
|
0, 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728, 2197, 2744, 3375, 4096, 4913, 5832, 6859, 8000, 9261, 10648, 12167, 13824, 15625, 17576, 19683, 21952, 24389, 27000, 29791, 32768, 35937, 39304, 42875, 46656, 50653, 54872, 59319, 64000, 68921, 74088, 79507
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,3
|
|
|
COMMENTS
|
a(n) is the sum of the next n odd numbers; i.e., group the odd numbers so that the n-th group contains n elements like this: (1), (3, 5), (7, 9, 11), (13, 15, 17, 19), (21, 23, 25, 27, 29), ...; then each group sum = n^3 = a(n). Also the median of each group = n^2 = mean. As the sum of first n odd numbers is n^2 this gives another proof of the fact that the n-th partial sum = (n(n + 1)/2)^2. - Amarnath Murthy, Sep 14 2002
Total number of triangles resulting from criss-crossing cevians within a triangle so that two of its sides are each n-partitioned. - Lekraj Beedassy, Jun 02 2004
Also structured triakis tetrahedral numbers (vertex structure 7) (cf. A100175 = alternate vertex); structured tetragonal prism numbers (vertex structure 7) (cf. A100177 = structured prisms); structured hexagonal diamond numbers (vertex structure 7) (cf. A100178 = alternate vertex; A000447 = structured diamonds); and structured trigonal anti-diamond numbers (vertex structure 7) (cf. A100188 = structured anti-diamonds). Cf. A100145 for more on structured polyhedral numbers. - James A. Record (james.record(AT)gmail.com), Nov 07 2004
Schlaefli symbol for this polyhedron: {4, 3}.
Least multiple of n such that every partial sum is a square. - Amarnath Murthy, Sep 09 2005
Draw a regular hexagon. Construct points on each side of the hexagon such that these points divide each side into equally sized segments (i.e., a midpoint on each side or two points on each side placed to divide each side into three equally sized segments or so on), do the same construction for every side of the hexagon so that each side is equally divided in the same way. Connect all such points to each other with lines that are parallel to at least one side of the polygon. The result is a triangular tiling of the hexagon and the creation of a number of smaller regular hexagons. The equation gives the total number of regular hexagons found where n = the number of points drawn + 1. For example, if 1 point is drawn on each side then n = 1 + 1 = 2 and a(n) = 2^3 = 8 so there are 8 regular hexagons in total. If 2 points are drawn on each side then n = 2 + 1 = 3 and a(n) = 3^3 = 27 so there are 27 regular hexagons in total. - Noah Priluck (npriluck(AT)gmail.com), May 02 2007
The solutions of the Diophantine equation: (X/Y)^2 - X*Y = 0 are of the form: (n^3, n) with n >= 1. The solutions of the Diophantine equation: (m^2)*(X/Y)^2k - XY = 0 are of the form: (m*n^(2k + 1), m*n^(2k - 1)) with m >= 1, k >= 1 and n >= 1. The solutions of the Diophantine equation: (m^2)*(X/Y)^(2k + 1) - XY = 0 are of the form: (m*n^(k + 1), m*n^k) with m >= 1, k >= 1 and n >= 1. - Mohamed Bouhamida, Oct 04 2007
Except for the first two terms, the sequence corresponds to the Wiener indices of C_{2n} i.e., the cycle on 2n vertices (n > 1). - K.V.Iyer, Mar 16 2009
Totally multiplicative sequence with a(p) = p^3 for prime p. - Jaroslav Krizek, Nov 01 2009
One of the 5 Platonic polyhedral (tetrahedral, cube, octahedral, dodecahedral and icosahedral) numbers (cf. A053012). - Daniel Forgues, May 14 2010
Numbers n for which order of torsion subgroup t of the elliptic curve y^2 = x^3 - n is t = 2. - Artur Jasinski, Jun 30 2010
The sequence with the lengths of the Pisano periods mod k is 1, 2, 3, 4, 5, 6, 7, 8, 3, 10, 11, 12, 13, 14, 15, 16, 17, 6, 19, 20, ... for k >= 1, apparently multiplicative and derived from A000027 by dividing every ninth term through 3. Cubic variant of A186646. - R. J. Mathar, Mar 10 2011
The number of atoms in a bcc (body-centered cubic) rhombic hexahedron with n atoms along one edge is n^3 (T. P. Martin, Shells of atoms, eq. (8)). - Brigitte Stepanov, Jul 02 2011
Twice the area of a triangle with vertices at (0, 0), (t(n - 1), t(n)), and (t(n), t(n - 1)), where t = A000217 are triangular numbers. - J. M. Bergot, Jun 25 2013
For n > 2, a(n) = twice the area of a triangle with vertices at points (binomial(n,3),binomial(n+2,3)), (binomial(n+1,3),binomial(n+1,3)), and (binomial(n+2,3),binomial(n,3)). - J. M. Bergot, Jun 14 2014
Determinants of the spiral knots S(4,k,(1,1,-1)). a(k) = det(S(4,k,(1,1,-1))). - Ryan Stees, Dec 14 2014
One of the oldest-known examples of this sequence is shown in the Senkereh tablet, BM 92698, which displays the first 32 terms in cuneiform. - Charles R Greathouse IV, Jan 21 2015
We construct a number triangle from the integers 1, 2, 3, ... 2*n-1 as follows. The first column contains all the integers 1, 2, 3, ... 2*n-1. Each succeeding column is the same as the previous column but without the first and last items. The last column contains only n. The sum of all the numbers in the triangle is n^3.
Here is the example for n = 4, where 1 + 2*2 + 3*3 + 4*4 + 3*5 + 2*6 + 7 = 64 = a(4):
1
2 2
3 3 3
4 4 4 4
5 5 5
6 6
7
(End)
For n > 0, a(n) is the number of compositions of n+11 into n parts avoiding parts 2 and 3. - Milan Janjic, Jan 07 2016
Does not satisfy Benford's law [Ross, 2012]. - N. J. A. Sloane, Feb 08 2017
Number of inequivalent face colorings of the cube using at most n colors such that each color appears at least twice. - David Nacin, Feb 22 2017
Consider A = {a,b,c} a set with three distinct members. The number of subsets of A is 8, including {a,b,c} and the empty set. The number of subsets from each of those 8 subsets is 27. If the number of such iterations is n, then the total number of subsets is a(n-1). - Gregory L. Simay, Jul 27 2018
By Fermat's Last Theorem, these are the integers of the form x^k with the least possible value of k such that x^k = y^k + z^k never has a solution in positive integers x, y, z for that k. - Felix Fröhlich, Jul 27 2018
|
|
|
REFERENCES
|
Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966. See p. 191.
R. L. Graham, D. E. Knuth, and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 255; 2nd. ed., p. 269. Worpitzky's identity (6.37).
T. Aaron Gulliver, "Sequences from cubes of integers", International Mathematical Journal, 4 (2003), no. 5, 439 - 445. See http://www.m-hikari.com/z2003.html for information about this journal. [I expanded the reference to make this easier to find. - N. J. A. Sloane, Feb 18 2019]
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
D. Wells, You Are A Mathematician, pp. 238-241, Penguin Books 1995.
|
|
|
LINKS
|
N. Brothers, S. Evans, L. Taalman, L. Van Wyk, D. Witczak, and C. Yarnall, Spiral knots, Missouri J. of Math. Sci., 22 (2010).
M. DeLong, M. Russell, and J. Schrock, Colorability and determinants of T(m,n,r,s) twisted torus knots for n equiv. +/-1(mod m), Involve, Vol. 8 (2015), No. 3, 361-384.
T. P. Martin, Shells of atoms, Phys. Reports, 273 (1996), 199-241, eq. (8).
Ed Pegg, Jr., Sequence Pictures, Math Games column, Dec 08 2003 [Cached copy, with permission (pdf only)]
|
|
|
FORMULA
|
G.f.: x*(1+4*x+x^2)/(1-x)^4. - Simon Plouffe in his 1992 dissertation
a(n) = lcm(n, (n - 1)^2) - (n - 1)^2. E.g.: lcm(1, (1 - 1)^2) - (1 - 1)^2 = 0, lcm(2, (2 - 1)^2) - (2 - 1)^2 = 1, lcm(3, (3 - 1)^2) - (3 - 1)^2 = 8, ... - Mats Granvik, Sep 24 2007
Starting (1, 8, 27, 64, 125, ...), = binomial transform of [1, 7, 12, 6, 0, 0, 0, ...]. - Gary W. Adamson, Nov 21 2007
a(n) = binomial(n+2,3) + 4*binomial(n+1,3) + binomial(n,3). [Worpitzky's identity for cubes. See. e.g., Graham et al., eq. (6.37). - Wolfdieter Lang, Jul 17 2019]
a(n) = n + 6*binomial(n+1,3) = binomial(n,1)+6*binomial(n+1,3). - Ron Knott, Jun 10 2019
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 6. - Ant King Apr 29 2013
a(k) = det(S(4,k,(1,1,-1))) = k*b(k)^2, where b(1)=1, b(2)=2, b(k) = 2*b(k-1) - b(k-2) = b(2)*b(k-1) - b(k-2). - Ryan Stees, Dec 14 2014
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Jon Tavasanis, Feb 21 2016
a(n) = n + Sum_{j=0..n-1} Sum_{k=1..2} binomial(3,k)*j^(3-k). - Patrick J. McNab, Mar 28 2016
a(n) = n*binomial(n+1, 2) + 2*binomial(n+1, 3) + binomial(n,3). - Tony Foster III, Nov 14 2017
Sum_{n>=1} 1/a(n) = zeta(3) (A002117).
Sum_{n>=1} (-1)^(n+1)/a(n) = 3*zeta(3)/4 (A197070). (End)
Product_{n>=1} (1 + 1/a(n)) = cosh(sqrt(3)*Pi/2)/Pi.
Product_{n>=2} (1 - 1/a(n)) = cosh(sqrt(3)*Pi/2)/(3*Pi). (End)
|
|
|
EXAMPLE
|
For k=3, b(3) = 2 b(2) - b(1) = 4-1 = 3, so det(S(4,3,(1,1,-1))) = 3*3^2 = 27.
For n=3, a(3) = 3 + (3*0^2 + 3*0 + 3*1^2 + 3*1 + 3*2^2 + 3*2) = 27. - Patrick J. McNab, Mar 28 2016
|
|
|
MAPLE
|
isA000578 := proc(r)
local p;
if r = 0 or r =1 then
true;
else
for p in ifactors(r)[2] do
if op(2, p) mod 3 <> 0 then
return false;
end if;
end do:
true ;
end if;
|
|
|
MATHEMATICA
|
CoefficientList[Series[x (1 + 4 x + x^2)/(1 - x)^4, {x, 0, 45}], x] (* Vincenzo Librandi, Jul 05 2014 *)
Accumulate[Table[3n^2+3n+1, {n, 0, 20}]] (* or *) LinearRecurrence[{4, -6, 4, -1}, {1, 8, 27, 64}, 20](* Harvey P. Dale, Aug 18 2018 *)
|
|
|
PROG
|
(Haskell)
a000578 = (^ 3)
a000578_list = 0 : 1 : 8 : zipWith (+)
(map (+ 6) a000578_list)
(map (* 3) $ tail $ zipWith (-) (tail a000578_list) a000578_list)
(Magma) I:=[0, 1, 8, 27]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..45]]; // Vincenzo Librandi, Jul 05 2014
(Python)
A000578_list, m = [], [6, -6, 1, 0]
for _ in range(10**2):
for i in range(3):
|
|
|
CROSSREFS
|
(1/12)*t*(n^3-n)+n for t = 2, 4, 6, ... gives A004006, A006527, A006003, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.
|
|
|
KEYWORD
|
nonn,core,easy,nice,mult,changed
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
Search completed in 1.155 seconds
|
