Search: a003242 -id:a003242
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A241902
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Decimal expansion of a constant related to Carlitz compositions (A003242).
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+20
9
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1, 7, 5, 0, 2, 4, 1, 2, 9, 1, 7, 1, 8, 3, 0, 9, 0, 3, 1, 2, 4, 9, 7, 3, 8, 6, 2, 4, 6, 3, 9, 8, 1, 5, 8, 7, 8, 7, 7, 8, 2, 0, 5, 8, 1, 8, 1, 3, 8, 1, 5, 9, 0, 5, 6, 1, 3, 1, 6, 5, 8, 6, 1, 3, 1, 7, 5, 1, 9, 3, 5, 1, 6, 7, 1, 5, 2, 0, 6, 0, 5, 0, 7, 7, 7, 4, 3, 8, 8, 7, 5, 6, 5, 7, 0, 9, 2, 4, 7, 1, 4, 1, 0, 0, 1
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OFFSET
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1,2
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LINKS
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A. Knopfmacher and H. Prodinger, On Carlitz compositions, European Journal of Combinatorics, Vol. 19, 1998, pp. 579-589.
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FORMULA
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Equals lim n -> infinity A003242(n)^(1/n).
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EXAMPLE
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1.7502412917183090312497386246398158787782...
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MATHEMATICA
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RealDigits[r /. FindRoot[Exp[QPolyGamma[0, 1 + Pi*I/Log[r], r]] == r^(3/2)/(1-r), {r, 3/2}, WorkingPrecision -> 120], 10, 110][[1]] (* Vaclav Kotesovec, Jun 19 2023 *)
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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A373420
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Number of Carlitz compositions of n (see A003242) such that the first and last parts are equal.
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+20
0
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1, 1, 1, 1, 2, 3, 2, 7, 11, 17, 26, 54, 86, 155, 272, 464, 816, 1447, 2507, 4400, 7706, 13456, 23570, 41293, 72212, 126394, 221282, 387219, 677714, 1186311, 2076170, 3633761, 6360219, 11131698, 19483066, 34100455, 59683664, 104460655, 182832044, 319999739
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OFFSET
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0,5
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LINKS
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FORMULA
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G.f.: 1 + Sum_{i>0} (x^i)*(C(x)*(x^i) + x^i + 1)/(1+x^i)^2 where C(x) is the g.f. for A003242.
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EXAMPLE
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a(7) = 7 counts: (1,2,1,2,1), (1,2,3,1), (1,3,2,1), (1,5,1), (2,3,2), (3,1,3), and (7).
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PROG
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(PARI)
C_x(N) = {my(g=1/(1-sum(k=1, N, x^k/(1+x^k)))); g}
A_x(i, N) = {my( x='x+O('x^N), f=(x^i)*(C_x(N)*(x^i)+x^i+1)/(1+x^i)^2); f}
D_x(N) = {my( x='x+O('x^N), f=1+sum(i=1, N, A_x(i, N))); Vec(f)}
D_x(40)
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CROSSREFS
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KEYWORD
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nonn,easy,new
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AUTHOR
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STATUS
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approved
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A000975
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a(2n) = 2*a(2n-1), a(2n+1) = 2*a(2n)+1 (also a(n) is the n-th number without consecutive equal binary digits).
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+10
294
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0, 1, 2, 5, 10, 21, 42, 85, 170, 341, 682, 1365, 2730, 5461, 10922, 21845, 43690, 87381, 174762, 349525, 699050, 1398101, 2796202, 5592405, 11184810, 22369621, 44739242, 89478485, 178956970, 357913941, 715827882, 1431655765, 2863311530, 5726623061, 11453246122
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OFFSET
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0,3
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COMMENTS
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Might be called the "Lichtenberg sequence" after Georg Christoph Lichtenberg, who discussed it in 1769 in connection with the Chinese Rings puzzle (baguenaudier). - Andreas M. Hinz, Feb 15 2017
Number of steps to change from a binary string of n 0's to n 1's using a Gray code. - Jon Stadler (jstadler(AT)coastal.edu)
Popular puzzles such as Spin-Out and The Brain Puzzler are based on the Gray binary system and require a(n) steps to complete for some number n.
Conjecture: {a(n)} also gives all j for which A048702(j) = A000217(j); i.e., if we take the a(n)-th triangular number (a(n)^2 + a(n))/2 and multiply it by 3, we get a(n)-th even-length binary palindrome A048701(a(n)) concatenated from a(n) and its reverse. E.g., a(4) = 10, which is 1010 in binary; the tenth triangular number is 55, and 55*3 = 165 = 10100101 in binary. - Antti Karttunen, circa 1999. (This has been now proved by Paul K. Stockmeyer in his arXiv:1608.08245 paper.) - Antti Karttunen, Aug 31 2016
Number of ways to tie a tie of n or fewer half turns, excluding mirror images. Also number of walks of length n or less on a triangular lattice with the following restrictions; given l, r and c as the lattice axes. 1. All steps are taken in the positive axis direction. 2. No two consecutive steps are taken on the same axis. 3. All walks begin with l. 4. All walks end with rlc or lrc. - Bill Blewett, Dec 21 2000
a(n) is the minimal number of vertices to be selected in a vertex-cover of the balanced tree B_n. - Sen-peng Eu, Jun 15 2002
Equivalently, numbers m whose binary representation is effectively, for some number k, both the lazy Fibonacci and the Zeckendorf representation of k (in which case k = A022290(m)). - Peter Munn, Oct 06 2022
a(n+1) gives row sums of Riordan array (1/(1-x), x(1+2x)). - Paul Barry, Jul 18 2005
Total number of initial 01's in all binary words of length n+1. Example: a(3) = 5 because the binary words of length 4 that start with 01 are (01)00, (01)(01), (01)10 and (01)11 and the total number of initial 01's is 5 (shown between parentheses). a(n) = Sum_{k >= 0} k*A119440(n+1, k). - Emeric Deutsch, May 19 2006
In Norway we call the 10-ring puzzle "strikketoy" or "knitwear" (see link). It takes 682 moves to free the two parts. - Hans Isdahl, Jan 07 2008
For n > 1, let B_n = the complete binary tree with vertex set V where |V| = 2^n - 1. If VC is a minimum-size vertex cover of B_n, Sen-Peng Eu points out that a(n) = |VC|. It also follows that if IS = V\VC, a(n+1) = |IS|. - K.V.Iyer, Apr 13 2009
This is the sequence A(0, 1; 1, 2; 1) of the family of sequences [a,b:c,d:k] considered by G. Detlefs, and treated as A(a,b;c,d;k) in the W. Lang link given below. - Wolfdieter Lang, Oct 18 2010
1) Denote by {n, k} the number of permutations of 1, ..., n with the up-down index k (for definition, see comment in A203827). Then max_k{n, k} = {n, a(n)} = A000111(n).
2) a(n) is the minimal number > a(n-1) with the Hamming distance d_H(a(n-1), a(n)) = n. Thus this sequence is the Hamming analog of triangular numbers 0, 1, 3, 6, 10, ... (End)
Represented in binary form each term after the second one contains every previous term as a substring.
The terms a(2) = 2 and a(3) = 5 are the only primes. Proof: For even n we get a(n) = 2*(2^(2*n) - 1)/3, which shows that a(n) is even, too, and obviously a(n) > 2 for all even n > 2. For odd n we have a(n) = (2^(n+1) - 1)/3 = (2^((n+1)/2) - 1) * (2^((n+1)/2) + 1)/3. Evidently, at least one of these factors is divisible by 3, both are greater than 6, provided n > 3. Hence it follows that a(n) is composite for all odd n > 3.
Represented as a binary number, a(n+1) has exactly n prime substrings. Proof: Evidently, a(1) = 1_2 has zero and a(2) = 10_2 has 1 prime substring. Let n > 1. Written in binary, a(n+1) is 101010101...01 (if n + 1 is odd) and is 101010101...10 (if n + 1 is even) with n + 1 digits. Only the 2- and 3-digits substrings 10_2 (=2) and 101_2 (=5) are prime substrings. All the other substrings are nonprime since every substring is a previous term and all terms unequal to 2 and 5 are nonprime. For even n + 1, the number of prime substrings equal to 2 = 10_2 is (n+1)/2, and the number of prime substrings equal to 5 = 101_2 is (n-1)/2, makes a sum of n. For odd n + 1 we get n/2 for both, the number of 2's and 5's prime substrings, in any case, the sum is n.
(End)
Also the number of different 3-colorings for the vertices of all triangulated planar polygons on a base with n+2 vertices if the colors of the two base vertices are fixed. - Patrick Labarque, Feb 09 2013
a(n) is the number of length n binary words containing at least one 1 and ending in an even number (possibly zero) of 0's. a(3) = 5 because we have: 001, 011, 100, 101, 111. - Geoffrey Critzer, Dec 15 2013
a(n) is the number of permutations of length n+1 having exactly one descent such that the first element of the permutation is an even number. - Ran Pan, Apr 18 2015
a(n) is the sequence of the last row of the Hadamard matrix H(2^n) obtained via Sylvester's construction: H(2) = [1,1;1,-1], H(2^n) = H(2^(n-1))*H(2), where * is the Kronecker product. - William P. Orrick, Jun 28 2015
Decimal representation of the x-axis, from the origin to the right edge, of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 131", based on the 5-celled von Neumann neighborhood. See A279053 for references and links. - Robert Price, Dec 05 2016
Conjecture: a(n+1) is the number of compositions of n with two kinds of parts, n and n', where the order of the 1 and 1' does not matter. For n=2, a(3) = 5 compositions, enumerated as follows: 2; 2'; 1,1; 1',1 = 1',1; 1',1'. - Gregory L. Simay, Sep 02 2017
Conjecture proved by recognizing the appropriate g.f. is x/(1 - x)(1 - x)(1 - 2*x^2 - 2x^3 - ...) = x/(1 - 2*x - x^2 + 2x^3). - Gregory L. Simay, Sep 10 2017
a(2*n) = n times the string [10] in binary representation, a(2*n+1) = n times the string [10] followed with [1] in binary representation. Example: a(7) = 85 = (1010101) in binary, a(8) = 170 = (10101010) in binary. - Ctibor O. Zizka, Nov 06 2018
Except for 0, these are the positive integers whose binary expansion has cuts-resistance 1. For the operation of shortening all runs by 1, cuts-resistance is the number of applications required to reach an empty word. Cuts-resistance 2 is A329862. - Gus Wiseman, Nov 27 2019
Let s(k) be the length of the Collatz orbit of k, e.g. s(1) = 1, s(2) = 2, s(3) = 5. Then s(a(n)) = n+3 for n >= 3. Proof by induction: s(a(3)) = s(5) = 6 = 3+3. For odd n >= 5 we have s(a(n)) = s(4*a(n-2)+1) = s(12*a(n-2)+4)+1 = s(3*a(n-2)+1)+3 = s(a(n-2))+2 = (n-2)+3+2 = n+3, and for even n >= 4 this gives s(a(n)) = s(2*a(n-1)) = s(a(n-1))+1 = (n-1)+3+1 = n+3.
Conjecture: For n >= 3, a(n) is the second largest natural number whose Collatz orbit has length n+3. (End)
With offset 1 the sequence equals the numbers of 1's from n = 1 to 3, 3 to 7, 7 to 15, ...; of A035263; as shown below:
..1 3 7 15...
..1 0 1 1 1 0 1 0 1 0 1 1 1 0 1...
..1.....2...........5......................10...; a(n) = Sum_(k=1..2n-1)A035263(k)
.....1...........2.......................5...; as to zeros.
..1's in the Tower of Hanoi game represent CW moves On disks in the pattern:
..0, 1, 2, 0, 1, 2, ... whereas even numbered disks move in the pattern:
..0, 2, 1, 0, 2, 1, ... (End)
Except for 0, numbers that are repunits in Gray-code representation (A014550). - Amiram Eldar, May 21 2021
Also the number of nonempty subsets of {1..n} with integer median, where the median of a multiset is the middle part in the odd-length case, and the average of the two middle parts in the even-length case. For example, the a(1) = 1 through a(4) = 10 subsets are:
{1} {1} {1} {1}
{2} {2} {2}
{3} {3}
{1,3} {4}
{1,2,3} {1,3}
{2,4}
{1,2,3}
{1,2,4}
{1,3,4}
{2,3,4}
The complement is counted by A005578.
For mean instead of median we have A051293, counting empty sets A327475.
(End)
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REFERENCES
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Thomas Fink and Yong Mao, The 85 Ways to Tie a Tie, Broadway Books, New York (1999), p. 138.
Clifford A. Pickover, The Math Book, From Pythagoras to the 57th Dimension, 250 Milestones in the History of Mathematics, Sterling Publ., NY, 2009.
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LINKS
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FORMULA
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a(n) = ceiling(2*(2^n-1)/3).
Alternating sum transform (PSumSIGN) of {2^n - 1} (A000225).
a(n) = a(n-1) + 2*a(n-2) + 1.
a(n) = 2*2^n/3 - 1/2 - (-1)^n/6.
a(n) = n + 2*Sum_{k = 1..n-2} a(k).
G.f.: x/((1+x)*(1-x)*(1-2*x)) = x/(1-2*x-x^2+2*x^3). - Paul Barry, Feb 11 2003
a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3). - Paul Barry, Feb 11 2003
a(n) = Sum_{k = 0..floor((n-1)/2)} 2^(n-2*k-1). - Paul Barry, Nov 11 2003
a(n+1) = Sum_{k=0..floor(n/2)} 2^(n-2*k); a(n+1) = Sum_{k = 0..n} Sum_{j = 0..k} (-1)^(j+k)*2^j. - Paul Barry, Nov 12 2003
(-1)^(n+1)*a(n) = Sum_{i = 0..n} Sum_{k = 1..i} k!*k* Stirling2(i, k)*(-1)^(k-1) = (1/3)*(-2)^(n+1)-(1/6)(3*(-1)^(n+1)-1). - Mario Catalani (mario.catalani(AT)unito.it), Dec 22 2003
a(n+1) = (n!/3)*Sum_{i - (-1)^i + j = n, i = 0..n, j = 0..n} 1/(i - (-1)^i)!/j!. - Benoit Cloitre, May 24 2004
a(n) = Sum_{k = 0..n} 2^(n-k-1)*(1-(-1)^k), row sums of A130125. - Paul Barry, Jul 28 2004
a(n) = Sum_{k = 0..n} binomial(k, n-k+1)2^(n-k); a(n) = Sum_{k = 0..floor(n/2)} binomial(n-k, k+1)2^k. - Paul Barry, Oct 07 2004
a(n) = floor(2^(n+1)/3) = ceiling(2^(n+1)/3) - 1 = A005578(n+1) - 1. - Paul Barry, Oct 08 2005
Convolution of "Number of fixed points in all 231-avoiding involutions in S_n." (A059570) with "1-n" (A024000), treating the result as if offset was 0. - Graeme McRae, Jul 12 2006
Starting (1, 2, 5, 10, 21, 42, ...), these are the row sums of triangle A135228. - Gary W. Adamson, Nov 23 2007
Let T = the 3 X 3 matrix [1,1,0; 1,0,1; 0,1,1]. Then T^n * [1,0,0] = [A005578(n), A001045(n), a(n-1)]. - Gary W. Adamson, Dec 25 2007
If we define f(m,j,x) = Sum_{k=j..m} binomial(m,k)*stirling2(k,j)*x^(m-k) then a(n-3) = (-1)^(n-1)*f(n,3,-2), (n >= 3). - Milan Janjic, Apr 26 2009
a(n) = round((2^(n+2)-3)/6) = floor((2^(n+1)-1)/3) = round((2^(n+1)-2)/3); a(n) = a(n-2) + 2^(n-1), n > 1. - Mircea Merca, Dec 27 2010
a(n) = binary XOR of 2^k-1 for k=0..n. - Paul D. Hanna, Nov 05 2011
E.g.f.: 2/3*exp(2*x) - 1/2*exp(x) - 1/6*exp(-x) = 2/3*U(0); U(k) = 1 - 3/(4*(2^k) - 4*(2^k)/(1+3*(-1)^k - 24*x*(2^k)/(8*x*(2^k)*(-1)^k - (k+1)/U(k+1)))); (continued fraction). - Sergei N. Gladkovskii, Nov 21 2011
a(n) = 2*(2^n - 1)/3, for even n; a(n) = (2^(n+1) - 1)/3 = (1/3)*(2^((n+1)/2) - 1)*(2^((n+1)/2) + 1), for odd n. - Hieronymus Fischer, Nov 22 2012
a(n) + a(n+1) = 2^(n+1) - 1. - Arie Bos, Apr 03 2013
G.f.: Q(0)/(3*(1-x)), where Q(k) = 1 - 1/(4^k - 2*x*16^k/(2*x*4^k - 1/(1 + 1/(2*4^k - 8*x*16^k/(4*x*4^k + 1/Q(k+1)))))); (continued fraction). - Sergei N. Gladkovskii, May 21 2013
a(n) = (2^(n+1) - 2 + (n mod 2))/3. - Paul Toms, Mar 18 2015
a(0) = 0, a(n) = 2*(a(n-1)) + (n mod 2). - Paul Toms, Mar 18 2015
Binary: a(n) = (a(n-1) shift left 1) + (a(n-1)) NOR (...11110). - Paul Toms, Mar 18 2015
a(n) = -(2^n)*a(-n) for even n; a(n) = -(2^(n+1))*a(-n) + 1 for odd n.
0 = +a(n)*(+2 +4*a(n) -4*a(n+1)) + a(n+1)*(-1 +a(n+1)) for all n in Z. (End)
a(4*k+d) = 2^(d+1)*a(4*k-1) + a(d), a(n+4) = a(n) + 2^n*10, a(0..3) = [0,1,2,5]. So the last digit is always 0,1,2,5 repeated. - Yuchun Ji, May 22 2023
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EXAMPLE
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a(4)=10 since 0001, 0011, 0010, 0110, 0111, 0101, 0100, 1100, 1101, 1111 are the 10 binary strings switching 0000 to 1111.
a(3) = 1 because "lrc" is the only way to tie a tie with 3 half turns, namely, pass the business end of the tie behind the standing part to the left, bring across the front to the right, then behind to the center. The final motion of tucking the loose end down the front behind the "lr" piece is not considered a "step".
a(4) = 2 because "lrlc" is the only way to tie a tie with 4 half turns. Note that since the number of moves is even, the first step is to go to the left in front of the tie, not behind it. This knot is the standard "four in hand", the most commonly known men's tie knot. By contrast, the second most well known tie knot, the Windsor, is represented by "lcrlcrlc".
a(n) = (2^0 - 1) XOR (2^1 - 1) XOR (2^2 - 1) XOR (2^3 - 1) XOR ... XOR (2^n - 1). - Paul D. Hanna, Nov 05 2011
G.f. = x + 2*x^2 + 5*x^3 + 10*x^4 + 21*x^5 + 42*x^6 + 85*x^7 + 170*x^8 + ...
a(9) = 341 = 2^8 + 2^6 + 2^4 + 2^2 + 2^0 = 4^4 + 4^3 + 4^2 + 4^1 + 4^0 = A002450(5). a(10) = 682 = 2*a(9) = 2*A002450(5). - Gregory L. Simay, Sep 27 2017
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MAPLE
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A000975 := proc(n) option remember; if n <= 1 then n else if n mod 2 = 0 then 2*A000975(n-1) else 2*A000975(n-1)+1 fi; fi; end;
f:=n-> if n mod 2 = 0 then (2^n-1)/3 else (2^n-2)/3; fi; [seq(f(n), n=0..40)]; # N. J. A. Sloane, Mar 21 2017
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MATHEMATICA
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Array[Ceiling[2(2^# - 1)/3] &, 41, 0]
RecurrenceTable[{a[0] == 0, a[1] == 1, a[n] == a[n - 1] + 2a[n - 2] + 1}, a, {n, 40}] (* or *)
LinearRecurrence[{2, 1, -2}, {0, 1, 2}, 40] (* Harvey P. Dale, Aug 10 2013 *)
f[n_] := Block[{exp = n - 2}, Sum[2^i, {i, exp, 0, -2}]]; Array[f, 33] (* Robert G. Wilson v, Oct 30 2015 *)
f[s_List] := Block[{a = s[[-1]]}, Append[s, If[OddQ@ Length@ s, 2a + 1, 2a]]]; Nest[f, {0}, 32] (* Robert G. Wilson v, Jul 20 2017 *)
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PROG
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(PARI) {a(n) = if( n<0, 0, 2 * 2^n \ 3)}; /* Michael Somos, Sep 04 2006 */
(PARI) a(n)=if(n<=0, 0, bitxor(a(n-1), 2^n-1)) \\ Paul D. Hanna, Nov 05 2011
(PARI) concat(0, Vec(x/(1-2*x-x^2+2*x^3) + O(x^100))) \\ Altug Alkan, Oct 30 2015
(PARI) {a(n) = (4*2^n - 3 - (-1)^n) / 6}; /* Michael Somos, Jul 23 2017 */
(Haskell)
a000975 n = a000975_list !! n
a000975_list = 0 : 1 : map (+ 1)
(zipWith (+) (tail a000975_list) (map (* 2) a000975_list))
(Magma) [(2^(n+1) - 2 + (n mod 2))/3: n in [0..40]]; // Vincenzo Librandi, Mar 18 2015
(GAP) List([0..35], n->(2^(n+1)-2+(n mod 2))/3); # Muniru A Asiru, Nov 01 2018
(Python)
def a(n): return (2**(n+1) - 2 + (n%2))//3
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CROSSREFS
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Cf. A000295, A005578, A015441, A043291, A053404, A059260, A077854, A119440, A127824, A130125, A135228, A155051, A179970, A264784.
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KEYWORD
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nonn,easy,nice
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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A032020
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Number of compositions (ordered partitions) of n into distinct parts.
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+10
214
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1, 1, 1, 3, 3, 5, 11, 13, 19, 27, 57, 65, 101, 133, 193, 351, 435, 617, 851, 1177, 1555, 2751, 3297, 4757, 6293, 8761, 11305, 15603, 24315, 30461, 41867, 55741, 74875, 98043, 130809, 168425, 257405, 315973, 431065, 558327, 751491, 958265, 1277867, 1621273
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OFFSET
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0,4
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COMMENTS
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a(n) = the number of different ways to run up a staircase with n steps, taking steps of distinct sizes where the order matters and there is no other restriction on the number or the size of each step taken. - Mohammad K. Azarian, May 21 2008
Compositions into distinct parts are equivalent to (1,1)-avoiding compositions. - Gus Wiseman, Jun 25 2020
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REFERENCES
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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.
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LINKS
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FORMULA
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"AGK" (ordered, elements, unlabeled) transform of 1, 1, 1, 1, ...
G.f.: Sum_{k>=0} k! * x^((k^2+k)/2) / Product_{j=1..k} (1-x^j). - David W. Wilson May 04 2000
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EXAMPLE
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a(6) = 11 because 6 = 5+1 = 4+2 = 3+2+1 = 3+1+2 = 2+4 = 2+3+1 = 2+1+3 = 1+5 = 1+3+2 = 1+2+3.
The a(0) = 1 through a(7) = 13 strict compositions:
() (1) (2) (3) (4) (5) (6) (7)
(1,2) (1,3) (1,4) (1,5) (1,6)
(2,1) (3,1) (2,3) (2,4) (2,5)
(3,2) (4,2) (3,4)
(4,1) (5,1) (4,3)
(1,2,3) (5,2)
(1,3,2) (6,1)
(2,1,3) (1,2,4)
(2,3,1) (1,4,2)
(3,1,2) (2,1,4)
(3,2,1) (2,4,1)
(4,1,2)
(4,2,1)
(End)
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MAPLE
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b:= proc(n, i) b(n, i):= `if`(n=0, [1], `if`(i<1, [], zip((x, y)
-> x+y, b(n, i-1), `if`(i>n, [], [0, b(n-i, i-1)[]]), 0))) end:
a:= proc(n) local l; l:=b(n, n): add((i-1)! *l[i], i=1..nops(l)) end:
# second Maple program:
T:= proc(n, k) option remember; `if`(k<0 or n<0, 0,
`if`(k=0, `if`(n=0, 1, 0), T(n-k, k) +k*T(n-k, k-1)))
end:
a:= n-> add(T(n, k), k=0..floor((sqrt(8*n+1)-1)/2)):
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MATHEMATICA
|
f[list_]:=Length[list]!; Table[Total[Map[f, Select[IntegerPartitions[n], Sort[#] == Union[#] &]]], {n, 0, 30}]
T[n_, k_] := T[n, k] = If[k<0 || n<0, 0, If[k==0, If[n==0, 1, 0], T[n-k, k] + k*T[n-k, k-1]]]; a[n_] := Sum[T[n, k], {k, 0, Floor[(Sqrt[8*n + 1] - 1) / 2]}]; Table[a[n], {n, 0, 60}] (* Jean-François Alcover, Sep 22 2015, after Alois P. Heinz *)
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PROG
|
(PARI)
N=66; q='q+O('q^N);
gf=sum(n=0, N, n!*q^(n*(n+1)/2) / prod(k=1, n, 1-q^k ) );
Vec(gf)
(PARI)
my(q = vector(N)); q[1] = [1, 0, 0, 0];
for (n = 2, N,
my(m = (sqrtint(8*n+1) - 1)\2);
q[n] = vector((1 + (m>>2)) << 2); q[n][1] = 1;
for (k = 2, m, q[n][k] = q[n-k][k] + q[n-k][k-1]));
return(q);
};
seq(N) = concat(1, apply(q -> sum(k = 1, #q, q[k] * k!), Q(N)));
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CROSSREFS
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Dominated by A003242 (anti-run compositions).
These compositions are ranked by A233564.
(1,1)-avoiding patterns are counted by A000142.
Numbers with strict prime signature are A130091.
(1,1,1)-avoiding compositions are counted by A232432.
(1,1)-matching compositions are counted by A261982.
Inseparable partitions are counted by A325535.
Patterns matched by compositions are counted by A335456.
Strict permutations of prime indices are counted by A335489.
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KEYWORD
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nonn,easy,nice
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AUTHOR
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STATUS
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approved
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A333489
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Numbers k such that the k-th composition in standard order is an anti-run (no adjacent equal parts).
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+10
213
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0, 1, 2, 4, 5, 6, 8, 9, 12, 13, 16, 17, 18, 20, 22, 24, 25, 32, 33, 34, 37, 38, 40, 41, 44, 45, 48, 49, 50, 52, 54, 64, 65, 66, 68, 69, 70, 72, 76, 77, 80, 81, 82, 88, 89, 96, 97, 98, 101, 102, 104, 105, 108, 109, 128, 129, 130, 132, 133, 134, 137, 140, 141
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OFFSET
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1,3
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COMMENTS
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A composition of n is a finite sequence of positive integers summing to n. The k-th composition in standard order (row k of A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again.
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LINKS
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EXAMPLE
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The sequence together with the corresponding compositions begins:
0: () 33: (5,1) 70: (4,1,2)
1: (1) 34: (4,2) 72: (3,4)
2: (2) 37: (3,2,1) 76: (3,1,3)
4: (3) 38: (3,1,2) 77: (3,1,2,1)
5: (2,1) 40: (2,4) 80: (2,5)
6: (1,2) 41: (2,3,1) 81: (2,4,1)
8: (4) 44: (2,1,3) 82: (2,3,2)
9: (3,1) 45: (2,1,2,1) 88: (2,1,4)
12: (1,3) 48: (1,5) 89: (2,1,3,1)
13: (1,2,1) 49: (1,4,1) 96: (1,6)
16: (5) 50: (1,3,2) 97: (1,5,1)
17: (4,1) 52: (1,2,3) 98: (1,4,2)
18: (3,2) 54: (1,2,1,2) 101: (1,3,2,1)
20: (2,3) 64: (7) 102: (1,3,1,2)
22: (2,1,2) 65: (6,1) 104: (1,2,4)
24: (1,4) 66: (5,2) 105: (1,2,3,1)
25: (1,3,1) 68: (4,3) 108: (1,2,1,3)
32: (6) 69: (4,2,1) 109: (1,2,1,2,1)
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MATHEMATICA
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stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n, 2]], 1], 0]]//Reverse;
Select[Range[0, 100], !MatchQ[stc[#], {___, x_, x_, ___}]&]
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CROSSREFS
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Anti-runs summing to n are counted by A003242(n).
A triangle counting maximal anti-runs of compositions is A106356.
A triangle counting maximal runs of compositions is A238279 or A238130.
Partitions whose first differences are an anti-run are A238424.
All of the following pertain to compositions in standard order (A066099):
- Adjacent equal pairs are counted by A124762.
- Weakly decreasing runs are counted by A124765.
- Weakly increasing runs are counted by A124766.
- Equal runs are counted by A124767.
- Strictly increasing runs are counted by A124768.
- Strictly decreasing runs are counted by A124769.
- Strict compositions are ranked by A233564.
- Constant compositions are ranked by A272919.
- Normal compositions are ranked by A333217.
- Anti-runs are counted by A333381.
- Adjacent unequal pairs are counted by A333382.
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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A088218
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Total number of leaves in all rooted ordered trees with n edges.
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+10
180
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1, 1, 3, 10, 35, 126, 462, 1716, 6435, 24310, 92378, 352716, 1352078, 5200300, 20058300, 77558760, 300540195, 1166803110, 4537567650, 17672631900, 68923264410, 269128937220, 1052049481860, 4116715363800, 16123801841550, 63205303218876, 247959266474052
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OFFSET
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0,3
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COMMENTS
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Essentially the same as A001700, which has more information.
Note that the unique rooted tree with no edges has no leaves, so a(0)=1 is by convention. - Michael Somos, Jul 30 2011
Number of ordered partitions of n into n parts, allowing zeros (cf. A097070) is binomial(2*n-1,n) = a(n) = essentially A001700. - Vladeta Jovovic, Sep 15 2004
Hankel transform is A000027; example: Det([1,1,3,10;1,3,10,35;3,10,35,126; 10,35,126,462]) = 4. - Philippe Deléham, Apr 13 2007
a(n) is the number of functions f:[n]->[n] such that for all x,y in [n] if x<y then f(x)<=f(y). So 2*a(n)-n = A045992(n). - Geoffrey Critzer, Apr 02 2009
Hankel transform of the aeration of this sequence is A000027 doubled: 1,1,2,2,3,3,... - Paul Barry, Sep 26 2009
The Fi1 and Fi2 triangle sums of A039599 are given by the terms of this sequence. For the definitions of these triangle sums see A180662. - Johannes W. Meijer, Apr 20 2011
Alternating row sums of Riordan triangle A094527. See the Philippe Deléham formula. - Wolfdieter Lang, Nov 22 2012
(-2)*a(n) is the Z-sequence for the Riordan triangle A110162. For the notion of Z- and A-sequences for Riordan arrays see the W. Lang link under A006232 with details and references. - Wolfdieter Lang, Nov 22 2012
Also the number of integer compositions of 2n with alternating (or reverse-alternating) sum 0 (ranked by A344619). This is equivalent to Ran Pan's comment at A001700. For example, the a(0) = 1 through a(3) = 10 compositions are:
() (11) (22) (33)
(121) (132)
(1111) (231)
(1122)
(1221)
(2112)
(2211)
(11121)
(12111)
(111111)
For n > 0, a(n) is also the number of integer compositions of 2n with alternating sum 2.
(End)
Number of terms in the expansion of (x_1+x_2+...+x_n)^n. - César Eliud Lozada, Jan 08 2022
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REFERENCES
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L. W. Shapiro and C. J. Wang, Generating identities via 2 X 2 matrices, Congressus Numerantium, 205 (2010), 33-46.
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LINKS
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FORMULA
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G.f.: (1 + 1 / sqrt(1 - 4*x)) / 2.
a(n) = binomial(2*n - 1, n).
a(n) = (n+1)*A000108(n)/2, n>=1. - B. Dubalski (dubalski(AT)atr.bydgoszcz.pl), Feb 05 2002 (in A060150)
a(n) = (0^n + C(2n, n))/2. - Paul Barry, May 21 2004
a(n) is the coefficient of x^n in 1 / (1 - x)^n and also the sum of the first n coefficients of 1 / (1 - x)^n. Given B(x) with the property that the coefficient of x^n in B(x)^n equals the sum of the first n coefficients of B(x)^n, then B(x) = B(0) / (1 - x).
G.f.: 1 / (2 - C(x)) = (1 - x*C(x))/sqrt(1-4*x) where C(x) is g.f. for Catalan numbers A000108. Second equation added by Wolfdieter Lang, Nov 22 2012.
a(n) = Sum_{k=0..n} binomial(2n, k)cos((n-k)*Pi)};
a(n) = Sum_{k=0..n} binomial(n, (n-k)/2)(1+(-1)^(n-k))cos(k*Pi/2)/2} (with interpolated zeros);
a(n) = Sum_{k=0..floor(n/2)} binomial(n, k)cos((n-2k)Pi/2)} (with interpolated zeros); (End)
G.f.: 1/(1-x/(1-2x/(1-(1/2)x/(1-(3/2)x/(1-(2/3)x/(1-(4/3)x/(1-(3/4)x/(1-(5/4)x/(1-... (continued fraction);
E.g.f.: (of aerated sequence) (1 + Bessel_I(0, 2*x))/2. (End)
E.g.f.: E(x) = 1+x/(G(0)-2*x) ; G(k) = (k+1)^2+2*x*(2*k+1)-2*x*(2*k+3)*((k+1)^2)/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Dec 21 2011
a(n) = Sum_{k=0..n}(-1)^k*binomial(2*n,n+k). - Mircea Merca, Jan 28 2012
a(n) = rf(n,n)/ff(n,n), where rf is the rising factorial and ff the falling factorial. - Peter Luschny, Nov 21 2012
D-finite with recurrence: n*a(n) +2*(-2*n+1)*a(n-1) = 0. - R. J. Mathar, Dec 04 2012
G.f.: 1 + x/W(0), where W(k) = 4*k+1 - (4*k+3)*x/(1 - (4*k+1)*x/(4*k+3 - (4*k+5)*x/(1 - (4*k+3)*x/W(k+1) ))) ; (continued fraction). - Sergei N. Gladkovskii, Nov 13 2014
E.g.f.: (1 + exp(2*x) * BesselI(0,2*x)) / 2. - Ilya Gutkovskiy, Nov 03 2021
Sum_{n>=0} 1/a(n) = 5/3 + 4*Pi/(9*sqrt(3)).
Sum_{n>=0} (-1)^n/a(n) = 3/5 - 8*log(phi)/(5*sqrt(5)), where phi is the golden ratio (A001622). (End)
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EXAMPLE
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G.f. = 1 + x + 3*x^2 + 10*x^3 + 35*x^4 + 126*x^5 + 462*x^6 + 1716*x^7 + ...
The five rooted ordered trees with 3 edges have 10 leaves.
..x........................
..o..x.x..x......x.........
..o...o...o.x..x.o..x.x.x..
..r...r....r....r.....r....
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MAPLE
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MATHEMATICA
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a[ n_] := SeriesCoefficient[(1 - x)^-n, {x, 0, n}];
c = (1 - (1 - 4 x)^(1/2))/(2 x); CoefficientList[Series[1/(1-(c-1)), {x, 0, 20}], x] (* Geoffrey Critzer, Dec 02 2010 *)
a[ n_] := If[ n < 0, 0, With[ {m = 2 n}, m! SeriesCoefficient[ (1 + BesselI[0, 2 x]) / 2, {x, 0, m}]]]; (* Michael Somos, Nov 22 2014 *)
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PROG
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(PARI) {a(n) = sum( i=0, n, binomial(n+i-2, i))};
(PARI) {a(n) = if( n<0, 0, polcoeff( (1 + 1 / sqrt(1 - 4*x + x * O(x^n))) / 2, n))};
(PARI) {a(n) = if( n<0, 0, polcoeff( 1 / (1 - x + x * O(x^n))^n, n))};
(PARI) {a(n) = if( n<0, 0, binomial( 2*n - 1, n))};
(PARI) {a(n) = if( n<1, n==0, polcoeff( subst((1 - x) / (1 - 2*x), x, serreverse( x - x^2 + x * O(x^n))), n))};
(Sage)
return rising_factorial(n, n)/falling_factorial(n, n)
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CROSSREFS
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Same as A001700 modulo initial term and offset.
A000041 counts partitions of 2n with alternating sum 0, ranked by A000290.
A003242 counts anti-run compositions.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A106356 counts compositions by number of maximal anti-runs.
A124754 gives the alternating sum of standard compositions.
A345197 counts compositions by sum, length, and alternating sum.
Compositions of n, 2n, or 2n+1 with alternating/reverse-alternating sum k:
Cf. A000027, A000070, A000097, A000108, A001622, A006232, A008965, A039599, A045992, A058696, A094527, A097070, A110162, A110555, A180662, A238279, A239830, A325534, A325535, A333213, A344607, A344611, A344617.
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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A106356
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Triangle T(n,k) 0<=k<n : Number of compositions of n with k adjacent equal parts.
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+10
179
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1, 1, 1, 3, 0, 1, 4, 3, 0, 1, 7, 6, 2, 0, 1, 14, 7, 8, 2, 0, 1, 23, 20, 10, 8, 2, 0, 1, 39, 42, 22, 13, 9, 2, 0, 1, 71, 72, 58, 28, 14, 10, 2, 0, 1, 124, 141, 112, 72, 33, 16, 11, 2, 0, 1, 214, 280, 219, 150, 92, 36, 18, 12, 2, 0, 1, 378, 516, 466, 311, 189, 112, 40, 20, 13, 2, 0, 1
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OFFSET
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1,4
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COMMENTS
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For n > 0, also the number of compositions of n with k + 1 maximal anti-runs (sequences without adjacent equal terms). - Gus Wiseman, Mar 23 2020
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LINKS
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A. Knopfmacher and H. Prodinger, On Carlitz compositions, European Journal of Combinatorics, Vol. 19, 1998, pp. 579-589.
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EXAMPLE
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T(4,1) = 3 because the compositions of 4 with 1 adjacent equal part are 1+1+2, 2+1+1, 2+2.
Triangle begins:
1;
1, 1;
3, 0, 1;
4, 3, 0, 1;
7, 6, 2, 0, 1;
14, 7, 8, 2, 0, 1;
23, 20, 10, 8, 2, 0, 1;
Row n = 6 counts the following compositions (empty column shown by dot):
(6) (33) (222) (11112) . (111111)
(15) (114) (1113) (21111)
(24) (411) (1122)
(42) (1131) (2211)
(51) (1221) (3111)
(123) (1311) (11121)
(132) (2112) (11211)
(141) (12111)
(213)
(231)
(312)
(321)
(1212)
(2121)
(End)
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MAPLE
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b:= proc(n, h, t) option remember;
if n=0 then `if`(t=0, 1, 0)
elif t<0 then 0
else add(b(n-j, j, `if`(j=h, t-1, t)), j=1..n)
fi
end:
T:= (n, k)-> b(n, -1, k):
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MATHEMATICA
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b[n_, h_, t_] := b[n, h, t] = If[n == 0, If[t == 0, 1, 0], If[t<0, 0, Sum[b[n-j, j, If [j == h, t-1, t]], {j, 1, n}]]]; T[n_, k_] := b[n, -1, k]; Table[Table[T[n, k], {k, 0, n-1}], {n, 1, 15}] // Flatten (* Jean-François Alcover, Feb 20 2015, after Alois P. Heinz *)
Table[Length[Select[Join@@Permutations/@IntegerPartitions[n], n==0||Length[Split[#, #1!=#2&]]==k+1&]], {n, 0, 12}, {k, 0, n}] (* Gus Wiseman, Mar 23 2020 *)
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CROSSREFS
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The version counting adjacent unequal parts is A238279.
The k-th composition in standard-order has A124762(k) adjacent equal parts and A333382(k) adjacent unequal parts.
The k-th composition in standard-order has A124767(k) maximal runs and A333381(k) maximal anti-runs.
The version for ascents/descents is A238343.
The version for weak ascents/descents is A333213.
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KEYWORD
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AUTHOR
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STATUS
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approved
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A005251
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a(0) = 0, a(1) = a(2) = a(3) = 1; thereafter, a(n) = a(n-1) + a(n-2) + a(n-4).
(Formerly M1059)
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+10
170
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0, 1, 1, 1, 2, 4, 7, 12, 21, 37, 65, 114, 200, 351, 616, 1081, 1897, 3329, 5842, 10252, 17991, 31572, 55405, 97229, 170625, 299426, 525456, 922111, 1618192, 2839729, 4983377, 8745217, 15346786, 26931732, 47261895, 82938844, 145547525, 255418101, 448227521
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OFFSET
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0,5
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COMMENTS
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a(n+3) is the number of n-bit sequences that avoid 010. Example: For n=4 the 12 sequences are all 4-bit sequences except 0100, 0101, 0010, 1010. - David Callan, Mar 25 2004
a(n+2) is the number of compositions (ordered partitions) of n where no two adjacent parts are != 1, see example. - Joerg Arndt, Jan 26 2013
a(n+1) is the number of compositions of n avoiding the part 2. - Joerg Arndt, Jul 13 2014
Number of different positive braids with n crossings of 3 strands.
This is a_2(n) in the Doroslovacki reference. Note that there is a typo in the paper in the formula for a_2(n): the upper bound in the inner sum should be "n-i" not "i-1". - Max Alekseyev, Jun 26 2007
a(n) is the number of peakless Motzkin paths of length n-1 with no UHH...HD's starting at level > 0 (here n > 0 and U=(1,1), H=(1,0), D=(1,-1)). Example: a(5)=7 because from all 8 peakless Motzkin paths of length 5 (see A004148) only UUHDD does not qualify. - Emeric Deutsch, Sep 13 2004
Equals the INVERT transform of (1, 0, 1, 1, 1, ...); equivalent to a(n) = a(n-1) + a(n-3) + a(n-4) + ... - Gary W. Adamson, Apr 27 2009
a(n) is the number of length n-1 words on {0,1} such that each string of 1's is followed by a string of at least two 0's. For example, a(5) = 4 because we have: 0000, 0100, 1000, and 1100. - Geoffrey Critzer, Aug 09 2013
a(n+1) is the top left entry of the n-th power of any of the 3 X 3 matrices [1, 1, 0; 0, 1, 1; 1, 0, 0] or [1, 0, 1; 1, 1, 0; 0, 1, 0] or [1, 1, 0; 0, 0, 1; 1, 0, 1] or [1, 0, 1; 1, 0, 0; 0, 1, 1]. - R. J. Mathar, Feb 03 2014
For n >= 2, a(n) is the number of (n-2)-length binary words with no isolated zeros. - Milan Janjic, Mar 07 2015
Apart from the first three terms, the total number of bargraphs of semiperimeter n of height at most two for n >= 2 starts 1, 2, 4, 7, 12, ... - Arnold Knopfmacher, Nov 02 2016
Number of DD-equivalence classes of Łukasiewicz paths. Łukasiewicz paths are DD-equivalent iff the positions of pattern DD are identical in these paths. - Sergey Kirgizov, Apr 08 2018
For n > 0, also the number of subsets of {1, ..., n - 3} such that if x and x + 2 are both in the subset, then so is x + 1. For example, the a(3) = 1 through a(7) = 12 subsets are:
{} {} {} {} {}
{1} {1} {1} {1}
{2} {2} {2}
{1,2} {3} {3}
{1,2} {4}
{2,3} {1,2}
{1,2,3} {1,4}
{2,3}
{3,4}
{1,2,3}
{2,3,4}
{1,2,3,4}
(End)
The two-dimensional version, which counts sets of pairs where, if two pairs are separated by graph-distance 2, then the intermediate pair or pairs are also in the set, is A329871. - Gus Wiseman, Nov 30 2019
a(n+1) is the number of ways to tile a strip of length n with squares, dominoes, and tetrominoes, where the first tile cannot be a domino. - Greg Dresden and Myanna Nash, Aug 18 2020
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REFERENCES
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S. Burckel, Efficient methods for three strand braids (submitted). [Apparently unpublished]
P. Chinn and S. Heubach, "Compositions of n with no occurrence of k", Congressus Numeratium, 2002, v. 162, pp. 33-51.
John H. Conway and R. K. Guy, The Book of Numbers, Copernicus Press, p. 205.
R. K. Guy, "Anyone for Twopins?" in D. A. Klarner, editor, The Mathematical Gardner. Prindle, Weber and Schmidt, Boston, 1981, pp. 2-15.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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R. K. Guy, Anyone for Twopins?, in D. A. Klarner, editor, The Mathematical Gardner. Prindle, Weber and Schmidt, Boston, 1981, pp. 2-15. [Annotated scanned copy, with permission]
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FORMULA
|
a(n) = 2*a(n-1) - a(n-2) + a(n-3).
23*a_n = 3*P_{2n+1} + 7*P_{2n} - 2*P_{2n-1}, where P_n are the Perrin numbers, A001608. - Don Knuth, Dec 09 2008
a(n) = (Sum_{j<n} a(j)) - a(n-2).
a(n+2) has g.f. (F_3(-x) + F_2(-x))/(F_4(-x) + F_3(-x)) = 1/(-x+1/(-x+1/(-x+1))) where F_n(x) is the n-th Fibonacci polynomial; see A011973. - Qiaochu Yuan (qchu(AT)mit.edu), Feb 19 2009
a(n) = hypergeom([(2-n)/3, 1-n/3, (1-n)/3], [1/2, -n+1], 27/4) for n > 1. - Peter Luschny, Apr 08 2018
G.f.: z/(1-z-z^3-z^4-z^5-...) for the compositions of n-1 avoiding 2. The g.f. for the number of compositions of n avoiding the part k is 1/(1-z-...-z^(k-1) - z^(k+1)-...). - Gregory L. Simay, Sep 09 2018
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EXAMPLE
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The a(5+2) = 12 compositions of 5 where no two adjacent parts are != 1 are
[ 1] [ 1 1 1 1 1 ]
[ 2] [ 1 1 1 2 ]
[ 3] [ 1 1 2 1 ]
[ 4] [ 1 1 3 ]
[ 5] [ 1 2 1 1 ]
[ 6] [ 1 3 1 ]
[ 7] [ 1 4 ]
[ 8] [ 2 1 1 1 ]
[ 9] [ 2 1 2 ]
[10] [ 3 1 1 ]
[11] [ 4 1 ]
[12] [ 5 ]
(End)
G.f. = x + x^2 + x^3 + 2*x^4 + 4*x^5 + 7*x^6 + 12*x^7 + 21*x^8 + 37*x^9 + ...
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MAPLE
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a := n -> `if`(n<=1, n, hypergeom([(2-n)/3, 1-n/3, (1-n)/3], [1/2, -n+1], 27/4)):
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MATHEMATICA
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LinearRecurrence[{2, -1, 1}, {0, 1, 1}, 40] (* Harvey P. Dale, May 05 2011 *)
a[ n_]:= If[n<0, SeriesCoefficient[ -x(1-x)/(1 -x + 2x^2 -x^3), {x, 0, -n}], SeriesCoefficient[ x(1-x)/(1 -2x +x^2 -x^3), {x, 0, n}]] (* Michael Somos, Dec 13 2013 *)
a[0] = 1; a[1] = a[2] = 0; a[n_] := a[n] = a[n-2] + a[n-3]; Table[a[2 n-1], {n, 1, 20}] (* Rigoberto Florez, Oct 15 2019 *)
Table[If[n==0, 0, Length[DeleteCases[Subsets[Range[n-3]], {___, x_, y_, ___}/; x+2==y]]], {n, 0, 10}] (* Gus Wiseman, Nov 25 2019 *)
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PROG
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(Haskell)
a005251 n = a005251_list !! n
a005251_list = 0 : 1 : 1 : 1 : zipWith (+) a005251_list
(drop 2 $ zipWith (+) a005251_list (tail a005251_list))
(PARI) {a(n) = if( n<0, polcoeff( -x*(1-x)/(1 -x +2*x^2 -x^3) + x*O(x^-n), -n), polcoeff( x*(1-x)/(1 -2*x +x^2 -x^3) + x*O(x^n), n))} /* Michael Somos, Dec 13 2013 */
(Magma) I:=[0, 1, 1, 1]; [n le 4 select I[n] else Self(n-1)+Self(n-2)+Self(n-4): n in [1..45]]; // Vincenzo Librandi, Nov 30 2018
(Magma) R<x>:=PowerSeriesRing(Integers(), 40); [0] cat Coefficients(R!( x*(1-x)/(1-2*x + x^2 - x^3) )); // Marius A. Burtea, Oct 24 2019
(SageMath) [sum( binomial(n-j-1, 2*j) for j in (0..floor((n-1)/3)) ) for n in (0..50)] # G. C. Greubel, Apr 13 2022
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CROSSREFS
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Bisection of Padovan sequence A000931.
Partial sums of A005314 shifted 3 times to the right, if we assume A005314(0) = 1.
Compositions without adjacent equal parts are A003242.
Compositions without isolated parts are A114901.
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KEYWORD
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nonn,nice,easy
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AUTHOR
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STATUS
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approved
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A025047
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Number of alternating compositions, i.e., compositions with alternating increases and decreases, starting with either an increase or a decrease.
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+10
157
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1, 1, 1, 3, 4, 7, 12, 19, 29, 48, 75, 118, 186, 293, 460, 725, 1139, 1789, 2814, 4422, 6949, 10924, 17168, 26979, 42404, 66644, 104737, 164610, 258707, 406588, 639009, 1004287, 1578363, 2480606, 3898599, 6127152, 9629623, 15134213, 23785388, 37381849, 58750468
(list;
graph;
refs;
listen;
history;
text;
internal format)
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OFFSET
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0,4
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COMMENTS
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Original name: Wiggly sums: number of sums adding to n in which terms alternately increase and decrease or vice versa.
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LINKS
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FORMULA
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a(n) = A025048(n) + A025049(n) - 1 = sum_k[A059881(n, k)] = sum_k[S(n, k) + T(n, k)] - 1 where if n>k>0 S(n, k) = sum_j[T(n - k, j)] over j>k and T(n, k) = sum_j[S(n - k, j)] over k>j (note reversal) and if n>0 S(n, n) = T(n, n) = 1; S(n, k) = A059882(n, k), T(n, k) = A059883(n, k). - Henry Bottomley, Feb 05 2001
a(n) ~ c * d^n, where d = 1.571630806607064114100138865739690782401305155950789062725..., c = 0.82222360450823867604750473815253345888526601460811483897... . - Vaclav Kotesovec, Sep 12 2014
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EXAMPLE
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There are a(7)=19 such compositions of 7:
[ 1] + [ 1 2 1 2 1 ]
[ 2] + [ 1 2 1 3 ]
[ 3] + [ 1 3 1 2 ]
[ 4] + [ 1 4 2 ]
[ 5] + [ 1 5 1 ]
[ 6] + [ 1 6 ]
[ 7] - [ 2 1 3 1 ]
[ 8] - [ 2 1 4 ]
[ 9] + [ 2 3 2 ]
[10] + [ 2 4 1 ]
[11] + [ 2 5 ]
[12] - [ 3 1 2 1 ]
[13] - [ 3 1 3 ]
[14] + [ 3 4 ]
[15] - [ 4 1 2 ]
[16] - [ 4 3 ]
[17] - [ 5 2 ]
[18] - [ 6 1 ]
[19] 0 [ 7 ]
For A025048(7)-1=10 of these the first two parts are increasing (marked by '+'),
and for A025049(7)-1=8 the first two parts are decreasing (marked by '-').
(End)
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MAPLE
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b:= proc(n, l, t) option remember; `if`(n=0, 1, add(
b(n-j, j, 1-t), j=`if`(t=1, 1..min(l-1, n), l+1..n)))
end:
a:= n-> 1+add(add(b(n-j, j, i), i=0..1), j=1..n-1):
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MATHEMATICA
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wigQ[y_]:=Or[Length[y]==0, Length[Split[y]]== Length[y]&&Length[Split[Sign[Differences[y]]]]==Length[y]-1];
Table[Length[Select[Join@@Permutations/@IntegerPartitions[n], wigQ]], {n, 0, 15}] (* Gus Wiseman, Jun 17 2021 *)
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PROG
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(PARI)
D(n, f)={my(M=matrix(n, n, j, k, k>=j), s=M[, n]); for(b=1, n, f=!f; M=matrix(n, n, j, k, if(k<j, if(f, if(k>1, M[j-k, k-1]), M[j-k, n]-M[j-k, k] ))); for(k=2, n, M[, k]+=M[, k-1]); s+=M[, n]); s~}
seq(n) = concat([1], D(n, 0) + D(n, 1) - vector(n, j, 1)) \\ Andrew Howroyd, Jan 31 2024
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CROSSREFS
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The version allowing pairs (x,x) is A344604.
A032020 counts strict compositions.
A106356 counts compositions by number of maximal anti-runs.
A114901 counts compositions where each part is adjacent to an equal part.
A274174 counts compositions with equal parts contiguous.
A345164 counts alternating permutations of prime indices.
A345165 counts partitions w/o alternating permutation, ranked by A345171.
A345170 counts partitions w/ alternating permutation, ranked by A345172.
Cf. A000070, A008965, A238279, A333755, A344606, A344614, A344653, A344740, A345163, A345166, A345169.
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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A000213
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Tribonacci numbers: a(n) = a(n-1) + a(n-2) + a(n-3) with a(0)=a(1)=a(2)=1.
(Formerly M2454 N0975)
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+10
147
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1, 1, 1, 3, 5, 9, 17, 31, 57, 105, 193, 355, 653, 1201, 2209, 4063, 7473, 13745, 25281, 46499, 85525, 157305, 289329, 532159, 978793, 1800281, 3311233, 6090307, 11201821, 20603361, 37895489, 69700671, 128199521, 235795681, 433695873, 797691075, 1467182629
(list;
graph;
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listen;
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internal format)
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OFFSET
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0,4
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COMMENTS
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The name "tribonacci number" is less well-defined than "Fibonacci number". The sequence A000073 (which begins 0, 0, 1) is probably the most important version, but the name has also been applied to A000213, A001590, and A081172. - N. J. A. Sloane, Jul 25 2024
Number of (n-1)-bit binary sequences with each one adjacent to a zero. - R. H. Hardin, Dec 24 2007
Equals INVERT transform of (1, 0, 2, 0, 2, 0, 2, ...). a(6) = 17 = (1, 1, 1, 3, 5, 9) dot (0, 2, 0, 2, 0, 1) = (0 + 2 + 0 + 6 + 0 + 9) = 17. - Gary W. Adamson, Apr 27 2009
Equals the number of tilings of a 2 X n grid using singletons and "S-shaped tetrominoes" (i.e., shapes of the form Polygon[{{0, 0}, {2, 0}, {2, 1}, {3, 1}, {3, 2}, {1, 2}, {1, 1}, {0, 1}}]).
Also equals the number of tilings of a 2 X n grid using singletons and "T-shaped tetrominoes" (i.e., shapes of the form Polygon[{{0, 0}, {3, 0}, {3, 1}, {2, 1}, {2, 2}, {1, 2}, {1, 1}, {0, 1}}]). (End)
Pisano period lengths: 1, 1, 13, 4, 31, 13, 48, 8, 39, 31, 110, 52, 168, 48, 403, 16, 96, 39, 360, 124, ... (differs from A106293). - R. J. Mathar, Aug 10 2012
a(n) is the number of compositions of n with no consecutive 1's. a(4) = 5 because we have: 4, 3+1, 1+3, 2+2, 1+2+1. Cf. A239791, A003242. - Geoffrey Critzer, Mar 27 2014
a(n+2) is the number of words of length n over alphabet {1,2,3} without having {11,12,22,23} as substrings. - Ran Pan, Sep 16 2015
a(n) is also the number of dominating sets on the (n-1)-path graph. - Eric W. Weisstein, Mar 31 2017
a(n) is also the number of maximal irredundant sets and minimal dominating sets in the (2n-3)-triangular snake graph. - Eric W. Weisstein, Jun 09 2019
a(n) is also the number of anti-palindromic compositions of n, where a composition (c(1), c(2),..., c(k)) is anti-palindromic if c(i) is not equal to c(k+1-i) whenever 1 <= i <= k/2. For instance, there are a(4) = 5 anti-palindromic compositions of 4: 4, 31, 13, 211, 112. - Jia Huang, Apr 08 2023
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REFERENCES
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Kenneth Edwards, Michael A. Allen, A new combinatorial interpretation of the Fibonacci numbers squared, Part II, Fib. Q., 58:2 (2020), 169-177.
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).
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LINKS
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George E. Andrews, Matthew Just, and Greg Simay, Anti-palindromic compositions, arXiv:2102.01613 [math.CO], 2021. Also Fib. Q., 60:2 (2022), 164-176. See Table 1.
B. G. Baumgart, Letter to the editor Part 1 Part 2 Part 3, Fib. Quart. 2 (1964), 260, 302.
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FORMULA
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G.f.: (1-x)*(1+x)/(1-x-x^2-x^3). - Ralf Stephan, Feb 11 2004
G.f.: 1 / (1 - x / (1 - 2*x^2 / (1 + x^2))). - Michael Somos, May 12 2012
a(n) = rightmost term of M^n * [1 1 1], where M is the 3 X 3 matrix [1 1 1 / 1 0 0 / 0 1 0]. M^n * [1 1 1] = [a(n+2) a(n+1) a(n)]. a(n)/a(n-1) tends to the tribonacci constant, 1.839286755...; an eigenvalue of M and a root of x^3 - x^2 - x - 1 = 0. - Gary W. Adamson, Dec 17 2004
a(n) = 2*a(n-1) - a(n-4), n > 3. - Gary Detlefs, Sep 13 2010
a(n) = Sum_{m=0..n/2} Sum_{i=0..m} binomial(n-2*m+1,m-i)*binomial(n-2*m+i, n-2*m). - Vladimir Kruchinin, Dec 17 2011
G.f.: 1+x/(U(0) - x) where U(k) = 1 - x^2/(1 - 1/(1 + 1/U(k+1))); (continued fraction, 3-step). - Sergei N. Gladkovskii, Nov 16 2012
G.f.: 1 + x + x^2/(G(0)-x) where G(k) = 1 - x*(2*k+1)/(1 - 1/(1 + (2*k+1)/G(k+1))); (continued fraction, 3-step). - Sergei N. Gladkovskii, Nov 17 2012
G.f.: (1+x)*(1-x)*(1 + x*(G(0)-1)/(x+1)) where G(k) = 1 + (1+x+x^2)/(1-x/(x+1/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jan 26 2013
G.f.: 1/(1+x-G(0)), where G(k) = 1 - 1/(1 - x/(x - 1/(1 + 1/(1 - x/(x + 1/G(k+1)))))); (continued fraction). - Sergei N. Gladkovskii, Jun 20 2013
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EXAMPLE
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G.f. = 1 + x + x^2 + 3*x^3 + 5*x^4 + 9*x^5+ 17*x^6 + 31*x^7 + 57*x^8 + ...
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MAPLE
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K:=(1-z^2)/(1-z-z^2-z^3): Kser:=series(K, z=0, 45): seq((coeff(Kser, z, n)), n= 0..34); # Zerinvary Lajos, Nov 08 2007
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MATHEMATICA
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LinearRecurrence[{1, 1, 1}, {1, 1, 1}, 45] (* Harvey P. Dale, May 23 2011 *)
Table[RootSum[-1 - # - #^2 + #^3 &, 2 #^n - 4 #^(n + 1) + 3 #^(n + 2) &]/11, {n, 0, 45}] (* Eric W. Weisstein, Apr 10 2018 *)
CoefficientList[Series[(1-x)(1+x)/(1-x-x^2-x^3), {x, 0, 45}], x] (* Eric W. Weisstein, Apr 10 2018 *)
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PROG
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(PARI) a(n)=tn=[1, 1, 1; 1, 0, 0; 0, 1, 0]^n; tn[3, 1]+tn[3, 2]+tn[3, 3] \\ Charles R Greathouse IV, Feb 18 2011
(Maxima) a(n):=sum(sum(binomial(n-2*m+1, m-i)*binomial(n-2*m+i, n-2*m), i, 0, m), m, 0, (n)/2); /* Vladimir Kruchinin, Dec 17 2011 */
(Haskell)
a000213 n = a000213_list !! n
a000213_list = 1 : 1 : 1 : zipWith (+) a000213_list
(tail $ zipWith (+) a000213_list (tail a000213_list))
(Magma) I:=[1, 1, 1]; [n le 3 select I[n] else Self(n-1) + Self(n-2) + Self(n-3): n in [1..45]]; // G. C. Greubel, Jun 09 2019
(Sage) ((1-x^2)/(1-x-x^2-x^3)).series(x, 45).coefficients(x, sparse=False) # G. C. Greubel, Jun 09 2019
(GAP) a:=[1, 1, 1];; for n in [4..45] do a[n]:=a[n-1]+a[n-2]+a[n-3]; od; a; # G. C. Greubel, Jun 09 2019
(Python)
alst = [1, 1, 1]
[alst.append(alst[n-1] + alst[n-2] + alst[n-3]) for n in range(3, 37)]
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CROSSREFS
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KEYWORD
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nonn,easy,nice
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AUTHOR
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STATUS
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approved
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