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a(n) = 5*a(n-1) - a(n-2) for n > 1, a(0) = 0, a(1) = 1.
(Formerly M3930)
+0
73
0, 1, 5, 24, 115, 551, 2640, 12649, 60605, 290376, 1391275, 6665999, 31938720, 153027601, 733199285, 3512968824, 16831644835, 80645255351, 386394631920, 1851327904249, 8870244889325, 42499896542376, 203629237822555, 975646292570399, 4674602225029440
COMMENTS
Nonnegative values of y satisfying x^2 - 21*y^2 = 4; values of x are in A003501. - Wolfdieter Lang, Nov 29 2002 a(n) is equal to the permanent of the (n-1) X (n-1) Hessenberg matrix with 5's along the main diagonal, i's along the superdiagonal and the subdiagonal (i is the imaginary unit), and 0's everywhere else. - John M. Campbell, Jun 09 2011 For n >= 1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,2,3,4}. - Milan Janjic, Jan 25 2015 For any three consecutive terms (x, y, z), y^2 - xz = 1 always applies.
a(n) = (t(i+2n) - t(i))/(t(i+n+1) - t(i+n-1)) where (t) is any recurrence t(k) = 4t(k-1) + 4t(k-2) - t(k-3) or t(k) = 5t(k-1) - t(k-2) without regard to initial valus.
In particular, if the recurrence (t) of the form (4,4,-1) has the same three initial values as the current sequence, a(n) = t(n) applies.
a(n) = (t(k+1)*t(k+n) - t(k)*t(k+n+1))/(y^2 - xz) where (t) is any recurrence of the current family with signature (5,-1) and (x, y, z) are any three consecutive terms of (t), for integer k >= 0. (End)
REFERENCES
F. A. Haight, On a generalization of Pythagoras' theorem, pp. 73-77 of J. C. Butcher, editor, A Spectrum of Mathematics. Auckland University Press, 1971.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
A. F. Horadam, Pell Identities, Fib. Quart., Vol. 9, No. 3, 1971, pp. 245-252.
FORMULA
G.f.: x/(1-5*x+x^2).
a(n) = ((5+sqrt(21))/2)^n-((5-sqrt(21))/2)^n)/sqrt(21). - Barry E. Williams, Aug 29 2000 a(n) = S(2*n-1, sqrt(7))/sqrt(7) = S(n-1, 5); S(n, x)=U(n, x/2), Chebyshev polynomials of 2nd kind, A049310. a(n) = Sum_{k=0..n-1} binomial(n+k, 2*k+1)*2^k. - Paul Barry, Nov 30 2004 a(n+1) = Sum_{k=0..n} Gegenbauer_C(n-k,k+1,2). - Paul Barry, Apr 21 2009 Product {n >= 1} (1 + 1/a(n)) = (1/3)*(3 + sqrt(21)). - Peter Bala, Dec 23 2012 Product {n >= 2} (1 - 1/a(n)) = (1/10)*(3 + sqrt(21)). - Peter Bala, Dec 23 2012 0 = -1 + a(n)*(+a(n) - 5*a(n+1)) + a(n+1)*(+a(n+1)) for all n in Z. - Michael Somos, Jan 22 2017 a(n) = 4(a(n-1) + a(n-2)) - a(n-3).
a(n) = 6(a(n-1) - a(n-2)) + a(n-3).
In general, for all sequences of the form U(n) = P*U(n-1) - U(n-2) the following applies:
U(n) = (P-1)*U(n-1) + (P-1)*U(n-2) - U(n-3).
U(n) = (P+1)*U(n-1) - (P+1)*U(n-2) + U(n-3). (End)
EXAMPLE
G.f. = x + 5*x^2 + 24*x^3 + 115*x^4 + 551*x^5 + 2640*x^6 + 12649*x^7 + ...
MATHEMATICA
a[n_]:=(MatrixPower[{{1, 3}, {1, 4}}, n].{{1}, {1}})[[2, 1]]; Table[a[n], {n, 0, 40}] (* Vladimir Joseph Stephan Orlovsky, Feb 19 2010 *) a[ n_] := ChebyshevU[2 n - 1, Sqrt[7]/2] / Sqrt[7]; (* Michael Somos, Jan 22 2017 *)
PROG
(PARI) {a(n) = subst(4*poltchebi(n+1) - 10*poltchebi(n), x, 5/2) / 21}; /* Michael Somos, Dec 04 2002 */ (PARI) {a(n) = imag((5 + quadgen(84))^n) / 2^(n-1)}; /* Michael Somos, Dec 04 2002 */ (PARI) {a(n) = polchebyshev(n - 1, 2, 5/2)}; /* Michael Somos, Jan 22 2017 */ (PARI) {a(n) = simplify( polchebyshev( 2*n - 1, 2, quadgen(28)/2) / quadgen(28))}; /* Michael Somos, Jan 22 2017 */ (Sage) [lucas_number1(n, 5, 1) for n in range(27)] # Zerinvary Lajos, Jun 25 2008 (Magma) [ n eq 1 select 0 else n eq 2 select 1 else 5*Self(n-1)-Self(n-2): n in [1..30] ]; // Vincenzo Librandi, Aug 19 2011
Binomial coefficient C(3n, n-3).
+0
6
1, 12, 105, 816, 5985, 42504, 296010, 2035800, 13884156, 94143280, 635745396, 4280561376, 28760021745, 192928249296, 1292706174900, 8654327655120, 57902201338905, 387221678682300, 2588713818544245
REFERENCES
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 828.
LINKS
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
FORMULA
E.g.f.: (1/6)*x^3*2F2(10/3,11/3; 5,11/2; 27*x/4).
a(n) ~ 3^(3*n+1/2)/(sqrt(Pi*n)*4^(n+2)). (End)
MAPLE
a:=n->sum(binomial(2*n-2, n+j)*binomial(n-1, n-j+1), j=0..n): seq(a(n), n=4..22); # Zerinvary Lajos, Jan 29 2007
PROG
(PARI) {a(n) = binomial(3*n, n-3)}; \\ G. C. Greubel, Mar 21 2019 (Magma) [Binomial(3*n, n-3): n in [3..30]]; // G. C. Greubel, Mar 21 2019 (Sage) [binomial(3*n, n-3) for n in (3..30)] # G. C. Greubel, Mar 21 2019 (GAP) List([3..30], n-> Binomial(3*n, n-3)) # G. C. Greubel, Mar 21 2019
Binomial coefficient C(4n,n-4).
+0
4
1, 20, 276, 3276, 35960, 376992, 3838380, 38320568, 377348994, 3679075400, 35607051480, 342700125300, 3284214703056, 31368725759168, 298824321028320, 2840671544105280, 26958221130508525, 255485622301674660
REFERENCES
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 828.
LINKS
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
FORMULA
E.g.f.: (1/24)*x^4*3F3(17/4,9/2,19/4; 17/3,6,19/3; 256*x/27).
a(n) ~ 2^(8*n+1/2)/(sqrt(Pi*n)*3^(3*n+9/2)). (End)
PROG
(Sage) [binomial(4*n, n-4) for n in (4..30)] # G. C. Greubel, Mar 21 2019 (GAP) List([4..30], n-> Binomial(4*n, n-4)) # G. C. Greubel, Mar 21 2019
Triangle read by rows: row n lists the first n positive integers in decreasing order.
+0
331
1, 2, 1, 3, 2, 1, 4, 3, 2, 1, 5, 4, 3, 2, 1, 6, 5, 4, 3, 2, 1, 7, 6, 5, 4, 3, 2, 1, 8, 7, 6, 5, 4, 3, 2, 1, 9, 8, 7, 6, 5, 4, 3, 2, 1, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1
COMMENTS
Old name: Triangle T(n,k) = n-k, n >= 1, 0 <= k < n. Fractal sequence formed by repeatedly appending strings m, m-1, ..., 2, 1.
The PARI functions t1 (this sequence), t2 ( A002260) can be used to read a square array T(n,k) (n >= 1, k >= 1) by antidiagonals upwards: n -> T(t1(n), t2(n)). - Michael Somos, Aug 23 2002, edited by M. F. Hasler, Mar 31 2020 A004736 is the mirror of the self-fission of the polynomial sequence (q(n,x)) given by q(n,x) = x^n+ x^(n-1) + ... + x + 1. See A193842 for the definition of fission. - Clark Kimberling, Aug 07 2011
Sequence B is called a reverse reluctant sequence of sequence A, if B is triangle array read by rows: row number k lists first k elements of the sequence A in reverse order. Sequence A004736 is the reverse reluctant sequence of sequence 1,2,3,... ( A000027). - Boris Putievskiy, Dec 13 2012 Riordan array (1/(1-x)^2,x). Call this array M and for k = 0,1,2,... define M(k) to be the lower unit triangular block array
/I_k 0\
\ 0 M/
having the k X k identity matrix I_k as the upper left block; in particular, M(0) = M. Then the infinite matrix product M(0)*M(1)*M(2)*... is equal to A078812. (End) T(n, k) gives the number of subsets of [n] := {1, 2, ..., n} with k consecutive numbers (consecutive k-subsets of [n]). - Wolfdieter Lang, May 30 2018 a(n) gives the distance from (n-1) to the smallest triangular number > (n-1). - Ctibor O. Zizka, Apr 09 2020 To construct the sequence, start from 1,2,,3,,,4,,,,5,,,,,6... where there are n commas after each "n". Then fill the empty places by the sequence itself. - Benoit Cloitre, Aug 17 2021 T(n,k) is the number of cycles of length 2*(k+1) in the (n+1)-ladder graph. There are no cycles of odd length. - Mohammed Yaseen, Jan 14 2023 The first 77 entries are also the signature sequence of log(3)= A002391. Then the two sequences start to differ. - R. J. Mathar, May 27 2024
REFERENCES
H. S. M. Coxeter, Regular Polytopes, 3rd ed., Dover, NY, 1973, pp 159-162.
LINKS
Glen Joyce C. Dulatre, Jamilah V. Alarcon, Vhenedict M. Florida, and Daisy Ann A. Disu, On Fractal Sequences, DMMMSU-CAS Science Monitor (2016-2017) Vol. 15 No. 2, 109-113.
FORMULA
a(n) = (2 - 2*n + round(sqrt(2*n)) + round(sqrt(2*n))^2)/2. - Brian Tenneson, Oct 11 2003 Recursion: e(n,k) = (e(n - 1, k)*e(n, k - 1) + 1)/e(n - 1, k - 1). - Roger L. Bagula, Mar 25 2009 a(n) = (t*t+3*t+4)/2-n, where t = floor((-1+sqrt(8*n-7))/2). - Boris Putievskiy, Dec 13 2012 T(n, k) = n - k + 1, n >= 1 and 1 <= k <= n.
T(n, k) = A002260(n+k-1, n-k+1). (End)
EXAMPLE
The triangle T(n, k) starts:
n\k 1 2 3 4 5 6 7 8 9 10 11 12 ...
1: 1
2: 2 1
3: 3 2 1
4: 4 3 2 1
5: 5 4 3 2 1
6: 6 5 4 3 2 1
7: 7 6 5 4 3 2 1
8: 8 7 6 5 4 3 2 1
9: 9 8 7 6 5 4 3 2 1
10: 10 9 8 7 6 5 4 3 2 1
11: 11 10 9 8 7 6 5 4 3 2 1
12: 12 11 10 9 8 7 6 5 4 3 2 1
T(6, 3) = 4 because the four consecutive 3-subsets of [6] = {1, 2, ..., 6} are {1, 2, 3}, {2, 3, 4}, {3, 4, 5} and {4, 5, 6}. - Wolfdieter Lang, May 30 2018
MAPLE
T := (n, k) -> n-k+1: seq(seq(T(n, k), k=1..n), n=1..13); # Johannes W. Meijer, Sep 28 2013
MATHEMATICA
Table[Range[n, 1, -1], {n, 20}]//Flatten (* Harvey P. Dale, May 27 2020 *)
PROG
(PARI) {a(n) = 1 + binomial(1 + floor(1/2 + sqrt(2*n)), 2) - n}
(PARI) {t1(n) = binomial( floor(3/2 + sqrt(2*n)), 2) - n + 1} /* A004736 */ (PARI) {t2(n) = n - binomial( floor(1/2 + sqrt(2*n)), 2)} /* A002260 */ (PARI) apply( A004736(n)=1-n+(n=sqrtint(8*n)\/2)*(n+1)\2, [1..99]) \\ M. F. Hasler, Mar 31 2020 (Excel) =if(row()>=column(); row()-column()+1; "") [ Mats Granvik, Jan 19 2009] (Haskell)
a004736 n k = n - k + 1
a004736_row n = a004736_tabl !! (n-1)
a004736_tabl = map reverse a002260_tabl
(Python)
def agen(rows):
for n in range(1, rows+1): yield from range(n, 0, -1)
CROSSREFS
Cf. A141419 (partial sums per row). See A001511 for definition of ordinal transform.
Concatenation of sequences (1,2,...,n-1,n,n-1,...,2) for n >= 2.
+0
12
1, 2, 1, 2, 3, 2, 1, 2, 3, 4, 3, 2, 1, 2, 3, 4, 5, 4, 3, 2, 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1, 2, 3, 4, 5, 6, 7, 6, 5, 4, 3, 2, 1, 2, 3, 4, 5, 6, 7, 8, 7, 6, 5, 4, 3, 2, 1, 2, 3, 4, 5, 6, 7, 8, 9, 8, 7, 6, 5, 4, 3, 2, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 10, 9
COMMENTS
Also concatenation of sequences n,n-1,...,2,1,2,...,n-1,n.
Table T(n,k) n, k > 0, T(n,k) = n-k+1, if n >= k, T(n,k) = k-n+1, if n < k. Table read by sides of squares from T(1,n) to T(n,n), then from T(n,n) to T(n,1). General case A209301. Let m be a natural number. The first column of the table T(n,1) is the sequence of the natural numbers A000027. In all columns with number k (k > 1) the segment with the length of (k-1): {m+k-2, m+k-3, ..., m} shifts the sequence A000027. For m=1 the result is A004739, for m=2 the result is A004738, for m=3 the result is A209301. - Boris Putievskiy, Jan 24 2013
REFERENCES
F. Smarandache, "Numerical Sequences", University of Craiova, 1975; [ See Arizona State University, Special Collection, Tempe, AZ, USA ].
FORMULA
a(n) = floor(sqrt(n) + 1/2) + 1 - abs(n - 1 - (floor(sqrt(n) + 1/2))^2). - Benoit Cloitre, Feb 08 2003 For the general case, a(n) = m*v + (2*v-1)*(t*t-n) + t, where t = floor((sqrt(n) - 1/2) + 1 and v = floor((n-1)/t) - t + 1.
For m=2, a(n) = 2*v + (2*v-1)*(t*t-n)+t, where t = floor((sqrt(n) - 1/2) + 1 and v = floor((n-1)/t) - t + 1. (End)
EXAMPLE
The start of the sequence as table:
1, 2, 3, 4, 5, 6, 7, ...
2, 1, 2, 3, 4, 5, 6, ...
3, 2, 1, 2, 3, 4, 5, ...
4, 3, 2, 1, 2, 3, 4, ...
5, 4, 3, 2, 1, 2, 3, ...
6, 5, 4, 3, 2, 1, 2, ...
7, 6, 5, 4, 3, 2, 1, ...
...
The start of the sequence as triangle array read by rows:
1;
2, 1, 2;
3, 2, 1, 2, 3;
4, 3, 2, 1, 2, 3, 4;
5, 4, 3, 2, 1, 2, 3, 4, 5;
6, 5, 4, 3, 2, 1, 2, 3, 4, 5, 6;
7, 6, 5, 4, 3, 2, 1, 2, 3, 4, 5, 6, 7;
...
Row number r contains 2*r - 1 numbers: r, r-1, ..., 1, 2, ..., r. (End)
MAPLE
local tri ;
tri := floor(sqrt(n)+1/2) ;
tri+1-abs(n-1-tri^2) ;
end proc:
MATHEMATICA
row[n_] := Range[n, 1, -1] ~Join~ Range[2, n];
PROG
(PARI) a(n)= floor(sqrt(n)+1/2)+1-abs(n-1-(floor(sqrt(n)+1/2)-1/2)^2)
Concatenation of sequences (1,2,2,...,n-1,n-1,n,n,n-1,n-1,...,2,2,1) for n >= 1.
+0
5
1, 1, 1, 2, 2, 1, 1, 2, 3, 3, 2, 1, 1, 2, 3, 4, 4, 3, 2, 1, 1, 2, 3, 4, 5, 5, 4, 3, 2, 1, 1, 2, 3, 4, 5, 6, 6, 5, 4, 3, 2, 1, 1, 2, 3, 4, 5, 6, 7, 7, 6, 5, 4, 3, 2, 1, 1, 2, 3, 4, 5, 6, 7, 8, 8, 7, 6, 5, 4, 3, 2, 1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 9, 8, 7, 6, 5, 4, 3, 2, 1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 10, 9, 8, 7
COMMENTS
Zeta(2, k/p) + Zeta(2, (p-k)/p) = (Pi/sin((Pi*a(n))/p))*2, where p=2,3,4, k=1..p-1.
This sequence is the odd subset of A003983 for odd p=3,5,7,9,.... Table T(n,k) n, k > 0, T(n,k) = n-k+1, if n >= k, T(n,k) = k-n, if n < k. Table read by sides of squares from T(1,n) to T(n,n), then from T(n,n) to T(n,1). General case A209301. Let m be a natural number. The first column of the table T(n,1) is the sequence of the natural numbers A000027. In all columns with number k (k > 1) the segment with the length of (k-1): {m+k-2, m+k-3, ..., m} shifts the sequence A000027. For m=1 the result is A004739, for m=2 the result is A004738, for m=3 the result is A209301. - Boris Putievskiy, Jan 24 2013
FORMULA
For the general case,
a(n) = m*v + (2*v-1)*(t*t-n) + t, where t = floor((sqrt(n) - 1/2) + 1 and v = floor((n-1)/t) - t + 1.
For m=1,
a(n) = v + (2*v-1)*(t*t-n) + t, where t = floor((sqrt(n) - 1/2) + 1 and v = floor((n-1)/t) - t + 1. (End)
EXAMPLE
The start of the sequence as table:
1, 1, 2, 3, 4, 5, 6, ...
2, 1, 1, 2, 3, 4, 5, ...
3, 2, 1, 1, 2, 3, 4, ...
4, 3, 2, 1, 1, 2, 3, ...
5, 4, 3, 2, 1, 1, 2, ...
6, 5, 4, 3, 2, 1, 1, ...
7, 6, 5, 4, 3, 2, 1, ...
...
The start of the sequence as triangle array read by rows:
1;
1, 1, 2;
2, 1, 1, 2, 3;
3, 2, 1, 1, 2, 3, 4;
4, 3, 2, 1, 1, 2, 3, 4, 5;
5, 4, 3, 2, 1, 1, 2, 3, 4, 5, 6;
6, 5, 4, 3, 2, 1, 1, 2, 3, 4, 5, 6, 7;
...
Row number r contains 2*r - 1 numbers: r-1, r-2, ..., 1, 1, 2, ..., r. (End)
MATHEMATICA
aa = {}; Do[Do[AppendTo[aa, (p/Pi) ArcSin[Sqrt[1/((1/Pi^2) (Zeta[2, k/p] + Zeta[2, (p - k)/p]))]]], {k, 1, p - 1}], {p, 3, 50, 2}]; Round[N[aa, 50]] (* Artur Jasinski, Mar 07 2010 *)
PROG
(Haskell)
a004739 n = a004739_list !! (n-1)
a004739_list = concat $ map (\n -> [1..n] ++ [n, n-1..1]) [1..]
The odd numbers: a(n) = 2*n + 1.
(Formerly M2400)
+0
1212
1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99, 101, 103, 105, 107, 109, 111, 113, 115, 117, 119, 121, 123, 125, 127, 129, 131
COMMENTS
Leibniz's series: Pi/4 = Sum_{n>=0} (-1)^n/(2n+1) (cf. A072172). Beginning of the ordering of the natural numbers used in Sharkovski's theorem - see the Cielsielski-Pogoda paper.
The Sharkovski ordering begins with the odd numbers >= 3, then twice these numbers, then 4 times them, then 8 times them, etc., ending with the powers of 2 in decreasing order, ending with 2^0 = 1.
Apart from initial term(s), dimension of the space of weight 2n cusp forms for Gamma_0(6).
a(1) = 1; a(n) is the smallest number such that a(n) + a(i) is composite for all i = 1 to n-1. - Amarnath Murthy, Jul 14 2003 Smallest number greater than n, not a multiple of n, but containing it in binary representation. - Reinhard Zumkeller, Oct 06 2003 Pi*sqrt(2)/4 = Sum_{n>=0} (-1)^floor(n/2)/(2n+1) = 1 + 1/3 - 1/5 - 1/7 + 1/9 + 1/11 ... [since periodic f(x)=x over -Pi < x < Pi = 2(sin(x)/1 - sin(2x)/2 + sin(3x)/3 - ...) using x = Pi/4 (Maor)]. - Gerald McGarvey, Feb 04 2005 a(n) = shortest side a of all integer-sided triangles with sides a <= b <= c and inradius n >= 1.
The odd numbers are the solution to the simplest recursion arising when assuming that the algorithm "merge sort" could merge in constant unit time, i.e., T(1):= 1, T(n):= T(floor(n/2)) + T(ceiling(n/2)) + 1. - Peter C. Heinig (algorithms(AT)gmx.de), Oct 14 2006
2n-5 counts the permutations in S_n which have zero occurrences of the pattern 312 and one occurrence of the pattern 123. - David Hoek (david.hok(AT)telia.com), Feb 28 2007
For n > 2, a(n-1) is the least integer not the sum of < n n-gonal numbers (0 allowed). - Jonathan Sondow, Jul 01 2007 Number of divisors of 4^(n-1) for n > 0. - J. Lowell, Aug 30 2008 Odd numbers(n) = 2*n+1 = square pyramidal number(3*n+1) / triangular number(3*n+1). - Pierre CAMI, Sep 27 2008 a(n) is also the maximum number of triangles that n+2 points in the same plane can determine. 3 points determine max 1 triangle; 4 points can give 3 triangles; 5 points can give 5; 6 points can give 7 etc. - Carmine Suriano, Jun 08 2009 Also the 3-rough numbers: positive integers that have no prime factors less than 3. - Michael B. Porter, Oct 08 2009 Given an L(2,1) labeling l of a graph G, let k be the maximum label assigned by l. The minimum k possible over all L(2,1) labelings of G is denoted by lambda(G). For n > 0, this sequence gives lambda(K_{n+1}) where K_{n+1} is the complete graph on n+1 vertices. - K.V.Iyer, Dec 19 2009 For n>0, continued fraction [1,1,n] = (n+1)/a(n); e.g., [1,1,7] = 8/15. - Gary W. Adamson, Jul 15 2010 Numbers that are the sum of two sequential integers. - Dominick Cancilla, Aug 09 2010 Cf. property described by Gary Detlefs in A113801: more generally, these numbers are of the form (2*h*n + (h-4)*(-1)^n - h)/4 (h and n in A000027), therefore ((2*h*n + (h-4)*(-1)^n - h)/4)^2 - 1 == 0 (mod h); in this case, a(n)^2 - 1 == 0 (mod 4). Also a(n)^2 - 1 == 0 (mod 8). - Bruno Berselli, Nov 17 2010 a(n) is the minimum number of tosses of a fair coin needed so that the probability of more than n heads is at least 1/2. In fact, Sum_{k=n+1..2n+1} Pr(k heads|2n+1 tosses) = 1/2. - Dennis P. Walsh, Apr 04 2012 1/N (i.e., 1/1, 1/2, 1/3, ...) = Sum_{j=1,3,5,...,infinity} k^j, where k is the infinite set of constants 1/exp.ArcSinh(N/2) = convergents to barover(N). The convergent to barover(1) or [1,1,1,...] = 1/phi = 0.6180339..., whereas c.f. barover(2) converges to 0.414213..., and so on. Thus, with k = 1/phi we obtain 1 = k^1 + k^3 + k^5 + ..., and with k = 0.414213... = (sqrt(2) - 1) we get 1/2 = k^1 + k^3 + k^5 + .... Likewise, with the convergent to barover(3) = 0.302775... = k, we get 1/3 = k^1 + k^3 + k^5 + ..., etc. - Gary W. Adamson, Jul 01 2012 Conjecture on primes with one coach ( A216371) relating to the odd integers: iff an integer is in A216371 (primes with one coach either of the form 4q-1 or 4q+1, (q > 0)); the top row of its coach is composed of a permutation of the first q odd integers. Example: prime 19 (q = 5), has 5 terms in each row of its coach: 19: [1, 9, 5, 7, 3] ... [1, 1, 1, 2, 4]. This is interpreted: (19 - 1) = (2^1 * 9), (19 - 9) = (2^1 * 5), (19 - 5) = (2^1 - 7), (19 - 7) = (2^2 * 3), (19 - 3) = (2^4 * 1). - Gary W. Adamson, Sep 09 2012 A005408 is the numerator 2n-1 of the term (1/m^2 - 1/n^2) = (2n-1)/(mn)^2, n = m+1, m > 0 in the Rydberg formula, while A035287 is the denominator (mn)^2. So the quotient a( A005408)/a( A035287) simulates the Hydrogen spectral series of all hydrogen-like elements. - Freimut Marschner, Aug 10 2013
This sequence has unique factorization. The primitive elements are the odd primes ( A065091). (Each term of the sequence can be expressed as a product of terms of the sequence. Primitive elements have only the trivial factorization. If the products of terms of the sequence are always in the sequence, and there is a unique factorization of each element into primitive elements, we say that the sequence has unique factorization. So, e.g., the composite numbers do not have unique factorization, because for example 36 = 4*9 = 6*6 has two distinct factorizations.) - Franklin T. Adams-Watters, Sep 28 2013 These are also numbers k such that (k^k+1)/(k+1) is an integer. - Derek Orr, May 22 2014 a(n-1) gives the number of distinct sums in the direct sum {1,2,3,..,n} + {1,2,3,..,n}. For example, {1} + {1} has only one possible sum so a(0) = 1. {1,2} + {1,2} has three distinct possible sums {2,3,4} so a(1) = 3. {1,2,3} + {1,2,3} has 5 distinct possible sums {2,3,4,5,6} so a(2) = 5. - Derek Orr, Nov 22 2014 The number of partitions of 4*n into at most 2 parts. - Colin Barker, Mar 31 2015 a(n) is representable as a sum of two but no fewer consecutive nonnegative integers, e.g., 1 = 0 + 1, 3 = 1 + 2, 5 = 2 + 3, etc. (see A138591). - Martin Renner, Mar 14 2016 Unique solution a( ) of the complementary equation a(n) = a(n-1)^2 - a(n-2)*b(n-1), where a(0) = 1, a(1) = 3, and a( ) and b( ) are increasing complementary sequences. - Clark Kimberling, Nov 21 2017 Also the number of maximal and maximum cliques in the n-centipede graph. - Eric W. Weisstein, Dec 01 2017 Lexicographically earliest sequence of distinct positive integers such that the average of any number of consecutive terms is always an integer. (For opposite property see A042963.) - Ivan Neretin, Dec 21 2017 Maximum number of non-intersecting line segments between vertices of a convex (n+2)-gon. - Christoph B. Kassir, Oct 21 2022 a(n) is the number of parking functions of size n+1 avoiding the patterns 123, 132, and 231. - Lara Pudwell, Apr 10 2023 a(n) is the maximum number of triangles in planar connected graphs of triangles with n+3 nodes. - Ya-Ping Lu, Jun 25 2024
REFERENCES
T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 2.
T. Dantzig, The Language of Science, 4th Edition (1954) page 276.
H. Doerrie, 100 Great Problems of Elementary Mathematics, Dover, NY, 1965, p. 73.
D. Hök, Parvisa mönster i permutationer [Swedish], (2007).
E. Maor, Trigonometric Delights, Princeton University Press, NJ, 1998, pp. 203-205.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
D. Applegate and J. C. Lagarias, The 3x+1 semigroup, Journal of Number Theory, Vol. 177, Issue 1, March 2006, pp. 146-159; see also the arXiv version, arXiv:math/0411140 [math.NT], 2004-2005.
FORMULA
a(n) = 2*n + 1. a(-1 - n) = -a(n). a(n+1) = a(n) + 2.
G.f.: (1 + x) / (1 - x)^2.
E.g.f.: (1 + 2*x) * exp(x).
G.f. with interpolated zeros: (x^3+x)/((1-x)^2 * (1+x)^2); e.g.f. with interpolated zeros: x*(exp(x)+exp(-x))/2. - Geoffrey Critzer, Aug 25 2012 Euler transform of length 2 sequence [3, -1]. - Michael Somos, Mar 30 2007 G.f. A(x) satisfies 0 = f(A(x), A(x^2)) where f(u, v) = v * (1 + 2*u) * (1 - 2*u + 16*v) - (u - 4*v)^2 * (1 + 2*u + 2*u^2). - Michael Somos, Mar 30 2007 a(n) = b(2*n + 1) where b(n) = n if n is odd is multiplicative. [This seems to say that A000027 is multiplicative? - R. J. Mathar, Sep 23 2011] a(n) = (n+1)^2 - n^2.
G.f. g(x) = Sum_{k>=0} x^floor(sqrt(k)) = Sum_{k>=0} x^ A000196(k). (End) a(n) = (n - 1) + n (sum of two sequential integers). - Dominick Cancilla, Aug 09 2010 n*a(2n+1)^2+1 = (n+1)*a(2n)^2; e.g., 3*15^2+1 = 4*13^2. - Charlie Marion, Dec 31 2010 arctanh(x) = Sum_{n>=0} x^(2n+1)/a(n). - R. J. Mathar, Sep 23 2011 a(n) = det(f(i-j+1))_{1<=i,j<=n}, where f(n) = A113311(n); for n < 0 we have f(n)=0. - Mircea Merca, Jun 23 2012 G.f.: Q(0), where Q(k) = 1 + 2*(k+1)*x/( 1 - 1/(1 + 2*(k+1)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 11 2013 a(n) = Product_{k=1..2*n} 2*sin(Pi*k/(2*n+1)) = Product_{k=1..n} (2*sin(Pi*k/(2*n+1)))^2, n >= 0 (undefined product = 1). See an Oct 09 2013 formula contribution in A000027 with a reference. - Wolfdieter Lang, Oct 10 2013 Noting that as n -> infinity, sqrt(n^2 + n) -> n + 1/2, let f(n) = n + 1/2 - sqrt(n^2 + n). Then for n > 0, a(n) = round(1/f(n))/4. - Richard R. Forberg, Feb 16 2014 a(n) = Sum_{k=0..n+1} binomial(2*n+1,2*k)*4^(k)*bernoulli(2*k). - Vladimir Kruchinin, Feb 24 2015 a(n) = Sum_{k=0..n} binomial(6*n+3, 6*k)*Bernoulli(6*k). - Michel Marcus, Jan 11 2016 O.g.f.: Sum_{n >= 1} phi(2*n-1)*x^(n-1)/(1 - x^(2*n-1)), where phi(n) is the Euler totient function A000010. - Peter Bala, Mar 22 2019 Sum_{n >= 1} (-1)^n/(a(n)*a(n+1)) = Pi/4 - 1/2 = 1/(3 + (1*3)/(4 + (3*5)/(4 + ... + (4*n^2 - 1)/(4 + ... )))). Cf. A016754. - Peter Bala, Mar 28 2024 a(n) = A055112(n)/oblong(n) = A193218(n+1)/Hex number(n). Compare to the Sep 27 2008 comment by Pierre CAMI. - Klaus Purath, Apr 23 2024 a(k*m) = k*a(m) - (k-1). - Ya-Ping Lu, Jun 25 2024
EXAMPLE
G.f. = q + 3*q^3 + 5*q^5 + 7*q^7 + 9*q^9 + 11*q^11 + 13*q^13 + 15*q^15 + ...
MATHEMATICA
CoefficientList[Series[(1 + x)/(-1 + x)^2, {x, 0, 20}], x] (* Eric W. Weisstein, Dec 01 2017 *)
PROG
(Magma) [ 2*n+1 : n in [0..100]];
(PARI) {a(n) = 2*n + 1}
(PARI) first(n) = Vec((1 + x)/(1 - x)^2 + O(x^n)) \\ Iain Fox, Dec 29 2017 (Haskell)
a005408 n = (+ 1) . (* 2)
(Maxima) makelist(2*n+1, n, 0, 30); /* Martin Ettl, Dec 11 2012 */
CROSSREFS
See A120062 for sequences related to integer-sided triangles with integer inradius n. Cf. A000754 (boustrophedon transform).
EXTENSIONS
Incorrect comment and example removed by Joerg Arndt, Mar 11 2010
a(n) = n*(n+2) = (n+1)^2 - 1.
(Formerly M2720)
+0
308
0, 3, 8, 15, 24, 35, 48, 63, 80, 99, 120, 143, 168, 195, 224, 255, 288, 323, 360, 399, 440, 483, 528, 575, 624, 675, 728, 783, 840, 899, 960, 1023, 1088, 1155, 1224, 1295, 1368, 1443, 1520, 1599, 1680, 1763, 1848, 1935, 2024, 2115, 2208, 2303, 2400, 2499, 2600
COMMENTS
Erdős conjectured that n^2 - 1 = k! has a solution if and only if n is 5, 11 or 71 (when k is 4, 5 or 7).
Second-order linear recurrences y(m) = 2y(m-1) + a(n)*y(m-2), y(0) = y(1) = 1, have closed form solutions involving only powers of integers. - Len Smiley, Dec 08 2001 Number of edges in the join of two cycle graphs, both of order n, C_n * C_n. - Roberto E. Martinez II, Jan 07 2002 Let k be a positive integer, M_n be the n X n matrix m_(i,j) = k^abs(i-j) then det(M_n) = (-1)^(n-1)*a(k-1)^(n-1). - Benoit Cloitre, May 28 2002 Also numbers k such that 4*k + 4 is a square. - Cino Hilliard, Dec 18 2003 For each term k, the function sqrt(x^2 + 1), starting with 1, produces an integer after k iterations. - Gerald McGarvey, Aug 19 2004 a(n) is the number of divisors of a(n+1) that are not greater than n. - Reinhard Zumkeller, Apr 09 2007 Nonnegative X values of solutions to the equation X^3 + X^2 = Y^2. To find Y values: b(n) = n(n+1)(n+2). - Mohamed Bouhamida, Nov 06 2007 Sequence allows us to find X values of the equation: X + (X + 1)^2 + (X + 2)^3 = Y^2. To prove that X = n^2 + 2n: Y^2 = X + (X + 1)^2 + (X + 2)^3 = X^3 + 7*X^2 + 15X + 9 = (X + 1)(X^2 + 6X + 9) = (X + 1)*(X + 3)^2 it means: (X + 1) must be a perfect square, so X = k^2 - 1 with k>=1. we can put: k = n + 1, which gives: X = n^2 + 2n and Y = (n + 1)(n^2 + 2n + 3). - Mohamed Bouhamida, Nov 12 2007 Toads and Frogs puzzle:
This is also the number of moves that it takes n frogs to swap places with n toads on a strip of 2n + 1 squares (or positions, or lily pads) where a move is a single slide or jump, illustrated for n = 2, a(n) = 8 by
T T - F F
T - T F F
T F T - F
T F T F -
T F - F T
- F T F T
F - T F T
F F T - T
F F - T T
I was alerted to this by the Holton article, but on consulting Singmaster's sources, I find that the puzzle goes back at least to 1867.
Probably the first to publish the number of moves for n of each animal was Edouard Lucas in 1883. (End)
Final digit belongs to a periodic sequence: 0, 3, 8, 5, 4, 5, 8, 3, 0, 9. - Mohamed Bouhamida, Sep 04 2009 [Comment edited by N. J. A. Sloane, Sep 24 2009] Let f(x) be a polynomial in x. Then f(x + n*f(x)) is congruent to 0 (mod(f(x)); here n belongs to N. There is nothing interesting in the quotients f(x + n*f(x))/f(x) when x belongs to Z. However, when x is irrational these quotients consist of two parts, a) rational integers and b) integer multiples of x. The present sequence represents the non-integer part when the polynomial is x^2 + x + 1 and x = sqrt(2), f(x+n*f(x))/f(x) = A056108(n) + a(n)*sqrt(2). - A.K. Devaraj, Sep 18 2009 For n >= 1, a(n) is the number for which 1/a(n) = 0.0101... ( A000035) in base (n+1). - Rick L. Shepherd, Sep 27 2009 For n > 0, continued fraction [n, 1, n] = (n+1)/a(n); e.g., [6, 1, 6] = 7/48. - Gary W. Adamson, Jul 15 2010 Starting (3, 8, 15, ...) = binomial transform of [3, 5, 2, 0, 0, 0, ...]; e.g., a(3) = 15 = (1*3 + 2*5 +1*2) = (3 + 10 + 2). - Gary W. Adamson, Jul 30 2010 a(n) is essentially the case 0 of the polygonal numbers. The polygonal numbers are defined as P_k(n) = Sum_{i=1..n} ((k-2)*i-(k-3)). Thus P_0(n) = 2*n-n^2 and a(n) = -P_0(n+2). See also A067998 and for the case k=1 A080956. - Peter Luschny, Jul 08 2011 a(n) is the maximal determinant of a 2 X 2 matrix with integer elements from {1, ..., n+1}, so the maximum determinant of a 2x2 matrix with integer elements from {1, ..., 5} = 5^2 - 1 = a(4) = 24. - Aldo González Lorenzo, Oct 12 2011 Using four consecutive triangular numbers t1, t2, t3 and t4, plot the points (0, 0), (t1, t2), and (t3, t4) to create a triangle. Twice the area of this triangle are the numbers in this sequence beginning with n = 1 to give 8. - J. M. Bergot, May 03 2012 Given a particle with spin S = n/2 (always a half-integer value), the quantum-mechanical expectation value of the square of the magnitude of its spin vector evaluates to <S^2> = S(S+1) = n(n+2)/4, i.e., one quarter of a(n) with n = 2S. This plays an important role in the theory of magnetism and magnetic resonance. - Stanislav Sykora, May 26 2012 Twice the harmonic mean [H(x, y) = (2*x*y)/(x + y)] of consecutive triangular numbers A000217(n) and A000217(n+1). - Raphie Frank, Sep 28 2012 Number m such that floor(sqrt(m)) = floor(m/floor(sqrt(m))) - 2 for m > 0. - Takumi Sato, Oct 10 2012 The solutions of equation 1/(i - sqrt(j)) = i + sqrt(j), when i = (n+1), j = a(n). For n = 1, 2 + sqrt(3) = 3.732050.. = A019973. For n = 2, 3 + sqrt(8) = 5.828427... = A156035. - Kival Ngaokrajang, Sep 07 2013 The integers in the closed form solution of a(n) = 2*a(n-1) + a(m-2)*a(n-2), n >= 2, a(0) = 0, a(1) = 1 mentioned by Len Smiley, Dec 08 2001, are m and -m + 2 where m >= 3 is a positive integer. - Felix P. Muga II, Mar 18 2014 Let m >= 3 be a positive integer. If a(n) = 2*a(n-1) + a(m-2) * a(n-2), n >= 2, a(0) = 0, a(1) = 1, then lim_{n->oo} a(n+1)/a(n) = m. - Felix P. Muga II, Mar 18 2014 For n >= 4 the Szeged index of the wheel graph W_n (with n + 1 vertices). In the Sarma et al. reference, Theorem 2.7 is incorrect. - Emeric Deutsch, Aug 07 2014 If P_{k}(n) is the n-th k-gonal number, then a(n) = t*P_{s}(n+2) - s*P_{t}(n+2) for s=t+1. - Bruno Berselli, Sep 04 2014 For n >= 1, a(n) is the dimension of the simple Lie algebra A_n. - Wolfdieter Lang, Oct 21 2015 Finding all positive integers (n, k) such that n^2 - 1 = k! is known as Brocard's problem, (see A085692). - David Covert, Jan 15 2016 For n > 0, a(n) mod (n+1) = a(n) / (n+1) = n. - Torlach Rush, Apr 04 2016 Conjecture: When using the Sieve of Eratosthenes and sieving (n+1..a(n)), with divisors (1..n) and n>0, there will be no more than a(n-1) composite numbers. - Fred Daniel Kline, Apr 08 2016 a(n) mod 8 is periodic with period 4 repeating (0,3,0,7), that is a(n) mod 8 = 5/2 - (5/2) cos(n*Pi) - sin(n*Pi/2) + sin(3*n*Pi/2). - Andres Cicuttin, Jun 02 2016 Also for n > 0, a(n) is the number of times that n-1 occurs among the first (n+1)! terms of A055881. - R. J. Cano, Dec 21 2016 The second diagonal of composites (the only prime is number 3) from the right on the Klauber triangle (see Kival Ngaokrajang link), which is formed by taking the positive integers and taking the first 1, the next 3, the following 5, and so on, each centered below the last. - Charles Kusniec, Jul 03 2017 Also the number of independent vertex sets in the n-barbell graph. - Eric W. Weisstein, Aug 16 2017 a(n) is the number of degrees of freedom in a triangular cell for a Raviart-Thomas or Nédélec first kind finite element space of order n. - Matthew Scroggs, Apr 22 2020 For n > 1, a(n-2) is the maximum number of elements in the second stage of the Quine-McCluskey algorithm whose minterms are not covered by the functions of n bits. At n=3, we have a(3-2) = a(1) = 1*(1+2) = 3 and f(A,B,C) = sigma(0,1,2,5,6,7).
.
0 1 2 5 6 7
+---------------
*(0,1)| X X
(0,2)| X X
(1,5)| X X
*(2,6)| X X
*(5,7)| X X
(6,7)| X X
.
*: represents the elements that are covered. (End)
1/a(n) is the ratio of the sum of the first k odd numbers and the sum of the next n*k odd numbers. - Melvin Peralta, Jul 15 2021 For n >= 1, the continued fraction expansion of sqrt(a(n)) is [n; {1, 2n}]. - Magus K. Chu, Sep 09 2022 Number of diagonals parallel to an edge in a regular (2*n+4)-gon (cf. A367204). - Paolo Xausa, Nov 21 2023
REFERENCES
E. R. Berlekamp, J. H. Conway and R. K. Guy, Winning Ways, Academic Press, NY, 2 vols., 1982, see index under Toads and Frogs Puzzle.
Martin Gardner, Perplexing Puzzles and Tantalizing Teasers, p. 21 (for "The Dime and Penny Switcheroo").
R. K. Guy, Unsolved Problems in Theory of Numbers, Section D25.
Derek Holton, Math in School, 37 #1 (Jan 2008) 20-22.
Edouard Lucas, Récréations Mathématiques, Gauthier-Villars, Vol. 2 (1883) 141-143.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
Spencer Daugherty, Pamela E. Harris, Ian Klein, and Matt McClinton, Metered Parking Functions, arXiv:2406.12941 [math.CO], 2024. See p. 22.
FORMULA
a(n) = (n! + (n+1)!)/(n-1)!, n > 0. - Gary Detlefs, Aug 10 2009 a(n) = floor(n^5/(n^3+1)) with offset 1 (a(1)=0). - Gary Detlefs, Feb 11 2010 a(n) = 2/(Integral_{x=0..Pi/2} (sin(x))^(n-1)*(cos(x))^3), for n > 0. - Francesco Daddi, Aug 02 2011 G.f.: U(0) where U(k) = -1 + (k+1)^2/(1 - x/(x + (k+1)^2/U(k+1))); (continued fraction, 3-step). - Sergei N. Gladkovskii, Oct 19 2012 a(n) = 15*C(n+4,3)*C(n+4,5)/(C(n+4,2)*C(n+4,4)). - Gary Detlefs, Aug 05 2013 For n >= 1, a(n^2 + n - 2) = a(n-1) * a(n). - Miko Labalan, Oct 15 2017 Sum_{n>=1} (-1)^(n+1)/a(n) = 1/4. - Amiram Eldar, Nov 04 2020 Product_{n>=1} (1 + 1/a(n)) = 2.
Product_{n>=1} (1 - 1/a(n)) = -sqrt(2)*sin(sqrt(2)*Pi)/Pi. (End)
EXAMPLE
G.f. = 3*x + 8*x^2 + 15*x^3 + 24*x^4 + 35*x^5 + 48*x^6 + 63*x^7 + 80*x^8 + ...
MATHEMATICA
ListCorrelate[{1, 2}, Range[-1, 50], {1, -1}, 0, Plus, Times] (* Harvey P. Dale, Aug 29 2015 *)
PROG
(PARI) concat(0, Vec(x*(3-x)/(1-x)^3 + O(x^90))) \\ Altug Alkan, Oct 22 2015 (Maxima) makelist(n*(n+2), n, 0, 56); /* Martin Ettl, Oct 15 2012 */ (Haskell)
a005563 n = n * (n + 2)
(SageMath) [n*(n+2) for n in range(61)] # G. C. Greubel, Mar 29 2024
CROSSREFS
a(n+1), n>=2, first column of triangle A120070.
Number of letters in the US English name of n, excluding spaces and hyphens.
(Formerly M2277)
+0
97
4, 3, 3, 5, 4, 4, 3, 5, 5, 4, 3, 6, 6, 8, 8, 7, 7, 9, 8, 8, 6, 9, 9, 11, 10, 10, 9, 11, 11, 10, 6, 9, 9, 11, 10, 10, 9, 11, 11, 10, 5, 8, 8, 10, 9, 9, 8, 10, 10, 9, 5, 8, 8, 10, 9, 9, 8, 10, 10, 9, 5, 8, 8, 10, 9, 9, 8, 10, 10, 9, 7, 10, 10, 12, 11, 11, 10, 12, 12, 11, 6, 9, 9, 11, 10, 10, 9, 11, 11, 10, 6, 9, 9, 11, 10, 10, 9, 11, 11, 10, 10, 13, 13, 15, 14, 14, 13, 15, 15, 14, 13, 16, 16, 18, 18, 17, 17, 19, 18, 18, 16
COMMENTS
Diane Karloff observes (Nov 27 2007) that repeatedly applying the map k-> A005589(k) to any starting value n always leads to 4 (cf. A016037, A133418). The above observation was previously made in 1972 by R. Schroeppel and R. W. Gosper in HAKMEM (Item 134). - Bartlomiej Pawlik, Jun 12 2023 For terms beyond a(100), this sequence uses the US English style, "one hundred one" (not "one hundred and one"), and the short scale (a billion = 10^9, not 10^12). - M. F. Hasler, Nov 03 2013 Explanation of Diane Karloff's observation above: In many languages there exists a number N, after which all numbers are written with fewer letters than the number itself. N is 4 in English, German and Bulgarian, and 11 in Russian. If in the interval [1,N] there are numbers equal to the number of their letters, then they are attractors. In English and German the only attractor is 4, in Bulgarian 3, in Russian there are two, 3 and 11. In the interval [1,N] there may also exist loops of numbers, for instance 4 and 6 in Bulgarian (6 and 4 letters respectively) or 4,5 and 6 in Russian (6,4 and 5 letters respectively). There are no loops in English, therefore the above observation is true. - Ivan N. Ianakiev, Sep 20 2014
REFERENCES
Problems Drive, Eureka, 37 (1974), 8-11 and 33.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
M. Beeler, R. W. Gosper and R. Schroeppel, Item 134, in Beeler, M., Gosper, R. W. and Schroeppel, R. HAKMEM. MIT AI Memo 239, Feb 29 1972. Eureka, Problems Drive, Eureka, 37 (1974), 8-11, 32-33, 24-27. (Annotated scanned copy) Eric Weisstein's World of Mathematics, Number.
EXAMPLE
Note that A052360(373373) = 64 whereas a(373373) = 56.
MATHEMATICA
inWords[n_] := Module[{r,
numNames = {"", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine"},
teenNames = {"ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen"},
tensNames = {"", "ten", "twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety"},
decimals = {"", "thousand", "million", "billion", "trillion", "quadrillion", "quintillion", "sextillion", "septillion", "octillion", "nonillion", "decillion", "undecillion", "duodecillion", "tredecillion", "quattuordecillion", "quindecillion", "sexdecillion", "septendecillion", "octodecillion", "novemdecillion", "vigintillion", "unvigintillion", "duovigintillion", "trevigintillion", "quattuorvigintillion", "quinvigintillion", "sexvigintillion", "septenvigintillion", "octovigintillion", "novemvigintillion", "trigintillion", "untrigintillion", "duotrigintillion"}},
r = If[# != 0, numNames[[# + 1]] <> "hundred"
(* <> If[#2 != 0||#3 != 0, " and", ""] *),
""] <> Switch[#2, 0, numNames[[#3 + 1]], 1, teenNames[[#3 + 1]], _, tensNames[[#2 + 1]] <> numNames[[#3 + 1]]] & @@@
(PadLeft[ FromDigits /@ Characters@ StringReverse@#, 3] & /@ StringCases[ StringReverse@ IntegerString@ n, RegularExpression["\\d{1, 3}"]]);
StringJoin@ Reverse@ MapThread[ If[# != "", StringJoin[##], ""] &, {r, Take[decimals, Length@ r]} ]]; (* modified for this sequence from what is presented in the link and good to 10^102 -1 *)
f[n_] := StringLength@ inWords@ n; f[0] = 4; Array[f, 84, 0]
a[n_] := StringLength[ StringReplace[ IntegerName[n, "Words"], ", " | " " | "\[Hyphen]" -> ""]]; a /@ Range[0, 83] (* Mma version >= 10, Giovanni Resta, Apr 10 2017 *)
PROG
(PARI) A005589(n, t=[10^12, #"trillion", 10^9, #"billion", 10^6, #"million", 1000, #"thousand", 100, #"hundred"])={ n>99 && forstep( i=1, #t, 2, n<t[i] && next; n=divrem(n, t[i]); n[1]>999 && error("n >= 10^", valuation(t[1], 10)+3, " requires extended 2nd argument"); return( A005589(n[1])+t[i+1]+if( n[2], A005589( n[2] )))); if( n<20, #(["zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen"][n+1]), #([ "twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety" ][n\10-1])+if( n%10, A005589(n%10)))} \\ Extension of 2nd arg to larger numbers is easy using the names listed in Mathematica section above. Only the string lengths are required, so it's easy to extend this list further without additional knowledge and without writing out the names. - M. F. Hasler, Jul 26 2011, minor edit on Jun 15 2021 (Python)
from num2words import num2words
def a(n):
x = num2words(n).replace(' and ', '')
l = [chr(i) for i in range(97, 123)]
return sum(1 for i in x if i in l)
EXTENSIONS
Corrected and extended by Larry Reeves (larryr(AT)acm.org) and Allan C. Wechsler, Mar 20 2000 More than the usual number of terms are shown in the DATA field to avoid confusion with A362123. - N. J. A. Sloane, Apr 20 2023
Numbers whose base-3 representation contains no 2.
(Formerly M2353)
+0
239
0, 1, 3, 4, 9, 10, 12, 13, 27, 28, 30, 31, 36, 37, 39, 40, 81, 82, 84, 85, 90, 91, 93, 94, 108, 109, 111, 112, 117, 118, 120, 121, 243, 244, 246, 247, 252, 253, 255, 256, 270, 271, 273, 274, 279, 280, 282, 283, 324, 325, 327, 328, 333, 334, 336, 337, 351, 352
COMMENTS
3 does not divide binomial(2s, s) if and only if s is a member of this sequence, where binomial(2s, s) = A000984(s) are the central binomial coefficients. This is the lexicographically earliest increasing sequence of nonnegative numbers that contains no arithmetic progression of length 3. - Robert Craigen (craigenr(AT)cc.umanitoba.ca), Jan 29 2001
Also final value of n - 1 written in base 2 and then read in base 3 and with finally the result translated in base 10. - Philippe LALLOUET (philip.lallouet(AT)wanadoo.fr), Jun 23 2007
Also numbers such that the balanced ternary representation is the same as the base 3 representation. - Alonso del Arte, Feb 25 2011 Fixed point of the morphism: 0 -> 01; 1 -> 34; 2 -> 67; ...; n -> (3n)(3n+1), starting from a(1) = 0. - Philippe Deléham, Oct 22 2011 It appears that this sequence lists the values of n which satisfy the condition sum(binomial(n, k)^(2*j), k = 0..n) mod 3 <> 0, for any j, with offset 0. See Maple code. - Gary Detlefs, Nov 28 2011 Also, it follows from the above comment by Philippe Lallouet that the sequence must be generated by the rules: a(1) = 0, and if m is in the sequence then so are 3*m and 3*m + 1. - L. Edson Jeffery, Nov 20 2015
REFERENCES
Richard K. Guy, Unsolved Problems in Number Theory, 3rd Edition, Springer, 2004, Section E10, pp. 317-323.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
Megumi Asada, Bruce Fang, Eva Fourakis, Sarah Manski, Nathan McNew, Steven J. Miller, Gwyneth Moreland, Ajmain Yamin, and Sindy Xin Zhang, Avoiding 3-Term Geometric Progressions in Hurwitz Quaternions, Williams College (2023). Matvey Borodin, Hannah Han, Kaylee Ji, Tanya Khovanova, Alexander Peng, David Sun, Isabel Tu, Jason Yang, William Yang, Kevin Zhang, and Kevin Zhao, Variants of Base 3 over 2, arXiv:1901.09818 [math.NT], 2019. Ben Chen, Richard Chen, Joshua Guo, Tanya Khovanova, Shane Lee, Neil Malur, Nastia Polina, Poonam Sahoo, Anuj Sakarda, Nathan Sheffield, and Armaan Tipirneni, On Base 3/2 and its Sequences, arXiv:1808.04304 [math.NT], 2018. P. Erdős, V. Lev, G. Rauzy, C. Sandor, and A. Sarkozy, Greedy algorithm, arithmetic progressions, subset sums and divisibility, Discrete Math., Vol. 200, No. 1-3 (1999), pp. 119-135 (see Table 1). alternate link. A. M. Odlyzko and R. P. Stanley, Some curious sequences constructed with the greedy algorithm, 1978, remark 1 ( PDF, PS, TeX).
FORMULA
Numbers n such that the coefficient of x^n is > 0 in prod (k >= 0, 1 + x^(3^k)). - Benoit Cloitre, Jul 29 2003 a(n+1) = Sum_{k=0..m} b(k)* 3^k and n = Sum( b(k)* 2^k ).
a(2n+1) = 3a(n+1), a(2n+2) = a(2n+1) + 1, a(0) = 0.
a(n+1) = 3*a(floor(n/2)) + n - 2*floor(n/2). - Ralf Stephan, Apr 27 2003 G.f.: (x/(1-x)) * Sum_{k>=0} 3^k*x^2^k/(1+x^2^k). - Ralf Stephan, Apr 27 2003 If the offset were changed to zero, then: a(0) = 0, a(n+1) = f(a(n) + 1, f(a(n)+1) where f(x, y) = if x < 3 and x <> 2 then y else if x mod 3 = 2 then f(y+1, y+1) else f(floor(x/3), y). (End)
We have liminf_{n->infinity} a(n)/n^(log(3)/log(2)) = 1/2 and limsup_{n->infinity} a(n)/n^(log(3)/log(2)) = 1. - Gheorghe Coserea, Sep 13 2015 a(2^k+m) = a(m) + 3^k with 1 <= m <= 2^k and 1 <= k, a(1)=0, a(2)=1. - Paul Weisenhorn, Mar 22 2020 Sum_{n>=2} 1/a(n) = 2.682853110966175430853916904584699374821677091415714815171756609672281184705... (calculated using Baillie and Schmelzer's kempnerSums.nb, see Links). - Amiram Eldar, Feb 12 2022 a(n) ≍ n^k, where k = log 3/log 2 = 1.5849625007. (I believe the constant varies from 1/2 to 1.) - Charles R Greathouse IV, Mar 29 2024
EXAMPLE
a(6) = 12 because 6 = 0*2^0 + 1*2^1 + 1*2^2 = 2+4 and 12 = 0*3^0 + 1*3^1 + 1*3^2 = 3 + 9.
This sequence regarded as a triangle with rows of lengths 1, 1, 2, 4, 8, 16, ...:
0
1
3, 4
9, 10, 12, 13
27, 28, 30, 31, 36, 37, 39, 40
81, 82, 84, 85, 90, 91, 93, 94, 108, 109, 111, 112, 117, 118, 120, 121
MAPLE
t := (j, n) -> add(binomial(n, k)^j, k=0..n):
for i from 1 to 400 do
if(t(4, i) mod 3 <>0) then print(i) fi
# alternative Maple program:
a:= proc(n) option remember: local k, m:
if n=1 then 0 elif n=2 then 1 elif n>2 then k:=floor(log[2](n-1)): m:=n-2^k: procname(m)+3^k: fi: end proc:
# third Maple program:
a:= n-> `if`(n=1, 0, irem(n-1, 2, 'q')+3*a(q+1)):
MATHEMATICA
Table[FromDigits[IntegerDigits[k, 2], 3], {k, 60}]
Select[Range[0, 400], DigitCount[#, 3, 2] == 0 &] (* Harvey P. Dale, Jan 04 2012 *) Join[{0}, Accumulate[Table[(3^IntegerExponent[n, 2] + 1)/2, {n, 57}]]] (* IWABUCHI Yu(u)ki, Aug 01 2012 *) FromDigits[#, 3]&/@Tuples[{0, 1}, 7] (* Harvey P. Dale, May 10 2019 *)
PROG
(Haskell)
a005836 n = a005836_list !! (n-1)
a005836_list = filter ((== 1) . a039966) [0..]
(Python)
return int(format(n-1, 'b'), 3) # Chai Wah Wu, Jan 04 2015 (Julia)
function a(n)
m, r, b = n, 0, 1
while m > 0
m, q = divrem(m, 2)
r += b * q
b *= 3
end
r end; [a(n) for n in 0:57] |> println # Peter Luschny, Jan 03 2021
CROSSREFS
Cf. A039966 (characteristic function). Cf. A002426, A004793, A005823, A007088, A007089, A032924, A033042- A033052, A054591, A055246, A062548, A065361, A074940, A081601, A081603, A081611, A083096, A089118, A121153, A170943, A185256. For generating functions Product_{k>=0} (1+a*x^(b^k)) for the following values of (a,b) see: (1,2) A000012 and A000027, (1,3) A039966 and A005836, (1,4) A151666 and A000695, (1,5) A151667 and A033042, (2,2) A001316, (2,3) A151668, (2,4) A151669, (2,5) A151670, (3,2) A048883, (3,3) A117940, (3,4) A151665, (3,5) A151671, (4,2) A102376, (4,3) A151672, (4,4) A151673, (4,5) A151674. Summary of increasing sequences avoiding arithmetic progressions of specified lengths (the second of each pair is obtained by adding 1 to the first):
EXTENSIONS
Edited by the Associate Editors of the OEIS, Apr 07 2009
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