OFFSET
0,3
COMMENTS
Weigh transform of the centered triangular numbers (A005448).
This sequence is obtained from the generalized Euler transform in A266964 by taking f(n) = -(3*n*(n-1)/2+1), g(n) = -1. - Seiichi Manyama, Nov 16 2017
LINKS
Seiichi Manyama, Table of n, a(n) for n = 0..10000
M. Bernstein and N. J. A. Sloane, Some canonical sequences of integers, Linear Alg. Applications, 226-228 (1995), 57-72; erratum 320 (2000), 210. [Link to arXiv version]
M. Bernstein and N. J. A. Sloane, Some canonical sequences of integers, Linear Alg. Applications, 226-228 (1995), 57-72; erratum 320 (2000), 210. [Link to Lin. Alg. Applic. version together with omitted figures]
N. J. A. Sloane, Transforms
Eric Weisstein's World of Mathematics, Centered Triangular Number
FORMULA
G.f.: Product_{k>=1} (1 + x^k)^A005448(k).
a(n) ~ exp(15*Zeta(3) / (28*Pi^2) - 6075*Zeta(3)^3 / (98*Pi^8) + (Pi/6 - 405*Zeta(3)^2 / (28*Pi^5)) * (5*n/7)^(1/4) - (9*sqrt(5/7) * Zeta(3) / (2*Pi^2)) * sqrt(n) + (2*Pi * (7/5)^(1/4)/3) * n^(3/4)) * 7^(1/8) / (2^(19/8) * 5^(1/8) * n^(5/8)). - Vaclav Kotesovec, Nov 16 2017
a(0) = 1 and a(n) = (1/n) * Sum_{k=1..n} b(k)*a(n-k) where b(n) = Sum_{d|n} d*(3*d*(d-1)/2+1)*(-1)^(1+n/d). - Seiichi Manyama, Nov 16 2017
MATHEMATICA
nmax = 33; CoefficientList[Series[Product[(1 + x^k)^(3 k (k - 1)/2 + 1), {k, 1, nmax}], {x, 0, nmax}], x]
a[n_] := a[n] = If[n == 0, 1, Sum[Sum[(-1)^(k/d + 1) d (3 d (d - 1)/2 + 1), {d, Divisors[k]}] a[n - k], {k, 1, n}]/n]; Table[a[n], {n, 0, 33}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Ilya Gutkovskiy, Nov 16 2017
STATUS
approved