OFFSET
1,8
COMMENTS
Also, the number of partitions of n such that (greatest part) = 2*(number of parts); hence, the number of partitions of n such that (rank + greatest part) = 0.
LINKS
Vaclav Kotesovec, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Seiichi Manyama)
Vaclav Kotesovec, Graph - the asymptotic ratio (20000 terms)
FORMULA
G.f.: Sum_{k>=1} x^(3*k-1) * Product_{j=1..k-1} (1-x^(2*k+j-1))/(1-x^j). - Seiichi Manyama, Jan 24 2022
a(n) ~ Pi^2 * exp(Pi*sqrt(2*n/3)) / (4 * 3^(3/2) * n^2). - Vaclav Kotesovec, Oct 17 2024
EXAMPLE
a(8) = 2 counts these partitions: 311111, 2222.
MATHEMATICA
z = 50; Table[Count[IntegerPartitions[n], p_ /; 2 Max[p] = = Length[p]], {n, z}]
(* or *)
nmax = 100; Rest[CoefficientList[Series[Sum[x^(3*k-1) * Product[(1 - x^(2*k+j-1)) / (1 - x^j), {j, 1, k-1}], {k, 1, nmax/3 + 1}], {x, 0, nmax}], x]] (* Vaclav Kotesovec, Oct 15 2024 *)
nmax = 100; p = x; s = x; Do[p = Normal[Series[p*x^3*(1 - x^(3*k - 1))*(1 - x^(3*k))*(1 - x^(3*k + 1))/((1 - x^(2*k + 1))*(1 - x^(2*k))*(1 - x^k)), {x, 0, nmax}]]; s += p; , {k, 1, nmax/3 + 1}]; Take[CoefficientList[s, x], nmax] (* Vaclav Kotesovec, Oct 16 2024 *)
PROG
(PARI) my(N=66, x='x+O('x^N)); concat(0, Vec(sum(k=1, N, x^(3*k-1)*prod(j=1, k-1, (1-x^(2*k+j-1))/(1-x^j))))) \\ Seiichi Manyama, Jan 24 2022
CROSSREFS
KEYWORD
nonn,easy,changed
AUTHOR
Clark Kimberling, Feb 13 2014
STATUS
approved