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A237753
Number of partitions of n such that 2*(greatest part) = (number of parts).
24
0, 1, 0, 0, 1, 1, 1, 2, 1, 2, 3, 4, 5, 7, 7, 9, 12, 15, 17, 23, 27, 34, 42, 50, 60, 75, 87, 106, 128, 154, 182, 222, 260, 311, 369, 437, 515, 613, 716, 845, 993, 1166, 1361, 1599, 1861, 2176, 2534, 2950, 3422, 3983, 4605, 5339, 6174, 7136, 8227, 9500, 10928
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)
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