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A237756
Number of partitions of n such that 3*(greatest part) = (number of parts).
6
0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 4, 4, 6, 7, 10, 10, 13, 14, 19, 21, 27, 31, 40, 45, 55, 64, 79, 91, 111, 127, 154, 177, 211, 243, 290, 333, 394, 455, 538, 618, 726, 834, 977, 1121, 1304, 1495, 1738, 1989, 2302, 2633, 3041, 3473, 3999, 4562, 5241
OFFSET
1,11
COMMENTS
Also, the number of partitions of n such that (greatest part) = 3*(number of parts).
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^(4*k-1) * Product_{j=1..k-1} (1-x^(3*k+j-1))/(1-x^j). - Seiichi Manyama, Jan 24 2022
a(n) ~ Pi^3 * exp(Pi*sqrt(2*n/3)) / (3*2^(5/2)*n^(5/2)). - Vaclav Kotesovec, Oct 17 2024
EXAMPLE
a(15) = 4 counts these partitions: [12,1,1,1], [9,5,1], [9,4,2], [9,3,3].
MATHEMATICA
z = 50; Table[Count[IntegerPartitions[n], p_ /; Max[p] = = 3 Length[p]], {n, z}]
(* or *)
nmax = 100; Rest[CoefficientList[Series[Sum[x^(4*k-1) * Product[(1 - x^(3*k+j-1)) / (1 - x^j), {j, 1, k-1}], {k, 1, nmax/4 + 1}], {x, 0, nmax}], x]] (* Vaclav Kotesovec, Oct 15 2024 *)
nmax = 100; p = x^2; s = x^2; Do[p = Normal[Series[p*x^4*(1 - x^(4*k - 1))*(1 - x^(4*k))*(1 - x^(4*k + 1))*(1 - x^(4*k + 2))/((1 - x^(3*k + 2))*(1 - x^(3*k + 1))*(1 - x^(3*k))*(1 - x^k)), {x, 0, nmax}]]; s += p; , {k, 1, nmax/4 + 1}]; Take[CoefficientList[s, x], nmax] (* Vaclav Kotesovec, Oct 16 2024 *)
PROG
(PARI) my(N=66, x='x+O('x^N)); concat([0, 0], Vec(sum(k=1, N, x^(4*k-1)*prod(j=1, k-1, (1-x^(3*k+j-1))/(1-x^j))))) \\ Seiichi Manyama, Jan 24 2022
CROSSREFS
Column 3 of A350879.
Sequence in context: A326115 A355746 A239485 * A010334 A010578 A328583
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Feb 13 2014
STATUS
approved