A237756 Number of partitions of n such that 3*(greatest part) = (number of parts).
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
Examples
a(15) = 4 counts these partitions: [12,1,1,1], [9,5,1], [9,4,2], [9,3,3].
Links
- Vaclav Kotesovec, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Seiichi Manyama)
Programs
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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 *) -
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
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
Comments