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A348164
Number of partitions of n such that 5*(greatest part) = (number of parts).
3
0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 2, 2, 3, 3, 3, 3, 5, 5, 7, 8, 10, 11, 15, 16, 20, 22, 26, 28, 35, 38, 46, 52, 62, 70, 85, 95, 112, 127, 148, 166, 195, 219, 254, 288, 332, 375, 435, 489, 562, 635, 726, 817, 936, 1051, 1198, 1348, 1531, 1721, 1957, 2196, 2489
OFFSET
1,17
COMMENTS
Also, the number of partitions of n such that (greatest part) = 5*(number of parts).
LINKS
FORMULA
G.f.: Sum_{k>=1} x^(6*k-1) * Product_{j=1..k-1} (1-x^(5*k+j-1))/(1-x^j).
a(n) ~ 5 * Pi^5 * exp(Pi*sqrt(2*n/3)) / (9 * 2^(3/2) * n^(7/2)). - Vaclav Kotesovec, Oct 17 2024
EXAMPLE
a(19) = 3 counts these partitions:
[3, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[3, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[2, 2, 2, 2, 2, 2, 2, 2, 2, 1].
MATHEMATICA
nmax = 100; Rest[CoefficientList[Series[Sum[x^(6*k-1) * Product[(1 - x^(5*k+j-1)) / (1 - x^j), {j, 1, k-1}], {k, 1, nmax/6 + 1}], {x, 0, nmax}], x]] (* Vaclav Kotesovec, Oct 15 2024 *)
nmax = 100; p = x^4; s = x^4; Do[p = Normal[Series[p*x^6*(1 - x^(6*k - 1))*(1 - x^(6*k))*(1 - x^(6*k + 1))*(1 - x^(6*k + 2))*(1 - x^(6*k + 3))*(1 - x^(6*k + 4))/((1 - x^(5*k + 4))*(1 - x^(5*k + 3))*(1 - x^(5*k + 2))*(1 - x^(5*k + 1))*(1 - x^(5*k))*(1 - x^k)), {x, 0, nmax}]]; s += p; , {k, 1, nmax/6 + 1}]; Take[CoefficientList[s, x], nmax] (* Vaclav Kotesovec, Oct 16 2024 *)
PROG
(PARI) my(N=66, x='x+O('x^N)); concat([0, 0, 0, 0], Vec(sum(k=1, N, x^(6*k-1)*prod(j=1, k-1, (1-x^(5*k+j-1))/(1-x^j)))))
CROSSREFS
Column 5 of A350879.
Sequence in context: A055377 A157524 A239496 * A299962 A340010 A128586
KEYWORD
nonn,changed
AUTHOR
Seiichi Manyama, Jan 25 2022
STATUS
approved