OFFSET
1,9
COMMENTS
In general, if k>=1 and g.f. = Sum_{m>0} (x^(k*m) / Product_{i>m} (1-x^i)), then a(n) ~ Pi * exp(Pi*sqrt(2*n/3)) / (3 * 2^(5/2) * n^(3/2)) * (1 - (3^(3/2)/(Pi*sqrt(2)) + (24*k - 35)*Pi/(24*sqrt(6)))/sqrt(n)). - Vaclav Kotesovec, Jul 05 2025
LINKS
Vaclav Kotesovec, Table of n, a(n) for n = 1..10000
FORMULA
G.f.: Sum_{m>0} (x^(5*m) / Product_{i>m} (1-x^i)). More generally, g.f. for number of partitions of n such that the least part occurs exactly k times is Sum_{m>0} (x^(k*m) / Product_{i>m} (1-x^i)). Vladeta Jovovic
From Vaclav Kotesovec, Jul 05 2025: (Start)
a(n) = -p(n) + 5*p(n+5) - p(n+6) - 2*p(n+7) - 2*p(n+8) - 3*p(n+9) + 3*p(n+10) + p(n+11) + 2*p(n+12) - p(n+13) - 2*p(n+14) + p(n+15), where p(n) = A000041(n).
a(n) ~ Pi * exp(Pi*sqrt(2*n/3)) / (3 * 2^(5/2) * n^(3/2)) * (1 - (3^(3/2)/(Pi*sqrt(2)) + 85*Pi/(24*sqrt(6)))/sqrt(n)). (End)
MATHEMATICA
f[n_] := Block[{p = IntegerPartitions[n], l = PartitionsP[n], c = 0, k = 1}, While[k < l + 1, q = PadLeft[ p[[k]], 6]; If[ q[[1]] != q[[6]] && q[[2]] == q[[6]], c++ ]; k++ ]; c]; Table[ f[n], {n, 54}]
Table[Count[IntegerPartitions[n], _?(Length[Split[#][[-1]]]==5&)], {n, 60}] (* Harvey P. Dale, Feb 07 2022 *)
nmax = 60; Rest[CoefficientList[Series[Sum[x^(5*m)/Product[1-x^k, {k, m+1, nmax}], {m, 1, nmax}], {x, 0, nmax}], x]] (* Vaclav Kotesovec, Jul 04 2025 *)
Table[-PartitionsP[n] + 5 PartitionsP[5 + n] - PartitionsP[6 + n] - 2 PartitionsP[7 + n] - 2 PartitionsP[8 + n] - 3 PartitionsP[9 + n] + 3 PartitionsP[10 + n] + PartitionsP[11 + n] + 2 PartitionsP[12 + n] - PartitionsP[13 + n] - 2 PartitionsP[14 + n] + PartitionsP[15 + n], {n, 1, 60}] (* Vaclav Kotesovec, Jul 05 2025 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Robert G. Wilson v, Jul 24 2004
STATUS
approved
