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A097092
Number of partitions of n such that the least part occurs exactly four times.
6
0, 0, 0, 1, 0, 1, 1, 3, 2, 4, 5, 9, 9, 14, 16, 26, 29, 40, 48, 67, 79, 105, 126, 165, 196, 253, 303, 385, 459, 572, 687, 852, 1014, 1244, 1482, 1807, 2145, 2595, 3075, 3701, 4375, 5231, 6170, 7350, 8641, 10247, 12025, 14201, 16620, 19557, 22839, 26790, 31209
OFFSET
1,8
COMMENTS
Number of partitions p of n such that 3*min(p) + (number of parts of p) is a part of p. - Clark Kimberling, Feb 28 2014
LINKS
FORMULA
G.f.: Sum_{m>0} (x^(4*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) + 4*p(n+4) - p(n+5) - 2*p(n+6) - 2*p(n+7) + p(n+8) + 2*p(n+9) - p(n+10), 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)) + 61*Pi/(24*sqrt(6)))/sqrt(n)). (End)
MATHEMATICA
a[n_] := Module[{p = IntegerPartitions[n], l = PartitionsP[n], c = 0, k = 1}, While[k < l + 1, q = PadLeft[ p[[k]], 5]; If[ q[[1]] != q[[5]] && q[[2]] == q[[5]], c++ ]; k++ ]; c]; Table[ a[n], {n, 53}]
Table[Count[IntegerPartitions[n], p_ /; MemberQ[p, Length[p] + 3*Min[p]]], {n, 50}] (* Clark Kimberling, Feb 28 2014 *)
Table[Count[IntegerPartitions[n], _?(Length[Split[#][[-1]]]==4&)], {n, 60}] (* Harvey P. Dale, Jan 18 2021 *)
nmax = 60; Rest[CoefficientList[Series[Sum[x^(4*m)/Product[1-x^k, {k, m+1, nmax}], {m, 1, nmax}], {x, 0, nmax}], x]] (* Vaclav Kotesovec, Jul 04 2025 *)
Table[-PartitionsP[n] + 4 PartitionsP[4 + n] - PartitionsP[5 + n] - 2 PartitionsP[6 + n] - 2 PartitionsP[7 + n] + PartitionsP[8 + n] + 2 PartitionsP[9 + n] - PartitionsP[10 + n], {n, 1, 60}] (* Vaclav Kotesovec, Jul 05 2025 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Robert G. Wilson v, Jul 24 2004
STATUS
approved