A237979 Number of strict partitions of n such that (least part) > number of parts.
0, 1, 1, 1, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 7, 7, 9, 10, 12, 13, 16, 17, 20, 22, 25, 28, 32, 35, 40, 45, 50, 56, 63, 70, 78, 87, 96, 107, 118, 131, 144, 160, 175, 194, 213, 235, 257, 284, 310, 342, 373, 410, 447, 491, 534, 585, 637, 696, 756, 826, 896, 977, 1060, 1153, 1250, 1359, 1471, 1597, 1729, 1874, 2026, 2195, 2371, 2565
Offset: 1
Examples
a(9) = 3 counts these partitions: 9, 63, 54.
Links
- Vaclav Kotesovec, Table of n, a(n) for n = 1..1000
Programs
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Mathematica
z = 50; q[n_] := q[n] = Select[IntegerPartitions[n], Max[Length /@ Split@#] == 1 &]; p1[p_] := p1[p] = DeleteDuplicates[p]; t[p_] := t[p] = Length[p1[p]] Table[Count[q[n], p_ /; Min[p] < t[p]], {n, z}] (* A237976 *) Table[Count[q[n], p_ /; Min[p] <= t[p]], {n, z}] (* A237977 *) Table[Count[q[n], p_ /; Min[p] == t[p]], {n, z}] (* A096401 *) Table[Count[q[n], p_ /; Min[p] > t[p]], {n, z}] (* A237979 *) Table[Count[q[n], p_ /; Min[p] >= t[p]], {n, z}] (* A025157 *)
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PARI
N=66; q='q+O('q^N); Vec(-1+sum(n=0, N, q^(n*(3*n+1)/2) / prod(k=1, n, 1-q^k ) )) \\ Joerg Arndt, Mar 09 2014
Formula
G.f. with a(0)=0: sum(n>=0, q^(n*(3*n+1)/2) / prod(k=1..n, 1-q^k ) ). [Joerg Arndt, Mar 09 2014]
a(n) ~ c^(1/4) * exp(2*sqrt(c*n)) / (2*sqrt(Pi*(1 + 3*r^2)) * n^(3/4)), where r = A263719 and c = 3*(log(r))^2/2 + polylog(2, 1-r). - Vaclav Kotesovec, Jan 15 2022
Comments