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A237827
Number of partitions of n such that 5*(least part) = greatest part.
9
0, 0, 0, 0, 0, 1, 1, 2, 3, 5, 7, 11, 13, 19, 24, 32, 39, 52, 61, 77, 93, 114, 133, 164, 188, 226, 261, 309, 353, 417, 471, 549, 622, 717, 808, 933, 1042, 1191, 1334, 1516, 1690, 1921, 2131, 2407, 2674, 3006, 3330, 3744, 4135, 4628, 5116, 5708, 6294, 7020
OFFSET
1,8
LINKS
David A. Corneth, Table of n, a(n) for n = 1..10000 (first 1000 terms from Alois P. Heinz)
FORMULA
G.f.: Sum_{k>=1} x^(6*k)/Product_{j=k..5*k} (1-x^j). - Seiichi Manyama, May 14 2023
EXAMPLE
a(8) = 2 counts these partitions: 521, 5111.
MATHEMATICA
z = 64; q[n_] := q[n] = IntegerPartitions[n];
Table[Count[q[n], p_ /; 3 Min[p] = = Max[p]], {n, z}] (* A237825*)
Table[Count[q[n], p_ /; 4 Min[p] = = Max[p]], {n, z}] (* A237826 *)
Table[Count[q[n], p_ /; 5 Min[p] = = Max[p]], {n, z}] (* A237827 *)
Table[Count[q[n], p_ /; 2 Min[p] + 1 = = Max[p]], {n, z}] (* A237828 *)
Table[Count[q[n], p_ /; 2 Min[p] - 1 = = Max[p]], {n, z}] (* A237829 *)
(* Second program: *)
kmax = 54;
Sum[x^(6 k)/Product[1 - x^j, {j, k, 5 k}], {k, 1, kmax}]/x + O[x]^kmax // CoefficientList[#, x]& (* Jean-François Alcover, May 30 2024, after Seiichi Manyama *)
PROG
(PARI) my(N=60, x='x+O('x^N)); concat([0, 0, 0, 0, 0], Vec(sum(k=1, N, x^(6*k)/prod(j=k, 5*k, 1-x^j)))) \\ Seiichi Manyama, May 14 2023
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Feb 16 2014
STATUS
approved