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%I #18 Dec 16 2020 08:56:20
%S 1,1,68,187,27763,108516,25958279,100664383,26388943467,109026138857,
%T 33100108402861,139752234469078,46498731704890104,200612215343574676,
%U 71799817534098086846,314741192906319529056
%N Number of set partitions of [k] into 5 blocks with equal element sum, where k is the n-th positive integer that allows such a partition.
%C k = 9, 10, 14, 15, 19, ... A047208(n+3) for n = 1, 2, 3, 4, 5, ... .
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Partition_of_a_set">Partition of a set</a>
%F a(n) = A275714(A047208(n+3),5).
%e a(1) = 1: 18|27|36|45|9 with k = 9.
%e a(2) = 1: 1(10)|29|38|47|56 with k = 10.
%p b:= proc() option remember; local i, j, t; `if`(args[1]=0,
%p `if`(nargs=2, 1, b(args[t] $t=2..nargs)), add(
%p `if`(args[j] -args[nargs]<0, 0, b(sort([seq(args[i]-
%p `if`(i=j, args[nargs], 0), i=1..nargs-1)])[],
%p args[nargs]-1)), j=1..nargs-1))
%p end:
%p a:= proc(n) option remember; (k-> (m->
%p b((m/5)$5, k)/5!)(k*(k+1)/2))(5+5*n/2+3/4*(1-(-1)^n))
%p end:
%p seq(a(n), n=1..8);
%t b[args_List] := b[args] = Module[{nargs = Length[args]}, If[args[[1]] == 0, If[nargs == 3, 1, b[args // Rest]], Sum[If[args[[j]] - Last[args] < 0, 0, b[Append[Sort[Flatten[Table[args[[i]] - If[i == j, Last[args], 0], {i, 1, nargs - 1}]]], Last[args] - 1]]], {j, 1, nargs - 1}]]];
%t a[n_] := a[n] = Function[k, Function[m, b[Append[Table[m/5, {5}], k]]/5!][k (k + 1)/2]][5 + 5n/2 + (3/4)(1 - (-1)^n)];
%t Table[Print[n, " ", a[n]]; a[n], {n, 1, 12}] (* _Jean-François Alcover_, Dec 16 2020, after _Alois P. Heinz_ *)
%Y Cf. A047208, A275714.
%K nonn
%O 1,3
%A _Alois P. Heinz_, Aug 07 2018
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