%I #33 Oct 23 2024 08:56:45
%S 1,1,1,1,1,1,3,1,3,1,3,1,3,1,3,1,3,4,3,1,6,1,3,4,3,1,6,1,3,4,3,1,6,5,
%T 3,4,3,5,6,1,3,8,3,1,6,5,3,4,3,5,6,1,3,8,8,1,6,5,3,9,3,5,6,1,8,8,3,1,
%U 6,10,3,4,3,5,11,1,3,8,3,6,12,5,3,4,8,5,12,1,3,13,3,1,12
%N a(n) is the total number of parts in all partitions of n into consecutive parts that differ by 5.
%C The one-part partition n = n is included in the count.
%C For the relation to heptagonal numbers see also A334540.
%H Seiichi Manyama, <a href="/A334732/b334732.txt">Table of n, a(n) for n = 1..10000</a>
%F G.f.: Sum_{n>=1} n*x^(n*(5*n-3)/2)/(1-x^n) (for proof see A330889).
%F Sum_{k=1..n} a(k) ~ 2^(3/2) * n^(3/2) / (3*sqrt(5)). - _Vaclav Kotesovec_, Oct 23 2024
%e For n = 27 there are three partitions of 27 into consecutive parts that differ by 5, including 27 as a valid partition. They are [27], [16, 11] and [14, 9, 4]. The number of parts of these partitions are 1, 2, 3 respectively and the total number of parts is 1 + 2 + 3 = 6, so the a(27) = 6.
%p A334732 := proc(n)
%p local a,k;
%p a := 0 ;
%p for k from 1 do
%p if n>= A000566(k) then
%p a := a+A334540(n,k);
%p else
%p return a;
%p end if;
%p end do:
%p end proc:
%p seq(A334732(n),n=1..120) ; # _R. J. Mathar_, Oct 02 2020
%t nmax = 100;
%t CoefficientList[Sum[n x^(n(5n-3)/2-1)/(1-x^n), {n, 1, nmax}]+O[x]^nmax, x] (* _Jean-François Alcover_, Nov 30 2020 *)
%t Table[Sum[If[n > 5*k*(k-1)/2 && IntegerQ[n/k - 5*(k-1)/2], k, 0], {k, Divisors[2*n]}], {n, 1, 100}] (* _Vaclav Kotesovec_, Oct 23 2024 *)
%o (PARI) my(N=66, x='x+O('x^N)); Vec(sum(k=1, N, k*x^(k*(5*k-3)/2)/(1-x^k))) \\ _Seiichi Manyama_, Dec 04 2020
%Y Row sums of A334540.
%Y Column k=5 of A334466.
%Y Sequences of the same family whose consecutive parts differs by k are: A000203 (k=0), A204217 (k=1), A066839 (k=2), A330889 (k=3), A334464 (k=4), this sequence (k=5), A334949 (k=6).
%Y Cf. A000566, A334465, A334540, A334541, A334733.
%K nonn,changed
%O 1,7
%A _Omar E. Pol_, May 09 2020
%E More terms from _R. J. Mathar_, Oct 02 2020