%I #30 Oct 23 2024 08:29:59
%S 1,1,1,1,1,1,1,3,1,3,1,3,1,3,1,3,1,3,1,3,4,3,1,6,1,3,4,3,1,6,1,3,4,3,
%T 1,6,1,3,4,7,1,6,1,7,4,3,1,10,1,3,4,7,1,6,1,7,4,3,1,10,1,3,4,7,6,6,1,
%U 7,4,8,1,10,1,3,9,7,1,6,1,12,4,3,1,10,6,3,4,7,1,11,1,7,4
%N a(n) is the total number of parts in all partitions of n into consecutive parts that differ by 6.
%C The one-part partition n = n is included in the count.
%C For the relation to the octagonal numbers see also A334947.
%H Seiichi Manyama, <a href="/A334949/b334949.txt">Table of n, a(n) for n = 1..10000</a>
%F G.f.: Sum_{n>=1} n*x^(n*(3*n-2))/(1-x^n). (For proof, see A330889. - _Omar E. Pol_, Nov 22 2020)
%F Sum_{k=1..n} a(k) ~ 2 * n^(3/2) / 3^(3/2). - _Vaclav Kotesovec_, Oct 23 2024
%e For n = 24 there are three partitions of 24 into consecutive parts that differ by 6, including 24 as a valid partition. They are [24], [15, 9] and [14, 8, 2]. There are 1, 2 and 3 parts respectively, hence the total number of parts is 1 + 2 + 3 = 6, so a(24) = 6.
%t nmax = 100;
%t CoefficientList[Sum[n x^(n(3n-2)-1)/(1-x^n), {n, 1, nmax}]+O[x]^nmax, x] (* _Jean-François Alcover_, Nov 30 2020 *)
%t Table[Sum[If[n > 3*k*(k-1), k, 0], {k, Divisors[n]}], {n, 1, 100}] (* _Vaclav Kotesovec_, Oct 22 2024 *)
%o (PARI) my(N=66, x='x+O('x^N)); Vec(sum(k=1, N, k*x^(k*(3*k-2))/(1-x^k))) \\ _Seiichi Manyama_, Dec 04 2020
%Y Row sums of A334947.
%Y Column k=6 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), A334732 (k=5), this sequence (k=6).
%Y Cf. A000567, A334946, A334947, A334948, A334953.
%K nonn,changed
%O 1,8
%A _Omar E. Pol_, May 27 2020