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%I #52 Dec 02 2020 18:08:39
%S 1,1,1,1,1,1,2,1,2,1,2,1,2,1,2,1,2,2,2,1,3,1,2,2,2,1,3,1,2,2,2,1,3,2,
%T 2,2,2,2,3,1,2,3,2,1,3,2,2,2,2,2,3,1,2,3,3,1,3,2,2,3,2,2,3,1,3,3,2,1,
%U 3,3,2,2,2,2,4,1,2,3,2,2,4,2,2,2,3,2,4,1,2,4,2,1,4,2,3,2,2,2,4,2,2,3,2,1,5
%N a(n) is the number of partitions of n into consecutive parts that differ by 5.
%C Note that all sequences of this family as A000005, A001227, A038548, A117277, A334461, etc. could be prepended with a(0) = 1 when they are interpreted as sequences of number of partitions, since A000041(0) = 1. However here a(0) is omitted in accordance with the mentioned members of the same family.
%C For the relation to heptagonal numbers see also A334465.
%H David A. Corneth, <a href="/A334541/b334541.txt">Table of n, a(n) for n = 1..10000</a>
%F The g.f. for "consecutive parts that differ by d" is Sum_{k>=1} x^(k*(d*k-d+2)/2) / (1-x^k); cf. A117277. - _Joerg Arndt_, Nov 30 2020
%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], so a(27) = 3.
%t first[n_] := Module[{res = Array[1&, n]}, For[i = 2, True, i++, start = i + 5 Binomial[i, 2]; If[start > n, Return[res]]; For[j = start, j <= n, j += i, res[[j]]++]]];
%t first[105] (* _Jean-François Alcover_, Nov 30 2020, after _David A. Corneth_ *)
%o (PARI) seq(N, d)=my(x='x+O('x^N)); Vec(sum(k=1, N, x^(k*(d*k-d+2)/2)/(1-x^k)));
%o seq(100, 5) \\ _Joerg Arndt_, May 06 2020
%o (PARI) first(n) = { my(res = vector(n, i, 1)); for(i = 2, oo, start = i + 5 * binomial(i, 2); if(start > n, return(res)); forstep(j = start, n, i, res[j]++ ) ); } \\ _David A. Corneth_, May 17 2020
%Y Row sums of A334465.
%Y Column k=5 of A323345.
%Y Sequences of this family whose consecutive parts differ by k are A000005 (k=0), A001227 (k=1), A038548 (k=2), A117277 (k=3), A334461 (k=4), this sequence (k=5).
%Y Cf. A000041, A000566, A303300.
%K nonn,easy
%O 1,7
%A _Omar E. Pol_, May 05 2020
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