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A117277 Number of partitions of n whose consecutive parts differ by 3.

%I A117277 #26 Oct 23 2024 09:27:24
%S A117277 1,1,1,1,2,1,2,1,2,1,2,2,2,1,3,1,2,2,2,1,3,2,2,2,2,2,3,1,2,3,2,1,3,2,
%T A117277 3,2,2,2,3,2,2,3,2,1,4,2,2,2,2,3,4,1,2,3,3,1,4,2,2,3,2,2,4,1,3,3,2,1,
%U A117277 4,4,2,2,2,2,5,1,3,3,2,2,4,2,2,3,3,2,4,1,2,4,3,2,4,2,3,2,2,3,4,3,2,3,2,1,6
%N A117277 Number of partitions of n whose consecutive parts differ by 3.
%C A117277 Also number of partitions of n such that if k is the largest part, then each of the parts 1,2,...,k-1 occurs exactly 3 times. Example: a(15)=3 because we have [3,3,2,2,2,1,1,1],[2,2,2,2,2,2,1,1,1] and [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1].
%C A117277 Row sums of A330887. - _Omar E. Pol_, May 07 2020
%C A117277 Column 3 of A323345. - _Omar E. Pol_, Dec 03 2020
%H A117277 Seiichi Manyama, <a href="/A117277/b117277.txt">Table of n, a(n) for n = 1..10000</a>
%F A117277 G.f.: Sum_{k>=1} x^((3*k^2-k)/2)/(1-x^k). In general, the generating function for the number of partitions in which consecutive parts differ by d is Sum_{k>=1} x^(k*(d*k-d+2)/2)/(1-x^k). For d=0, 1 and 2 one obtains A000005, A001227 and A038548, respectively.
%e A117277 a(15) = 3 because we have [15], [9,6] and [8,5,2].
%p A117277 g:=sum(x^((3*k^2-k)/2)/(1-x^k),k=1..10): gser:=series(g,x=0,140): seq(coeff(gser,x^n),n=1..135);
%t A117277 Table[Sum[If[n > 3*k*(k-1)/2 && IntegerQ[n/k - 3*(k-1)/2], 1, 0], {k, Divisors[2*n]}], {n, 1, 100}] (* _Vaclav Kotesovec_, Oct 23 2024 *)
%o A117277 (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 A117277 seq(100,3) \\ _Joerg Arndt_, May 05 2020
%Y A117277 Cf. A000005, A001227, A038548, A323345, A330887, A338731.
%K A117277 nonn
%O A117277 1,5
%A A117277 _Emeric Deutsch_, Mar 07 2006