A334461 - OEIS

%I #29 Oct 22 2024 12:54:16

%S 1,1,1,1,1,2,1,2,1,2,1,2,1,2,2,2,1,3,1,2,2,2,1,3,1,2,2,3,1,3,1,3,2,2,

%T 1,4,1,2,2,3,1,3,1,3,3,2,1,4,1,3,2,3,1,3,2,3,2,2,1,5,1,2,2,3,2,4,1,3,

%U 2,3,1,5,1,2,3,3,1,4,1,4,2,2,1,5,2,2,2,3,1,5,2,3,2,2,2,5,1,3,2,4,1,4,1,3,4

%N a(n) is the number of partitions of n into consecutive parts that differ by 4.

%H Seiichi Manyama, <a href="/A334461/b334461.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 = 28 there are three partitions of 28 into consecutive parts that differ by 4, including 28 as a valid partition. They are [28], [16, 12] and [13, 9, 5, 1]. So a(28) = 3.

%t nmax = 105;

%t col[k_] := col[k] = CoefficientList[Sum[x^(n(k n - k + 2)/2 - 1)/(1 - x^n), {n, 1, nmax}] + O[x]^nmax, x];

%t a[n_] := col[4][[n]];

%t Array[a, nmax] (* _Jean-François Alcover_, Nov 30 2020 *)

%t Table[Sum[If[n > 2*k*(k-1), 1, 0], {k, Divisors[n]}], {n, 1, 100}] (* _Vaclav Kotesovec_, Oct 22 2024 *)

%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, 4) \\ _Joerg Arndt_, May 05 2020

%Y Row sums of A334460.

%Y Column k=4 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), this sequence (k=4), A334541 (k=5), A334948 (k=6).

%K nonn

%O 1,6

%A _Omar E. Pol_, May 01 2020