A334464 - OEIS

%I #29 Dec 04 2020 11:54:59

%S 1,1,1,1,1,3,1,3,1,3,1,3,1,3,4,3,1,6,1,3,4,3,1,6,1,3,4,7,1,6,1,7,4,3,

%T 1,10,1,3,4,7,1,6,1,7,9,3,1,10,1,8,4,7,1,6,6,7,4,3,1,15,1,3,4,7,6,12,

%U 1,7,4,8,1,16,1,3,9,7,1,12,1,12,4,3,1,16,6,3,4,7,1,17,8,7,4

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

%C The one-part partition n = n is included in the count.

%C For the relation to hexagonal numbers see also A334462.

%H Seiichi Manyama, <a href="/A334464/b334464.txt">Table of n, a(n) for n = 1..10000</a>

%F G.f.: Sum_{n>=1} n*x^(n*(2*n-1))/(1-x^n). (For proof, see A330889. - _N. J. A. Sloane_, Nov 21 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]. The number of parts of these partitions are 1, 2, 4 respectively. The total number of parts is 1 + 2 + 4 = 7, so a(28) = 7.

%t nmax = 100;

%t CoefficientList[Sum[n x^(n(2n-1)-1)/(1-x^n), {n, 1, nmax}]+O[x]^nmax, x] (* _Jean-François Alcover_, Nov 30 2020 *)

%o (PARI) my(N=66, x='x+O('x^N)); Vec(sum(k=1, N, k*x^(k*(2*k-1))/(1-x^k))) \\ _Seiichi Manyama_, Dec 04 2020

%Y Row sums of A334462.

%Y Column k=4 of A334466.

%Y Cf. A000384.

%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), this sequence (k=4), A334732 (k=5), A334949 (k=6).

%Y Cf. A334461.

%K nonn

%O 1,6

%A _Omar E. Pol_, May 05 2020