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A330889 a(n) is the total number of parts in all partitions of n into consecutive parts that differ by 3.

%I A330889 #56 Oct 23 2024 08:56:13
%S A330889 1,1,1,1,3,1,3,1,3,1,3,4,3,1,6,1,3,4,3,1,6,5,3,4,3,5,6,1,3,8,3,1,6,5,
%T A330889 8,4,3,5,6,6,3,8,3,1,11,5,3,4,3,10,12,1,3,8,8,1,12,5,3,9,3,5,12,1,8,8,
%U A330889 3,1,12,17,3,4,3,5,17,1,10,8,3,6,12,5,3,11,8,5,12,1,3,13
%N A330889 a(n) is the total number of parts in all partitions of n into consecutive parts that differ by 3.
%C A330889 The one-part partition n = n is included in the count.
%C A330889 For the relation to pentagonal numbers see also A330888.
%H A330889 Seiichi Manyama, <a href="/A330889/b330889.txt">Table of n, a(n) for n = 1..10000</a>
%F A330889 Conjecture: G.f.: Sum_{n>=1} n*x^(n*(3*n-1)/2)/(1-x^n).
%F A330889 Proof from _Matthew C. Russell_, Nov 21 2020:
%F A330889 The summation variable n runs over the number of parts in the partition.
%F A330889 For fixed n, the smallest such partition is:
%F A330889 1 + 4 + 7 + ... + (3n-2).
%F A330889 The above sum is equal to n * (3*n-1) / 2. That's where the x^(n*(3*n-1)/2) factor comes from.
%F A330889 Then we want to (add 1 to every part), (add 2 to every part), etc. to get 2 + 5 + 8 + ..., 3 + 6 + 9 + ..., which corresponds to adding n, 2*n, 3*n, etc. to the base partition. So we divide by (1 - x^n).
%F A330889 Multiply by n (to count the total number of parts) and we are done. QED
%F A330889 Sum_{k=1..n} a(k) ~ (2/3)^(3/2) * n^(3/2). - _Vaclav Kotesovec_, Oct 23 2024
%e A330889 For n = 21 there are three partitions of 21 into consecutive parts that differ by 3, including 21 as a partition. They are [21], [12, 9] and [10, 7, 4]. The number of parts of these partitions are 1, 2 and 3 respectively. The total number of parts is 1 + 2 + 3 = 6, so a(27) = 6.
%p A330889 A330889 := proc(n)
%p A330889     local a;
%p A330889     a := 0 ;
%p A330889     for k from 1 do
%p A330889         if n>= A000325(k) then
%p A330889             a := a+A330888(n,k);
%p A330889         else
%p A330889             return a;
%p A330889         end if;
%p A330889     end do:
%p A330889 end proc: # _R. J. Mathar_, Oct 02 2020
%t A330889 nmax = 100;
%t A330889 CoefficientList[Sum[n x^(n(3n-1)/2-1)/(1-x^n), {n, 1, nmax}]+O[x]^nmax, x] (* _Jean-François Alcover_, Nov 30 2020 *)
%t A330889 Table[Sum[If[n > 3*k*(k-1)/2 && IntegerQ[n/k - 3*(k-1)/2], k, 0], {k, Divisors[2*n]}], {n, 1, 100}] (* _Vaclav Kotesovec_, Oct 23 2024 *)
%o A330889 (PARI) my(N=66, x='x+O('x^N)); Vec(sum(k=1, N, k*x^(k*(3*k-1)/2)/(1-x^k))) \\ _Seiichi Manyama_, Dec 04 2020
%Y A330889 Row sums of A330888.
%Y A330889 Column k=3 of A334466.
%Y A330889 Sequences of the same family whose consecutive parts differs by k are: A000203 (k=0), A204217 (k=1), A066839 (k=2), this sequence (k=3), A334464 (k=4), A334732 (k=5), A334949 (k=6).
%Y A330889 Cf. A338730.
%K A330889 nonn,easy
%O A330889 1,5
%A A330889 _Omar E. Pol_, Apr 30 2020
%E A330889 More terms from _R. J. Mathar_, Oct 02 2020