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

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, 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, 3, 1, 12, 17, 3, 4, 3, 5, 17, 1, 10, 8, 3, 6, 12, 5, 3, 11, 8, 5, 12, 1, 3, 13

Offset: 1

Author

Omar E. Pol, Apr 30 2020

Keywords

• nonn
• easy

Comments

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

For the relation to pentagonal numbers see also A330888.

Examples

			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.
		

Links

• Seiichi Manyama, Table of n, a(n) for n = 1..10000

Crossrefs

Row sums of A330888.

Column k=3 of A334466.

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).

Cf. A338730.

Programs

Maple

A330889 := proc(n)
    local a;
    a := 0 ;
    for k from 1 do
        if n>= A000325(k) then
            a := a+A330888(n,k);
        else
            return a;
        end if;
    end do:
end proc: # R. J. Mathar, Oct 02 2020

Mathematica

nmax = 100;
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 *)
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 *)

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

Formula

Conjecture: G.f.: Sum_{n>=1} n*x^(n*(3*n-1)/2)/(1-x^n).

Proof from Matthew C. Russell, Nov 21 2020:

The summation variable n runs over the number of parts in the partition.

For fixed n, the smallest such partition is:

1 + 4 + 7 + ... + (3n-2).

The above sum is equal to n * (3*n-1) / 2. That's where the x^(n*(3*n-1)/2) factor comes from.

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).

Multiply by n (to count the total number of parts) and we are done. QED

Sum_{k=1..n} a(k) ~ (2/3)^(3/2) * n^(3/2). - Vaclav Kotesovec, Oct 23 2024

Extensions

More terms from R. J. Mathar, Oct 02 2020