A334732 a(n) is the total number of parts in all partitions of n into consecutive parts that differ by 5.
1, 1, 1, 1, 1, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 4, 3, 1, 6, 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, 3, 4, 3, 5, 6, 1, 3, 8, 8, 1, 6, 5, 3, 9, 3, 5, 6, 1, 8, 8, 3, 1, 6, 10, 3, 4, 3, 5, 11, 1, 3, 8, 3, 6, 12, 5, 3, 4, 8, 5, 12, 1, 3, 13, 3, 1, 12
Offset: 1
Keywords
Examples
For n = 27 there are three partitions of 27 into consecutive parts that differ by 5, including 27 as a valid partition. They are [27], [16, 11] and [14, 9, 4]. The number of parts of these partitions are 1, 2, 3 respectively and the total number of parts is 1 + 2 + 3 = 6, so the a(27) = 6.
Links
- Seiichi Manyama, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
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Maple
A334732 := proc(n) local a,k; a := 0 ; for k from 1 do if n>= A000566(k) then a := a+A334540(n,k); else return a; end if; end do: end proc: seq(A334732(n),n=1..120) ; # R. J. Mathar, Oct 02 2020
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Mathematica
nmax = 100; CoefficientList[Sum[n x^(n(5n-3)/2-1)/(1-x^n), {n, 1, nmax}]+O[x]^nmax, x] (* Jean-François Alcover, Nov 30 2020 *) Table[Sum[If[n > 5*k*(k-1)/2 && IntegerQ[n/k - 5*(k-1)/2], k, 0], {k, Divisors[2*n]}], {n, 1, 100}] (* Vaclav Kotesovec, Oct 23 2024 *)
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PARI
my(N=66, x='x+O('x^N)); Vec(sum(k=1, N, k*x^(k*(5*k-3)/2)/(1-x^k))) \\ Seiichi Manyama, Dec 04 2020
Formula
G.f.: Sum_{n>=1} n*x^(n*(5*n-3)/2)/(1-x^n) (for proof see A330889).
Sum_{k=1..n} a(k) ~ 2^(3/2) * n^(3/2) / (3*sqrt(5)). - Vaclav Kotesovec, Oct 23 2024
Extensions
More terms from R. J. Mathar, Oct 02 2020
Comments