A334948 a(n) is the number of partitions of n into consecutive parts that differ by 6.
1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 3, 1, 3, 1, 3, 2, 2, 1, 4, 1, 2, 2, 3, 1, 3, 1, 3, 2, 2, 1, 4, 1, 2, 2, 3, 2, 3, 1, 3, 2, 3, 1, 4, 1, 2, 3, 3, 1, 3, 1, 4, 2, 2, 1, 4, 2, 2, 2, 3, 1, 4, 1, 3, 2, 2, 2, 5, 1, 2, 2, 4, 1, 4, 1, 3, 3
Offset: 1
Keywords
Examples
For n = 24 there are three partitions of 24 into consecutive parts that differ by 6, including 24 as a valid partition. They are [24], [15, 9] and [14, 8, 2], so a(24) = 3.
Links
- Seiichi Manyama, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
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Mathematica
nmax = 105; col[k_] := col[k] = CoefficientList[Sum[x^(n(k n - k + 2)/2 - 1)/(1 - x^n), {n, 1, nmax}] + O[x]^nmax, x]; a[n_] := col[6][[n]]; Array[a, nmax] (* Jean-François Alcover, Nov 30 2020 *) Table[Count[IntegerPartitions[n],?(Union[Abs[Differences[#]]]=={6}&)]+1,{n,110}] (* _Harvey P. Dale, Dec 07 2020 *) Table[Sum[If[n > 3*k*(k-1), 1, 0], {k, Divisors[n]}], {n, 1, 100}] (* Vaclav Kotesovec, Oct 22 2024 *)
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PARI
my(N=66, x='x+O('x^N)); Vec(sum(k=1, N, x^(k*(3*k-2))/(1-x^k))) \\ Seiichi Manyama, Dec 04 2020
Formula
G.f.: Sum_{k>=1} x^(k*(3*k - 2)) / (1 - x^k). - Ilya Gutkovskiy, Nov 23 2020
Comments