A316996 Consider a partition of n into distinct parts with the summands written in binary notation. a(n) is the number of such partitions whose binary representation has an odd number of binary ones.

0, 1, 1, 0, 2, 1, 1, 5, 2, 2, 7, 6, 6, 12, 10, 8, 23, 19, 15, 37, 27, 32, 60, 42, 54, 87, 74, 88, 130, 116, 134, 206, 173, 203, 305, 256, 325, 437, 375, 485, 624, 574, 700, 879, 836, 1008, 1268, 1190, 1433, 1773, 1688, 2059, 2443, 2376, 2883, 3362, 3356, 3978

Offset: 0

Author

David S. Newman, Jul 18 2018

Keywords

• nonn

Examples

			For n = 5 there are 3 partitions to be examined: 5, 4+1, and 3+2. In binary these are 101, 100+1, and 11+10, which have 2, 2, and 3 binary ones respectively, so a(5) = 1.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A000009, A000120, A102437, A317239.

Programs

Maple

h:= proc(n) option remember; `if`(n=0, 0, irem(n, 2, 'q')+h(q)) end:
b:= proc(n, i, t) option remember; `if`(i*(i+1)/2 b(n$2, 0):
seq(a(n), n=0..100);  # Alois P. Heinz, Jul 24 2018

Mathematica

h[n_] := h[n] = If[n == 0, 0, Mod[n, 2] + h[Quotient[n, 2]]];
b[n_, i_, t_] := b[n, i, t] = If[i*(i + 1)/2 < n, 0, If[n == 0, t, b[n, i - 1, t] + b[n - i, Min[n - i, i - 1], Mod[t + h[i], 2]]]];
a[n_] := b[n, n, 0];
a /@ Range[0, 100] (* Jean-François Alcover, Sep 16 2019, after Alois P. Heinz *)

PARI

seq(n)={apply(t->polcoeff(lift(t), 1), Vec(prod(i=1, n, 1 + x^i*Mod( y^hammingweight(i), y^2-1 ) + O(x*x^n))))} \\ Andrew Howroyd, Jul 23 2018

Formula

a(n) + A317239(n) = A000009(n).

a(n) ~ exp(Pi*sqrt(n/3)) / (8 * 3^(1/4) * n^(3/4)). - Vaclav Kotesovec, Oct 09 2018