A316996 -id:A316996 - cp's OEIS frontend

A102437 Let pi be an unrestricted partition of n 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, 1, 3, 3, 5, 9, 10, 14, 22, 28, 37, 53, 66, 85, 120, 147, 188, 252, 308, 394, 509, 621, 783, 990, 1210, 1500, 1872, 2272, 2793, 3447, 4152, 5064, 6184, 7414, 8984, 10856, 12964, 15592, 18711, 22250, 26576, 31690, 37520, 44565, 52856, 62292

Offset: 0

David S. Newman, Feb 23 2005

nonn

			a(5) = 3 because there are 3 partitions of 5 with an odd number of binary ones in their binary representation, namely: 11+10, 10+10+1 and 1+1+1+1+1.

Alois P. Heinz, Table of n, a(n) for n = 0..1000

Cf. A000041, A000120, A102425, A316996.

p:= proc(n) option remember; local c, m;
      c:= 0; m:= n;
      while m>0 do c:= c +irem(m, 2, 'm') od;
      c
    end:
b:= proc(n, i, t) option remember;
      if n<0 then 0
    elif n=0 then t
    elif i=0 then 0
    else b(n, i-1, t) +b(n-i, i, irem(p(i)+t, 2))
      fi
    end:
a:= n-> b(n, n, 0):
seq(a(n), n=0..60);  # Alois P. Heinz, Feb 21 2011

Table[Length[Select[Map[Apply[Join,#]&,Map[IntegerDigits[#,2]&,Partitions[n]]],OddQ[Count[#,1]]&]],{n,0,40}] (* Geoffrey Critzer, Sep 28 2013 *)

seq(n)={apply(t->polcoeff(lift(t), 1), Vec(prod(i=1, n, 1/(1 - x^i*Mod( y^hammingweight(i), y^2-1 )) + O(x*x^n))))} \\ Andrew Howroyd, Jul 20 2018

More terms from Vladeta Jovovic, Feb 23 2005

A317239 Number of partitions of n into distinct parts with an even sum of Hamming weights.

Original entry on oeis.org

1, 0, 0, 2, 0, 2, 3, 0, 4, 6, 3, 6, 9, 6, 12, 19, 9, 19, 31, 17, 37, 44, 29, 62, 68, 55, 91, 104, 92, 140, 162, 134, 217, 245, 207, 329, 343, 323, 489, 497, 489, 686, 726, 731, 980, 1040, 1036, 1400, 1477, 1491, 1970, 2038, 2139, 2744, 2835, 3016, 3752, 3939

Offset: 0

Alois P. Heinz, Jul 24 2018

nonn

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

Cf. A000009, A000120, A316996.

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, 1):
seq(a(n), n=0..100);

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, 1];
a /@ Range[0, 100](* Jean-François Alcover, Dec 12 2020, after Alois P. Heinz *)

a(n) + A316996(n) = A000009(n).

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

cp's OEIS Frontend

A102437 Let pi be an unrestricted partition of n 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.

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