%I #25 Jul 25 2018 10:58:13

%S 0,1,1,1,3,3,5,9,10,14,22,28,37,53,66,85,120,147,188,252,308,394,509,

%T 621,783,990,1210,1500,1872,2272,2793,3447,4152,5064,6184,7414,8984,

%U 10856,12964,15592,18711,22250,26576,31690,37520,44565,52856,62292

%N 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.

%H Alois P. Heinz, <a href="/A102437/b102437.txt">Table of n, a(n) for n = 0..1000</a>

%e 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.

%p p:= proc(n) option remember; local c, m;

%p c:= 0; m:= n;

%p while m>0 do c:= c +irem(m, 2, 'm') od;

%p c

%p end:

%p b:= proc(n, i, t) option remember;

%p if n<0 then 0

%p elif n=0 then t

%p elif i=0 then 0

%p else b(n, i-1, t) +b(n-i, i, irem(p(i)+t, 2))

%p fi

%p end:

%p a:= n-> b(n, n, 0):

%p seq(a(n), n=0..60); # _Alois P. Heinz_, Feb 21 2011

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

%o (PARI) 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

%Y Cf. A000041, A000120, A102425, A316996.

%K nonn

%O 0,5

%A _David S. Newman_, Feb 23 2005

%E More terms from _Vladeta Jovovic_, Feb 23 2005