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%I #19 Jul 22 2018 20:36:23
%S 1,0,1,2,2,4,6,6,12,16,20,28,40,48,69,91,111,150,197,238,319,398,493,
%T 634,792,968,1226,1510,1846,2293,2811,3395,4197,5079,6126,7469,8993,
%U 10781,13051,15593,18627,22333,26598,31571,37655,44569,52702,62462
%N Let pi be an unrestricted partition of n with the summands written as binary numbers; a(n) is the number of such partitions with an even number of binary ones.
%H Alois P. Heinz, <a href="/A102425/b102425.txt">Table of n, a(n) for n = 0..1000</a>
%e a(5) = 4 because there are 4 partitions of 5 whose binary representations have an even number of binary ones, namely 101, 100+1, 11+1+1, 10+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 1-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]]],EvenQ[Count[#,1]]&]],{n,0,40}] (* _Geoffrey Critzer_, Sep 28 2013 *)
%o (PARI) seq(n)={apply(t->polcoeff(lift(t), 0), 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, A102437.
%K nonn
%O 0,4
%A _David S. Newman_, Feb 23 2005
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