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A372325 Numbers whose binary expansion has an even number of 1's among positions listed in this sequence.

%I A372325 #28 May 28 2024 18:27:32
%S A372325 0,2,5,7,8,10,13,15,16,18,21,23,24,26,29,31,33,35,36,38,41,43,44,46,
%T A372325 49,51,52,54,57,59,60,62,64,66,69,71,72,74,77,79,80,82,85,87,88,90,93,
%U A372325 95,97,99,100,102,105,107,108,110,113,115,116,118,121,123,124
%N A372325 Numbers whose binary expansion has an even number of 1's among positions listed in this sequence.
%H A372325 Robert Israel, <a href="/A372325/b372325.txt">Table of n, a(n) for n = 1..10000</a>
%e A372325 118 is in the sequence because 118 = 2^6 + 2^5 + 2^4 + 2^2 + 2^1, and an even number of the exponents 6,5,4,2,1 (namely 2,5) are in the sequence.
%e A372325 8192 is not in the sequence because 8192 = 2^13, and 13 is in the sequence.
%p A372325 R:= 0: RL:= [1]: nextp:= 2: m:= 1: count:= 0:
%p A372325 for i from 1 while count < 100 do
%p A372325   L:= convert(i,base,2);
%p A372325   if i = nextp then
%p A372325     nextp:= 2*nextp;
%p A372325     if R[1+nops(RL)] = m then RL:= [op(RL),m+1] fi;
%p A372325     m:= m+1;
%p A372325   fi;
%p A372325   if convert(L[RL],`+`)::even
%p A372325   then R:= R,i; count:= count+1
%p A372325   fi
%p A372325 od:
%p A372325 R; # _Robert Israel_, May 28 2024
%o A372325 (Python)
%o A372325 from itertools import count, islice
%o A372325 def agen():  # generator of terms
%o A372325     aset = 0 # stored as a bitmask
%o A372325     for k in count(0):
%o A372325         if (k&aset).bit_count()%2 == 0:
%o A372325             yield k
%o A372325             aset += (1<<k)
%o A372325 print(list(islice(agen(), 63))) # _Michael S. Branicky_, Apr 28 2024
%Y A372325 Cf. A133457, A272011.
%K A372325 nonn,easy,base
%O A372325 1,2
%A A372325 _David A. Madore_, Apr 27 2024