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%I #24 Sep 06 2017 04:08:07
%S 1,1,2,0,3,1,1,0,5,2,2,0,2,0,0,0,8,2,2,0,2,0,0,0,2,0,0,0,0,0,0,0,13,2,
%T 2,1,2,1,1,0,2,1,1,0,1,0,0,0,2,1,1,0,1,0,0,0,1,0,0,0,0,0,0,0,28,2,2,2,
%U 2,2,2,0,2,2,2,0,2,0,0,0,2,2,2,0,2,0,0,0,2,0,0,0,0,0,0,0,2,2,2,0,2,0,0,0,2
%N a(0) = 1. For n >= 1, a(n) = number of earlier terms of the sequence that have the same number of ones in their binary representations as n.
%C A115211(n) = A000120(a(n)) = number of ones in binary representation of a(n) for n>0; record values: A115212(n) = a(A115213(n)). - _Reinhard Zumkeller_, Jan 17 2006
%H Michael De Vlieger, <a href="/A113504/b113504.txt">Table of n, a(n) for n = 0..16384</a>
%H Michael De Vlieger, <a href="/A113504/a113504.txt">Records and First Positions of Records of A113504</a>.
%e The first 8 terms (terms 0 through 7) written in binary are [1,1,10,0,11,1,1,0]. Term 8 gives the number of earlier terms with the same number of 1's in their binary representation as 8 (which is 1000 in binary, for one 1). a(8) = 5 because there are five terms among the earlier terms with one binary 1 (terms with one 1: 1, 1, 2, 1 and 1).
%t Block[{a = {{1}}}, Do[AppendTo[a, IntegerDigits[#, 2]] &@ Count[a, k_ /; Count[k, 1] == DigitCount[i, 2, 1]], {i, 104}]; FromDigits[#, 2] & /@ a] (* _Michael De Vlieger_, Sep 05 2017 *)
%o (PARI) lista(nn) = {vh = [1]; print1(1, ", "); for (n=1, nn, nb = #select(x->(x==hammingweight(n)), vh); print1(nb, ", "); vh = concat(vh, hammingweight(nb)););} \\ _Michel Marcus_, Sep 06 2017
%Y Cf. A000120, A113503, A115212, A115213.
%K nonn,base
%O 0,3
%A _Leroy Quet_, Jan 10 2006
%E More terms from _Reinhard Zumkeller_, Jan 17 2006
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