%I #25 Aug 20 2017 10:54:56
%S 0,1,1,3,1,4,5,11,1,5,3,7,7,19,13,25,1,6,7,17,8,23,31,61,9,29,19,39,
%T 47,107,77,137,1,7,2,5,1,3,1,2,5,17,11,23,3,7,5,9,11,41,13,28,7,17,6,
%U 11,37,97,67,127,19,39,29,49,1,8,9,23,10,31,41,83,11,39,25,53,61,145,103,187
%N If n = Sum_{m>=1} 2^(m-1) * b(n,m), where each b(n,m) is 0 or 1 and the sum is a finite sum, then a(n) = numerator of Sum_{m>=1} b(n,m)/m.
%H Peter Kagey, <a href="/A116416/b116416.txt">Table of n, a(n) for n = 0..10000</a>
%e 13 in binary is 1101. So a(13) is the numerator of 1/4 + 1/3 + 1 = 19/12, since the binary digits at positions (from right to left) 1, 3 and 4 are each 1 and the other digits are 0.
%t Table[Numerator@ Total@ MapIndexed[#1/ First@ #2 &, Reverse@ IntegerDigits[n, 2]], {n, 0, 79}] (* _Michael De Vlieger_, Aug 19 2017 *)
%o (PARI) a(n) = {my(b = Vecrev(binary(n))); numerator(sum(k=1, #b, b[k]/k));} \\ _Michel Marcus_, Apr 18 2016
%Y Cf. A007088, A116417.
%K easy,frac,nonn,look
%O 0,4
%A _Leroy Quet_, Feb 13 2006
%E More terms from _Joshua Zucker_, May 03 2006