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a(n) = numerator of Sum_{k>=1} floor(n/k)/2^k.
1

%I #20 Sep 08 2022 08:45:54

%S 0,1,5,15,43,103,263,591,1391,3103,7007,15039,33983,72063,156543,

%T 334591,722687,1510911,3255807,6773759,14433279,30193663,63535103,

%U 131264511,278589439,575004671,1200349183,2484846591,5189910527,10648256511,22287450111,45648642047

%N a(n) = numerator of Sum_{k>=1} floor(n/k)/2^k.

%H Jinyuan Wang, <a href="/A178965/b178965.txt">Table of n, a(n) for n = 0..1000</a>

%F a(n) = Sum_{i=1..n} 2^(n-i)*floor(n/i). - _Ridouane Oudra_, Jul 30 2019

%e a(3)=15 because Sum_{k>=1} floor(3/k)/2^k = 15/8.

%p seq(add(2^(n-i)*floor(n/i), i=1..n), n=0..60); # _Ridouane Oudra_, Jul 30 2019

%t Table[Numerator[Sum[Floor[n/k]/2^k, {k, 1, Infinity}]], {n, 0, 25}]

%o (Magma) [0] cat [&+[2^(n-i)*Floor(n/i):i in [1..n]]:n in [1..25]]; // _Marius A. Burtea_, Jul 30 2019

%o (PARI) a(n) = numerator(sum(k=1,n,floor(n/k)/2^k)); \\ _Jinyuan Wang_, Jul 31 2019

%K nonn,frac

%O 0,3

%A _Vladimir Reshetnikov_, Dec 31 2010