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A066767 a(n) = Sum_{k=1..n} sigma(k)*2^(n-k) where sigma(k) = A000203(k) is the sum of divisors of k.

%I A066767 #36 Apr 29 2022 16:16:11
%S A066767 1,5,14,35,76,164,336,687,1387,2792,5596,11220,22454,44932,89888,
%T A066767 179807,359632,719303,1438626,2877294,5754620,11509276,23018576,
%U A066767 46037212,92074455,184148952,368297944,736595944,1473191918,2946383908
%N A066767 a(n) = Sum_{k=1..n} sigma(k)*2^(n-k) where sigma(k) = A000203(k) is the sum of divisors of k.
%C A066767 a(n) is the numerator of the unreduced fraction of the n-th partial sum of Sum_{k>=1} sigma(k)/2^k where the denominator of that unreduced fraction is 2^n. The partial sums converge to A066766 = 2.744033...
%D A066767 Steven R. Finch, Mathematical Constants, Cambridge, 2003, pp. 354-361.
%H A066767 Steven R. Finch, <a href="http://www.people.fas.harvard.edu/~sfinch/constant/dig/dig.html">Digital Search Tree Constants</a> [Broken link]
%H A066767 Steven R. Finch, <a href="http://web.archive.org/web/20010413214937/http://www.mathsoft.com:80/asolve/constant/dig/dig.html">Digital Search Tree Constants</a> [From the Wayback machine]
%e A066767 a(1) = 2*(1/2);
%e A066767 a(2) = 4*(1/2 + (1+2)/4) since sigma(1) = 1 and sigma(2) = 1 + 2 = 3;
%e A066767 a(3) = 8*(1/2 + (1+2)/4 + (1+3)/8);
%e A066767 a(4) = 16*(1/2 + (1+2)/4 + (1+3)/8 + (1+2+4)/16).
%o A066767 (PARI) smv(v)= s=0; for(i=1,matsize(v)[2],s=s+v[i]); s
%o A066767 a(n)= sm=0; for(j=1,n,sm=sm+smv(divisors(j)/2^j)); sm*2^n
%o A066767 (PARI) a(n) = 2^n*(sum(k=1, n, sigma(k)/2^k)); \\ _Michel Marcus_, Apr 25 2022
%Y A066767 Cf. A000203, A066766.
%K A066767 nonn
%O A066767 1,2
%A A066767 _Randall L Rathbun_, Jan 16 2002