%I #11 Mar 15 2015 20:22:54

%S 2,7,115,1366,13561,135178,1351546,13546438,135481777,1354575715

%N Sum of digits of (2^(10^n)).

%F a(n)= sum_{m=0..floor(log(2^(10^n)))} floor(10*((2^(10^n))/(10^(((floor(log(2^(10^n)))+1))-m)) - floor ((2^(10^n))/(10^(((floor(log(2^(10^n)))+1))-m))))))

%F Limit a(n)/10^n, as n -> inf., is 1.35463...=4.5*log(2). For large m, mean value of digits of 2^m is 4.5, according to the uniform probability distribution of digits 0..9 in 2^m. Also, number of decimal digits in 2^m is log(2)*m, hence the formula for limit a(n)/10^n. - _Zak Seidov_

%e a(4)=sum(m=0,floor(log(2^(10^4))),floor(10*((2^(10^4))/(10^(((floor(log(2^(10^4)))+1))-m)) - floor ((2^(10^4))/(10^(((floor(log(2^(10^4)))+1))-m))))))=13561.

%t f[n_] := Plus @@ IntegerDigits[2^(10^n)]; Table[ f[n], {n, 0, 7}] (* _Robert G. Wilson v_, Nov 05 2004 *)

%t f[n_] := Plus @@ IntegerDigits[2^(10^n)]; Table[ f[n], {n, 0, 7}] (* _Robert G. Wilson v_, Nov 05 2004 *) (* Or *)

%t g[n_] := Sum[ Floor[10*((2^(10^n))/(10^(((Floor[ Log[10, 2^(10^n)]] + 1)) - m)) - Floor[(2^(10^n))/(10^(((Floor[ Log[10, 2^(10^n)]] + 1)) - m))])], {m, 0, Floor[ Log[10, 2^(10^n)]]}]; Table[ g[n], {n, 0, 6}]

%K nonn,base

%O 0,1

%A _Yalcin Aktar_, Nov 05 2004

%E a(5)-a(7) from _Robert G. Wilson v_, Nov 05 2004

%E a(8) and a(9) from _Zak Seidov_, Nov 23 2007