0,1

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))))))

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

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.

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

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

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}]

Sequence in context: A326964 A034902 A382839 * A270749 A206151 A070521

Adjacent sequences: A101426 A101427 A101428 * A101430 A101431 A101432

nonn,base

Yalcin Aktar, Nov 05 2004

a(5)-a(7) from Robert G. Wilson v, Nov 05 2004

a(8) and a(9) from Zak Seidov, Nov 23 2007

approved