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A101429
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Sum of digits of (2^(10^n)).
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0
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OFFSET
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0,1
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LINKS
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FORMULA
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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
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EXAMPLE
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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.
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MATHEMATICA
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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}]
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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