%I #31 Dec 11 2016 20:41:42
%S 1,2,4,9,9,9,8,19,9,8,17,27,27,27,28,17,26,35,45,45,46,37,25,44,53,65,
%T 42,72,74,52,70,90,92,74,53,62,72,70,93,61,81,80,89,100,91,80,91,79,
%U 99,99,99,98,107,117,118,106,130,86,123,155,137,117,118,105,136
%N Left-concatenate zeros to 2^(n-1) such that it has n digits. In the regular array formed by listing the found powers, a(n) is the sum of (nonzero) digits in column n.
%C After carries, this is the decimal expansion of Sum_{i>=0} 0.2^i = 1.25. For n > 2, the 10^0's digit of a(n) + the 10^1's digit of a(n+1) + ... + the 10^m's digit of a(n+m) = 9 for some finite m.
%C Conjecture: a(n) ~ c*n where c ~= 1.93.
%C Conjecture: lim_{n->infinity} a(n)/n = (9/2)*log_5(2) =
%C 1.93804... - _Jon E. Schoenfield_, Dec 09 2016
%H Robert G. Wilson v, <a href="/A279035/b279035.txt">Table of n, a(n) for n = 1..10000</a>
%e 1
%e .2
%e . 4
%e . .8
%e . .16
%e . . 32
%e . . .64
%e . . .128
%e . . . 256
%e . . . .512
%e . . . .1024
%e The sum of digits of the first column is 1. Therefore, a(1) = 1.
%e The sum of digits in column 4 is 8 + 1 = 9. Therefore, a(4) = 9.
%e With the powers of 2 listed above, we can find n up to n = 7. For n > 8, some digits from 2^m compose a(n) for m > 10.
%t f[n_, b_] := Block[{k = n}, While[k < n + Floor[ k*Log10[b]], k++]; Plus @@ Mod[ Quotient[ Table[ b^j*10^(k - j), {j, n -1, k}], 10^(k - n +1)], 10]]; Table[f[n, 2], {n, 65}]
%t (* _Robert G. Wilson v_, Dec 03 2016 *)
%Y Cf. A000079, A016134.
%K nonn,base
%O 1,2
%A _David A. Corneth_, Dec 03 2016