

A279035


Leftconcatenate zeros to 2^(n1) 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.


1



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, 42, 72, 74, 52, 70, 90, 92, 74, 53, 62, 72, 70, 93, 61, 81, 80, 89, 100, 91, 80, 91, 79, 99, 99, 99, 98, 107, 117, 118, 106, 130, 86, 123, 155, 137, 117, 118, 105, 136
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OFFSET

1,2


COMMENTS

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.
Conjecture: a(n) ~ c*n where c ~= 1.93.
Conjecture: lim_{n>infinity} a(n)/n = (9/2)*log_5(2) =


LINKS



EXAMPLE

1
.2
. 4
. .8
. .16
. . 32
. . .64
. . .128
. . . 256
. . . .512
. . . .1024
The sum of digits of the first column is 1. Therefore, a(1) = 1.
The sum of digits in column 4 is 8 + 1 = 9. Therefore, a(4) = 9.
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.


MATHEMATICA

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}]


CROSSREFS



KEYWORD

nonn,base


AUTHOR



STATUS

approved



