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%I #15 May 20 2021 00:22:45
%S 1,13,67,94,139,220,805
%N Numbers k such that k divides the sum of digits of 7^k.
%C There are no other terms < 3000. - _Stefan Steinerberger_, Mar 14 2006
%C No more terms < 50000. - _David Wasserman_, May 30 2008
%C From _Jon E. Schoenfield_, May 29 2010: (Start)
%C No more terms < 100000. It is nearly certain that no terms exist beyond 805.
%C Let f(k) be the sum of digits of 7^k. Let d be the number of digits, i.e., d=ceiling(log_10(7^k)).
%C Let s(m) be the sum of m random digits (each drawn independently from a uniform distribution over the integers 0 through 9).
%C As k increases, the behavior of f(k)/k becomes increasingly similar to that of s(d)/k.
%C The mean and variance of s(d)/k are 4.5*d/k and 28.5*d/k^2, respectively.
%C For large values of k, the distribution of s(d)/k approaches a standard normal distribution with mean 4.5*log_10(7) (approximately 3.80294) and variance 28.5*log_10(7)/k.
%C The probability P(k) that s(d)/k departs from the mean by an amount at least sufficient to reach the nearest higher or lower integer (so that k divides the sum of digits) becomes vanishingly small (e.g., P(50000) < 10^-18, P(100000) < 10^-36, P(150000) < 10^-54), and the same is true of the sum of P(i) for all i >= k (this sum is less than 10^-33 at k=100000). (End)
%e 13 divides the sum of digits of 7^13 (i.e., 9 + 6 + 8 + 8 + 9 + 0 + 1 + 0 + 4 + 0 + 7 = 52), so 13 is in the sequence.
%t For[n = 1, n < 2000, n++, a := DigitCount[7^n]; If[IntegerQ[Sum[a[[i]]*i, {i, 1, 9}]/n], Print[n]]] (* _Stefan Steinerberger_, Mar 14 2006 *)
%Y Cf. A062927, A067862, A067864.
%K hard,more,nonn,base
%O 1,2
%A _Shyam Sunder Gupta_ and _Amarnath Murthy_, Feb 16 2002
%E Edited by _Jon E. Schoenfield_, May 29 2010
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