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%I #22 Nov 29 2012 12:18:30
%S 2,4,5,7,9,9,9,11,12,13,13,15,15,16,16,18,18,20,23,21
%N Exponents for the terms in A217368: least number which taken to the a(n)-th power has exactly n copies of each decimal digit.
%C This sequence gives the exponents a(n) such that A217368(n)^a(n) has n copies of each digit 0-9.
%C In the limit of large k, the probability of a uniformly selected 10k-digit number having k copies of each base-10 digit is C*k^(-4.5), where C is approximately 8.09451*10^(-4) (by the use of Stirling's approximation to the factorial function applied to the multinomial corresponding to the number of such 10k-digit numbers divided by the total number of 10k-digit numbers). Also, the number of n-th powers of this length is very nearly equal to (1-10^(-1/n))*10^(10k/n) as long as n is not too large. That is, the former probability is reciprocal polynomial in k, while the number of n-th powers for a given n is exponential in k as long as k is large enough. Then, under the assumption that the digits of powers are randomly distributed, this sequence will increase without bound. A217378(n+1) < A217378(n) for the first time for n=19.
%e A217368(3) = 643905 raised to the 5th power has exactly 3 copies of each digit in its decimal representation, and no number smaller than 643905 has a power of the same nature. Therefore a(3)=5.
%Y Cf. A217368 and references therein.
%K nonn,base
%O 1,1
%A _James G. Merickel_, Oct 01 2012
%E Edited by _M. F. Hasler_, Oct 05 2012
%E a(13) and a(14) added by _James G. Merickel_, Oct 06 2012 and Oct 08 2012
%E a(15)-a(19) added by _James G. Merickel_, Oct 19 2012
%E a(20) added by _James G. Merickel_, Nov 28 2012