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We exclude powers of 10 and numbers of the form 11...111 in which the number of 1's is a power of 10. Then a(n) is the smallest number (not excluded) whose trajectory under iteration of "x -> sum of n-th powers of digits of x" reaches 1.
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%I #40 Sep 15 2023 04:14:29

%S 19,7,112,11123,1111222,111111245666689,1111133333333335,

%T 1111122333333333333333333346677777777888,

%U 22222222222222222226666668888888,233444445555555555555555555555555555555555555555555577,1222222222233333333333333444444444455555555555555556666666666666666666666677778888889

%N We exclude powers of 10 and numbers of the form 11...111 in which the number of 1's is a power of 10. Then a(n) is the smallest number (not excluded) whose trajectory under iteration of "x -> sum of n-th powers of digits of x" reaches 1.

%C For n!=2, it appears that the first step in the trajectory is always to a power of 10, so that the task would be to find the shortest and lexicographically smallest partition of a power of 10 into parts 1^n,...,9^n.

%e a(1) = 19 since 1^1 + 9^1 = 10 and 1^1 + 0^1 = 1.

%e a(3) = 112 since 1^3 + 1^3 + 2^3 = 10 and 1^3 + 0^3 = 1.

%Y Cf. A007770, A035497, A046519.

%K nonn,base

%O 1,1

%A _Simon R Blow_, Aug 07 2023

%E a(6), a(8), and a(9) corrected by, and a(10) and a(11) from _Jon E. Schoenfield_, Aug 10 2023

%E Definition clarified by _N. J. A. Sloane_, Sep 15 2023