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a(n) is the smallest k > 1 such that the sum of digits of n^k is a power of n (or -1 if no such k exists).
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%I #38 Jan 24 2024 08:00:43

%S 2,2,2,18,8,7,4,3,2,2,45741764,4216,32,537,39,44,3,3,1187,13,67,4

%N a(n) is the smallest k > 1 such that the sum of digits of n^k is a power of n (or -1 if no such k exists).

%C a(11), if it is not -1, seems likely to exceed 4*10^6.

%C Additional known terms: a(12)..a(22) = {4216, 32, 537, 39, 44, 3, 3, 1187, 13, 67, 4}; a(24)..a(28) = {88, 4, 3, 3, 4}; a(30) = 399, a(31) = 7, a(33)..a(55) = {159, 7, 5, 4, 191, 188, 228, 13, 389332, 236, 7, 11543, 6, 5, 302, 292, 15405, 788, 337, 18213, 7, 6, 21248}; a(57)..a(60) = {413, 7, 25683, 1044}; a(62) = 476.

%C a(10^m) = 2 for all m >= 0, since the sum of digits of (10^m)^2 is 1.

%C If n is not divisible by 10, then a(n) tends to be fairly close to a number x such that the number of digits of n^x is (2/9)*n^j for some positive integer j, i.e., log_10(n^x) ~ (2/9)*n^j, so a(n) ~ (2/9)*n^j/log_10(n) for some integer j. E.g., a(12) = 4216 ~ 4269.90... = (2/9)*12^4/log_10(12). For n = 11, such numbers x are (2/9)*11^j/log_10(11) = 0.213389... * 11^j, which, for j = 1..7, round to 2, 26, 284, 3124, 34366, 378032, and 4158357. (By exhaustive search, a(n) > 10^5 (or a(n) = -1) for n = 23, 29, 32, 56, and 61, and (if a(11) != -1) a(11) > 4*10^5, so a(11) seems very likely to be either in the general vicinity of 4.16*10^6 or > 4.5*10^7.)

%C a(32) = 4950773; a(61) = 1722427. - _Martin Ehrenstein_, Nov 25 2022

%C 3*10^7 < a(23) <= 555650815. - _Martin Ehrenstein_, Nov 28 2022

%e The sum of digits of 1^2 = 1 is 1, which is a power of 1, so a(1) = 2.

%e The sum of digits of 2^2 = 4 is 4, which is a power of 2, so a(2) = 2.

%e For k = 2..17, the sum of digits of 4^k is {7, 10, 13, 7, 19, 22, 25, 19, 31, 25, 37, 40, 43, 37, 58, 61}, none of which is a power of 4, but the sum of digits of 4^18 = 68719476736 is 6+8+7+1+9+4+7+6+7+3+6 = 64 = 4^3, so a(4) = 18.

%o (PARI) isok(k,n) = my(s=sumdigits(n^k), x, y); if (s==1, return(2)); (s==n) || ((ispower(s,,&x)) && ((x==n) || (ispower(n,,&y) && (y==x))));

%o a(n) = my(k=2); while (!isok(k,n), k++); k; \\ _Michel Marcus_, Nov 25 2022

%Y Cf. A095412, A118872.

%Y Cf. A066005 (sum of digits of 11^n).

%K sign,base,more

%O 1,1

%A _Jon E. Schoenfield_, Nov 24 2022

%E a(11)-a(22) from _Martin Ehrenstein_, Nov 26 2022