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A175398 Sequence of resulting numbers after iterations of {((((D_1^D_2)^D_3)^D_4)^... )^D_k, where D_k is the k-th digit D of the number r and k is the digit number of the number r in the decimal expansion of r (A055642)} needed to reach a one-digit number starting at r = n.

%I A175398 #15 Jan 14 2021 21:15:38
%S A175398 0,1,2,3,4,5,6,7,8,9,1,1,1,1,1,1,1,1,1,1,1,2,4,8,1,9,1,1,1,9,1,3,9,1,
%T A175398 8,1,1,1,1,1,1,4,1,1,1,1,1,1,1,1,1
%N A175398 Sequence of resulting numbers after iterations of {((((D_1^D_2)^D_3)^D_4)^... )^D_k, where D_k is the k-th digit D of the number r and k is the digit number of the number r in the decimal expansion of r (A055642)} needed to reach a one-digit number starting at r = n.
%C A175398 a(n) = 1 - 9 for infinitely many n.
%C A175398 E.g., a(n) = b (b = 1, 2, ..., 9) for numbers n = b*10^k + A002275(k), where k >= 1.
%C A175398 a(n) = 1 for numbers n such that A055642(A133500(n)) = 1 for n >= 1, e.g., if the number n starts with a digit 1 or contains a digit 0 or for n >= 1.
%C A175398 Sequences after k steps of defined iteration (k >= 0):
%C A175398 0th step: A001477: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, ...
%C A175398 1st step: A133500: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 1, 4, 16, 64, 256, 1024, 4096, 16384, 65536, 262144, 1, ...
%C A175398 2nd step: A175399: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 4, 8, 1, 9, 1296, 1, 1073741824, 25, 1, 3, 9, 128, 8, 4096, 1628413597910449, 72057594037927936, 221073919720733357899776, 1, 1, 4, 1, 1296, 1073741824, 1, 1, 1, ...
%C A175398 3rd step: A175400: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 4, 8, 1, 9, 1, 1, 1, 32, 1, 3, 9, 1, 8, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
%C A175398 4th step: A175401: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 4, 8, 1, 9, 1, 1, 1, 9, 1, 3, 9, 1, 8, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
%C A175398 See A175402 and A175403.
%e A175398 For n = 29: a(29) = 9 because for the number 29 there are 4 steps of defined iteration: {2^9 = 512}, {(5^1)^2 = 25}, {2^5 = 32}, {3^2 = 9}. Resulting number is 9.
%K A175398 nonn,base
%O A175398 0,3
%A A175398 _Jaroslav Krizek_, May 01 2010