%I
%S 0,1,2,3,4,5,6,7,8,9,0,1,2,3,4,5,6,7,8,9,0,1,4,9,6,1,1,1,1,1,0,1,8,
%T 1,1,1,1,1,8,1,0,1,6,1,1,1,1,1,1,1,0,1,8,1,1,1,1,1,1,1,0,1,1,1,1,1,
%U 1,1,1,1,0,1,1,1,1,1,1,1,1,1,0,1,1,1,1,1,1,1,1,1,0,1,8,1,1,1,1,1,1,1,1
%N The singledigit number obtained by iterated mapping of r (starting with n) to a powertower of its digits, or 1 if such a singledigit number is never reached.
%C Define a map r>A175420(r) which takes the base10 digits of r = sum_{i>=0} d_i*10^i and assigns the powertower (d_0^d_1)^d_2)^d^3... to the result. There are A055642(r)1 exponentiations in this expression. Singledigit numbers are fixed points of the map.
%C Starting with n, this map is iterated as often as needed to result in a singledigit number, which becomes a(n). In case the iteration does not reach a singledigit number (i.e., enters cycles with only multidigit numbers), a(n)= 1.
%C The entries 1 to 9 appear infinitely often in the sequence.
%C The entry 1 appears infinitely often in the sequence, see A175426.
%C After k iterations (k >= 0) we reach the following sequences:
%C 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 1st step: A175420: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 0, 1, 8, 27, 64, 125, 216, 343, 512, 729, 0, 1, 16, 81, 256, 625, 1296, 2401, 4096, 6561, 0, ...
%C 2nd step: A175421: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 4, 9, 6, 25, 216, 6561, 4096, 1, 0, 1, 8, 49, 4096, 25, 36, 531441, 32, 22876792454961 0, 1, 6, 1, 60466176, 244140625, 101559956668416 1, 1, 1, 0, ...
%C 3rd step: A175422: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 4, 9, 6, 25, 36, 1, 1, 1, 0, 1, 8, 6561, 1, 25, 216, 1, 8, 1, 0, 1, 6, 1, 1, 1, 1, 1, 1, 1, 0, ...
%C 4th step: A175423: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 4, 9, 6, 25, 216, 1, 1, 1, 0, 1, 8, 1, 1, 25, 36, 1, 8, 1, 0, 1, 6, 1, 1, 1, 1, 1, 1, 1, 0, ...
%e For n = 33: a(33) = 1 because starting with 33 we reach a singledigit 1 after 4 iterations: 3^3 = 27, 7^2 = 49, 9^4 = 6561, ((1^6)^5)^6 = 1.
%e For n = 25: a(25) = 1 because starting with 25 the iteration enters a loop of 2digit numbers: 5^2 = 25, 5^2 = 25, ...
%Y Cf. A175420  A175427.
%K sign,base
%O 0,3
%A _Jaroslav Krizek_, May 09 2010
