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%I #33 Mar 09 2024 11:33:10
%S 0,1,6,87,96,8203125,8656,2343,56,289,0,6611,16,53,36,765380859375,
%T 5616,777,76,179,0,2421,6,47,4976,908447265625,84203776,83,96,669,0,
%U 6431,4176,713,16,8046875,7136,917,6,759,0,7641,6,1107,56,48828125,9696,23,36
%N Numbers formed by the rightmost decimal digits of n^(n^n) that are the same as those of n^(n^(n^n)).
%C The common digits might include leading 0's (such as at n = 5) and they are discarded (in particular, a(0) = 0 indicates that the corresponding zero digit term results in a 0 integer entry).
%C a(k*10) = 0 for every positive integer k, since (k*10)^((k*10)^(k*10)) and (k*10)^((k*10)^((k*10)^(k*10))) have in common only their rightmost (k*10)^(k*10) digits.
%H Jorge Jiménez Urroz and José Luis Andrés Yebra, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL12/Yebra/yebra4.html">On the Equation a^x == x (mod b^n)</a>, Journal of Integer Sequences, Article 09.8.8, 2009.
%H Marco Ripà, <a href="https://arxiv.org/abs/2402.07929">Congruence speed of tetration bases ending with 0</a>, arXiv:2402.07929 [math.NT], 2024.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/JoyceSequence.html">Joyce Sequence</a>.
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Knuth's_up-arrow_notation">Knuth's up-arrow notation</a>.
%F a(n) = A002488(n) (mod 10^k), where k is such that n^(n^n) == n^(n^(n^n)) (mod 10^k) and n^(n^n) <> n^(n^(n^n)) (mod 10^(k+1)).
%e For n = 3, 3^(3^3) = 7625597484987 and 3^(3^(3^3)) == 387 (mod 1000) so there are two common final digits a(3) = 87.
%Y Cf. A002488, A317824, A317903, A317905, A349425.
%K sign,base
%O 0,3
%A _Marco Ripà_, Jan 27 2024