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%I #19 Dec 12 2024 23:24:18
%S 8,64,2486,5,4268,8426,2,4,4862,46,82,6248,6842,8624,2684,28,6,9,7139,
%T 3179,19,1397,1793,91,3971,7931,9713,9317
%N Positive integers in A376446 sorted according to their appearance in that sequence.
%C Since a(28) = A376446(700001) = 9317, the present sequence has at least 28 terms.
%C If we merge A376446(n) and A377124(n*10), taking A376446(n) if and only if n is not a multiple of 10 and A376446(n*10) otherwise, we should get the sequence: 8, 64, 2486, 5, 4268, 8426, 2, 1, 4, 4862, 46, 82, 6248, 6, 6842, 8624, 2684, 28, 9, 7139, 3179, 19, 1397, 1793, 91, 3971, 7931, 9713, 9317 (which the author conjectures to be complete, as the present one).
%C Moreover, by construction, each term of this sequence is necessarily a circular permutation of the digits of one term of A376842 (e.g., a(4) = 2486 since A376842(4) = 6248).
%H Marco Ripà, <a href="https://arxiv.org/abs/2411.00015">Graham's number stable digits: an exact solution</a>, arXiv:2411.00015 [math.GM], 2024.
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Graham's_number">Graham's Number</a>.
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Tetration">Tetration</a>.
%F a(1) = 8, a(2) = 64, ..., a(28) = 9317 (and a(28) is the last term of the present sequence - conjectured).
%e a(2) = 64 since A376446(2) = 64 (which is different from A376446(1) = 8).
%Y Cf. A317905, A370211, A372490, A373387, A376446, A376842, A377124.
%Y Cf. A000007, A018247, A018248, A063006, A091661, A091663, A091664, A120817, A120818, A290372, A290373, A290374, A290375.
%K nonn,fini,more,new
%O 2,1
%A _Marco Ripà_, Nov 25 2024