%I
%S 1,4,7,10,13,19,22,25,28,31,34,37,40,55,58,61,64,67,70,73,76,79,82,85,
%T 88,91,94,97,100,103,106,109,112,115,118,121,163,166,169,172,175,178,
%U 181,184,187,190,193,196,199,202,205,208,211,214,217,220,223,226,229,232
%N Numbers congruent to 1 (mod 3) that are the quotient of two Cantor numbers (A005823).
%C Let C be the Cantor numbers (A005823), and let A be the set of integers congruent to 1 (mod 3) representable as the quotient of two nonzero elements of C. It is easy to see that if (3/2)*3^i < n < 2*3^i for some i, then n cannot be in A. Initial empirical data suggested that these are the only integers congruent to 1 (mod 3) not in A. However, there are additional "sporadic" counterexamples enumerated by A339636, whose structure is not well understood.
%C A simple automatonbased (or breadthfirst search) algorithm can establish in O(n) time whether n is in A or not.
%H J. S. Athreya, B. Reznick, and J. T. Tyson, <a href="https://doi.org/10.1080/00029890.2019.1528121">Cantor set arithmetic</a>, Amer. Math. Monthly 126 (2019), 417.
%H James Haoyu Bai, Joseph Meleshko, Samin Riasat, and Jeffrey Shallit, <a href="https://doi.org/10.48550/arXiv.2202.13694">Quotients of Palindromic and Antipalindromic Numbers</a>, arXiv:2202.13694 [math.NT], 2022.
%e 106 is in the sequence, because 106=1462376/13796, and 1462376 in base 3 is 2202022000002, and 13796 in base 3 is 200220222.
%Y Cf. A005823, A339636.
%K nonn
%O 1,2
%A _Jeffrey Shallit_, Dec 11 2020
