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%I #24 Jul 04 2021 13:44:46
%S 1,6,8,12,14,16,20,22,27,30,32,34,38,40,45,46,51,53,56,58,63,64,69,71,
%T 75,77,79,84,86,88,92,94,99,100,105,107,110,112,117,118,123,125,129,
%U 131,133,136,141,143,147,149,151,155,157,162
%N Numbers k such that A026600(k) = 1.
%C It appears that a(n) gives the position of its own n-th 1 modulo 3 term, the n-th 2 modulo 3 term in A026602, and the n-th multiple of 3 in A026603. A026602 and A026603 appear to have analogous indexical properties. - _Matthew Vandermast_, Oct 06 2010
%C This follows directly from the generating morphism for A026600: a 1 in position k creates a 1 in position 3k-2, a 2 in position 3k-1, and a 3 in position 3k. Since each block of three terms in A026600 is a permutation of {1,2,3}, these created terms are the k-th terms of their respective index sequences. The proof for the other index sequences is similar. - _Charlie Neder_, Mar 10 2019
%H Amiram Eldar, <a href="/A026601/b026601.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = A079498(n) + 1.
%Y Cf. A026600, A026602, A026603, A079498.
%K nonn
%O 1,2
%A _Clark Kimberling_
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