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If any power of 2 ends with k 3's and 6's, they must be the first k terms of this sequence in reverse order.
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%I #14 Jan 27 2022 22:46:14

%S 6,3,3,6,6,3,6,3,6,3,3,3,3,3,6,3,3,3,6,3,3,3,3,3,6,6,3,3,6,3,3,6,3,6,

%T 3,3,3,6,3,6,3,6,6,3,3,6,3,6,6,6,3,3,3,3,6,6,3,6,3,3,3,6,6,6,3,6,3,3,

%U 6,6,6,3,6,6,3,3,6,3,3,3,6,3,3,6,3,6,3,3,6,6,3,3,6,3,3,6,3,3,6,3,6,3,6,3,6,3,3,3

%N If any power of 2 ends with k 3's and 6's, they must be the first k terms of this sequence in reverse order.

%H Ray Chandler, <a href="/A023407/b023407.txt">Table of n, a(n) for n = 1..10000</a>

%H Milan Janjic, <a href="http://www.pmfbl.org/janjic/">Enumerative Formulas for Some Functions on Finite Sets</a>

%F a(n) = 3*A023396(n).

%Y Cf. A023396.

%K nonn,base

%O 1,1

%A _David W. Wilson_