%I #18 Jan 27 2022 22:48:04

%S 2,3,2,3,2,2,2,3,2,3,2,3,3,2,3,3,3,3,3,3,2,3,2,2,2,3,3,3,3,3,3,3,3,2,

%T 3,3,2,3,3,3,3,3,2,2,2,2,3,3,2,3,2,2,3,2,2,3,2,2,3,2,2,2,2,3,2,2,3,2,

%U 3,3,3,2,2,3,3,3,2,3,3,3,2,2,2,2,2,3,2,2,3,3,3,2,3,2,3,3,2,3,2,3,3,2,2,2,2,2,3,2

%N In base 10, if any power of 2 ends with k 2's and 3's, they must be the first k terms of this sequence in reverse order.

%H Ray Chandler, <a href="/A023397/b023397.txt">Table of n, a(n) for n = 1..10000</a>

%e No power of 2 ends with 3, so the first term is 2.

%e No power of 2 is == 22 (mod 100), since 4 does not divide 22, so the next term is 3 (and 4 does divide 32).

%e No power of 2 is == 332 (mod 1000), since 8 does not divide 332, so the next term is 2 (and 8 does divide 232). And so on.

%Y Cf. A023410, A053316.

%K nonn,base

%O 1,1

%A _David W. Wilson_