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If any power of 2 ends with k 2's and 7's, they must be the first k terms of this sequence in reverse order.
3

%I #13 Jan 27 2022 22:44:38

%S 2,7,2,2,2,2,7,2,7,2,7,7,7,7,2,2,7,2,7,2,7,7,2,7,7,2,7,2,7,7,2,2,2,2,

%T 7,2,7,2,7,2,2,2,7,7,7,7,2,2,7,7,7,2,7,2,2,7,7,7,7,2,7,7,2,7,7,2,7,7,

%U 7,7,7,2,2,7,2,7,7,7,7,7,2,2,7,7,2,2,7,7,2,7,2,7,2,7,2,7,2,7,2,7,2,2,7,2,7,7,2,7

%N If any power of 2 ends with k 2's and 7's, they must be the first k terms of this sequence in reverse order.

%H Ray Chandler, <a href="/A023399/b023399.txt">Table of n, a(n) for n = 1..10000</a>

%K nonn,base

%O 1,1

%A _David W. Wilson_