A023409 - OEIS

A023409

If any power of 2 ends with k 6's and 7's, they must be the first k terms of this sequence in reverse order.

6, 7, 7, 7, 6, 6, 6, 6, 7, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 6, 7, 7, 6, 6, 7, 7, 7, 6, 6, 6, 6, 7, 6, 7, 7, 7, 7, 7, 6, 6, 6, 6, 7, 6, 7, 7, 6, 6, 6, 6, 7, 6, 7, 6, 6, 7, 6, 7, 7, 7, 6, 6, 6, 7, 6, 7, 7, 7, 6, 6, 6, 6, 6, 7, 6, 6, 6, 7, 7, 6, 7, 7, 6, 7, 7, 6, 7, 6, 6, 7, 7, 6, 7, 6, 7, 7, 6, 6, 7, 7, 6, 6, 7

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OFFSET

0,1

COMMENTS

From Robert Israel, Mar 30 2018: (Start)

a(0)=6. If the concatenation 6a(n)...a(0) (as a decimal number) is divisible by 2^(n+2) then a(n+1)=6, otherwise a(n+1)=7.

Pomerance (see link) shows the sequence is not eventually periodic. (End)

LINKS

Robert Israel, Table of n, a(n) for n = 0..10000

C. Pomerance, Sixes and sevens, Missouri J. Math. Sci. 6 (1994), 62-63.

MAPLE

a[0]:= 6: v:= 6: