

A023409


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


2



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, else 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), 6263.


MAPLE

a[0]:= 6: v:= 6:
for n from 1 to 100 do
if 6*10^n+v mod 2^(n+1)=0 then a[n]:= 6 else a[n]:= 7 fi;
v:= v + a[n]*10^n
od:
seq(a[i], i=0..100); # Robert Israel, Mar 30 2018


CROSSREFS

Cf. A023396, A023397, A023398, A023399, A023400, A023401, A023402, A023403, A023404, A023405, A023406, A023407, A023408, A023410, A023411, A023412, A023413, A023414, A023415.
Sequence in context: A084256 A197692 A322415 * A008938 A256425 A021600
Adjacent sequences: A023406 A023407 A023408 * A023410 A023411 A023412


KEYWORD

nonn,base


AUTHOR

David W. Wilson


STATUS

approved



