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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.

%I A023409 #23 Jan 27 2022 22:48:11
%S A023409 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,
%T A023409 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,
%U A023409 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
%N 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.
%C A023409 From _Robert Israel_, Mar 30 2018: (Start)
%C A023409 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.
%C A023409 Pomerance (see link) shows the sequence is not eventually periodic. (End)
%H A023409 Robert Israel, <a href="/A023409/b023409.txt">Table of n, a(n) for n = 0..10000</a>
%H A023409 C. Pomerance, <a href="https://www.math.dartmouth.edu/~carlp/sixes.pdf">Sixes and sevens</a>, Missouri J. Math. Sci. 6 (1994), 62-63.
%p A023409 a[0]:= 6: v:= 6:
%p A023409 for n from 1 to 100 do
%p A023409   if 6*10^n+v mod 2^(n+1)=0 then a[n]:= 6 else a[n]:= 7 fi;
%p A023409   v:= v + a[n]*10^n
%p A023409 od:
%p A023409 seq(a[i],i=0..100); # _Robert Israel_, Mar 30 2018
%Y A023409 Cf. A023396, A023397, A023398, A023399, A023400, A023401, A023402, A023403, A023404, A023405, A023406, A023407, A023408, A023410, A023411, A023412, A023413, A023414, A023415.
%K A023409 nonn,base
%O A023409 0,1
%A A023409 _David W. Wilson_