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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.
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%I #23 Jan 27 2022 22:48:11

%S 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 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 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 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 From _Robert Israel_, Mar 30 2018: (Start)

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

%H Robert Israel, <a href="/A023409/b023409.txt">Table of n, a(n) for n = 0..10000</a>

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

%p for n from 1 to 100 do

%p if 6*10^n+v mod 2^(n+1)=0 then a[n]:= 6 else a[n]:= 7 fi;

%p v:= v + a[n]*10^n

%p od:

%p seq(a[i],i=0..100); # _Robert Israel_, Mar 30 2018

%Y Cf. A023396, A023397, A023398, A023399, A023400, A023401, A023402, A023403, A023404, A023405, A023406, A023407, A023408, A023410, A023411, A023412, A023413, A023414, A023415.

%K nonn,base

%O 0,1

%A _David W. Wilson_