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:
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
KEYWORD
nonn,base
AUTHOR
STATUS
approved