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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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.
Crossrefs
Programs
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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
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