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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. 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 (list; graph; refs; listen; history; text; internal format)
 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 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 * A349991 A351784 A008938 Adjacent sequences: A023406 A023407 A023408 * A023410 A023411 A023412 KEYWORD nonn,base AUTHOR David W. Wilson STATUS approved

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Last modified August 7 20:39 EDT 2024. Contains 375017 sequences. (Running on oeis4.)