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A023409 If any power of 2 ends with k 6's and 7's, they must be the first k elements 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, else 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 * A008938 A256425 A021600

Adjacent sequences:  A023406 A023407 A023408 * A023410 A023411 A023412

KEYWORD

nonn,base

AUTHOR

David W. Wilson

STATUS

approved

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Last modified June 21 15:37 EDT 2021. Contains 345364 sequences. (Running on oeis4.)