A023410 -id:A023410 - cp's OEIS frontend

A023397 In base 10, if any power of 2 ends with k 2's and 3's, they must be the first k terms of this sequence in reverse order.

2, 3, 2, 3, 2, 2, 2, 3, 2, 3, 2, 3, 3, 2, 3, 3, 3, 3, 3, 3, 2, 3, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 2, 3, 3, 2, 3, 3, 3, 3, 3, 2, 2, 2, 2, 3, 3, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 2, 2, 3, 2, 2, 3, 2, 3, 3, 3, 2, 2, 3, 3, 3, 2, 3, 3, 3, 2, 2, 2, 2, 2, 3, 2, 2, 3, 3, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3, 3, 2, 2, 2, 2, 2, 3, 2

Offset: 1

David W. Wilson

			No power of 2 ends with 3, so the first term is 2.
No power of 2 is == 22 (mod 100), since 4 does not divide 22, so the next term is 3 (and 4 does divide 32).
No power of 2 is == 332 (mod 1000), since 8 does not divide 332, so the next term is 2 (and 8 does divide 232). And so on.

Ray Chandler, Table of n, a(n) for n = 1..10000

Cf. A023410, A053316.

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.

Original entry on oeis.org

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

Offset: 0

David W. Wilson

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)

Robert Israel, Table of n, a(n) for n = 0..10000
C. Pomerance, Sixes and sevens, Missouri J. Math. Sci. 6 (1994), 62-63.

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

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

A053338 a(n) contains n digits (either '6' or '9') and is divisible by 2^n.

Original entry on oeis.org

6, 96, 696, 9696, 69696, 669696, 6669696, 96669696, 696669696, 9696669696, 69696669696, 969696669696, 9969696669696, 69969696669696, 969969696669696, 9969969696669696, 99969969696669696, 999969969696669696

Offset: 1

Henry Bottomley, Mar 06 2000

Ray Chandler, Table of n, a(n) for n = 1..1000

Cf. A023410, A050621, A050622, A035014.

Table[Select[FromDigits/@Tuples[{6,9},n],Mod[#,2^n]==0&],{n,20}]//Flatten (* Harvey P. Dale, Sep 15 2023 *)

a(n)=a(n-1)+10^(n-1)*(6+3*[a(n-1)/2^(n-1) mod 2]) i.e. a(n) ends with a(n-1); if (n-1)-th term is divisible by 2^n then n-th term begins with a 6, if not then n-th term begins with a 9.

cp's OEIS Frontend

A023397 In base 10, if any power of 2 ends with k 2's and 3's, they must be the first k terms of this sequence in reverse order.

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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.

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A053338 a(n) contains n digits (either '6' or '9') and is divisible by 2^n.

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