A023401 -id:A023401 - cp's OEIS frontend

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

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

A053314 a(n) contains n digits (either '1' or '4') and is divisible by 2^n.

Original entry on oeis.org

4, 44, 144, 4144, 14144, 414144, 1414144, 41414144, 441414144, 1441414144, 11441414144, 411441414144, 4411441414144, 44411441414144, 444411441414144, 1444411441414144, 41444411441414144, 441444411441414144

Offset: 1

Henry Bottomley, Mar 06 2000

Robert Israel, Table of n, a(n) for n = 1..999

Cf. A023401, A050621, A050622, A035014.

A[1]:= 4:
for n from 2 to 100 do
   if A[n-1] mod 2^n = 0 then A[n]:= A[n-1]+4*10^(n-1)
   else A[n]:= A[n-1]+10^(n-1)
fi
od:
seq(A[i],i=1..100); # Robert Israel, Oct 27 2019

nxt[{n_,a_}]:={n+1,If[Divisible[a,2^(n+1)],4*10^IntegerLength[a]+ a, 10^IntegerLength[ a]+a]}; NestList[nxt,{1,4},20][[All,2]] (* Harvey P. Dale, Oct 30 2022 *)

a(n) = a(n-1) + 10^(n-1)*(4 - 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 4, if not then n-th term begins with a 1.

Formula corrected by Robert Israel, Oct 27 2019

cp's OEIS Frontend

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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