cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A069871 Numbers m that divide the concatenation of m-1 and m+1.

Original entry on oeis.org

3, 9, 11, 33, 111, 333, 1111, 3333, 11111, 33333, 111111, 142857, 333333, 1111111, 3333333, 11111111, 33333333, 111111111, 333333333, 1111111111, 3333333333, 11111111111, 33333333333, 111111111111, 142857142857, 333333333333, 1111111111111
Offset: 1

Views

Author

Amarnath Murthy, Apr 24 2002

Keywords

Comments

All the numbers of the form (10^k - 1)/3 and (10^k - 1)/9 are terms.
These (i.e., (10^k - 1)/3 for k >= 1, (10^k - 1)/9 for k >= 2, and (10^(6*k) - 1)/7 for k >= 1) are all the terms of the sequence, apart from 9. This is because if 10^(i-1) <= x+1 < 10^i, x | 10^i*(x-1) + x + 1 iff x | 10^i - 1, and then 1 < d = (10^i - 1)/x <= (10^i - 1)/(10^(i-1)-1) < 10. Since 2,4,5,6,8 can't divide 10^i-1, d must be 3, 7 or 9. - Robert Israel, Nov 04 2014
The terms of the sequence satisfy the condition that both m-1 and m+1 must be greater than 0. If m-1=0 were admitted then 1 would also be part of the sequence. - Michel Marcus, Nov 05 2014

Examples

			3 belongs to this sequence since 3 divides 24; 11 belongs to this sequence since 11 divides 1012.
9 belongs to this sequence since 9 divides the concatenation of 8 and 10, i.e., 810.
142857 belongs to this sequence since 142857 divides the concatenation of 142856 and 142858, i.e., 142856142858/142857 = 999994.
		

Crossrefs

Programs

  • Maple
    N:= 10: # to get all terms with at most N digits
    3,9, seq(seq((10^k-1)/d, d = `if`(k mod 6 = 0, [9,7,3],[9,3])), k = 2 .. N); # Robert Israel, Nov 04 2014
  • Mathematica
    Select[ Range[10^8], Mod[ FromDigits[ Join[ IntegerDigits[ # - 1], IntegerDigits[ # + 1]]], # ] == 0 & ]
  • PARI
    isok(n) = eval(concat(Str(n-1), Str(n+1))) % n == 0; \\ Michel Marcus, Nov 04 2014

Formula

From Robert Israel, Nov 04 2014: (Start)
a(1+2*j + 13*k) = (10^(1+j+6*k)-1)/9, j=0..5, k >= 0 (except for j=k=0).
a(2*j + 13*k) = (10^(j+6*k)-1)/3, j=0..5, k >= 0 (except for j=k=0 and j=1,k=0).
a(13*k - 1) = (10^(6*k)-1)/7, k >= 1.
(End)

Extensions

More terms from Sascha Kurz, Feb 10 2003
Missing a(12) added by Paolo P. Lava and missing a(25) added by Alois P. Heinz, Nov 03 2014