A069871 Numbers m that divide the concatenation of m-1 and m+1.
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
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.
Programs
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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
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Mathematica
Select[ Range[10^8], Mod[ FromDigits[ Join[ IntegerDigits[ # - 1], IntegerDigits[ # + 1]]], # ] == 0 & ]
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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
Comments