cp's OEIS Frontend

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

%I A069871 #47 Mar 16 2024 15:19:55
%S A069871 3,9,11,33,111,333,1111,3333,11111,33333,111111,142857,333333,1111111,
%T A069871 3333333,11111111,33333333,111111111,333333333,1111111111,3333333333,
%U A069871 11111111111,33333333333,111111111111,142857142857,333333333333,1111111111111
%N A069871 Numbers m that divide the concatenation of m-1 and m+1.
%C A069871 All the numbers of the form (10^k - 1)/3 and (10^k - 1)/9 are terms.
%C A069871 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
%C A069871 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
%F A069871 From _Robert Israel_, Nov 04 2014: (Start)
%F A069871 a(1+2*j + 13*k) = (10^(1+j+6*k)-1)/9, j=0..5, k >= 0 (except for j=k=0).
%F A069871 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).
%F A069871 a(13*k - 1) = (10^(6*k)-1)/7, k >= 1.
%F A069871 (End)
%e A069871 3 belongs to this sequence since 3 divides 24; 11 belongs to this sequence since 11 divides 1012.
%e A069871 9 belongs to this sequence since 9 divides the concatenation of 8 and 10, i.e., 810.
%e A069871 142857 belongs to this sequence since 142857 divides the concatenation of 142856 and 142858, i.e., 142856142858/142857 = 999994.
%p A069871 N:= 10: # to get all terms with at most N digits
%p A069871 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
%t A069871 Select[ Range[10^8], Mod[ FromDigits[ Join[ IntegerDigits[ # - 1], IntegerDigits[ # + 1]]], # ] == 0 & ]
%o A069871 (PARI) isok(n) = eval(concat(Str(n-1), Str(n+1))) % n == 0; \\ _Michel Marcus_, Nov 04 2014
%Y A069871 Cf. A069860, A069862, A088797, A088798.
%Y A069871 Cf. A077192.
%K A069871 nonn,base
%O A069871 1,1
%A A069871 _Amarnath Murthy_, Apr 24 2002
%E A069871 More terms from _Sascha Kurz_, Feb 10 2003
%E A069871 Missing a(12) added by _Paolo P. Lava_ and missing a(25) added by _Alois P. Heinz_, Nov 03 2014