A067934 - OEIS

A067934

Let rep(k) = (10^k - 1)/9 be the k-th repunit number = 11111..1111 with k 1 digits, then sequence gives values of k such that rep(k) == 1 (mod k).

1, 2, 5, 7, 10, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 55, 59, 61, 67, 71, 73, 79, 83, 89, 91, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 259

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OFFSET

1,2

COMMENTS

Due to Fermat's little theorem, all prime numbers except 3 are in the sequence. E.g., rep(17) = 1 + 17*653594771241830.

Numbers n such that 10^n == 10 (mod 9n). The number (10^n - 1)/9 is a term if and only if n is a term. - Thomas Ordowski, Apr 28 2018

Generally, the repunit theorem: Let integer b <> 1 and n be a positive integer. Define R_b(n) = (b^n-1)/(b-1) = N. Then R_b(N) == 1 (mod N) if and only if N == 1 (mod n). - Thomas Ordowski, Apr 28 2018

Proof: (b^N-1)/(b-1)-1 = (b^N-b)/(b-1) is divisible by N if and only if b^N-b is divisible by b^n-1. Since b^N-b == b^(N mod n)-b (mod b^n-1), we have that b^N-b is divisible by b^n-1 if and only if N == 1 (mod n). QED. - Max Alekseyev, Apr 28 2018

Terms which are not prime are 1 U A303608. - Robert G. Wilson v, Jun 13 2018

No multiples of 3 are in this sequence. - Eric Chen, Jun 13 2018

A005939 is subsequence. - Eric Chen, Jun 13 2018

LINKS

Michael De Vlieger, Table of n, a(n) for n = 1..10000

EXAMPLE

(10^11 - 1)/9 = 11111111111 == 1 (mod 11), so 11 is a term.

We also have the congruence 10^11 == 10 (mod 9*11).

MATHEMATICA

{1}~Join~Select[Range[260], Mod[#2, #1] == 1 & @@ {#, (10^# - 1)/9} &] (* Michael De Vlieger, May 06 2018 *)

fQ[n_] := PowerMod[10, n, 9 n] == 10; fQ[1] = True; Select[Range@260, fQ] (* Robert G. Wilson v, Jun 13 2018 *)

PROG