A000864 -id:A000864 - cp's OEIS frontend

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

Offset: 1

Benoit Cloitre, Mar 05 2002

nonn

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

			(10^11 - 1)/9 = 11111111111 == 1 (mod 11), so 11 is a term.
We also have the congruence 10^11 == 10 (mod 9*11).

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

Cf. A000864, A015919, A303608, A005939.

{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 *)

is(n)=n==1 || ((10^n-1)/9)%n==1 \\ Eric Chen, Jun 13 2018

A303608 Repunit pseudoprimes: composite numbers k such that (10^k - 1)/9 == 1 (mod k).

Original entry on oeis.org

10, 55, 91, 259, 370, 385, 451, 481, 505, 703, 715, 1045, 1105, 1729, 2035, 2465, 2821, 2981, 3367, 4141, 4187, 5005, 5461, 6533, 6541, 6565, 6601, 7471, 7777, 8149, 8401, 8695, 8905, 8911, 10001, 10585, 11111, 12403, 13366, 13981, 14245, 14645, 14701, 14911, 15211, 15841, 18685

Offset: 1

Thomas Ordowski, Apr 27 2018

nonn

Composite numbers k such that 10^k == 10 (mod 9k).
If k is a term, then so is (10^k - 1)/9. Thus, the sequence is infinite.
No terms are divisible by 3. - Robert Israel, May 28 2018

			(10^10 - 1)/9 = 1111111111 == 1 (mod 10), so the composite 10 is a term.
Equivalently, we have the congruence 10^10 == 10 (mod 9*10).

Robert Israel, Table of n, a(n) for n = 1..4130

A000864 is a subsequence.

Composite numbers in A067934. - Michel Marcus, Apr 27 2018

filter:=  n -> n mod 3 <> 0 and (10&^n - 10) mod n = 0\ and not isprime(n):
select(filter,[$4..10^5]); # Robert Israel, May 28 2018

Select[Range@ 20000, ! PrimeQ@# && PowerMod[10, #, 9 #] == 10 &] (* Robert G. Wilson v, Apr 28 2018 *)

isok(n) = (n>1) && !isprime(n) && Mod(10, 9*n)^n == 10; \\ Michel Marcus, Apr 28 2018

a(4) onward from Robert G. Wilson v, Apr 27 2018

cp's OEIS Frontend

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).

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