A095250 -id:A095250 - cp's OEIS frontend

A094918 a(n) = (3^n-1)/2 mod n.

0, 0, 1, 0, 1, 4, 1, 0, 4, 4, 1, 4, 1, 4, 13, 0, 1, 4, 1, 0, 13, 4, 1, 16, 21, 4, 13, 12, 1, 4, 1, 0, 13, 4, 23, 4, 1, 4, 13, 0, 1, 28, 1, 40, 31, 4, 1, 16, 15, 24, 13, 40, 1, 40, 33, 32, 13, 4, 1, 40, 1, 4, 13, 0, 56, 34, 1, 40, 13, 54, 1, 40, 1, 4, 28, 40, 37, 52, 1, 0, 40, 4, 1, 28, 36, 4, 13, 24

Offset: 1

N. J. A. Sloane, Jun 18 2004

Antti Karttunen, Table of n, a(n) for n = 1..16384
Antti Karttunen, Data supplement: n, a(n) computed for n = 1..65537

(k^n-1)/(k-1) mod n for k = 1, ..., 10 gives A082495, this sequence, A094919, A094920, A094921, A094922, A094923, A094924 and A095250.

Cf. A003462.

A094918(n) = lift(Mod((3^n-1)/2,n)); \\ Antti Karttunen, May 19 2020

A067935 Let rep(n) be the n-th repunit number, sequence gives values of n such that : rep(n)==rep(2) (mod n).

Original entry on oeis.org

1, 2, 4, 5, 10, 14, 20, 22, 25, 26, 34, 38, 44, 46, 50, 58, 62, 74, 82, 86, 94, 100, 106, 110, 118, 122, 134, 140, 142, 146, 158, 166, 178, 182, 185, 194, 202, 206, 214, 218, 220, 226, 254, 260, 262, 274, 278, 298, 302, 308, 314, 326, 334, 346, 350, 358, 362

Offset: 1

Benoit Cloitre, Mar 05 2002

nonn

(10^5-1)/9 = 11+5*2220, hence 5 is in the sequence.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A002275, A095250 (rep(n) mod n).

filter:= n -> 10&^n -100 mod (9*n) = 0:
select(filter, [$1..400]); # Robert Israel, Jul 02 2019

A172372 Least number k such that the n-th prime not dividing 10 (A004139(n)) divides the repunit (10^k-1)/9.

Original entry on oeis.org

3, 6, 2, 6, 16, 18, 22, 28, 15, 3, 5, 21, 46, 13, 58, 60, 33, 35, 8, 13, 41, 44, 96, 4, 34, 53, 108, 112, 42, 130, 8, 46, 148, 75, 78, 81, 166, 43, 178, 180, 95, 192, 98, 99, 30, 222, 113, 228, 232, 7, 30, 50, 256, 262, 268, 5, 69, 28, 141, 146, 153, 155, 312, 79, 110

Offset: 1

Michel Lagneau, Feb 01 2010

nonn

If p is an odd prime different from 5, then p divides an infinite number of terms of the sequence of repunits {1, 11, 111, 1111, ... }. The proof is elementary: let p be such a prime. If p = 3, then 3 divides (10^3-1)/9 = 111. Otherwise, take k = (10^p - 1)/9; by the Fermat theorem, 10^(p-1) == 1 (mod p), so p divides (10^(p-1)-1); since p is relatively prime to 9, it divides k. Trivially, if p divides any k digit repunit, it divides the k*m digit repunit as well.

Essentially the same as A002371. - T. D. Noe, Apr 11 2012

			3 divides 111, but not 1 or 11, so a(1) = 3.
7 divides 111111 but not 1, 11, 111, 1111, or 11111, so a(2) = 6.

David Wells, The Factors of the Repunits 11 through R_40, The Penguin Dictionary of Curious and Interesting Numbers, p. 219 Penguin 1986.
David Wells, The Penguin Dictionary of Curious and Interesting Numbers, Penguin Books, 1997.
David Wells, Curious and Interesting Numbers (Revised), Penguin Books, page 114.

Project Euler, Problem 129: Repunit divisibility
S. S. Wagstaff, Jr., The Cunningham Project
Eric Weisstein's World of Mathematics, Repunit.

Cf. A002275 (repunits), A002371 (period of decimal expansion of 1/prime(n)), A004139 (odd primes excluding 5), A095250 (11111111... (n times) mod n).

a(n) = {k=1; p = if(n>1, prime(n+2), 3); while((10^k-1)/9 % p, k++); k;} \\ Michel Marcus, May 25 2014

Corrected and edited by Michel Lagneau, Apr 25 2010

Term 6 between terms 44 and 96 doesn't belong to the sequence. The same for term 43 between terms 43 and 178. Corrected and edited by Krzysztof Wojtas, May 25 2014

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A094918 a(n) = (3^n-1)/2 mod n.

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A067935 Let rep(n) be the n-th repunit number, sequence gives values of n such that : rep(n)==rep(2) (mod n).

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A172372 Least number k such that the n-th prime not dividing 10 (A004139(n)) divides the repunit (10^k-1)/9.

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