

A172372


Least number k such that the nth prime not dividing 10 (A004139(n)) divides the repunit (10^k1)/9.


0



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
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,1


COMMENTS

If p is a 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 (10^31)/9 = 111. Otherwise, take k = (10^p  1)/9; by the Fermat theorem, 10^(p1) == 1 (mod p), so p divides (10^(p1)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


REFERENCES

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


LINKS

Table of n, a(n) for n=1..65.
Project Euler, Problem 129: Repunit divisibility
S. S. Wagstaff, Jr.,The Cunningham Project
Eric Weisstein's World of Mathematics, Repunit.


EXAMPLE

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.


PROG

(PARI) a(n) = {k=1; p = if(n>1, prime(n+2), 3); while((10^k1)/9 % p, k++); k; } \\ Michel Marcus, May 25 2014


CROSSREFS

Cf. A002275 (repunits), A095250 (11111111... (n times) mod n).
Sequence in context: A259498 A118453 A021969 * A046901 A169751 A105332
Adjacent sequences: A172369 A172370 A172371 * A172373 A172374 A172375


KEYWORD

nonn


AUTHOR

Michel Lagneau, Feb 01 2010


EXTENSIONS

Added punctuation to the examples. 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


STATUS

approved



