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A172372
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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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0
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3, 6, 2, 6, 16, 18, 22, 28, 15, 3, 5, 21, 46, 13, 58, 60, 33, 35, 8, 13, 41, 44, 6, 96, 4, 34, 53, 108, 112, 42, 130, 8, 46, 148, 75, 78, 81, 166, 43, 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
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OFFSET
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1,1
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COMMENTS
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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 (10^3-1)/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
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REFERENCES
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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.
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LINKS
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Table of n, a(n) for n=1..67.
S. S. Wagstaff, Jr.,The Cunningham Project
Eric Weisstein's World of Mathematics, Repunit.
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EXAMPLE
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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.
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CROSSREFS
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A002275 A002275, Repunits: (10^n - 1)/9. A095250 a(n) = 11111111... (n times) = (10^n-1)/9 reduced mod n
Sequence in context: A194033 A118453 A021969 * A046901 A169751 A105332
Adjacent sequences: A172369 A172370 A172371 * A172373 A172374 A172375
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KEYWORD
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nonn
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AUTHOR
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Michel Lagneau, Feb 01 2010
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EXTENSIONS
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Added punctuation to the examples. Corrected and edited by Michel Lagneau, Apr 25 2010
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STATUS
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approved
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