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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, 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
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OFFSET
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1,1
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COMMENTS
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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.
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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.
David 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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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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PROG
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(PARI) 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
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CROSSREFS
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Cf. A002275 (repunits), A002371 (period of decimal expansion of 1/prime(n)), A004139 (odd primes excluding 5), A095250 (11111111... (n times) mod n).
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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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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STATUS
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approved
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