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A084740
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Least k such that (n^k-1)/(n-1) is prime, or 0 if no such prime exists.
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16
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2, 3, 2, 3, 2, 5, 3, 0, 2, 17, 2, 5, 3, 3, 2, 3, 2, 19, 3, 3, 2, 5, 3, 0, 7, 3, 2, 5, 2, 7, 0, 3, 13, 313, 2, 13, 3, 349, 2, 3, 2, 5, 5, 19, 2, 127, 19, 0, 3, 4229, 2, 11, 3, 17, 7, 3, 2, 3, 2, 7, 3, 5, 0, 19, 2, 19, 5, 3, 2, 3, 2, 5, 5, 3, 41, 3, 2, 5, 3, 0, 2, 5, 17, 5, 11, 7, 2, 3, 3, 4421, 439, 7, 5, 7, 2, 17, 13, 3, 2, 3, 2, 19, 97, 3, 2, 17, 2, 17, 3, 3, 2, 23, 29, 7, 59
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OFFSET
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2,1
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COMMENTS
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When (n^k-1)/(n-1) is prime, k must be prime. As mentioned by Dubner, when n is a perfect power, then (n^k-1)/(n-1) will usually be composite for all k, which is the case for n = 9, 25, 32, 49, 64, 81, 121, 125, 144, 169, 216, 225, 243, 289, 324, 343, ... - T. D. Noe, Jan 30 2004
a(152) > prime(1100) or 0. - Derek Orr, Nov 29 2014
a(n)=2 if and only if n=p-1, where p is an odd prime; that is, n belongs to A006093, except 2. - Thomas Ordowski, Sep 19 2015
Probably a(152) = 270217 since (152^270217-1)/(152-1) has been shown to be probably prime. - Michael Stocker, Jan 24 2019
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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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a(7) = 5 as (7^5 - 1 )/(7 - 1) = 2801 = 1 + 7 + 7^2 + 7^3 + 7^4 is a prime but no smaller partial sum yields a prime.
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PROG
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(PARI) a(n) = {l=List([9, 25, 32, 49, 64, 81, 121, 125, 144, 169, 216, 225, 243, 289, 324, 343]); for(q=1, #l, if(n==l[q], return(0))); k=1; while(k, s=(n^prime(k)-1)/(n-1); if(ispseudoprime(s), return(prime(k))); k++)}
n=2; while(n<361, print1(a(n), ", "); n++) \\ Derek Orr, Jul 13 2014
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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Amarnath Murthy and Meenakshi Srikanth (menakan_s(AT)yahoo.com), Jun 15 2003
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EXTENSIONS
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STATUS
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approved
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