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Search: id:A039678
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| A039678 |
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Smallest a>1 such that a^(p-1)-1 is divisible by p^2, p = n-th prime. |
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+0
5
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| 5, 8, 7, 18, 3, 19, 38, 28, 28, 14, 115, 18, 51, 19, 53, 338, 53, 264, 143, 11, 306, 31, 99, 184, 53, 181, 43, 164, 96, 68, 38, 58, 19, 328, 313, 78, 226, 65, 253, 259, 532, 78, 176, 276, 143, 174, 165, 69, 330, 44, 33, 332, 94, 263, 48, 79, 171, 747, 731, 20, 147, 91, 40
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OFFSET
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1,1
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COMMENT
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Using Fermat's little theorem twice, it is easy to see that a=p^2-1 solves this problem for all odd primes p. In fact, there appear to be exactly p-1 values of a with 1 <= a <= p^2 for which a^(p-1)=1 (mod p^2). See A096082 for the related open problem. - T. D. Noe (noe(AT)sspectra.com), Aug 24 2008
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REFERENCES
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P. Ribenboim, The New Book of Prime Number Records, Springer, 1996, 345-349.
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LINKS
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T. D. Noe, Table of n, a(n) for n = 1..10000
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EXAMPLE
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For n=3, p=5 is 3rd prime and 5^2 = 25 divides 7^4 - 1 = 2401.
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CROSSREFS
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Adjacent sequences: A039675 A039676 A039677 this_sequence A039679 A039680 A039681
Sequence in context: A165909 A019845 A053787 this_sequence A131040 A007450 A068470
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KEYWORD
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nonn,nice
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AUTHOR
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Felice Russo (felice.russo(AT)katamail.com)
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EXTENSIONS
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More terms from David W. Wilson (davidwwilson(AT)comcast.net)
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