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Search: id:A134307
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| A134307 |
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Primes such that A^(p-1)==1 (mod P^2) for some A where 1<A<=p-1. |
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+0
2
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| 11, 29, 37, 43, 59, 71, 79, 97, 103, 109, 113, 127, 131, 137, 151, 163, 181, 191, 197, 199, 211, 223, 229, 233, 241, 257, 263, 269, 281, 283, 293, 307, 313, 331, 347, 349, 353, 359, 367, 373, 379, 397, 401, 419, 421, 433, 439, 449, 461, 463, 487, 499, 509
(list; graph; listen)
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OFFSET
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1,1
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COMMENT
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It's worth observing that there are p-1 elements of order dividing p-1 modulo
p^2 that are of the form r^(k*p) mod p^2 where r is a primitive
element modulo p and k=0,1,...,p-2. Heuristically, one can expect that
at least one of them belongs to the interval [2,p-1] with the
probability about 1 - (1 - 1/p)^(p-1) ~= 1 - 1/e.
Numerically, among primes below 1000 (of the total number
pi(1000)=168) there 103 terms of your sequence and the ratio 103/168
~= 0.613 which is already somewhat close to 1-1/e ~= 0.632.
As of replacing p^2 with p^3, heuristically it likely that the
sequence is finite (since 1 - (1 - 1/p^2)^(p-1) tends to 0 as p
grows). [Max Alekseyev]
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REFERENCES
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L.E.Dickson: "History of the theory of numbers", vol.1, p.105
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LINKS
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T. D. Noe, Table of n, a(n) for n=1..1000
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EXAMPLE
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Examples (pairs [p, A]):
[11, 3]
[11, 9]
[29, 14]
[37, 18]
[43, 19]
[59, 53]
[71, 11]
[71, 26]
[79, 31]
[97, 53]
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PROGRAM
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(PARI) { forprime (p=2, 1000, for (a=2, p-1, p2 = p^2; q = ( Mod(a, p2)^(p-1) == Mod(1, p2) ); if ( q, print1(p, ", "); break() ); ); ); }
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CROSSREFS
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Cf. A001220, A055578, A039678, A143548.
Sequence in context: A124110 A153768 A092194 this_sequence A087693 A106017 A106065
Adjacent sequences: A134304 A134305 A134306 this_sequence A134308 A134309 A134310
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KEYWORD
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nonn
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AUTHOR
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Joerg Arndt (arndt(AT)jjj.de), Aug 27 2008
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