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A134307
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Primes such that A^(p-1) == 1 (mod p^2) for some A in the range 2 <= A <= p-1.
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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; refs; listen; history; internal format)
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OFFSET
| 1,1
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COMMENTS
| 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 probability about 1 - (1 - 1/p)^(p-1) ~= 1 - 1/e.
Numerically, among the primes below 1000 (out of the total number pi(1000)=168) there are 103 terms of the sequence, and the ratio 103/168 = 0.613 which is already somewhat close to 1-1/e ~= 0.632.
If we replace p^2 with p^3, heuristically it is likely that the sequence is finite (since 1 - (1 - 1/p^2)^(p-1) tends to 0 as p grows). [Max Alekseyev]
Replacing p^2 with p^3 gives just the one term (113) for p < 10^6. - Joerg Arndt, Jan 07 2011
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REFERENCES
| L. E. Dickson: "History of the theory of numbers", vol. 1, p. 105.
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LINKS
| T. D. Noe, Table of n, a(n) for n=1..1000
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EXAMPLE
| 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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PROG
| (PARI)
{ forprime (p=2, 1000,
for (a=2, p-1, p2 = p^2;
if( Mod(a, p2)^(p-1) == Mod(1, p2), print1(p, ", ") ; break() );
); ); }
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CROSSREFS
| Cf. A001220, A055578, A039678, A143548.
Sequence in context: A124110 A153768 A092194 * A087693 A106017 A106065
Adjacent sequences: A134304 A134305 A134306 * A134308 A134309 A134310
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KEYWORD
| nonn,changed
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AUTHOR
| Joerg Arndt (arndt(AT)jjj.de), Aug 27 2008
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