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A319040
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Numbers k > 1 such that Pell(k) == 1 (mod k).
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3
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7, 17, 23, 31, 35, 41, 47, 71, 73, 79, 89, 97, 103, 113, 127, 137, 151, 167, 169, 191, 193, 199, 223, 233, 239, 241, 257, 263, 271, 281, 311, 313, 337, 353, 359, 367, 383, 385, 401, 409, 431, 433, 439, 449, 457, 463, 479, 487, 503, 521, 569, 577, 593, 599
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OFFSET
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1,1
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COMMENTS
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It appears that most of the terms of this sequence are primes. The composite terms are 35, 169, 385, 899, 961, 1121, ... (A319042).
The primes in the sequence give A001132 (primes == +-1 (mod 8)), since for primes p we have Pell(p) == (2/p) (mod p) where (2/p) is the Legendre symbol. - Jianing Song, Sep 10 2018
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LINKS
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EXAMPLE
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k = 7 is in the sequence since Pell(7) = 169 = 7 * 24 + 1 == 1 (mod 7).
k = 11 is not in the sequence: Pell(11) = 5741 = 11 * 522 - 1 !== 1 (mod 11).
k = 35 is in the sequence: Pell(35) = 8822750406821 = 35 * 252078583052 + 1 == 1 (mod 35).
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MAPLE
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isA319040 := k -> simplify(2^(k-1)*hypergeom([1-k/2, (1-k)/2], [1-k], -1)) mod k = 1: A319040List := b -> select(isA319040, [$1..b]):
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MATHEMATICA
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Select[Range[500], Mod[Fibonacci[#, 2], #] == 1 &] (* Alonso del Arte, Sep 08 2018 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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