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A309587
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Primes p with 2 zeros in a fundamental period of A006190 mod p.
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12
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7, 11, 17, 19, 31, 47, 59, 67, 71, 83, 113, 151, 163, 167, 223, 227, 239, 257, 271, 307, 313, 331, 337, 359, 379, 383, 431, 463, 479, 487, 499, 521, 587, 601, 619, 631, 641, 643, 673, 683, 691, 739, 743, 787, 809, 811, 827, 839, 863, 947, 967, 983
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OFFSET
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1,1
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COMMENTS
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For p > 2, p is in this sequence if and only if (all these conditions are equivalent):
(b) 8 divides ord(p,(3+sqrt(13))/2), where ord(p,u) is the smallest integer k > 0 such that (u^k - 1)/p is an algebraic integer;
(c) 4 divides ord(p,(11+3*sqrt(13))/2);
(e) 4 divides ord(p,-(11+3*sqrt(13))/2).
In general, let {x(n)} be a sequence defined by x(0) = 0, x(1) = 1, x(n+2) = m*x(n+1) + x(n). Let pi(k) be the Pisano period of {x(n)} modulo k, i.e., pi(k) = min{l > 0 : x(n+l) == x(n) (mod k) for all n}, r(k) = min{l > 0 : k divides x(l)} and w(k) be the number of zeros in a fundamental period of {x(n)} modulo k. Let u = (m + sqrt(m^2+4))/2, p be an odd prime, then these conditions are equivalent:
(1) w(p) = 2;
(2) 8 divides pi(p);
(3) 8 divides ord(p,u);
(4) 4 divides ord(p,u^2);
(5) 4 divides r(p);
(6) 4 divides ord(p,-u^2).
This can be shown by noting that pi(p) = p^c*ord(p,u) and r(p) = p^c*ord(p,-u^2) for some c (if p does not divide m^2 + 4 then c = 0, otherwise c = 1). Also, Pi(p) is always even, so ord(p,u) is always even.
This sequence contains all primes congruent to 7, 11, 15, 19, 31, 47 modulo 52.
Conjecturely, this sequence has density 1/3 in the primes.
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LINKS
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PROG
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(PARI) forprime(p=2, 1000, if(A322906(p)==2, print1(p, ", ")))
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CROSSREFS
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Let {x(n)} be the sequence defined in the comment section.
| m=1 | m=2 | m=3
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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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