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A309586
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Primes p with 1 zero in a fundamental period of A006190 mod p.
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12
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2, 3, 23, 43, 53, 61, 79, 101, 103, 107, 127, 131, 139, 173, 179, 191, 199, 211, 251, 263, 277, 283, 311, 347, 367, 419, 433, 439, 443, 467, 491, 503, 523, 547, 563, 569, 571, 599, 607, 647, 659, 677, 719, 727, 751, 757, 823, 829, 859, 881, 883, 887, 907
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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) ord(p,(3+sqrt(13))/2) == 2 (mod 4), where ord(p,u) is the smallest integer k > 0 such that (u^k - 1)/p is an algebraic integer;
(c) ord(p,(11+3*sqrt(13))/2) is odd;
(e) ord(p,-(11+3*sqrt(13))/2) == 2 (mod 4).
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) = 1;
(2) pi(p) == 2 (mod 4);
(3) ord(p,u) == 2 (mod 4);
(4) ord(p,u^2) is odd;
(5) r(p) == 2 (mod 4);
(6) ord(p,-u^2) == 2 (mod 4).
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 3, 23, 27, 35, 43, 51 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, 900, if(A322906(p)==1, 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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