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Revision History for A308795 (Bold, blue-underlined text is an addition; faded, red-underlined text is a deletion.)

Showing entries 1-10 | older changes
Primes p such that A001177(p) = (p-1)/2.
(history; published version)
#13 by Peter Luschny at Fri Jul 05 16:01:40 EDT 2019
STATUS

proposed

approved

#12 by Jean-François Alcover at Fri Jul 05 10:19:22 EDT 2019
STATUS

editing

proposed

#11 by Jean-François Alcover at Fri Jul 05 10:19:16 EDT 2019
MATHEMATICA

pn[n_] := For[k = 1, True, k++, If[Mod[Fibonacci[k], n] == 0, Return[k]]];

Reap[For[p = 2, p < 3600, p = NextPrime[p], If[pn[p] == (p - 1)/2, Print[p]; Sow[p]]]][[2, 1]] (* Jean-François Alcover, Jul 05 2019 *)

STATUS

proposed

editing

#10 by Jianing Song at Thu Jul 04 11:16:28 EDT 2019
STATUS

editing

proposed

#9 by Jianing Song at Thu Jul 04 11:14:01 EDT 2019
PROG

(PARI) Entry_for_decomposing_prime(p) = my(k=1, M=[k, 1; 1, 0], Id=[1, 0; 0, 1]); if(isprime(p)&&kronecker(k^2+4, p)==1, my(v=divisors(p-1)); for(d=1, #v, if((Mod(M, p)^v[d])[2, 1]==0, return(v[d]))))

STATUS

proposed

editing

#8 by Jianing Song at Wed Jul 03 06:19:11 EDT 2019
STATUS

editing

proposed

#7 by Jianing Song at Tue Jul 02 11:42:50 EDT 2019
COMMENTS

For an odd rational prime p:

This Here k = 1, and this sequence gives primes such that (a) holds and s = 2. For even s, all terms are congruent to 1 modulo 4.

Number of terms below 10^N:

N | Number | Decomposing primes*

3 | 15 | 78

4 | 115 | 609

5 | 839 | 4777

6 | 6913 | 39210

7 | 58891 | 332136

8 | 510784 | 2880484

* Here "Decomposing primes" means primes such that Legendre(5,p) = 1, i.e., p == 1, 4 (mod 5).

#6 by Jianing Song at Sun Jun 30 11:07:47 EDT 2019
DATA

29, 41, 101, 181, 229, 241, 349, 409, 449, 509, 569, 601, 641, 929, 941, 1021, 1061, 1109, 1129, 1201, 1229, 1321, 1481, 1489, 1549, 1609, 1621, 1669, 1709, 1741, 1789, 1801, 1861, 1889, 2029, 2069, 2129, 2609, 2621, 2861, 3209, 3301, 3361, 3389, 3449, 3461, 3581, 3761, 3769, 3821, 3889, 3989, 4129, 4229, 4241

PROG

(PARI) Entry_for_decomposing_prime(p) = my(k=1, M=[k, 1; 1, 0], Id=[1, 0; 0, 1]); if(isprime(p)&&kronecker(k^2+4, p)==1, my(v=divisors(p-1)); for(d=1, #v, if((Mod(M, p)^v[d])[2, 1]==0, return(v[d]))))

forprime(p=2, 3600, if(Entry_for_decomposing_prime(p)==(p-1)/2, print1(p, ", ")))

#5 by Jianing Song at Tue Jun 25 07:20:40 EDT 2019
COMMENTS

Primes p such that ord(-(13+sqrt(5))/2,p) = 2*(p+-1)/3, 2, where ord(z,p) is the smallest integer k > 0 such that (z^k-1)/p is an algebraic integer.

#4 by Jianing Song at Tue Jun 25 07:16:48 EDT 2019
COMMENTS

Primes p such that ord((1+sqrt(5))/2,p) = 2*(p+1)/3, where ord(z,p) is the smallest integer k > 0 such that (z^k-1)/p is an algebraic integer.

Let {T(n)} be a sequence defined by T(0) = 0, T(1) = 1, T(n) = k*T(n-1) + T(n-2), K be the quadratic field Q[sqrt(k^2+4)], O_K be the ring of integer of K, u = (k+sqrt(k^2+4))/2. For a prime p, not dividing k^2 + 4, the entry point Pisano period of {T(n)} modulo p (that is, the smallest m > 0 such that T(n+m) == 0 T(n) (mod p) for all n) is the multiplicative order of -ord(u^2 modulo ,p. Here ); the multiplicative order entry point of u {T(n)} modulo p (that is defined as , the smallest positive integer k m > 0 such that T(u^k-1m)/ == 0 (mod p )) is an algebraic integerord(-u^2,p).