A053032 -id:A053032 - cp's OEIS frontend

A122869 Primes p that divide Lucas((p-1)/2), where Lucas is A000032.

11, 19, 31, 59, 71, 79, 131, 139, 151, 179, 191, 199, 211, 239, 251, 271, 311, 331, 359, 379, 419, 431, 439, 479, 491, 499, 571, 599, 619, 631, 659, 691, 719, 739, 751, 811, 839, 859, 911, 919, 971, 991, 1019, 1031, 1039, 1051, 1091, 1151, 1171, 1231, 1259

Offset: 1

Alexander Adamchuk, Sep 16 2006

nonn

Final digit of a(n) is 1 or 9.
A002145 is the union of this sequence and A122870, Primes p that divide Lucas((p+1)/2).
Conjecture: This sequence is just the primes congruent to 11 or 19 mod 20. - Charles R Greathouse IV, May 25 2011 [The conjecture is correct. - Jianing Song, Jun 20 2025]
Note that F(p-1) = F((p-1)/2)*Lucas((p-1)/2), where F = A000045. Since gcd(F(n),Lucas(n)) = 1 or 2 (because Lucas(n)^2 - 5*F(n)^2 = 4*(-1)^n), this sequence lists primes p such that p divides F(p-1) but does not divides F((p-1)/2). By Propositions 1.1 and 1.2 (the k = 3 case) of my link below, this is primes p == 11, 19 (mod 20). - Jianing Song, Jun 20 2025

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Vincenzo Librandi)
Jianing Song, Lucas sequences and entry point modulo p
Eric Weisstein's World of Mathematics, Gaussian Prime.
Eric Weisstein's World of Mathematics, Lucas Number.

Cf. A000032, A000045, A053032, A076518, A122870.

Subsequence of A002145, A003626, A040105, A040147 and A064739.

Select[Prime[Range[1000]],IntegerQ[(Fibonacci[(#1-1)/2-1]+Fibonacci[(#1-1)/2+1])/#1]&]

lista(kmax) = {my(lucas1 = 1, lucas2 = 3, lucas3, p); for(k = 3, kmax, lucas3 = lucas1 + lucas2; p = 2*k + 1; if(isprime(p) && !(lucas3 % p), print1(p, ", ")); lucas1 = lucas2; lucas2 = lucas3);} \\ Amiram Eldar, Jun 06 2024

A086598 Number of distinct prime factors in Lucas(n).

Original entry on oeis.org

1, 0, 1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 3, 1, 2, 3, 1, 1, 3, 1, 2, 3, 3, 2, 3, 3, 2, 3, 2, 2, 4, 1, 2, 3, 3, 4, 4, 1, 2, 4, 3, 1, 5, 2, 4, 6, 3, 1, 4, 2, 4, 4, 3, 1, 4, 4, 2, 4, 3, 3, 6, 1, 2, 6, 2, 5, 5, 2, 2, 5, 4, 1, 4, 2, 3, 7, 2, 4, 4, 1, 2, 5, 4, 2, 6, 4, 2, 5, 3, 2, 6, 3, 3, 4, 4, 5, 4, 2, 4, 7, 4, 3, 6, 3, 4, 9

Offset: 0

T. D. Noe, Jul 24 2003

Interestingly, the Lucas numbers separate the primes into three disjoint sets: (A053028) primes that do not divide any Lucas number, (A053027) primes that divide Lucas numbers of even index and (A053032) primes that divide Lucas numbers of odd index.

Max Alekseyev, Table of n, a(n) for n = 0..1411 (first 1000 terms from T. D. Noe)
Blair Kelly, Fibonacci and Lucas Factorizations
Eric Weisstein's World of Mathematics, Lucas Number

Cf. A000204 (Lucas numbers), A086599 (number of prime factors, counting multiplicity), A086600 (number of primitive prime factors).

Cf. A053027, A053028, A053032.

[#PrimeDivisors(Lucas(n)): n in [1..100]]; // Vincenzo Librandi, Jul 26 2017

Lucas[n_] := Fibonacci[n+1] + Fibonacci[n-1]; Table[Length[FactorInteger[Lucas[n]]], {n, 150}]

a(n)=omega(fibonacci(n-1)+fibonacci(n+1)) \\ Charles R Greathouse IV, Sep 14 2015

a(n) = Sum{d|n and n/d odd} A086600(d) + 1 if 6|n, a Mobius-like transform

a(0)=1 prepended by Max Alekseyev, Jun 15 2025

A065156 Numbers n such that some Lucas number (A000204) is divisible by n.

Original entry on oeis.org

1, 2, 3, 4, 6, 7, 9, 11, 14, 18, 19, 22, 23, 27, 29, 31, 38, 41, 43, 44, 46, 47, 49, 54, 58, 59, 62, 67, 71, 76, 79, 81, 82, 83, 86, 94, 98, 101, 103, 107, 116, 118, 121, 123, 124, 127, 129, 131, 134, 139, 142, 151, 158, 161, 162, 163, 166, 167, 179, 181, 191, 199

Offset: 1

Dean Hickerson, Oct 18 2001

From A.H.M. Smeets, Sep 20 2020 (Start)
For the Fibonacci numbers, each natural number divides some Fibonacci number (see A001177).
If, for some number m, m divides some Lucas number L_i (=A000204(i)), then, the smallest i satisfies i <= m. (End)

A.H.M. Smeets, Table of n, a(n) for n = 1..20000 (terms 1..1000 from T. D. Noe)
B. Avila and T. Khovanova, Free Fibonacci Sequences, arXiv preprint arXiv:1403.4614 [math.NT], 2014 and J. Int. Seq. 17 (2014) # 14.8.5

Complement of A064362. Cf. A000204.

Cf. A001177, A053032, A140409.

test[ n_ ] := For[ a=1; b=3, True, t=b; b=Mod[ a+b, n ]; a=t, If[ b==0, Return[ True ] ]; If[ a==2&&b==1, Return[ False ] ] ]; Select[ Range[ 200 ], test ]
Take[Flatten[Divisors/@LucasL[Range[200]]]//Union,70] (* Harvey P. Dale, Jun 07 2020 *)

a, n = 0, 0
while n < 1000:
    a, f0, f1, i = a+1, 1, 2, 1
    if f1%a == 0:
        n = n+1
        print(n,a)
    else:
        while f0%a != 0 and i <= a:
            f0, f1, i = f0+f1, f0, i+1
        if i <= a:
            n = n+1
            print(n,a) # A.H.M. Smeets, Sep 20 2020

Equals {1,2,4} union {p^e | p in A140409 and e > 0} union {2*p^e | p in A140409 and e > 0} union {4*p | p in A053032} union {4*p*q | p, q in A053032}. - A.H.M. Smeets, Sep 20 2020

A106306 Primes that yield a simple orbit structure in 2-step recursions.

Original entry on oeis.org

2, 3, 7, 13, 17, 23, 37, 41, 43, 47, 53, 61, 67, 73, 83, 89, 97, 103, 107, 109, 113, 127, 137, 149, 157, 163, 167, 173, 193, 197, 223, 227, 233, 241, 257, 263, 269, 277, 281, 283, 293, 307, 313, 317, 337, 347, 353, 367, 373, 383, 389, 397, 401, 409, 421, 433

Offset: 1

T. D. Noe, May 02 2005

nonn

Consider the 2-step recursion x(k)=x(k-1)+x(k-2) mod n. For any of the n^2 initial conditions x(1) and x(2) in Zn, the recursion has a finite period. When n is a prime in this sequence, all of the orbits, except the one containing (0,0), have the same length.
Except for 5, this appears to be the complement of A053032, odd primes p with one 0 in Fibonacci numbers mod p. - T. D. Noe, May 03 2005
A prime p is in this sequence if either (1) the polynomial x^2-x-1 mod p has no zeros for x in [0,p-1] (see A086937) or (2) the polynomial has zeros, but none is a root of unity mod p. The first few primes in the second category are 41, 61, 89 and 109. - T. D. Noe, May 12 2005

Eric Weisstein's World of Mathematics, Fibonacci n-Step Number.

Cf. A015134 (orbits of 2-step sequences).

A140409 Prime factors of Lucas numbers.

Original entry on oeis.org

2, 3, 7, 11, 19, 23, 29, 31, 41, 43, 47, 59, 67, 71, 79, 83, 101, 103, 107, 127, 131, 139, 151, 163, 167, 179, 181, 191, 199, 211, 223, 227, 229, 239, 241, 251, 263, 271, 281, 283, 307, 311, 331, 347, 349, 359, 367, 379, 383, 401, 409, 419, 431, 439, 443, 449

Offset: 1

Tanya Khovanova, Jun 16 2008

nonn

a(n) is A096362(n) in ascending order.

C. Smyth, The terms in Lucas Sequences divisible by their indices, JIS 13 (2010) #10.2.4.

Cf. A096362 Order in which prime factors first occur in the Lucas sequence.

Union of 2, A053027 and A053032 - T. D. Noe, Jun 21 2008

More terms from T. D. Noe, Jun 21 2008

A189767 Least number k such that the set of numbers {Fibonacci(i) mod n, i=0..k-1} contains all possible residues of Fibonacci(i) mod n.

Original entry on oeis.org

1, 2, 4, 5, 10, 10, 13, 11, 17, 22, 9, 23, 19, 37, 20, 23, 25, 19, 17, 53, 15, 25, 37, 23, 50, 61, 53, 45, 13, 58, 29, 47, 39, 25, 77, 23, 55, 17, 47, 59, 31, 37, 65, 29, 93, 37, 25, 23, 81, 148, 67, 75, 77, 53, 19, 45, 71, 37, 57, 119, 43, 29, 45, 95, 103

Offset: 1

T. D. Noe, May 10 2011

nonn

Sequence A066853 gives the number of possible residues of the sequence Fibonacci(i) mod n for i=0,1,2,.... Here we compute the smallest k required to find all A066853(n) residues in the first k terms (starting at i=0) of Fibonacci sequence (mod n). We know that k is at most A001175(n), the period of Fibonacci(i) mod n. It appears that when n is a prime in A053032, then a(n)=n-1.

			Consider n=8. The Fibonacci numbers mod 8 have period 12: 0, 1, 1, 2, 3, 5, 0, 5, 5, 2, 7, 1. The set of residues is {0, 1, 2, 3, 5, 7}. How long does it take to find all 6 residues in the sequence Fibonacci(i) mod n? The answer is 11 because 7 finally appears as Fibonacci(10) mod 8.

T. D. Noe, Table of n, a(n) for n = 1..1000

Cf. A000045 (Fibonacci numbers), A001175, A053032, A066853, A189768 (residues).

F:= proc(n)
 local r, k, a,ap, t, V;
ap:= 0: a:= 1; r:= 1;
V:= Array(0..n-1);
V[0]:= 1;
V[1]:= 1;
for k from 2 do
     t:= a + ap mod n;
     ap:= a;
     a:= t;
     if ap = 0 and a = 1 then return r +1 fi;
     if V[t] = 0 then
        r:=k;
        V[t]:= 1;
     fi
od:
end proc:
F(1):= 1:
seq(F(n),n=1..100); # Robert Israel, Dec 23 2015

pisano[n_] := Module[{a={1,0},a0,k=0,s}, If[n==1, 1, a0=a; While[k++; s=Mod[Total[a],n]; a[[1]]=a[[2]]; a[[2]]=s; a != a0]; k]]; Table[p=pisano[n]; f=Mod[Fibonacci[Range[0,p]],n]; u=Union[f]; k=1; While[Union[Take[f,k]] != u, k++]; k, {n,100}]

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A122869 Primes p that divide Lucas((p-1)/2), where Lucas is A000032.

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A086598 Number of distinct prime factors in Lucas(n).

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A065156 Numbers n such that some Lucas number (A000204) is divisible by n.

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A106306 Primes that yield a simple orbit structure in 2-step recursions.

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A140409 Prime factors of Lucas numbers.

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A189767 Least number k such that the set of numbers {Fibonacci(i) mod n, i=0..k-1} contains all possible residues of Fibonacci(i) mod n.

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