A106535 -id:A106535 - cp's OEIS frontend

A001177 Fibonacci entry points: a(n) = least k >= 1 such that n divides Fibonacci number F_k (=A000045(k)).

1, 3, 4, 6, 5, 12, 8, 6, 12, 15, 10, 12, 7, 24, 20, 12, 9, 12, 18, 30, 8, 30, 24, 12, 25, 21, 36, 24, 14, 60, 30, 24, 20, 9, 40, 12, 19, 18, 28, 30, 20, 24, 44, 30, 60, 24, 16, 12, 56, 75, 36, 42, 27, 36, 10, 24, 36, 42, 58, 60, 15, 30, 24, 48, 35, 60, 68, 18, 24, 120

Offset: 1

N. J. A. Sloane

nonn

In the formula, the relation a(p^e) = p^(e-1)*a(p) is called Wall's conjecture, which has been verified for primes up to 10^14. See A060305. Primes for which this relation fails are called Wall-Sun-Sun primes. - T. D. Noe, Mar 03 2009
All solutions to F_m == 0 (mod n) are given by m == 0 (mod a(n)). For a proof see, e.g., Vajda, p. 73. [Old comment changed by Wolfdieter Lang, Jan 19 2015]
If p is a prime of the form 10n +- 1 then a(p) is a divisor of p-1. If q is a prime of the form 10n +- 3 then a(q) is a divisor of q+1. - Robert G. Wilson v, Jul 07 2007
Definition 1 in Riasat (2011) calls this k(n), or sometimes just k. Corollary 1 in the same paper, "every positive integer divides infinitely many Fibonacci numbers," demonstrates that this sequence is infinite. - Alonso del Arte, Jul 27 2013
If p is a prime then a(p) <= p+1. This is because if p is a prime then exactly one of the following Fibonacci numbers is a multiple of p: F(p-1), F(p) or F(p+1). - Dmitry Kamenetsky, Jul 23 2015
From Renault 1996:
  1. a(lcm(n,m)) = lcm(a(n), a(m)).
  2. if n|m then a(n)|a(m).
  3. if m has prime factorization m=p1^e1 * p2^e2 * ... * pn^en then a(m) = lcm(a(p1^e1), a(p2^e2), ..., a(pn^en)). - Dmitry Kamenetsky, Jul 23 2015
a(n)=n if and only if n=5^k or n=12*5^k for some k >= 0 (see Marques 2012). - Dmitry Kamenetsky, Aug 08 2015
Every positive integer (except 2) eventually appears in this sequence. This is because every Fibonacci number bigger than 1 (except Fibonacci(6)=8 and Fibonacci(12)=144) has at least one prime factor that is not a factor of any earlier Fibonacci number (see Knott reference). Let f(n) be such a prime factor for Fibonacci(n); then a(f(n))=n. - Dmitry Kamenetsky, Aug 08 2015
We can reconstruct the Fibonacci numbers from this sequence using the formula Fibonacci(n+2) = 1 + Sum_{i: a(i) <= n} phi(i)*floor(n/a(i)), where phi(n) is Euler's totient function A000010 (see the Stroinski link). For example F(6) = 1 + phi(1)*floor(4/a(1)) + phi(2)*floor(4/a(2)) + phi(3)*floor(4/a(4)) = 1 + 1*4 + 1*1 + 2*1 = 8. - Peter Bala, Sep 10 2015
Conjecture: Sum_{d|n} phi(d)*a(d) = A232656(n). - Logan J. Kleinwaks, Oct 28 2017
a(F_m) = m for all m > 1. Indeed, let (b(j)) be defined by b(1)=b(2)=1, and b(j+2) = (b(j) + b(j+1)) mod n. Then a(n) equals the index of the first occurrence of 0 in (b(j)). Example: if n=4 then b = A079343 = 1,1,2,3,1,0,1,1,..., so a(4)=6. If n is a Fibonacci number n=F_m, then obviously a(n)=m. Note that this gives a simple proof of the fact that all integers larger than 2 occur in (a(n)). - Michel Dekking, Nov 10 2017

			a(4) = 6 because the smallest Fibonacci number that 4 divides is F(6) = 8.
a(5) = 5 because the smallest Fibonacci number that 5 divides is F(5) = 5.
a(6) = 12 because the smallest Fibonacci number that 6 divides is F(12) = 144.
From _Wolfdieter Lang_, Jan 19 2015: (Start)
a(2) = 3, hence 2 | F(m) iff m = 2*k, for k >= 0;
a(3) = 4, hence 3 | F(m) iff m = 4*k, for k >= 0;
etc. See a comment above with the Vajda reference.
(End)

A. Brousseau, Fibonacci and Related Number Theoretic Tables. Fibonacci Association, San Jose, CA, 1972, p. 25.
B. H. Hannon and W. L. Morris, Tables of Arithmetical Functions Related to the Fibonacci Numbers. Report ORNL-4261, Oak Ridge National Laboratory, Oak Ridge, Tennessee, June 1968.
Alfred S. Posamentier & Ingmar Lehmann, The (Fabulous) Fibonacci Numbers, Afterword by Herbert A. Hauptman, Nobel Laureate, 2. 'The Minor Modulus m(n)', Prometheus Books, NY, 2007, page 329-342.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
S. Vajda, Fibonacci and Lucas numbers and the Golden Section, Ellis Horwood Ltd., Chichester, 1989.
N. N. Vorob'ev, Fibonacci numbers, Blaisdell, NY, 1961.

T. D. Noe, Table of n, a(n) for n = 1..10000
A. Allard and P. Lecomte, Periods and entry points in Fibonacci sequence, Fib. Quart. 17 (1) (1979) 51-57.
R. C. Archibald (?), Review of B. H. Hannon and W. L. Morris, Tables of arithmetical functions related to the Fibonacci numbers, Math. Comp., 23 (1969), 459-460.
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.
Alfred Brousseau, Fibonacci and Related Number Theoretic Tables, Fibonacci Association, San Jose, CA, 1972. See p. 25.
J. D. Fulton and W. L. Morris, On arithmetical functions related to the Fibonacci numbers, Acta Arithmetica, 16 (1969), 105-110.
Molly FitzGibbons, Steven J. Miller, and Amanda Verga, Dynamics of the Fibonacci Order of Appearance Map, arXiv:2309.14501 [math.NT], 2023.
Ramon Glez-Regueral, An entry-point algorithm for high-speed factorization, Thirteenth Internat. Conf. Fibonacci Numbers Applications, Patras, Greece, 2008.
B. H. Hannon and W. L. Morris, Tables of Arithmetical Functions Related to the Fibonacci Numbers [Annotated and scanned copy]
Ron Knott, The first 300 Fibonacci numbers factorized
Paolo Leonetti and Carlo Sanna, On the greatest common divisor of n and the nth Fibonacci number, arXiv:1704.00151 [math.NT], 2017. See z(n).
Diego Marques, Fixed points of the order of appearance in the Fibonacci sequence, Fibonacci Quart. 50:4 (2012), pp. 346-352.
Diego Marques, The order of appearance of the product of consecutive Lucas numbers, Fibonacci Quarterly, 51 (2013), 38-43.
Diego Marques, Sharper upper bounds for the order of appearance in the fibonacci sequence, Fibonacci Quarterly, 51 (2013), pp. 233-238.
Zuzana Masáková and Edita Pelantová, Midy's Theorem in non-integer bases and divisibility of Fibonacci numbers, arXiv:2401.03874 [math.NT], 2024. See page 9.
Marc Renault, The Fibonacci sequence under various moduli, Masters Thesis, Wake Forest University, 1996.
Samin Riasat, Z[phi] and the Fibonacci Sequence Modulo n, Mathematical Reflections 1 (2011): 1 - 7.
H. J. A. Salle, A Maximum Value for the Rank of Apparition of Integers in Recursive Sequences , Fibonacci Quart. 13.2 (1975) 159-161.
U. Stroinski, Connection between Euler's totient function and Fibonacci numbers, Mathematics Stack Exchange, Feb 17 2015.
D. D. Wall, Fibonacci series modulo m, Amer. Math. Monthly 67 (6) (1960) 525-532.
Eric Weisstein, MathWorld: Wall-Sun-Sun Prime

Cf. A000045, A001175, A001176, A060383, A001602. First occurrence of k is given in A131401. A233281 gives such k that a(k) is a prime.
From Antti Karttunen, Dec 21 2013: (Start)
Various derived sequences:
A047930(n) = A000045(a(n)).
A037943(n) = A000045(a(n))/n.
A217036(n) = A000045(a(n)-1) mod n.
A132632(n) = a(n^2).
A132633(n) = a(n^3).
A214528(n) = a(n!).
A215011(n) = a(A000217(n)).
A215453(n) = a(n^n).
Analogous sequence for the tribonacci numbers: A046737, for Lucas numbers: A223486, for Pell numbers: A214028.
Cf. also A000057, A106535, A120255, A120256, A175026, A213648, A214031, A214781, A214783, A230359, A233283, A233285, A233287. (End)

a001177 n = head [k | k <- [1..], a000045 k `mod` n == 0]
-- Reinhard Zumkeller, Jan 15 2014

A001177 := proc(n)
        for k from 1 do
                if combinat[fibonacci](k) mod n = 0 then
                        return k;
                end if;
        end do:
end proc: # R. J. Mathar, Jul 09 2012
N:= 1000: # to get a(1) to a(N)
L:= ilcm($1..N):
count:= 0:
for n from 1 while count < N do
  fn:= igcd(L,combinat:-fibonacci(n));
  divs:= select(`<=`,numtheory:-divisors(fn),N);
  for d in divs do if not assigned(A[d]) then count:= count+1; A[d]:= n fi od:
od:
seq(A[n],n=1..N); # Robert Israel, Oct 14 2015

fibEntry[n_] := Block[{k = 1}, While[ Mod[ Fibonacci@k, n] != 0, k++ ]; k]; Array[fibEntry, 74] (* Robert G. Wilson v, Jul 04 2007 *)

a(n)=if(n<0,0,s=1;while(fibonacci(s)%n>0,s++);s) \\ Benoit Cloitre, Feb 10 2007

ap(p)=my(k=p+[0, -1, 1, 1, -1][p%5+1], f=factor(k)); for(i=1, #f[, 1], for(j=1, f[i, 2], if((Mod([1, 1; 1, 0], p)^(k/f[i, 1]))[1, 2], break); k/=f[i, 1])); k
a(n)=if(n==1,return(1)); my(f=factor(n), v); v=vector(#f~, i, if(f[i,1]>1e14,ap(f[i,1]^f[i,2]), ap(f[i,1])*f[i,1]^(f[i,2]-1))); if(f[1,1]==2&&f[1,2]>1, v[1]=3<Charles R Greathouse IV, May 08 2017

(define (A001177 n) (let loop ((k 1)) (cond ((zero? (modulo (A000045 k) n)) k) (else (loop (+ k 1)))))) ;; Antti Karttunen, Dec 21 2013

A001175(n) = A001176(n) * a(n) for n >= 1.
a(n) = n if and only if n is of form 5^k or 12*5^k (proved in Marques paper), a(n) = n - 1 if and only if n is in A106535, a(n) = n + 1 if and only if n is in A000057, a(n) = n + 5 if and only if n is in 5*A000057, ... - Benoit Cloitre, Feb 10 2007
a(1) = 1, a(2) = 3, a(4) = 6 and for e > 2, a(2^e) = 3*2^(e-2); a(5^e) = 5^e; and if p is an odd prime not 5, then a(p^e) = p^max(0, e-s)*a(p) where s = valuation(A000045(a(p)), p) (Wall's conjecture states that s = 1 for all p). If (m, n) = 1 then a(m*n) = lcm(a(m), a(n)). See Posamentier & Lahmann. - Robert G. Wilson v, Jul 07 2007; corrected by Max Alekseyev, Oct 19 2007, Jun 24 2011
Apparently a(n) = A213648(n) + 1 for n >= 2. - Art DuPre, Jul 01 2012
a(n) < n^2. [Vorob'ev]. - Zak Seidov, Jan 07 2016
a(n) < n^2 - 3n + 6. - Jinyuan Wang, Oct 13 2018
a(n) <= 2n [Salle]. - Jon Maiga, Apr 25 2019

Definition corrected by Wolfdieter Lang, Jan 19 2015

A000057 Primes dividing all Fibonacci sequences.

Original entry on oeis.org

2, 3, 7, 23, 43, 67, 83, 103, 127, 163, 167, 223, 227, 283, 367, 383, 443, 463, 467, 487, 503, 523, 547, 587, 607, 643, 647, 683, 727, 787, 823, 827, 863, 883, 887, 907, 947, 983, 1063, 1123, 1163, 1187, 1283, 1303, 1327, 1367, 1423, 1447, 1487, 1543, 1567, 1583

Offset: 1

N. J. A. Sloane

nonn

Here a Fibonacci sequence is a sequence which begins with any two integers and continues using the rule s(n+2) = s(n+1) + s(n). These primes divide at least one number in each such sequence. - Don Reble, Dec 15 2006
Primes p such that the smallest positive m for which Fibonacci(m) == 0 (mod p) is m = p + 1. In other words, the n-th prime p is in this sequence iff A001602(n) = p + 1. - Max Alekseyev, Nov 23 2007
Cubre and Rouse comment that this sequence is not known to be infinite. - Charles R Greathouse IV, Jan 02 2013
Number of terms up to 10^n: 3, 7, 38, 249, 1894, 15456, 130824, 1134404, 10007875, 89562047, .... - Charles R Greathouse IV, Nov 19 2014
These are also the fixed points of sequence A213648 which gives the minimal number of 1's such that n*[n; 1,..., 1, n] = [x; ..., x], where [...] denotes simple continued fractions. - M. F. Hasler, Sep 15 2015
It appears that for n >= 2, all first differences are congruent to 0 (mod 4). - Christopher Hohl, Dec 28 2018
The comment above is equivalent to a(n) == 3 (mod 4) for n >= 2. This is indeed correct. Actually it can be proved that a(n) == 3, 7 (mod 20) for n >= 2. Let p != 2, 5 be a prime, then: A001175(p) divides (p - 1)/2 if p == 1, 9 (mod 20); p - 1 if p == 11, 19 (mod 20); (p + 1)/2 if p == 13, 17 (mod 20). So the remaining cases are p == 3, 7 (mod 20). - Jianing Song, Dec 29 2018

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Christian G. Bower and T. D. Noe, Table of n, a(n) for n = 1..1000
U. Alfred, Primes which are factors of all Fibonacci sequences, Fib. Quart., 2 (1964), 33-38.
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.
D. M. Bloom, On periodicity in generalized Fibonacci sequences, Am. Math. Monthly 72 (8) (1965) 856-861.
H. E. A. Campbell and David L. Wehlau, Zigzag polynomials, Artin's conjecture and trinomials, Finite Fields and Their Applications (2023) Vol. 89, 102198.
Paul Cubre and Jeremy Rouse, Divisibility properties of the Fibonacci entry point, arXiv:1212.6221 [math.NT], 2012.
Ron Knott, General Fibonacci Series.
Rishi Kumar, Kepler Sets of Second-Order Linear Recurrence Sequences Over Q_p, arXiv:2406.05890 [math.NT], 2024. See pp. 2, 7.
N. J. A. Sloane, Families of Essentially Identical Sequences, Mar 24 2021 (Includes this sequence).

Subsequence of A064414.

Cf. A001175, A001602, A079346, A106535, A213648.

Select[Prime[Range[1000]], Function[p, a=0; b=1; n=1; While[b != 0, t=b; b = Mod[(a+b), p]; a=t; n++]; n>p]] (* Jean-François Alcover, Aug 05 2018, after Charles R Greathouse IV *)

select(p->my(a=0,b=1,n=1,t);while(b,t=b;b=(a+b)%p; a=t; n++); n>p, primes(1000)) \\ Charles R Greathouse IV, Jan 02 2013

is(p)=fordiv(p-1,d,if(((Mod([1,1;1,0],p))^d)[1,2]==0,return(0)));fordiv(p+1,d,if(((Mod([1,1;1,0],p))^d)[1,2]==0,return(d==p+1 && isprime(p)))) \\ Charles R Greathouse IV, Jan 02 2013

is(p)=if((p-2)%5>1, return(0)); my(f=factor(p+1)); for(i=1, #f~, if((Mod([1, 1; 1, 0], p)^((p+1)/f[i, 1]))[1, 2]==0, return(0))); isprime(p) \\ Charles R Greathouse IV, Nov 19 2014

More terms from Don Reble, Nov 14 2006

A003147 Primes p with a Fibonacci primitive root g, i.e., such that g^2 = g + 1 (mod p).

Original entry on oeis.org

5, 11, 19, 31, 41, 59, 61, 71, 79, 109, 131, 149, 179, 191, 239, 241, 251, 269, 271, 311, 359, 379, 389, 409, 419, 431, 439, 449, 479, 491, 499, 569, 571, 599, 601, 631, 641, 659, 701, 719, 739, 751, 821, 839, 929, 971, 1019, 1039, 1051, 1091, 1129, 1171, 1181, 1201, 1259, 1301

Offset: 1

N. J. A. Sloane

Primes p with a primitive root g such that g^2 = g + 1 (mod p).
Not the same as primes with a Fibonacci number as primitive root; cf. A083701. - Jonathan Sondow, Feb 17 2013
For all except the initial term 5, these are numbers such that the Pisano period equals 1 less than the Pisano number, i.e., where A001175(n) = n-1. - Matthew Goers, Sep 20 2013
As shown in the paper by Brison, these are also the primes p such that there is a Fibonacci-type sequence (mod p) that begins with (1,b) and encounters all numbers less than p in the first p-1 iterations (for some b). - T. D. Noe, Feb 26 2014
Shanks (1972) conjectured that the relative asymptotic density of this sequence in the sequence of primes is 27*c/38 = 0.2657054465..., where c is Artin's constant (A005596). The conjecture was proved on the assumption of a generalized Riemann hypothesis by Lenstra (1977) and Sander (1990). - Amiram Eldar, Jan 22 2022

			3 is a primitive root mod 5, and 3^2 = 3 + 1 mod 5, so 5 is a member. - _Jonathan Sondow_, Feb 17 2013

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000 (first 1000 terms from Noe)
Bob Bastasz, Lyndon words of a second-order recurrence, Fibonacci Quarterly, Vol. 58, No. 5 (2020), pp. 25-29.
Owen J. Brison, Complete Fibonacci sequences in finite fields, Fibonacci Quarterly, Vol. 30, No. 4 (1992), pp. 295-304.
Alexandru Gica, Quadratic Residues in Fibonacci Sequences, Fibonacci Quart., Vol. 46/47, No. 1 (2008/2009), pp. 68-72. See Theorem 5.1.
Liang-Chung Hsia, Hua-Chieh Li, and Wei-Liang Sun, Certain Diagonal Equations and Conflict-Avoiding Codes of Prime Lengths, arXiv:2302.00920 [math.NT], 2023.
H. W. Lenstra, Jr., On Artin's conjecture and Euclid's algorithm in global fields, Invent. Math., Vol. 42 (1977), pp. 202-224; alternative link.
J. W. Sander, On Fibonacci primitive roots, Fibonacci Quarterly, Vol. 28, No. 1 (1990), pp. 79-80.
Daniel Shanks, Fibonacci primitive roots, end of article, Fibonacci Quarterly, Vol. 10, No. 2 (1972), pp. 163-168, 181.
Daniel Shanks and Larry Taylor, An Observation of Fibonacci Primitive Roots, Fibonacci Quarterly, Vol. 11, No. 2 (1973), pp. 159-160.
Index entries for primes by primitive root.

Subsequence of A038872.
Cf. A001175, A005596, A083701.
See also A106535.

filter:=proc(n) local g,r;
if not isprime(n) then return false fi;
r:= [msolve(g^2 -g - 1, n)][1];
numtheory:-order(rhs(op(r)),n) = n-1
end proc:
select(filter, [5,seq(seq(10*i+j,j=[1,9]),i=1..1000)]); # Robert Israel, May 22 2015

okQ[p_] := AnyTrue[PrimitiveRootList[p], Mod[#^2, p] == Mod[#+1, p]&]; Select[Prime[Range[300]], okQ] (* Jean-François Alcover, Jan 04 2016 *)

is(n)=if(kronecker(5,n)<1||!isprime(n), return(n==5)); my(s=sqrt(Mod(5,n))); znorder((1+s)/2)==n-1 || znorder((1-s)/2)==n-1 \\ Charles R Greathouse IV, May 22 2015

More terms from David W. Wilson
Cross-reference from Charles R Greathouse IV, Nov 05 2009
Definition clarified by M. F. Hasler, Jun 05 2018

A001578 Smallest primitive prime factor of Fibonacci number F(n), or 1 if F(n) has no primitive prime factor.

Original entry on oeis.org

1, 1, 2, 3, 5, 1, 13, 7, 17, 11, 89, 1, 233, 29, 61, 47, 1597, 19, 37, 41, 421, 199, 28657, 23, 3001, 521, 53, 281, 514229, 31, 557, 2207, 19801, 3571, 141961, 107, 73, 9349, 135721, 2161, 2789, 211, 433494437, 43, 109441, 139, 2971215073, 1103, 97, 101

Offset: 1

N. J. A. Sloane

nonn

A prime factor of F(n) is called primitive if it does not divide F(r) for any r < n.
A Fibonacci number can have more than one primitive factor; the primitive factors of F(19) are 37 and 113.
From Robert Israel, Oct 13 2015: (Start)
Since gcd(F(n),F(k)) = F(gcd(n,k)), the non-primitive prime factors of F(n) are factors of F(k) for some proper divisors k of n.
Since prime p divides F(p-1) if p == 1 or 4 (mod 5), F(p+1) if p == 2 or 3 mod 5, F(p) if p = 5, we have a(n) >= n-1 if a(n) > 1.
a(n) = n-1 iff n=2 or n-1 is in A000057.
a(n) = n+1 iff n+1 is a prime in A106535. (End)

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Max Alekseyev, Table of n, a(n) for n = 1..1452 (terms 1..1000 from Blair Kelly and T. D. Noe, terms 1409..1422 from Tyler Busby)
Mansur S. Boase, A Result About the Primes Dividing Fibonacci Numbers, The Fibonacci Quarterly, 39.5 (2001) 386.
R. D. Carmichael, On the numerical factors of the arithmetic forms α^n ± β^n, Annals of Math., 15 (1/4) (1913), 30-70.
D. Jarden, On the greatest primitive divisors of Fibonacci and Lucas numbers with prime-power subscripts, Fib. Quart. 1(#3) (1963), 15-31.
Blair Kelly, Fibonacci and Lucas Factorizations

Cf. A000045, A000057, A106535, A086597 (number of primitive prime factors in F(n)), A061488 (1's omitted), A262341 (largest primitive prime factor of F(n)).

for n from 1 to 350 do
  f:= combinat:-fibonacci(n);
  if not isprime(n) then
    for k in map(t -> n/t, numtheory:-factorset(n)) do
       fk:= combinat:-fibonacci(k);
       g:= igcd(f,fk);
       while g > 1 do
         f:= f/g;
         g:= igcd(f,fk);
       od
    od
  fi;
  if f = 1 then A[n]:= 1; next fi;
  F:= map(t -> t[1],ifactors(f,easy)[2]);
  p:= select(type, F,integer);
  if nops(p) >= 1 then A[n]:= min(p); next fi;
  A[n]:= min(numtheory:-factorset(f));
od:
seq(A[i],i=1..350); # Robert Israel, Oct 13 2015

prms={}; Table[f=First/@FactorInteger[Fibonacci[n]]; p=Complement[f, prms]; prms=Join[prms, p]; If[p=={}, 1, First[p]], {n, 50}]

a(n) = {my(v = vector(n, k, fibonacci(k))); my(vf = vector(n, k, factor(v[k])[,1]~)); for (k=1, n-1, vf[n] = setminus(vf[n], vf[k]);); if (#vf[n], vecmin(vf[n]), 1);} \\ Michel Marcus, May 11 2021

a(n) = 1 if and only if n = 1, 2, 6, or 12, by Carmichael's theorem. - Jonathan Sondow, Dec 07 2017

Edited by T. D. Noe, Apr 15 2004

Definition clarified at the suggestion of Joerg Arndt by Jonathan Sondow, Oct 13 2015

A308800 Primes p such that A001177(p) = (p-1)/7.

Original entry on oeis.org

2731, 3739, 4831, 6091, 11159, 13679, 14771, 16871, 19559, 20399, 24179, 26111, 29191, 31039, 33811, 34511, 34679, 35911, 40111, 41651, 49211, 55259, 56099, 60859, 62819, 69539, 71191, 71359, 71471, 73291, 74831, 85751, 87991, 96979, 97231, 97931, 104959, 108179

Offset: 1

Jianing Song, Jun 25 2019

nonn

Primes p such that ord(-(3+sqrt(5))/2,p) = (p-1)/7, 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 Pisano period of {T(n)} modulo p (that is, the smallest m > 0 such that T(n+m) == T(n) (mod p) for all n) is ord(u,p); the entry point of {T(n)} modulo p (that is, the smallest m > 0 such that T(m) == 0 (mod p)) is ord(-u^2,p).
For an odd prime p:
(a) if p decomposes in K, then (O_K/pO_K)* (the multiplicative group of O_K modulo p) is congruent to C_(p-1) X C_(p-1), so the entry point of {T(n)} modulo p is equal to (p-1)/s, s = 1, 2, 3, 4, ...;
(b) if p is inert in K, then u^(p+1) == -1 (mod p), (-u^2)^(p+1) == 1 (mod p), so the entry point of {T(n)} modulo p is equal to (p+1)/s, s = 1, 2, 3, 4, ...
Here k = 1, and this sequence gives primes such that (a) holds and s = 7. For odd s, all terms are congruent to 3 modulo 4.
Number of terms below 10^N:
  N | Number | Decomposing primes*
  3 |      0 |            78
  4 |      4 |           609
  5 |     36 |          4777
  6 |    347 |         39210
  7 |   2801 |        332136
  8 |  24291 |       2880484
  * Here "Decomposing primes" means primes such that Legendre(5,p) = 1, i.e., p == 1, 4 (mod 5).

Amiram Eldar, Table of n, a(n) for n = 1..10000

Similar sequences that give primes such that (a) holds: A106535 (s=1), A308795 (s=2), A308796 (s=3), A308797 (s=4), A308798 (s=5), A308799 (s=6), this sequence (s=7), A308801 (s=8), A308802 (s=9).

pn[n_] := For[k = 1, True, k++, If[Mod[Fibonacci[k], n] == 0, Return[k]]];
Reap[For[p = 2, p < 50000, p = NextPrime[p], If[Mod[p, 7] == 1, If[pn[p] == (p - 1)/7, Print[p]; Sow[p]]]]][[2, 1]] (* Jean-François Alcover, Jul 05 2019 *)

Entry_for_decomposing_prime(p) = my(k=1, M=[k, 1; 1, 0]); 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, 109000, if(Entry_for_decomposing_prime(p)==(p-1)/7, print1(p, ", ")))

A308795 Primes p such that A001177(p) = (p-1)/2.

Original entry on oeis.org

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

Offset: 1

Jianing Song, Jun 25 2019

nonn

Primes p such that ord(-(3+sqrt(5))/2,p) = (p-1)/2, 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 Pisano period of {T(n)} modulo p (that is, the smallest m > 0 such that T(n+m) == T(n) (mod p) for all n) is ord(u,p); the entry point of {T(n)} modulo p (that is, the smallest m > 0 such that T(m) == 0 (mod p)) is ord(-u^2,p).
For an odd prime p:
(a) if p decomposes in K, then (O_K/pO_K)* (the multiplicative group of O_K modulo p) is congruent to C_(p-1) X C_(p-1), so the entry point of {T(n)} modulo p is equal to (p-1)/s, s = 1, 2, 3, 4, ...;
(b) if p is inert in K, then u^(p+1) == -1 (mod p), (-u^2)^(p+1) == 1 (mod p), so the entry point of {T(n)} modulo p is equal to (p+1)/s, s = 1, 2, 3, 4, ...
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).

Similar sequences that give primes such that (a) holds: A106535 (s=1), this sequence (s=2), A308796 (s=3), A308797 (s=4), A308798 (s=5), A308799 (s=6), A308800 (s=7), A308801 (s=8), A308802 (s=9).

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 *)

Entry_for_decomposing_prime(p) = my(k=1, M=[k, 1; 1, 0]); 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, ", ")))

A308796 Primes p such that A001177(p) = (p-1)/3.

Original entry on oeis.org

139, 151, 331, 619, 811, 1231, 1279, 1291, 1471, 1579, 1699, 1999, 2239, 2251, 2371, 2659, 3271, 3331, 3391, 3499, 3631, 3919, 4051, 4159, 4231, 4759, 5059, 5839, 6079, 6619, 6691, 6991, 7219, 7639, 8059, 8599, 8731, 8971, 9151, 9319, 9679, 9739, 10099, 10459, 10771

Offset: 1

Jianing Song, Jun 25 2019

nonn

Primes p such that ord(-(3+sqrt(5))/2,p) = (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 Pisano period of {T(n)} modulo p (that is, the smallest m > 0 such that T(n+m) == T(n) (mod p) for all n) is ord(u,p); the entry point of {T(n)} modulo p (that is, the smallest m > 0 such that T(m) == 0 (mod p)) is ord(-u^2,p).
For an odd prime p:
(a) if p decomposes in K, then (O_K/pO_K)* (the multiplicative group of O_K modulo p) is congruent to C_(p-1) X C_(p-1), so the entry point of {T(n)} modulo p is equal to (p-1)/s, s = 1, 2, 3, 4, ...;
(b) if p is inert in K, then u^(p+1) == -1 (mod p), (-u^2)^(p+1) == 1 (mod p), so the entry point of {T(n)} modulo p is equal to (p+1)/s, s = 1, 2, 3, 4, ...
Here k = 1, and this sequence gives primes such that (a) holds and s = 3. For odd s, all terms are congruent to 3 modulo 4.
Number of terms below 10^N:
  N | Number | Decomposing primes*
  3 |      5 |            78
  4 |     42 |           609
  5 |    312 |          4777
  6 |   2490 |         39210
  7 |  20958 |        332136
  8 | 181493 |       2880484
  * Here "Decomposing primes" means primes such that Legendre(5,p) = 1, i.e., p == 1, 4 (mod 5).

Bob Bastasz, Lyndon words of a second-order recurrence, Fibonacci Quarterly (2020) Vol. 58, No. 5, 25-29.

Similar sequences that give primes such that (a) holds: A106535 (s=1), A308795 (s=2), this sequence (s=3), A308797 (s=4), A308798 (s=5), A308799 (s=6), A308800 (s=7), A308801 (s=8), A308802 (s=9).

pn[n_] := For[k = 1, True, k++, If[Mod[Fibonacci[k], n] == 0, Return[k]]];
Reap[For[p = 2, p < 10000, p = NextPrime[p], If[Mod[p, 3] == 1, If[pn[p] == (p - 1)/3, Print[p]; Sow[p]]]]][[2, 1]] (* Jean-François Alcover, Jul 05 2019 *)

Entry_for_decomposing_prime(p) = my(k=1, M=[k, 1; 1, 0]); 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, 11000, if(Entry_for_decomposing_prime(p)==(p-1)/3, print1(p, ", ")))

A308797 Primes p such that A001177(p) = (p-1)/4.

Original entry on oeis.org

61, 109, 149, 269, 389, 401, 701, 809, 821, 1049, 1181, 1249, 1289, 1301, 1361, 1409, 1429, 1721, 1901, 1949, 2141, 2309, 2341, 2381, 2549, 2729, 2741, 2801, 2909, 3049, 3061, 3089, 3109, 3169, 3181, 3221, 3229, 3541, 3701, 3709, 3929, 4001, 4049, 4349, 4421, 4649

Offset: 1

Jianing Song, Jun 25 2019

nonn

Primes p such that ord(-(3+sqrt(5))/2,p) = (p-1)/4, 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 Pisano period of {T(n)} modulo p (that is, the smallest m > 0 such that T(n+m) == T(n) (mod p) for all n) is ord(u,p); the entry point of {T(n)} modulo p (that is, the smallest m > 0 such that T(m) == 0 (mod p)) is ord(-u^2,p).
For an odd prime p:
(a) if p decomposes in K, then (O_K/pO_K)* (the multiplicative group of O_K modulo p) is congruent to C_(p-1) X C_(p-1), so the entry point of {T(n)} modulo p is equal to (p-1)/s, s = 1, 2, 3, 4, ...;
(b) if p is inert in K, then u^(p+1) == -1 (mod p), (-u^2)^(p+1) == 1 (mod p), so the entry point of {T(n)} modulo p is equal to (p+1)/s, s = 1, 2, 3, 4, ...
Here k = 1, and this sequence gives primes such that (a) holds and s = 4.
Number of terms below 10^N:
  N | Number | Decomposing primes*
  3 |      9 |            78
  4 |     81 |           609
  5 |    651 |          4777
  6 |   5268 |         39210
  7 |  44188 |        332136
  8 | 383224 |       2880484
  * Here "Decomposing primes" means primes such that Legendre(5,p) = 1, i.e., p == 1, 4 (mod 5).

Similar sequences that give primes such that (a) holds: A106535 (s=1), A308795 (s=2), A308796 (s=3), this sequence (s=4), A308798 (s=5), A308799 (s=6), A308800 (s=7), A308801 (s=8), A308802 (s=9).

pn[n_] := For[k = 1, True, k++, If[Mod[Fibonacci[k], n] == 0, Return[k]]];
Reap[For[p = 2, p < 6000, p = NextPrime[p], If[Mod[p, 4] == 1, If[pn[p] == (p - 1)/4, Print[p]; Sow[p]]]]][[2, 1]] (* Jean-François Alcover, Jul 05 2019 *)

Entry_for_decomposing_prime(p) = my(k=1, M=[k, 1; 1, 0]); 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, 5000, if(Entry_for_decomposing_prime(p)==(p-1)/4, print1(p, ", ")))

A308798 Primes p such that A001177(p) = (p-1)/5.

Original entry on oeis.org

211, 691, 991, 1031, 1151, 1511, 1871, 1951, 2591, 3251, 3851, 4391, 4651, 4691, 4751, 4871, 5531, 5591, 6011, 6211, 6271, 6491, 7211, 7451, 8011, 8171, 8831, 9011, 9091, 9371, 9431, 9931, 10391, 10531, 10691, 10891, 11071, 12011, 12071, 12911, 14051, 14251, 14591

Offset: 1

Jianing Song, Jun 25 2019

nonn

Primes p such that ord(-(3+sqrt(5))/2,p) = (p-1)/5, 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 Pisano period of {T(n)} modulo p (that is, the smallest m > 0 such that T(n+m) == T(n) (mod p) for all n) is ord(u,p); the entry point of {T(n)} modulo p (that is, the smallest m > 0 such that T(m) == 0 (mod p)) is ord(-u^2,p).
For an odd prime p:
(a) if p decomposes in K, then (O_K/pO_K)* (the multiplicative group of O_K modulo p) is congruent to C_(p-1) X C_(p-1), so the entry point of {T(n)} modulo p is equal to (p-1)/s, s = 1, 2, 3, 4, ...;
(b) if p is inert in K, then u^(p+1) == -1 (mod p), (-u^2)^(p+1) == 1 (mod p), so the entry point of {T(n)} modulo p is equal to (p+1)/s, s = 1, 2, 3, 4, ...
Here k = 1, and this sequence gives primes such that (a) holds and s = 5. For odd s, all terms are congruent to 3 modulo 4.
Number of terms below 10^N:
  N | Number | Decomposing primes*
  3 |      3 |            78
  4 |     32 |           609
  5 |    192 |          4777
  6 |   1521 |         39210
  7 |  12542 |        332136
  8 | 109034 |       2880484
  * Here "Decomposing primes" means primes such that Legendre(5,p) = 1, i.e., p == 1, 4 (mod 5).

Similar sequences that give primes such that (a) holds: A106535 (s=1), A308795 (s=2), A308796 (s=3), A308797 (s=4), this sequence (s=5), A308799 (s=6), A308800 (s=7), A308801 (s=8), A308802 (s=9).

pn[n_] := For[k = 1, True, k++, If[Mod[Fibonacci[k], n] == 0, Return[k]]];
Reap[For[p = 2, p < 15000, p = NextPrime[p], If[Mod[p, 5] == 1, If[pn[p] == (p - 1)/5, Print[p]; Sow[p]]]]][[2, 1]] (* Jean-François Alcover, Jul 05 2019 *)

Entry_for_decomposing_prime(p) = my(k=1, M=[k, 1; 1, 0]); 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, 15000, if(Entry_for_decomposing_prime(p)==(p-1)/5, print1(p, ", ")))

cp's OEIS Frontend

A214029 The union of the disjoint prime sequences A000057 and A106535.

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A001177 Fibonacci entry points: a(n) = least k >= 1 such that n divides Fibonacci number F_k (=A000045(k)).

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A000057 Primes dividing all Fibonacci sequences.

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A003147 Primes p with a Fibonacci primitive root g, i.e., such that g^2 = g + 1 (mod p).

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A001578 Smallest primitive prime factor of Fibonacci number F(n), or 1 if F(n) has no primitive prime factor.

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A308800 Primes p such that A001177(p) = (p-1)/7.

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A308795 Primes p such that A001177(p) = (p-1)/2.