A053028 -id:A053028 - cp's OEIS frontend

A000032 Lucas numbers beginning at 2: L(n) = L(n-1) + L(n-2), L(0) = 2, L(1) = 1.

2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 521, 843, 1364, 2207, 3571, 5778, 9349, 15127, 24476, 39603, 64079, 103682, 167761, 271443, 439204, 710647, 1149851, 1860498, 3010349, 4870847, 7881196, 12752043, 20633239, 33385282, 54018521, 87403803

Offset: 0

N. J. A. Sloane, May 24 1994

Cf. A000204 for Lucas numbers beginning with 1.
Also the number of independent vertex sets and vertex covers for the cycle graph C_n for n >= 2. - Eric W. Weisstein, Jan 04 2014
Also the number of matchings in the n-cycle graph C_n for n >= 3. - Eric W. Weisstein, Oct 01 2017
Also the number of maximal independent vertex sets (and maximal vertex covers) for the n-helm graph for n >= 3. - Eric W. Weisstein, May 27 2017
Also the number of maximal independent vertex sets (and maximal vertex covers) for the n-sunlet graph for n >= 3. - Eric W. Weisstein, Aug 07 2017
This is also the Horadam sequence (2, 1, 1, 1). - Ross La Haye, Aug 18 2003
For distinct primes p, q, L(p) is congruent to 1 mod p, L(2p) is congruent to 3 mod p and L(pq) is congruent 1 + q(L(q) - 1) mod p. Also, L(m) divides F(2km) and L((2k + 1)m), k, m >= 0.
a(n) = Sum_{k=0..ceiling((n - 1)/2)} P(3; n - 1 - k, k), n >= 1, with a(0) = 2. These are the sums over the SW-NE diagonals in P(3; n, k), the (3, 1) Pascal triangle A093560. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs. Also SW-NE diagonal sums of the (1, 2) Pascal triangle A029635 (with T(0, 0) replaced by 2).
Suppose psi = log(phi) = A002390. We get the representation L(n) = 2*cosh(n*psi) if n is even; L(n) = 2*sinh(n*psi) if n is odd. There is a similar representation for Fibonacci numbers (A000045). Many Lucas formulas now easily follow from appropriate sinh- and cosh-formulas. For example: the identity cosh^2(x) - sinh^2(x) = 1 implies L(n)^2 - 5*F(n)^2 = 4*(-1)^n (setting x = n*psi). - Hieronymus Fischer, Apr 18 2007
From John Blythe Dobson, Oct 02 2007, Oct 11 2007: (Start)
The parity of L(n) follows easily from its definition, which shows that L(n) is even when n is a multiple of 3 and odd otherwise.
The first six multiplication formulas are:
L(2n) = L(n)^2 - 2*(-1)^n;
L(3n) = L(n)^3 - 3*(-1)^n*L(n);
L(4n) = L(n)^4 - 4*(-1)^n*L(n)^2 + 2;
L(5n) = L(n)^5 - 5*(-1)^n*L(n)^3 + 5*L(n);
L(6n) = L(n)^6 - 6*(-1)^n*L(n)^4 + 9*L(n)^2 - 2*(-1)^n.
Generally, L(n) | L(mn) if and only if m is odd.
In the expansion of L(mn), where m represents the multiplier and n the index of a known value of L(n), the absolute values of the coefficients are the terms in the m-th row of the triangle A034807. When m = 1 and n = 1, L(n) = 1 and all the terms are positive and so the row sums of A034807 are simply the Lucas numbers. (End)
From John Blythe Dobson, Nov 15 2007: (Start)
The comments submitted by Miklos Kristof on Mar 19 2007 for the Fibonacci numbers (A000045) contain four important identities that have close analogs in the Lucas numbers:
For a >= b and odd  b, L(a + b) + L(a - b) = 5*F(a)*F(b).
For a >= b and even b, L(a + b) + L(a - b) = L(a)*L(b).
For a >= b and odd  b, L(a + b) - L(a - b) = L(a)*L(b).
For a >= b and even b, L(a + b) - L(a - b) = 5*F(a)*F(b).
A particularly interesting instance of the difference identity for even b is L(a + 30) - L(a - 30) = 5*F(a)*832040, since 5*832040 is divisible by 100, proving that the last two digits of Lucas numbers repeat in a cycle of length 60 (see A106291(100)). (End)
From John Blythe Dobson, Nov 15 2007: (Start)
The Lucas numbers satisfy remarkable difference equations, in some cases best expressed using Fibonacci numbers, of which representative examples are the following:
L(n) - L(n - 3)  = 2*L(n - 2);
L(n) - L(n - 4)  = 5*F(n - 2);
L(n) - L(n - 6)  = 4*L(n - 3);
L(n) - L(n - 12) = 40*F(n - 6);
L(n) - L(n - 60) = 4160200*F(n - 30).
These formulas establish, respectively, that the Lucas numbers form a cyclic residue system of length 3 (mod 2), of length 4 (mod 5), of length 6 (mod 4), of length 12 (mod 40) and of length 60 (mod 4160200). The divisibility of the last modulus by 100 accounts for the fact that the last two digits of the Lucas numbers begin to repeat at L(60).
The divisibility properties of the Lucas numbers are very complex and still not fully understood, but several important criteria are established in Zhi-Hong Sun's 2003 survey of congruences for Fibonacci numbers. (End)
Sum_{n>0} a(n)/(n*2^n) = 2*log(2). - Jaume Oliver Lafont, Oct 11 2009
A010888(a(n)) = A030133(n). - Reinhard Zumkeller, Aug 20 2011
The powers of phi, the golden ratio, approach the values of the Lucas numbers, the odd powers from above and the even powers from below. - Geoffrey Caveney, Apr 18 2014
Inverse binomial transform is (-1)^n * a(n). - Michael Somos, Jun 03 2014
Lucas numbers are invariant to the following transformation for all values of the integers j and n, including negative values, thus: L(n) = (L(j+n) + (-1)^n * L(j-n))/L(j). The same transformation applied to all sequences of the form G(n+1) = m * G(n) + G(n-1) yields Lucas numbers for m = 1, except where G(j) = 0, regardless of initial values which may be nonintegers. The corresponding sequences for other values of m are: for m = 2, 2*A001333; for m = 3, A006497; for m = 4, 2*A001077; for m = 5, A087130; for m = 6, 2*A005667; for m = 7, A086902. The invariant ones all have G(0) = 2, G(1) = m. A related family of sequences is discussed at A059100. - Richard R. Forberg, Nov 23 2014
If x=a(n), y=a(n+1), z=a(n+2), then -x^2 - z*x - 3*y*x - y^2 + y*z + z^2 = 5*(-1)^(n+1). - Alexander Samokrutov, Jul 04 2015
A conjecture on the divisibility of infinite subsequences of Lucas numbers by prime(n)^m, m >= 1, is given in A266587, together with the prime "entry points". - Richard R. Forberg, Dec 31 2015
A trapezoid has three lengths of sides in order L(n-1), L(n+1), L(n-1). For increasing n a very close approximation to the maximum area will have the fourth side equal to 2*L(n). For a trapezoid with sides L(n-1), L(n-3), L(n-1), the fourth side will be L(n). - J. M. Bergot, Mar 17 2016
Satisfies Benford's law [Brown-Duncan, 1970; Berger-Hill, 2017]. - N. J. A. Sloane, Feb 08 2017
Lucas numbers L(n) and Fibonacci numbers F(n), being related by the formulas F(n) = (F(n-1) + L(n-1))/2 and L(n) = 2 F(n+1) - F(n), are a typical pair of "autosequences" (see the link to OEIS Wiki). - Jean-François Alcover, Jun 09 2017
For n >= 3, the Lucas number L(n) is the dimension of a commutative Hecke algebra of affine type A_n with independent parameters. See Theorem 1.4, Corollary 1.5, and the table on page 524 in the link "Hecke algebras with independent parameters". - Jia Huang, Jan 20 2019
From Klaus Purath, Apr 19 2019: (Start)
While all prime numbers appear as factors in the Fibonacci numbers, this is not the case with the Lucas numbers. For example, L(n) is never divisible by the following prime numbers < 150: 5, 13, 17, 37, 53, 61, 73, 89, 97, 109, 113, 137, 149 ... See A053028. Conjecture: Three properties can be determined for these prime numbers:
First observation: The prime factors > 3 occur in the Fibonacci numbers with an odd index.
Second observation: These are the prime numbers p congruent to 2, 3 (modulo 5), which occur both in Fibonacci(p+1) and in Fibonacci((p+1)/2) as prime factors, or the prime numbers p congruent to 1, 4 (modulo 5), which occur both in Fibonacci((p-1)/2) and in Fibonacci((p-1)/(2^k)) with k >= 2.
Third observation: The Pisano period lengths of these prime numbers, given in A001175, are always divisible by 4, but not by 8. In contrast, those of the prime factors of Lucas numbers are divisible either by 2, but not by 4, or by 8. (See also comment in A053028 by N. J. A. Sloane, Feb 21 2004). (End)
L(n) is the sum of 4*k consecutive terms of the Fibonacci sequence (A000045) divided by Fibonacci(2*k): (Sum_{i=0..4*k-1, k>=1} F(n+i))/F(2*k) = L(n+2*k+1). Sequences extended to negative indices, following the rule a(n-1) = a(n+1) - a(n). - Klaus Purath, Sep 15 2019
If one forms a sequence (A) of the Fibonacci type with the initial values A(0) = A022095(n) and A(1) = A000285(n), then A(n+1) = L(n+1)^2 always applies. - Klaus Purath, Sep 29 2019
From Kai Wang, Dec 18 2019: (Start)
L((2*m+1)k)/L(k) = Sum_{i=0..m-1} (-1)^(i*(k+1))*L((2*m-2*i)*k) + (-1)^(m*k).
Example: k=5, m=2, L(5)=11, L(10)=123, L(20)=15127, L(25)=167761. L(25)/L(5) = 15251, L(20) + L(10) + 1 = 15127 + 123 + 1 = 15251. (End)
From Peter Bala, Dec 23 2021: (Start)
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) hold for all prime p and positive integers n and k.
For a positive integer k, the sequence (a(n))n>=1 taken modulo k becomes a purely periodic sequence. For example, taken modulo 11, the sequence becomes [1, 3, 4, 7, 0, 7, 7, 3, 10, 2, 1, 3, 4, 7, 0, 7, 7, 3, 10, 2, ...], a periodic sequence with period 10. (End)
For any sequence with recurrence relation b(n) = b(n-1) + b(n-2), it can be shown that the recurrence relation for every k-th term is given by: b(n) = A000032(k) * b(n-k) + (-1)^(k+1) * b(n-2k), extending to negative indices as necessary. - Nick Hobson, Jan 19 2024
For n >= 3, L(n) is the number of (n-1)-digit numbers where all consecutive pairs of digits have a difference of at least 8. - Edwin Hermann, Apr 19 2025

			G.f. = 2 + x + 3*x^2 + 4*x^3 + 7*x^4 + 11*x^5 + 18*x^6 + 29*x^7 + ...

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Index entries for sequences related to Chebyshev polynomials..
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2).
Index entries for linear recurrences with constant coefficients, signature (1,1).
Index entries for two-way infinite sequences.
Index entries for "core" sequences.
Index entries for sequences related to Benford's law.

Cf. A000204. A000045(n) = (2*L(n + 1) - L(n))/5.
First row of array A103324.
a(n) = A101220(2, 0, n), for n > 0.
a(k) = A090888(1, k) = A109754(2, k) = A118654(2, k - 1), for k > 0.
Cf. A131774, A001622, A002878 (L(2n+1)), A005248 (L(2n)), A006497, A080039, A049684 (summation of Fibonacci(4n+2)), A106291 (Pisano periods), A057854 (complement), A354265 (generalized Lucas numbers).
Cf. sequences with formula Fibonacci(n+k)+Fibonacci(n-k) listed in A280154.
Subsequence of A047201.

a000032 n = a000032_list !! n
a000032_list = 2 : 1 : zipWith (+) a000032_list (tail a000032_list)
-- Reinhard Zumkeller, Aug 20 2011

Magma
```
[Lucas(n): n in [0..120]];
```

with(combinat): A000032 := n->fibonacci(n+1)+fibonacci(n-1);
seq(simplify(2^n*(cos(Pi/5)^n+cos(3*Pi/5)^n)), n=0..36)

a[0] := 2; a[n] := Nest[{Last[#], First[#] + Last[#]} &, {2, 1}, n] // Last
Array[2 Fibonacci[# + 1] - Fibonacci[#] &, 50, 0] (* Joseph Biberstine (jrbibers(AT)indiana.edu), Dec 26 2006 *)
Table[LucasL[n], {n, 0, 36}] (* Zerinvary Lajos, Jul 09 2009 *)
LinearRecurrence[{1, 1}, {2, 1}, 40] (* Harvey P. Dale, Sep 07 2013 *)
LucasL[Range[0, 20]] (* Eric W. Weisstein, Aug 07 2017 *)
CoefficientList[Series[(-2 + x)/(-1 + x + x^2), {x, 0, 20}], x] (* Eric W. Weisstein, Sep 21 2017 *)

{a(n) = if(n<0, (-1)^n * a(-n), if( n<2, 2-n, a(n-1) + a(n-2)))};

{a(n) = if(n<0, (-1)^n * a(-n), polsym(x^2 - x - 1, n)[n+1])};

{a(n) = real((2 + quadgen(5)) * quadgen(5)^n)};

a(n)=fibonacci(n+1)+fibonacci(n-1) \\ Charles R Greathouse IV, Jun 11 2011

polsym(1+x-x^2, 50) \\ Charles R Greathouse IV, Jun 11 2011

def A000032_gen(): # generator of terms
    a, b = 2, 1
    while True:
        yield a
        a, b = b, a+b
it = A000032_gen()
A000032_list = [next(it) for  in range(50)] # _Cole Dykstra, Aug 02 2022

from sympy import lucas
def A000032(n): return lucas(n) # Chai Wah Wu, Sep 23 2023

[(i:=3)+(j:=-1)] + [(j:=i+j)+(i:=j-i) for  in range(100)] # _Jwalin Bhatt, Apr 02 2025

[lucas_number2(n,1,-1) for n in range(37)] # Zerinvary Lajos, Jun 25 2008

G.f.: (2 - x)/(1 - x - x^2).
L(n) = ((1 + sqrt(5))/2)^n + ((1 - sqrt(5))/2)^n = phi^n + (1-phi)^n.
L(n) = L(n - 1) + L(n - 2) = (-1)^n * L( - n).
L(n) = Fibonacci(2*n)/Fibonacci(n) for n > 0. - Jeff Burch, Dec 11 1999
E.g.f.: 2*exp(x/2)*cosh(sqrt(5)*x/2). - Len Smiley, Nov 30 2001
L(n) = F(n) + 2*F(n - 1) = F(n + 1) + F(n - 1). - Henry Bottomley, Apr 12 2000
a(n) = sqrt(F(n)^2 + 4*F(n + 1)*F(n - 1)). - Benoit Cloitre, Jan 06 2003 [Corrected by Gary Detlefs, Jan 21 2011]
a(n) = 2^(1 - n)*Sum_{k=0..floor(n/2)} C(n, 2k)*5^k. a(n) = 2T(n, i/2)( - i)^n with T(n, x) Chebyshev's polynomials of the first kind (see A053120) and i^2 = - 1. - Paul Barry, Nov 15 2003
L(n) = 2*F(n + 1) - F(n). - Paul Barry, Mar 22 2004
a(n) = (phi)^n + ( - phi)^( - n). - Paul Barry, Mar 12 2005
From Miklos Kristof, Mar 19 2007: (Start)
Let F(n) = A000045 = Fibonacci numbers, L(n) = a(n) = Lucas numbers:
L(n + m) + (-1)^m*L(n - m) = L(n)*L(m).
L(n + m) - (-1)^m*L(n - m) = 8*F(n)*F(m).
L(n + m + k) + (-1)^k*L(n + m - k) + (-1)^m*(L(n - m + k) + (-1)^k*L(n - m - k)) = L(n)*L(m)*L(k).
L(n + m + k) - (-1)^k*L(n + m - k) + (-1)^m*(L(n - m + k) - (-1)^k*L(n - m - k)) = 5*F(n)*L(m)*F(k).
L(n + m + k) + (-1)^k*L(n + m - k) - (-1)^m*(L(n - m + k) + (-1)^k*L(n - m - k)) = 5*F(n)*F(m)*L(k).
L(n + m + k) - (-1)^k*L(n + m - k) - (-1)^m*(L(n - m + k) - (-1)^k*L(n - m - k)) = 5*L(n)*F(m)*F(k). (End)
Inverse: floor(log_phi(a(n)) + 1/2) = n, for n>1. Also for n >= 0, floor((1/2)*log_phi(a(n)*a(n+1))) = n. Extension valid for all integers n: floor((1/2)*sign(a(n)*a(n+1))*log_phi|a(n)*a(n+1)|) = n {where sign(x) = sign of x}. - Hieronymus Fischer, May 02 2007
Let f(n) = phi^n + phi^(-n), then L(2n) = f(2n) and L(2n + 1) = f(2n + 1) - 2*Sum_{k>=0} C(k)/f(2n + 1)^(2k + 1) where C(n) are Catalan numbers (A000108). - Gerald McGarvey, Dec 21 2007, modified by Davide Colazingari, Jul 01 2016
Starting (1, 3, 4, 7, 11, ...) = row sums of triangle A131774. - Gary W. Adamson, Jul 14 2007
a(n) = trace of the 2 X 2 matrix [0,1; 1,1]^n. - Gary W. Adamson, Mar 02 2008
From Hieronymus Fischer, Jan 02 2009: (Start)
For odd n: a(n) = floor(1/(fract(phi^n))); for even n>0: a(n) = ceiling(1/(1 - fract(phi^n))). This follows from the basic property of the golden ratio phi, which is phi - phi^(-1) = 1 (see general formula described in A001622).
a(n) = round(1/min(fract(phi^n), 1 - fract(phi^n))), for n>1, where fract(x) = x - floor(x). (End)
E.g.f.: exp(phi*x) + exp(-x/phi) with phi: = (1 + sqrt(5))/2 (golden section). 1/phi = phi - 1. See another form given in the Smiley e.g.f. comment. - Wolfdieter Lang, May 15 2010
L(n)/L(n - 1) -> A001622. - Vincenzo Librandi, Jul 17 2010
a(n) = 2*a(n-2) + a(n-3), n>2. - Gary Detlefs, Sep 09 2010
L(n) = floor(1/fract(Fibonacci(n)*phi)), for n odd. - Hieronymus Fischer, Oct 20 2010
L(n) = ceiling(1/(1 - fract(Fibonacci(n)*phi))), for n even. - Hieronymus Fischer, Oct 20 2010
L(n) = 2^n * (cos(Pi/5)^n + cos(3*Pi/5)^n). - Gary Detlefs, Nov 29 2010
L(n) = (Fibonacci(2*n - 1)*Fibonacci(2*n + 1) - 1)/(Fibonacci(n)*Fibonacci(2*n)), n != 0. - Gary Detlefs, Dec 13 2010
L(n) = sqrt(A001254(n)) = sqrt(5*Fibonacci(n)^2 - 4*(-1)^(n+1)). - Gary Detlefs, Dec 26 2010
L(n) = floor(phi^n) + ((-1)^n + 1)/2 = A014217(n) +((-1)^n+1)/2, where phi = A001622. - Gary Detlefs, Jan 20 2011
L(n) = Fibonacci(n + 6) mod Fibonacci(n + 2), n>2. - Gary Detlefs, May 19 2011
For n >= 2, a(n) = round(phi^n) where phi is the golden ratio. - Arkadiusz Wesolowski, Jul 20 2012
a(p*k) == a(k) (mod p) for primes p. a(2^s*n) == a(n)^(2^s) (mod 2) for s = 0,1,2.. a(2^k) ==  - 1 (mod 2^k). a(p^2*k) == a(k) (mod p) for primes p and s = 0,1,2,3.. [Hoggatt and Bicknell]. - R. J. Mathar, Jul 24 2012
From Gary Detlefs, Dec 21 2012: (Start)
L(k*n) = (F(k)*phi + F(k - 1))^n + (F(k + 1) - F(k)*phi)^n.
L(k*n) = (F(n)*phi + F(n - 1))^k + (F(n + 1) - F(n)*phi)^k.
where phi = (1 + sqrt(5))/2, F(n) = A000045(n).
(End)
L(n) = n * Sum_{k=0..floor(n/2)} binomial(n - k,k)/(n - k), n>0 [H. W. Gould]. - Gary Detlefs, Jan 20 2013
G.f.: G(0), where G(k) = 1 + 1/(1 - (x*(5*k-1))/((x*(5*k+4)) - 2/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 15 2013
L(n) = F(n) + F(n-1) + F(n-2) + F(n-3). - Bob Selcoe, Jun 17 2013
L(n) = round(sqrt(L(2n-1) + L(2n-2))). - Richard R. Forberg, Jun 24 2014
L(n) = (F(n+1)^2 - F(n-1)^2)/F(n) for n>0. - Richard R. Forberg, Nov 17 2014
L(n+2) = 1 + A001610(n+1) = 1 + Sum_{k=0..n} L(k). - Tom Edgar, Apr 15 2015
L(i+j+1) = L(i)*F(j) + L(i+1)*F(j+1) with F(n)=A000045(n). - J. M. Bergot, Feb 12 2016
a(n) = (L(n+1)^2 + 5*(-1)^n)/L(n+2). - J. M. Bergot, Apr 06 2016
Dirichlet g.f.: PolyLog(s,-1/phi) + PolyLog(s,phi), where phi is the golden ratio. - Ilya Gutkovskiy, Jul 01 2016
L(n) = F(n+2) - F(n-2). - Yuchun Ji, Feb 14 2016
L(n+1) = A087131(n+1)/2^(n+1) = 2^(-n)*Sum_{k=0..n} binomial(n,k)*5^floor((k+1)/2). - Tony Foster III, Oct 14 2017
L(2*n) = (F(k+2*n) + F(k-2*n))/F(k); n >= 1, k >= 2*n. - David James Sycamore, May 04 2018
From Greg Dresden and Shaoxiong Yuan, Jul 16 2019: (Start)
L(3n + 4)/L(3n + 1) has continued fraction: n 4's followed by a single 7.
L(3n + 3)/L(3n) has continued fraction: n 4's followed by a single 2.
L(3n + 2)/L(3n - 1) has continued fraction: n 4's followed by a single -3. (End)
From Klaus Purath, Sep 15 2019: (Start)
All involved sequences extended to negative indices, following the rule a(n-1) = a(n+1) - a(n).
L(n) = (2*L(n+2) - L(n-3))/5.
L(n) = (2*L(n-2) + L(n+3))/5.
L(n) = F(n-3) + 2*F(n).
L(n) = 2*F(n+2) - 3*F(n).
L(n) = (3*F(n-1) + F(n+2))/2.
L(n) = 3*F(n-3) + 4*F(n-2).
L(n) = 4*F(n+1) - F(n+3).
L(n) = (F(n-k) + F(n+k))/F(k) with odd k>0.
L(n) = (F(n+k) - F(n-k))/F(k) with even k>0.
L(n) = A001060(n-1) - F(n+1).
L(n) = (A022121(n-1) - F(n+1))/2.
L(n) = (A022131(n-1) - F(n+1))/3.
L(n) = (A022139(n-1) - F(n+1))/4.
L(n) = (A166025(n-1) - F(n+1))/5.
The following two formulas apply for all sequences of the Fibonacci type.
(a(n-2*k) + a(n+2*k))/a(n) = L(2*k).
(a(n+2*k+1) - a(n-2*k-1))/a(n) = L(2*k+1). (End)
L(n) = F(n-k)*L(k+1) + F(n-k-1)*L(k), for all k >= 0, where F(n) = A000045(n). - Michael Tulskikh, Dec 06 2019
F(n+2*m) = L(m)*F(n+m) + (-1)^(m-1)*F(n) for all n >= 0 and m >= 0. - Alexander Burstein, Mar 31 2022
a(n) = i^(n-1)*cos(n*c)/cos(c) = i^(n-1)*cos(c*n)*sec(c), where c = Pi/2 + i*arccsch(2). - Peter Luschny, May 23 2022
From Yike Li and Greg Dresden, Aug 25 2022: (Start)
L(2*n) = 5*binomial(2*n-1,n) - 2^(2*n-1) + 5*Sum_{j=1..n/5} binomial(2*n,n+5*j) for n>0.
L(2*n+1) = 2^(2n) - 5*Sum_{j=0..n/5} binomial(2*n+1,n+5*j+3). (End)
From Andrea Pinos, Jul 04 2023: (Start)
L(n) ~ Gamma(1/phi^n) + gamma.
L(n) = Re(phi^n + e^(i*Pi*n)/phi^n). (End)
L(n) = ((Sum_{i=0..n-1} L(i)^2) - 2)/L(n-1). - Jules Beauchamp, May 03 2025
From Peter Bala, Jul 09 2025: (Start)
The following series telescope:
For  k >= 1, Sum_{n >= 1} (-1)^((k+1)*(n+1)) * a(2*n*k)/(a((2*n-1)*k)*a((2*n+1)*k)) = 1/a(k)^2.
For positive even k, Sum_{n >= 1} 1/(a(k*n) - (a(k) + 2)/a(k*n)) = 1/(a(k) - 2) and
Sum_{n >= 1} (-1)^(n+1)/(a(k*n) + (a(k) - 2)/a(k*n)) = 1/(a(k) + 2).
For positive odd k, Sum_{n >= 1} 1/(a(k*n) - (-1)^n*(a(2*k) + 2)/a(k*n)) = (a(k) + 2)/(2*(a(2*k) - 2)) and
Sum_{n >= 1} (-1)^(n+1)/(a(k*n) - (-1)^n*(a(2*k) + 2)/a(k*n)) = (a(k) - 2)/(2*(a(2*k) - 2)). (End)

A001176 Number of zeros in fundamental period of Fibonacci numbers mod n.

Original entry on oeis.org

1, 1, 2, 1, 4, 2, 2, 2, 2, 4, 1, 2, 4, 2, 2, 2, 4, 2, 1, 2, 2, 1, 2, 2, 4, 4, 2, 2, 1, 2, 1, 2, 2, 4, 2, 2, 4, 1, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 4, 2, 2, 4, 2, 2, 2, 2, 1, 1, 2, 4, 1, 2, 2, 4, 2, 2, 2, 2, 2, 1, 2, 4, 4, 2, 1, 2, 2, 1, 2, 2, 2, 2, 2, 4, 2, 2, 2, 4, 2, 2, 2, 2, 2, 2, 2, 4, 2, 2, 2, 1, 2, 2, 2, 2

Offset: 1

N. J. A. Sloane

If the Fibonacci numbers are indexed so that 3 is the fourth number, then if the modulo base is a Fibonacci number (>= 3) with an even index, the period has 2 zeros. If the base is a Fibonacci number (>= 5) with an odd index, the period has 4 zeros. - Kerry Mitchell, Dec 11 2005

For a proof that A001177(n) divides the period length A001175(n) for n >= 1, see, e.g., the Vajda reference, p. 73. This comment refers to the present first formula. - Wolfdieter Lang, Jan 19 2015

			{F(n) mod 1} has fundamental period (0) with 1 zero.
{F(n) mod 2} has fundamental period (0,1,1) with 1 zero.
{F(n) mod 3} has fundamental period (0,1,1,2,0,2,2,1) with 2 zeros.
{F(n) mod 4} has fundamental period (0,1,1,2,3,1), with 1 zero.
{F(n) mod 5} has fundamental period (0,1,1,2,3,0,3,3,1,4,0,4,4,3,2,0,2,2,4,1) with 4 zeros.

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, Jun 1968.
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.

T. D. Noe, Table of n, a(n) for n = 1..1000
Brennan Benfield and Michelle Manes, The Fibonacci Sequence is Normal Base 10, arXiv:2202.08986 [math.NT], 2022.
J. D. Fulton and W. L. Morris, On arithmetical functions related to the Fibonacci numbers, Acta Arithmetica, 16 (1969), 105-110.
B. H. Hannon and W. L. Morris, Tables of Arithmetical Functions Related to the Fibonacci Numbers [Annotated and scanned copy]
M. Renault, Fibonacci sequence modulo m
Review of B. H. Hannon and W. L. Morris tables, Math. Comp., 23 (1969), 459-460.

Cf. A001175, A001177, A053027, A053028, A053029, A053030, A053031, A053032.

Cf. A235715.

a001176 1 = 1
a001176 n = f 1 ps 0 where
   f 0 (1 : xs) z = z
   f _ (x : xs) z = f x xs (z + 0 ^ x)
   ps = 1 : 1 : zipWith (\u v -> (u + v) `mod` n) (tail ps) ps
-- Reinhard Zumkeller, Jan 15 2014

With[{fibs=Fibonacci[Range[2000]]},Table[Count[FindTransientRepeat[ Mod[ fibs, n], 3][[2]],0],{n,110}]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Dec 26 2016 *)

a(n) = A001175(n)/A001177(n) for n >= 1.
a(n) = ord(n, fibonacci(A001177(n) + 1)), where ord(n, a) is the multiplicative order of a modulo n. - Mircea Merca, Jan 03 2011
a(n) = A128924(n,1). - Reinhard Zumkeller, Jan 17 2014
From Isaac Saffold, Aug 30 2018: (Start)
With the sole exception of a(8) = 2,
  a(p^k) = 1 if A007814(A001175(p^k)) < 2.
  a(p^k) = 4 if A007814(A001175(p^k)) = 2.
  a(p^k) = 2 if A007814(A001175(p^k)) > 2. (End)
From Jianing Song, Sep 01 2018: (Start)
a(2^e) = 1 if e <= 2, otherwise 2. For odd primes p, a(p^e) = 4 if A001177(p) is odd; 1 if A001177(p) is even but not divisible by 4; 2 if A001177(p) is divisible by 4.
a(n) = 2 for n == 0, 3, 7, 8, 12, 15 (mod 20). a(p^e) = 1 if primes p == 11, 19 (mod 20); 4 if p == 13, 17 (mod 20). Conjecture: 1/6 of the primes congruent to 1 or 9 mod 40 satisfy a(p^e) = 1, 2/3 of them satisfy a(p^e) = 2 and 1/6 of them satisfy a(p^e) = 4; also, 1/2 of the primes congruent to 21 or 29 mod 40 satisfy a(p^e) = 1 and 1/2 of them satisfy a(p^e) = 4. (End)

Better description and more terms from Henry Bottomley, Feb 01 2000
Examples from David W. Wilson, Jan 05 2005
Replaced the old Renault link with a working one. - Wolfdieter Lang, Jan 17 2015

A022307 Number of distinct prime factors of n-th Fibonacci number.

Original entry on oeis.org

0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 1, 2, 1, 2, 3, 3, 1, 3, 2, 4, 3, 2, 1, 4, 2, 2, 4, 4, 1, 5, 2, 4, 3, 2, 3, 5, 3, 3, 3, 6, 2, 5, 1, 5, 5, 3, 1, 6, 3, 5, 3, 4, 2, 6, 4, 6, 5, 3, 2, 8, 2, 3, 5, 6, 3, 5, 3, 5, 5, 7, 2, 8, 2, 4, 5, 5, 4, 6, 2, 9, 7, 3, 1, 9, 4, 3, 4, 9, 2, 10, 4, 6, 4, 2, 6, 9, 4, 5, 6

Offset: 0

Clark Kimberling

nonn

Although every prime divides some Fibonacci number, this is not true for the Lucas numbers. Exactly 1/3 of all primes do not divide any Lucas number. See Lagarias and Moree for more details. - Jonathan Vos Post, Dec 06 2006
First occurrence of k: 0, 3, 8, 15, 20, 30, 40, 70, 60, 80, 90, 140, 176, 120, 168, 180, 324, 252, 240, 378, ..., . - Robert G. Wilson v, Dec 10 2006 [Other than 0, this is sequence A060320. - Jon E. Schoenfield, Dec 30 2016]
Row lengths of table A060442. - Reinhard Zumkeller, Aug 30 2014
If k properly divides n then a(n) >= a(k) + 1, except for a(6) = a(3) = 1. - Robert Israel, Aug 18 2015

Alfred Brousseau, Fibonacci and Related Number Theoretic Tables, The Fibonacci Association, 1972, pages 1-8.

Max Alekseyev, Table of n, a(n) for n = 0..1422 (terms 0..1000 and 1001..1408 from T. D. Noe and Amiram Eldar, respectively).
Blair Kelly, Fibonacci and Lucas Factorizations
J. C. Lagarias, The set of primes dividing the Lucas numbers has density 2/3, Pacific J. Math., 118. No. 2, (1985), 449-461.
J. C. Lagarias, Errata to: The set of primes dividing the Lucas numbers has density 2/3, Pacific J. Math., 162, No. 2, (1994), 393-396.
Hisanori Mishima, WIFC (World Integer Factorization Center), Fibonacci numbers (n = 1 to 100, n = 101 to 200, n = 201 to 300, n = 301 to 400, n = 401 to 480).
Pieter Moree, Counting Divisors of Lucas Numbers, Pacific J. Math, Vol. 186, No. 2, 1998, pp. 267-284.
Eric Weisstein's World of Mathematics, Fibonacci Number

Cf. A038575 (number of prime factors, counting multiplicity), A086597 (number of primitive prime factors).
Cf. A000032, A000040, A000045, A001221, A053028.
Cf. A060442, A086598 (omega(Lucas(n))).
Cf. A060320. - Jon E. Schoenfield, Dec 30 2016

a022307 n = if n == 0 then 0 else a001221 $ a000045 n
-- Reinhard Zumkeller, Aug 30 2014

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

Table[Length[FactorInteger[Fibonacci[n]]], {n, 150}]

a(n)=omega(fibonacci(n)) \\ Charles R Greathouse IV, Feb 03 2014

a(n) = Sum{d|n} A086597(d), Mobius transform of A086597.

a(n) = A001221(A000045(n)) = omega(F(n)). - Jonathan Vos Post, Dec 06 2006

A053032 Odd primes p with one zero in Fibonacci numbers mod p.

Original entry on oeis.org

11, 19, 29, 31, 59, 71, 79, 101, 131, 139, 151, 179, 181, 191, 199, 211, 229, 239, 251, 271, 311, 331, 349, 359, 379, 419, 431, 439, 461, 479, 491, 499, 509, 521, 541, 571, 599, 619, 631, 659, 691, 709, 719, 739, 751, 809, 811, 839, 859, 911, 919, 941, 971

Offset: 1

Henry Bottomley, Feb 23 2000

nonn

Also, odd primes that divide Lucas numbers of odd index. - T. D. Noe, Jul 25 2003
From Charles R Greathouse IV, Dec 14 2016: (Start)
It seems that this sequence contains about 1/3 of the primes. In particular, members of this sequence constitute:
        35 of the first 10^2 primes
       330 of the first 10^3 primes
      3328 of the first 10^4 primes
     33371 of the first 10^5 primes
    333329 of the first 10^6 primes
   3333720 of the first 10^7 primes
  33333463 of the first 10^8 primes
etc. (End)
Of the Fibonacci-like sequences modulo a prime p that are not A000004, one of them has a period length less than A001175(p) if and only if p = 5 or p is in this sequence. - Isaac Saffold, Dec 18 2018
Odd primes in A053031. - Jianing Song, Jun 19 2019

			From _Michael B. Porter_, Jan 25 2019: (Start)
The Fibonacci numbers (mod 7) repeat the pattern 0, 1, 1, 2, 3, 5, 1, 6, 0, 6, 6, 5, 4, 2, 6, 1. Since there are two zeros, 7 is not in the sequence.
The Fibonacci numbers (mod 11) repeat the pattern 0, 1, 1, 2, 3, 5, 8, 2, 10, 1 which has only one zero, so 11 is in the sequence.
(End)

T. D. Noe, Table of n, a(n) for n = 1..1000
C. Ballot and M. Elia, Rank and period of primes in the Fibonacci sequence; a trichotomy, Fib. Quart., 45 (No. 1, 2007), 56-63 (The sequence B1).
Nicholas Bragman and Eric Rowland, Limiting density of the Fibonacci sequence modulo powers of p, arXiv:2202.00704 [math.NT], 2022.
M. Renault, Fibonacci sequence modulo m

Cf. A001175, A001177. See A112860 for another version.
Cf. A000204 (Lucas numbers), A001602 (index of the smallest Fibonacci number divisible by prime(n)).
Let {x(n)} be a sequence defined by x(0) = 0, x(1) = 1, x(n+2) = m*x(n+1) + x(n). Let w(k) be the number of zeros in a fundamental period of {x(n)} modulo k.
                             |    m=1     |   m=2   |   m=3
-----------------------------+------------+---------+---------
The sequence {x(n)}          | A000045    | A000129 | A006190
The sequence {w(k)}          | A001176    | A214027 | A322906
Primes p such that w(p) = 1  | A112860*   | A309580 | A309586
Primes p such that w(p) = 2  | A053027**  | A309581 | A309587
Primes p such that w(p) = 4  | A053028*** | A261580 | A309588
Numbers k such that w(k) = 1 | A053031    | A309583 | A309591
Numbers k such that w(k) = 2 | A053030    | A309584 | A309592
Numbers k such that w(k) = 4 | A053029    | A309585 | A309593
* and also this sequence U {2}
** also primes dividing Lucas numbers of even index
*** also primes dividing no Lucas number

Prime@ Rest@ Position[Table[Count[Drop[NestWhile[Append[#, Mod[Total@ Take[#, -2], n]] &, {1, 1}, If[Length@ # < 3, True, Take[#, -2] != {1, 1}] &], -2], 0], {n, Prime@ Range@ 168}], 1][[All, 1]] (* Michael De Vlieger, Aug 08 2018 *)

fibmod(n,m)=(Mod([1, 1; 1, 0], m)^n)[1, 2]
is(n)=my(k=n+[0, -1, 1, 1, -1][n%5+1]); k>>=valuation(k,2)-1; fibmod(k,n)==0 && fibmod(k/2,n) && isprime(n) \\ Charles R Greathouse IV, Dec 14 2016

A prime p = prime(i) is in this sequence if p > 2 and A001602(i)/2 is odd. - T. D. Noe, Jul 25 2003

A053027 Odd primes p with 2 zeros in Fibonacci numbers mod p.

Original entry on oeis.org

3, 7, 23, 41, 43, 47, 67, 83, 103, 107, 127, 163, 167, 223, 227, 241, 263, 281, 283, 307, 347, 367, 383, 401, 409, 443, 449, 463, 467, 487, 503, 523, 547, 563, 569, 587, 601, 607, 641, 643, 647, 683, 727, 743, 769, 787, 823, 827, 863, 881, 883, 887, 907, 929

Offset: 1

Henry Bottomley, Feb 23 2000

nonn

Also, odd primes that divide Lucas numbers of even index. - T. D. Noe, Jul 25 2003
Primes in A053030. - Jianing Song, Jun 19 2019
From Jianing Song, Jun 16 2024: (Start)
Primes p such that A001176(p) = 2.
For p > 2, p is in this sequence if and only if 8 divides of A001175(p), and if and only if 4 divides A001177(p). For a proof of the equivalence between A001176(p) = 2 and 4 dividing A001177(p), see Section 2 of my link below.
This sequence contains all primes congruent to 3, 7 (mod 20). This corresponds to case (2) for k = 3 in the Conclusion of Section 1 of my link below.
Conjecturely, this sequence has density 1/3 in the primes. (End) [Comment rewritten by Jianing Song, Jun 16 2024 and Jun 25 2024]

T. D. Noe, Table of n, a(n) for n=1..1000
C. Ballot and M. Elia, Rank and period of primes in the Fibonacci sequence; a trichotomy, Fib. Quart., 45 (No. 1, 2007), 56-63 (The sequence B3).
Nicholas Bragman and Eric Rowland, Limiting density of the Fibonacci sequence modulo powers of p, arXiv:2202.00704 [math.NT], 2022.
M. Renault, Fibonacci sequence modulo m
Jianing Song, Lucas sequences and entry point modulo p

Cf. A001175, A001177.
Cf. A000204 (Lucas numbers), A001602 (index of the smallest Fibonacci number divisible by prime(n)).
Let {x(n)} be a sequence defined by x(0) = 0, x(1) = 1, x(n+2) = m*x(n+1) + x(n). Let w(k) be the number of zeros in a fundamental period of {x(n)} modulo k.
                             |    m=1    |   m=2   |   m=3
-----------------------------+-----------+---------+---------
The sequence {x(n)}          | A000045   | A000129 | A006190
The sequence {w(k)}          | A001176   | A214027 | A322906
Primes p such that w(p) = 1  | A112860*  | A309580 | A309586
Primes p such that w(p) = 2  | this seq  | A309581 | A309587
Primes p such that w(p) = 4  | A053028** | A261580 | A309588
Numbers k such that w(k) = 1 | A053031   | A309583 | A309591
Numbers k such that w(k) = 2 | A053030   | A309584 | A309592
Numbers k such that w(k) = 4 | A053029   | A309585 | A309593
* and also A053032 (primes dividing Lucas numbers of odd index) U {2}
** also primes dividing no Lucas number

A prime p = prime(i) is in this sequence if p > 2 and A001602(i)/2 is even. - T. D. Noe, Jul 25 2003

A053029 Numbers with 4 zeros in Fibonacci numbers mod m.

Original entry on oeis.org

5, 10, 13, 17, 25, 26, 34, 37, 50, 53, 61, 65, 73, 74, 85, 89, 97, 106, 109, 113, 122, 125, 130, 137, 146, 149, 157, 169, 170, 173, 178, 185, 193, 194, 197, 218, 221, 226, 233, 250, 257, 265, 269, 274, 277, 289, 293, 298, 305, 313, 314, 317, 325, 337, 338, 346

Offset: 1

Henry Bottomley, Feb 23 2000

nonn

Conjecture: m is on this list iff m is an odd number all of whose factors are on this list or m is twice such an odd number.

A001176(a(n)) = A128924(a(n),1) = 4. - Reinhard Zumkeller, Jan 17 2014

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
Brennan Benfield and Michelle Manes, The Fibonacci Sequence is Normal Base 10, arXiv:2202.08986 [math.NT], 2022.
Brennan Benfield and Oliver Lippard, Connecting Zeros in Pisano Periods to Prime Factors of K-Fibonacci Numbers, arXiv:2407.20048 [math.NT], 2024.
M. Renault, Fibonacci sequence modulo m

Let {x(n)} be a sequence defined by x(0) = 0, x(1) = 1, x(n+2) = m*x(n+1) + x(n). Let w(k) be the number of zeros in a fundamental period of {x(n)} modulo k.
                             |   m=1    |   m=2   |   m=3
-----------------------------+----------+---------+---------
The sequence {x(n)}          | A000045  | A000129 | A006190
The sequence {w(k)}          | A001176  | A214027 | A322906
Primes p such that w(p) = 1  | A112860* | A309580 | A309586
Primes p such that w(p) = 2  | A053027  | A309581 | A309587
Primes p such that w(p) = 4  | A053028  | A261580 | A309588
Numbers k such that w(k) = 1 | A053031  | A309583 | A309591
Numbers k such that w(k) = 2 | A053030  | A309584 | A309592
Numbers k such that w(k) = 4 | this seq | A309585 | A309593
* and also A053032 U {2}

a053029 n = a053029_list !! (n-1)
a053029_list = filter ((== 4) . a001176) [1..]
-- Reinhard Zumkeller, Jan 17 2014

A053030 Numbers with 2 zeros in Fibonacci numbers mod m.

Original entry on oeis.org

3, 6, 7, 8, 9, 12, 14, 15, 16, 18, 20, 21, 23, 24, 27, 28, 30, 32, 33, 35, 36, 39, 40, 41, 42, 43, 45, 46, 47, 48, 49, 51, 52, 54, 55, 56, 57, 60, 63, 64, 66, 67, 68, 69, 70, 72, 75, 77, 78, 80, 81, 82, 83, 84, 86, 87, 88, 90, 91, 92, 93, 94, 95, 96, 98, 99, 100, 102, 103, 104

Offset: 1

Henry Bottomley, Feb 23 2000

nonn

m is on this list iff m does not have 1 or 4 zeros in the Fibonacci sequence modulo m.

A001176(a(n)) = A128924(a(n),1) = 2. - Reinhard Zumkeller, Jan 17 2014

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Brennan Benfield and Oliver Lippard, Connecting Zeros in Pisano Periods to Prime Factors of K-Fibonacci Numbers, arXiv:2407.20048 [math.NT], 2024. See p. 2.
Brennan Benfield and Michelle Manes, The Fibonacci Sequence is Normal Base 10, arXiv:2202.08986 [math.NT], 2022.
M. Renault, Fibonacci sequence modulo m

Let {x(n)} be a sequence defined by x(0) = 0, x(1) = 1, x(n+2) = m*x(n+1) + x(n). Let w(k) be the number of zeros in a fundamental period of {x(n)} modulo k.
                             |   m=1    |   m=2   |   m=3
-----------------------------+----------+---------+---------
The sequence {x(n)}          | A000045  | A000129 | A006190
The sequence {w(k)}          | A001176  | A214027 | A322906
Primes p such that w(p) = 1  | A112860* | A309580 | A309586
Primes p such that w(p) = 2  | A053027  | A309581 | A309587
Primes p such that w(p) = 4  | A053028  | A261580 | A309588
Numbers k such that w(k) = 1 | A053031  | A309583 | A309591
Numbers k such that w(k) = 2 | this seq | A309584 | A309592
Numbers k such that w(k) = 4 | A053029  | A309585 | A309593
* and also A053032 U {2}

a053030 n = a053030_list !! (n-1)
a053030_list = filter ((== 2) . a001176) [1..]
-- Reinhard Zumkeller, Jan 17 2014

A053031 Numbers with 1 zero in Fibonacci numbers mod m.

Original entry on oeis.org

1, 2, 4, 11, 19, 22, 29, 31, 38, 44, 58, 59, 62, 71, 76, 79, 101, 116, 118, 121, 124, 131, 139, 142, 151, 158, 179, 181, 191, 199, 202, 209, 211, 229, 236, 239, 242, 251, 262, 271, 278, 284, 302, 311, 316, 319, 331, 341, 349, 358, 359, 361, 362, 379, 382, 398

Offset: 1

Henry Bottomley, Feb 23 2000

nonn

Conjecture: m is on this list iff m is an odd number all of whose factors are on this list or m is 2 or 4 times such an odd number.
A001176(a(n)) = A128924(a(n),1) = 1. - Reinhard Zumkeller, Jan 16 2014
Also numbers n such that A001175(n) = A001177(n). - Daniel Suteu, Aug 08 2018

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
Brennan Benfield and Michelle Manes, The Fibonacci Sequence is Normal Base 10, arXiv:2202.08986 [math.NT], 2022.
Brennan Benfield and Oliver Lippard, Connecting Zeros in Pisano Periods to Prime Factors of K-Fibonacci Numbers, arXiv:2407.20048 [math.NT], 2024.
M. Renault, Fibonacci sequence modulo m

Let {x(n)} be a sequence defined by x(0) = 0, x(1) = 1, x(n+2) = m*x(n+1) + x(n). Let w(k) be the number of zeros in a fundamental period of {x(n)} modulo k.
                             |   m=1    |   m=2   |   m=3
-----------------------------+----------+---------+---------
The sequence {x(n)}          | A000045  | A000129 | A006190
The sequence {w(k)}          | A001176  | A214027 | A322906
Primes p such that w(p) = 1  | A112860* | A309580 | A309586
Primes p such that w(p) = 2  | A053027  | A309581 | A309587
Primes p such that w(p) = 4  | A053028  | A261580 | A309588
Numbers k such that w(k) = 1 | this seq | A309583 | A309591
Numbers k such that w(k) = 2 | A053030  | A309584 | A309592
Numbers k such that w(k) = 4 | A053029  | A309585 | A309593
* and also A053032 U {2}

a053031 n = a053031_list !! (n-1)
a053031_list = filter ((== 1) . a001176) [1..]
-- Reinhard Zumkeller, Jan 16 2014

With[{s = {1}~Join~Table[Count[Drop[NestWhile[Append[#, Mod[Total@ Take[#, -2], n]] &, {1, 1}, If[Length@ # < 3, True, Take[#, -2] != {1, 1}] &], -2], 0], {n, 2, 400}]}, Position[s, 1][[All, 1]] ] (* Michael De Vlieger, Aug 08 2018 *)

entryp(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
entry(n)=if(n==1, return(1)); my(f=factor(n), v); v=vector(#f~, i, if(f[i, 1]>1e14, entryp(f[i, 1]^f[i, 2]), entryp(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, Dec 14 2016

A112860 2 together with A053032.

Original entry on oeis.org

2, 11, 19, 29, 31, 59, 71, 79, 101, 131, 139, 151, 179, 181, 191, 199, 211, 229, 239, 251, 271, 311, 331, 349, 359, 379, 419, 431, 439, 461, 479, 491, 499, 509, 521, 541, 571, 599, 619, 631, 659, 691, 709, 719, 739, 751, 809, 811, 839, 859, 911, 919, 941, 971

Offset: 1

N. J. A. Sloane, Nov 30 2007

nonn

Consists of the primes that are in neither A053027 nor A053028.
From Jianing Song, Jun 16 2024: (Start)
Primes p such that A001176(p) = 1.
For p > 2, p is in this sequence if and only if A001175(p) == 2 (mod 4), and if and only if A001177(p) == 2 (mod 4). For a proof of the equivalence between A001176(p) = 1 and A001177(p) == 2 (mod 4), see Section 2 of my link below.
This sequence contains all primes congruent to 11, 19 (mod 20). This corresponds to case (3) for k = 3 in the Conclusion of Section 1 of my link below.
Conjecturely, this sequence has density 1/3 in the primes. (End) [Comment rewritten by Jianing Song, Jun 16 2024 and Jun 25 2024]

T. D. Noe, Table of n, a(n) for n=1..1001
C. Ballot and M. Elia, Rank and period of primes in the Fibonacci sequence; a trichotomy, Fib. Quart., 45 (No. 1, 2007), 56-63 (The sequence B1).
Jianing Song, Lucas sequences and entry point modulo p

Cf. A001175, A001177.
Let {x(n)} be a sequence defined by x(0) = 0, x(1) = 1, x(n+2) = m*x(n+1) + x(n). Let w(k) be the number of zeros in a fundamental period of {x(n)} modulo k.
                             |   m=1     |   m=2   |   m=3
-----------------------------+-----------+---------+---------
The sequence {x(n)}          | A000045   | A000129 | A006190
The sequence {w(k)}          | A001176   | A214027 | A322906
Primes p such that w(p) = 1  | this seq* | A309580 | A309586
Primes p such that w(p) = 2  | A053027   | A309581 | A309587
Primes p such that w(p) = 4  | A053028   | A261580 | A309588
Numbers k such that w(k) = 1 | A053031   | A309583 | A309591
Numbers k such that w(k) = 2 | A053030   | A309584 | A309592
Numbers k such that w(k) = 4 | A053029   | A309585 | A309593
* and also A053032 U {2}

A261580 Primes p such that A214028(p) is odd.

Original entry on oeis.org

5, 13, 29, 37, 53, 61, 101, 109, 137, 149, 157, 173, 181, 197, 229, 269, 277, 293, 317, 349, 373, 389, 397, 421, 461, 509, 521, 541, 557, 569, 593, 613, 653, 661, 677, 701, 709, 733, 757, 773, 797, 821, 829, 853, 857, 877, 941, 953, 997, 1013, 1021, 1061, 1069

Offset: 1

Michel Marcus, Aug 25 2015

nonn

From Jianing Song, Aug 13 2019: (Start)
Primes p with 4 zeros in a fundamental period of A000129 mod p, that is, primes p such that A214027(p) = 4. For a proof of the equivalence between A214027(p) = 4 and A214028(p) being odd, see Section 2 of my link below.
For p > 2, p is in this sequence if and only if A175181(p) == 4 (mod 8).
This sequence contains all primes congruent to 5 modulo 8. This corresponds to case (1) for k = 6 in the Conclusion of Section 1 of my link below.
Conjecturely, since (k+2)/2 = 4 is a square, this sequence has density 7/24 in the primes; see the end of Section 1 of my link. (End) [Comment rewritten by Jianing Song, Jun 16 2024 and Jun 20 2024]
The conjecture above is an analog of Hasse's result that the set {p prime : ord(2,p) is odd} has density 7/24 in the primes, where ord(a,m) is the multiplicative order of a modulo m; see A014663. - Jianing Song, Jun 26 2025

			The smallest Pell number divisible by the prime 5 has index 3, which is odd, so 5 is in the sequence.

Jianing Song, Table of n, a(n) for n = 1..1280
Bernadette Faye and Florian Luca, Pell Numbers whose Euler Function is a Pell Number, arXiv:1508.05714 [math.NT], 2015.
Jianing Song, Lucas sequences and entry point modulo p

Cf. A214028, A261581.
Cf. also A175181.
Let {x(n)} be a sequence defined by x(0) = 0, x(1) = 1, x(n+2) = m*x(n+1) + x(n). Let w(k) be the number of zeros in a fundamental period of {x(n)} modulo k.
                             |   m=1    |   m=2    |   m=3
-----------------------------+----------+----------+---------
The sequence {x(n)}          | A000045  | A000129  | A006190
The sequence {w(k)}          | A001176  | A214027  | A322906
Primes p such that w(p) = 1  | A112860* | A309580  | A309586
Primes p such that w(p) = 2  | A053027  | A309581  | A309587
Primes p such that w(p) = 4  | A053028  | this seq | A309588
Numbers k such that w(k) = 1 | A053031  | A309583  | A309591
Numbers k such that w(k) = 2 | A053030  | A309584  | A309592
Numbers k such that w(k) = 4 | A053029  | A309585  | A309593
* and also A053032 U {2}

f[n_] := Block[{k = 1}, While[Mod[Simplify[((1 + Sqrt@ 2)^k - (1 - Sqrt@ 2)^k)/(2 Sqrt@ 2)], n] != 0, k++]; k]; Select[Prime@ Range@ 180, OddQ@ f@ # &] (* Michael De Vlieger, Aug 25 2015 *)

pell(n) = polcoeff(Vec(x/(1-2*x-x^2) + O(x^(n+1))), n);
z(n) = {k=1; while (pell(k) % n, k++); k;}
lista(nn) = {forprime(p=2, nn, if (z(p) % 2, print1(p, ", ")););}

forprime(p=2, 1100, if(A214027(p)==4, print1(p, ", "))) \\ Jianing Song, Aug 13 2019

cp's OEIS Frontend

A000032 Lucas numbers beginning at 2: L(n) = L(n-1) + L(n-2), L(0) = 2, L(1) = 1.

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A001176 Number of zeros in fundamental period of Fibonacci numbers mod n.

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A022307 Number of distinct prime factors of n-th Fibonacci number.

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A053032 Odd primes p with one zero in Fibonacci numbers mod p.

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A053027 Odd primes p with 2 zeros in Fibonacci numbers mod p.

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A053029 Numbers with 4 zeros in Fibonacci numbers mod m.

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