A001077 -id:A001077 - 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)

A001076 Denominators of continued fraction convergents to sqrt(5).

Original entry on oeis.org

0, 1, 4, 17, 72, 305, 1292, 5473, 23184, 98209, 416020, 1762289, 7465176, 31622993, 133957148, 567451585, 2403763488, 10182505537, 43133785636, 182717648081, 774004377960, 3278735159921, 13888945017644, 58834515230497, 249227005939632, 1055742538989025

Offset: 0

N. J. A. Sloane

a(2*n+1) with b(2*n+1) := A001077(2*n+1), n >= 0, give all (positive integer) solutions to Pell equation b^2 - 5*a^2 = -1, a(2*n) with b(2*n) := A001077(2*n), n >= 1, give all (positive integer) solutions to Pell equation b^2 - 5*a^2 = +1 (cf. Emerson reference).
Bisection: a(2*n+1) = T(2*n+1, sqrt(5))/sqrt(5) = A007805(n), n >= 0 and a(2*n) = 4*S(n-1,18), n >= 0, with T(n,x), resp. S(n,x), Chebyshev's polynomials of the first, resp. second kind. S(-1,x)=0. See A053120, resp. A049310. S(n,18)=A049660(n+1). - Wolfdieter Lang, Jan 10 2003
Apart from initial terms, this is the Pisot sequence E(4,17), a(n) = floor(a(n-1)^2/a(n-2) + 1/2).
This is also the Horadam sequence (0,1,1,4), having the recurrence relation a(n) = s*a(n-1) + r*a(n-2); for n > 1, where a(0) = 0, a(1) = 1, s = 4, r = 1. a(n) / a(n-1) converges to 5^1/2 + 2 as n approaches infinity. 5^(1/2) + 2 can also be written as (2 * Phi) + 1 and Phi^2 + Phi. - Ross La Haye, Aug 18 2003
Numerators of continued fraction [4, 4, 4, ...], where the convergents to [4, 4, 4, ...] = (4/1, 17/4, 72/17, ...). Let X = the 2 X 2 matrix [0, 1; 1, 4]; then X^n = [a(n-1), a(n); a(n), a(n+1)]; e.g., X^3 = [4, 17; 17, 72]. Let C = the limit of a(n)/a(n-1) = 2 + sqrt(5) = 4.236067977...; then C^n = a(n+1) + (1/C)*a(n), where (1/C) = 0.236067977... . Example: C^3 = 76.01315556..., = 72 + 17*(0.2360679...). - Gary W. Adamson, Dec 15 2007, corrected by Greg Dresden, Sep 16 2019, corrected by Alex Mark, Jul 21 2020
Sqrt(5) = 4/2 + 4/17 + 4/(17*305) + 4/(305*5473) + 4/(5473*98209) + ... . - Gary W. Adamson, Dec 15 2007
a(p) == 20^((p-1)/2) (mod p) for odd primes p. - Gary W. Adamson, Feb 22 2009
a(n) = A167808(3*n). - Reinhard Zumkeller, Nov 12 2009
For n >= 2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 4's along the main diagonal and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
Moreover, a(n) is the second binomial transform of (0,1,0,5,0,25,...) (see also A033887). This fact can be proved similarly like the proof of Paul Barry's remark in A033887 by using the following scaling identity for delta-Fibonacci numbers: y^n b(n;x/y) = Sum_{k=0..n} binomial(n,k) (y-1)^(n-k) b(k;x) and the fact that b(n;2) = (1-(-1)^n) 5^floor(n/2). - Roman Witula, Jul 12 2012
Binomial transform of 0, 1, 2, 8, 24, 80, 256, ... (A063727 with offset 1). - R. J. Mathar, Feb 05 2014
For n >= 1, a(n) equals the number of words of length n-1 on alphabet {0,1,...,4} avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015
With offset 1 is the INVERT transform of A006190: (1, 3, 10, 33, 109, 360, ...). - Gary W. Adamson, Jul 24 2015
From Rogério Serôdio, Mar 30 2018: (Start)
This is a divisibility sequence (i.e., if n|m then a(n)|a(m)).
gcd(a(n),a(n+k)) = a(gcd(n, k)) for all positive integers n and k. (End)
The initial 0 of this sequence is in contradiction with the fact that 0 is no valid denominator and according to all standard references, the first convergent of a continued fraction is p(0)/q(0) = b(0)/1 where b(0) is the first term of the continued fraction, given by the integer part of the number. One may artificially define q(-1) = 0 to have a recurrent relation q(n) = b(n)*q(n-1) + q(n-2), n >= 1, but then its index should be -1. - M. F. Hasler, Nov 01 2019
Number of 4-compositions of n restricted to odd parts (and allowed zeros); see Hopkins & Ouvry reference. - Brian Hopkins, Aug 17 2020
From Michael A. Allen, Feb 15 2023: (Start)
Also called the 4-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence.
a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are 4 kinds of squares available. (End)
a(n) is the smallest nonnegative integer that is the sum of n, but no fewer, Fibonacci numbers including negative-index Fibonacci numbers (A039834), with that sum being a(n) = Sum_{i=0..n-1} A000045(3*i+1). a(n) is also the smallest nonnegative integer that is the sum of n, but no fewer, terms each of which is either a Fibonacci number or the negative of a Fibonacci number. (See A027941 for negatives disallowed.) - Mike Speciner, Oct 08 2023
From Enrique Navarrete, Dec 16 2023: (Start)
a(n) is the number of compositions of n when there are P(k) sorts of parts k, with k,n > = 1, where P(k) = A006190(k) is the k-th 3-metallonacci number (see example below).
In general, the number of compositions with k-metallonacci number of parts is counted by the (k+1)-st metallonacci sequence (note k=1 and k=2 are the Fibonacci and the Pell numbers, respectively). (End).
a(n) is the number of tilings of a 2 X n rectangle missing the top right 1 X 1 cell, using 1 X 1 squares, dominoes and right trominoes. Compare to A110679 which is the same problem but without the missing top right cell. - Greg Dresden and Yilin Zhu, Jul 10 2025

			1 2 9 38 161 (A001077)
-,-,-,--,---, ...
0 1 4 17 72 (A001076)
G.f. = x + 4*x^2 + 17*x^3 + 72*x^4 + 305*x^5 + 1292*x^6 + 5473*x^7 + 23184*x^8 + ...
From _Enrique Navarrete_, Dec 16 2023: (Start)
From the comment on compositions with 3-metallonacci sorts of parts, A006190(k), there are A006190(1)=1 type of 1, A006190(2)=3 types of 2, A006190(3)=10 types of 3, A006190(4)=33 types of 4, A006190(5)=109 types of 5 and A006190(6)=360 types of 6. The following table gives the number of compositions of n=6:
Composition, number of such compositions, number of compositions of this type:
 6,              1,      360;
 5+1,            2,      218;
 4+2,            2,      198;
 3+3,            1,      100;
 4+1+1,          3,       99;
 3+2+1,          6,      180;
 2+2+2,          1,       27;
 3+1+1+1,        4,       40;
 2+2+1+1,        6,       54;
 2+1+1+1+1,      5,       15;
 1+1+1+1+1+1,    1,        1;
for a total of a(6)=1292 compositions of n=6. (End)

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 23.
S. Koshkin, Non-classical linear divisibility sequences ..., Fib. Q., 57 (No. 1, 2019), 68-80. See Table 1.
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).
V. Thébault, Les Récréations Mathématiques. Gauthier-Villars, Paris, 1952, p. 282.

G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0..200 from T. D. Noe)
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.
Paraskevas K. Alvanos and Konstantinos A. Draziotis, Integer Solutions of the Equation y^2 = Ax^4 + B, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.4.
Elwyn Berlekamp and Richard K. Guy, Fibonacci Plays Billiards (2003), arXiv:2002.03705 [math.HO], 2020. See p. 5.
D. Birmajer, J. B. Gil, and M. D. Weiner, On the Enumeration of Restricted Words over a Finite Alphabet, J. Int. Seq. 19 (2016) # 16.1.3, Example 8.
D. W. Boyd, Some integer sequences related to the Pisot sequences, Acta Arithmetica, 34 (1979), 295-305.
D. W. Boyd, Linear recurrence relations for some generalized Pisot sequences, Advances in Number Theory ( Kingston ON, 1991) 333-340, Oxford Sci. Publ., Oxford Univ. Press, New York, 1993.
Latham Boyle and Paul J. Steinhardt, Self-Similar One-Dimensional Quasilattices, arXiv preprint arXiv:1608.08220 [math-ph], 2016.
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242, Thm. 1, p. 233.
M. C. Firengiz and A. Dil, Generalized Euler-Seidel method for second order recurrence relations, Notes on Number Theory and Discrete Mathematics, Vol. 20, 2014, No. 4, 21-32.
Yixing Fu, E. J. König, J. H. Wilson, Yang-Zhi Chou, and J. H. Pixley, Magic-angle semimetals, arXiv:1809.04604 [cond-mat.str-el], 2018.
Juan B. Gil and Aaron Worley, Generalized metallic means, arXiv:1901.02619 [math.NT], 2019.
Brian Hopkins and Stéphane Ouvry, Combinatorics of Multicompositions, arXiv:2008.04937 [math.CO], 2020.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 398
Milan Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
Tanya Khovanova, Recursive Sequences
C. Pita, On s-Fibonomials, J. Int. Seq. 14 (2011) # 11.3.7.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Kai Wang, On k-Fibonacci Sequences And Infinite Series List of Results and Examples, 2020.
Shaoxiong (Steven) Yuan, Generalized Identities of Certain Continued Fractions, arXiv:1907.12459 [math.NT], 2019.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (4,1).

Row n=4 of A073133, A172236 and A352361.
Cf. A000045, A001077, A015448, A175183 (Pisano periods).
Cf. A049660, A007805, A110679.
Partial sums of A033887. First differences of A049652. Bisection of A059973.
Third column of array A028412.

a:=[0,1];; for n in [3..30] do a[n]:=4*a[n-1]+a[n-2]; od; a; # Muniru A Asiru, Mar 31 2018

I:=[0,1]; [n le 2 select I[n] else 4*Self(n-1) + Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 24 2018

A001076:=-1/(-1+4*z+z**2); # conjectured by Simon Plouffe in his 1992 dissertation

Join[{0}, Denominator[Convergents[Sqrt[5], 30]]] (* Harvey P. Dale, Dec 10 2011 *)
a[ n_] := Fibonacci[3*n] / 2; (* Michael Somos, Feb 23 2014 *)
a[ n_] := ((2 + Sqrt[5])^n - (2 - Sqrt[5])^n) /(2 Sqrt[5]) // Simplify; (* Michael Somos, Feb 23 2014 *)
LinearRecurrence[{4, 1}, {0, 1}, 26] (* Jean-François Alcover, Sep 23 2017 *)
a[ n_] := Fibonacci[n, 4]; (* Michael Somos, Nov 02 2021 *)

a(n):=sum(4^(n-1-2*k)*binomial(n-k-1,n-2*k-1),k,0,floor((n)/2));/* Vladimir Kruchinin, Oct 02 2022 */

numlib::fibonacci(3*n)/2 $ n = 0..30; // Zerinvary Lajos, May 09 2008

{a(n) = fibonacci(3*n) / 2}; /* Michael Somos, Aug 11 2009 */

{a(n) = imag( (2 + quadgen(20))^n )}; /* Michael Somos, Feb 23 2014 */

{a(n) = polchebyshev(n-1, 2, 2*I)/I^(n-1)}; /* Michael Somos, Nov 02 2021 */

[lucas_number1(n,4,-1) for n in range(23)] # Zerinvary Lajos, Apr 23 2009

[fibonacci(3*n)/2 for n in range(23)] # Zerinvary Lajos, May 15 2009

a(n) = 4*a(n-1) + a(n-2), n > 1. a(0)=0, a(1)=1.
G.f.: x/(1 - 4*x - x^2).
a(n) = ((2+sqrt(5))^n - (2-sqrt(5))^n)/(2*sqrt(5)).
a(n) = A014445(n)/2 = F(3n)/2.
a(n) = ((-i)^(n-1))*S(n-1, 4*i), with i^2 = -1 and S(n, x) := U(n, x/2) Chebyshev's polynomials of the second kind. See A049310. S(-1, x) = 0.
a(n) = Sum_{i=0..n} Sum_{j=0..n} Fibonacci(i+j)*n!/(i!j!(n-i-j)!)/2. - Paul Barry, Feb 06 2004
E.g.f.: exp(2*x)*sinh(sqrt(5)*x)/sqrt(5). - Vladeta Jovovic, Sep 01 2004
a(n) = F(1) + F(4) + F(7) + ... + F(3n-2), for n > 0.
Conjecture: 2a(n+1) = a(n+2) - A001077(n+1). - Creighton Dement, Nov 28 2004
a(n) = Sum_{k=0..n} Sum_{j=0..n} C(n, j)*C(j, k)*F(j)/2. - Paul Barry, Feb 14 2005
a(n) = A048876(n) - A048875(n). - Creighton Dement, Mar 19 2005
Let M = {{0, 1}, {1, 4}}, v[1] = {0, 1}, v[n] = M.v[n - 1]; then a(n) = v[n][[1]]. - Roger L. Bagula, May 29 2005
a(n) = F(n, 4), the n-th Fibonacci polynomial evaluated at x=4. - T. D. Noe, Jan 19 2006
[A015448(n), a(n)] = [1,4; 1,3]^n * [1,0]. - Gary W. Adamson, Mar 21 2008
a(n) = (Sum_{k=0..n} Fibonacci(3*k-2)) + 1. - Gary Detlefs, Dec 26 2010
a(n) = (3*(-1)^n*F(n) + 5*F(n)^3)/2, n >= 0. See the general D. Jennings formula given in a comment on triangle A111125, where also the reference is given. Here the second (k=1) row [3,1] applies. - Wolfdieter Lang, Sep 01 2012
Sum_{k>=1} (-1)^(k-1)/(a(k)*a(k+1)) = (Sum_{k>=1} (-1)^(k-1)/(F_k*F_(k+1)))^3 = phi^(-3), where F_n is the n-th Fibonacci numbers (A000045) and phi is golden ratio (A001622). - Vladimir Shevelev, Feb 23 2013
G.f.: Q(0)*x/(2-4*x), where Q(k) = 1 + 1/(1 - x*(5*k-4)/(x*(5*k+1) - 2/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Oct 11 2013
a(-n) = -(-1)^n * a(n). - Michael Somos, Feb 23 2014
The o.g.f. A(x) = x/(1 - 4*x - x^2) satisfies A(x) + A(-x) + 8*A(x)*A(-x) = 0 or equivalently (1 + 8*A(x))*(1 + 8*A(-x)) = 1. The o.g.f. for A049660 equals -A(sqrt(x))*A(-sqrt(x)). - Peter Bala, Apr 02 2015
From Rogério Serôdio, Mar 30 2018: (Start)
Some properties:
(1) a(n)*a(n+1) = 4*Sum_{k=1..n} a(k)^2;
(2) a(n)^2 + a(n+1)^2 = a(2*n+1);
(3) a(n)^2 - a(n-2)^2 = 4*a(n-1)*(a(n) + a(n-2));
(4) a(m*(p+1)) = a(m*p)*a(m+1) + a(m*p-1)*a(m);
(5) a(n-k)*a(n+k) = a(n)^2 + (-1)^(n+k+1)*a(k)^2;
(6) a(n-1)*a(n+1) = a(n)^2 + (-1)^n (particular case of (5)!);
(7) a(2*n) = 2*a(n)*(2*a(n) + a(n-1));
(8) 3*Sum_{k=2..n+1} a(k)*a(k-1) is equal to a(n+1)^2 if n odd, and is equal to a(n+1)^2 - 1 if n is even;
(9) a(n) - a(n-2*k+1) = alpha(k)*a(n-2*k+1) + a(n-4*k+2), where alpha(k) = (2+sqrt(5))^(2*k-1) + (2-sqrt(5))^(2*k-1);
(10) 31|Sum_{k=n..n+9} a(k), for all positive n. (End)
O.g.f.: x*exp(Sum_{n >= 1} Lucas(3*n)*x^n/n) = x + 4*x^2 + 17*x^3 + .... - Peter Bala, Oct 11 2019
a(n) = Sum_{k=0..floor(n/2)} 4^(n-2*k-1)*C(n-k-1,n-2*k-1). - Vladimir Kruchinin, Oct 02 2022
a(n) = i^(n-1)*S(n-1, -4*i), with i = sqrt(-1), and the Chebyshev S-polynomials (see A049310) with S(n, -1) = 0. - Gary Detlefs and Wolfdieter Lang, Mar 06 2023
G.f.: x/(1 - 4*x - x^2) = Sum_{n >= 0} x^(n+1) * ( Product_{k = 1..n} (m*k + 4 - m + x)/(1 + m*k*x) ) for arbitrary m (a telescoping series). - Peter Bala, May 08 2024
a(n) = 4^(n-1)*hypergeom([(1-n)/2, 1-n/2], [1-n], -1/4) for n > 0. - Peter Luschny, Mar 30 2025
a(n) = a(n-1) + A110679(n-1) + A110679(n-2) = a(n-1) + Fibonacci(3*n-2). - Greg Dresden and Yilin Zhu, Jul 10 2025

A049660 a(n) = Fibonacci(6*n)/8.

Original entry on oeis.org

0, 1, 18, 323, 5796, 104005, 1866294, 33489287, 600940872, 10783446409, 193501094490, 3472236254411, 62306751484908, 1118049290473933, 20062580477045886, 360008399296352015, 6460088606857290384, 115921586524134874897, 2080128468827570457762

Offset: 0

Clark Kimberling

For n >= 2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 18's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011
For n >= 2, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,17}. - Milan Janjic, Jan 25 2015
10*a(n)^2 = Tri(4)*S(n-1, 18)^2 is the triangular number Tri((T(n, 9) - 1)/2), with Tri, S and T given in A000217, A049310 and A053120. This is instance k = 4 of the k-family of identities given in a comment on A001109. - Wolfdieter Lang, Feb 01 2016
Possible solutions for y in Pell equation x^2 - 80*y^2 = 1. The values for x are given in A023039. - Herbert Kociemba, Jun 05 2022

			a(3) = F(6 * 3) / 8 = F(18) / 8 = 2584 / 8 = 323. - _Indranil Ghosh_, Feb 06 2017

Indranil Ghosh, Table of n, a(n) for n = 0..796 (terms 0..200 from Vincenzo Librandi)
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
R. Flórez, R. A. Higuita, and A. Mukherjee, Alternating Sums in the Hosoya Polynomial Triangle, Article 14.9.5 Journal of Integer Sequences, Vol. 17 (2014).
Tanya Khovanova, Recursive Sequences
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (18,-1).

Column m=6 of array A028412.
Partial sums of A007805.
Cf. A000045, A001076, A001077, A014445, A014448, A015448, A099843, A134492, A155179.

[Fibonacci(6*n)/8: n in [0..30]]; // G. C. Greubel, Dec 02 2017

with (combinat):seq(fibonacci(2*n,4)/4, n=0..16); # Zerinvary Lajos, Apr 20 2008

Fibonacci[6*Range[0,20]]/8 (* Harvey P. Dale, Nov 23 2011 *)
LinearRecurrence[{18,-1}, {0,1}, 30] (* G. C. Greubel, Dec 02 2017 *)
Table[ChebyshevU[-1 + n, 9], {n, 0, 18}] (* Herbert Kociemba, Jun 05 2022 *)

numlib::fibonacci(6*n)/8 $ n = 0..25; // Zerinvary Lajos, May 09 2008

a(n)=fibonacci(6*n)/8  \\ Charles R Greathouse IV, Apr 17 2012

[lucas_number1(n,18,1) for n in range(0,20)] # Zerinvary Lajos, Jun 25 2008

[fibonacci(6*n)/8 for n in range(0, 17)] # Zerinvary Lajos, May 15 2009

G.f.: x/(1 - 18*x + x^2).
a(n) = A134492(n)/8.
a(n) ~ (1/40)*sqrt(5)*(sqrt(5) + 2)^(2*n). - Joe Keane (jgk(AT)jgk.org), May 15 2002
For all terms k of the sequence, 80*k^2 + 1 is a square. Limit_{n->oo} a(n)/a(n-1) = 8*phi + 5 = 9 + 4*sqrt(5). - Gregory V. Richardson, Oct 14 2002
a(n) = S(n-1, 18) with S(n, x) := U(n, x/2), Chebyshev's polynomials of the second kind. S(-1, x) := 0. See A049310.
a(n) = (((9 + 4*sqrt(5))^n - (9 - 4*sqrt(5))^n))/(8*sqrt(5)).
a(n) = sqrt((A023039(n)^2 - 1)/80) (cf. Richardson comment).
a(n) = 18*a(n-1) - a(n-2). - Gregory V. Richardson, Oct 14 2002
a(n) = A001076(2n)/4.
a(n) = 17*(a(n-1) + a(n-2)) - a(n-3) = 19*(a(n-1) - a(n-2)) + a(n-3). - Mohamed Bouhamida, May 26 2007
a(n+1) = Sum_{k=0..n} A101950(n,k)*17^k. - Philippe Deléham, Feb 10 2012
Product_{n>=1} (1 + 1/a(n)) = (1/2)*(2 + sqrt(5)). - Peter Bala, Dec 23 2012
Product_{n>=2} (1 - 1/a(n)) = (2/9)*(2 + sqrt(5)). - Peter Bala, Dec 23 2012
a(n) = (1/32)*(F(6*n + 3) - F(6*n - 3)).
Sum_{n>=1} 1/(4*a(n) + 1/(4*a(n))) = 1/4. Compare with A001906 and A049670. - Peter Bala, Nov 29 2013
From Peter Bala, Apr 02 2015: (Start)
Sum_{n >= 1} a(n)*x^(2*n) = -G(x)*G(-x), where G(x) = Sum_{n >= 1} A001076(n)*x^n.
1 + 4*Sum_{n >= 1} a(n)*x^(2*n) = (1 + F(x))*(1 + F(-x)) = (1 + 2*x*G(x))*(1 - 2*x*G(-x)), where F(x) = Sum_{n >= 1} Fibonacci(3*n + 3)*x^n.
1 + 7*Sum_{n >= 1} a(n)*x^(2*n) = (1 + G(x))*(1 + G(-x)) = (1 + 7*G(x))*(1 + 7*G(-x)).
1 + 12*Sum_{n >= 1} a(n)*x^(2*n) = (1 + 2*G(x))*(1 + 2*G(-x)) = (1 + 6*G(x))*(1 + 6*G(-x)) = (1 + A(x))*(1 + A(-x)), where A(x) = Sum_{n >= 1} Fibonacci(3*n)*x^n is the o.g.f for A014445.
1 + 15*Sum_{n >= 1} a(n)*x^(2*n) = (1 + 5*G(x))*(1 + 5*G(-x)) = (1 + 3*G(x))*(1 + 3*G(-x)) = H(x)*H(-x), where H(x) = Sum_{n >= 0} A155179(n)*x^n.
1 + 16*Sum_{n >= 1} a(n)*x^(2*n) = (1 + 4*G(x))*(1 + 4*G(-x)) = (1 + 2* Sum_{n >= 1} Fibonacci(3*n - 1)*x^n)*(1 + 2* Sum_{n >= 1} Fibonacci(3*n - 1)*(-x)^n) = (1 + 2* Sum_{n >= 1} Fibonacci(3*n + 1)*x^n)*(1 + 2* Sum_{n >= 1} Fibonacci(3*n + 1)*(-x)^n).
1 + 20*Sum_{n >= 1} a(n)*x^(2*n) = (1 + Sum_{n >= 1} Lucas(3*n)*x^n)*(1 + Sum_{n >= 1} Lucas(3*n)*(-x)^n).
1 - 5*Sum_{n >= 1} a(n)*x^(2*n) = (1 + Sum_{n >= 1} A001077(n+1)*x^n)*(1 + Sum_{n >= 1} A001077(n+1)*(-x)^n).
1 - 9*Sum_{n >= 1} a(n)*x^(2*n) = (1 - G(x))*(1 - G(-x)) = (1 + 9*G(x))*(1 + 9*G(-x)).
1 - 16*Sum_{n >= 1} a(n)*x^(2*n) = (1 + 2*Sum_{n >= 1} A099843(n)*x^n)*(1 + 2*Sum_{n >= 1} A099843(n)*(-x)^n).
1 - 20*Sum_{n >= 1} a(n)*x^(2*n) = (1 - 2*G(x))*(1 - 2*G(-x)) = (1 + 10*G(x))*(1 + 10*G(-x)).
(End)

Chebyshev and other comments from Wolfdieter Lang, Nov 08 2002

A014448 Even Lucas numbers: L(3n).

Original entry on oeis.org

2, 4, 18, 76, 322, 1364, 5778, 24476, 103682, 439204, 1860498, 7881196, 33385282, 141422324, 599074578, 2537720636, 10749957122, 45537549124, 192900153618, 817138163596, 3461452808002, 14662949395604, 62113250390418

Offset: 0

Mohammad K. Azarian

This is the Lucas sequence V(4,-1). - Bruno Berselli, Jan 08 2013

			a(4) = L(3 * 4) = L(12) = 322. - _Indranil Ghosh_, Feb 05 2017

Indranil Ghosh, Table of n, a(n) for n = 0..1591
Pooja Bhadouria, Deepika Jhala and Bijendra Singh, Binomial Transforms of the k-Lucas Sequences and its Properties, The Journal of Mathematics and Computer Science (JMCS), Volume 8, Issue 1 (2014), pp. 81-92; sequence L_{4,n}.
H. H. Ferns, Problem B-115, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 5, No. 2 (1967), p. 202; Identities for F_{kn} and L{kn}, Solution to Problem B-115 by Stanley Rabinowitz, ibid., Vol. 6, No. 1 (1968), pp. 92-93.
Tanya Khovanova, Recursive Sequences.
Michael Z. Spivey and Laura L. Steil, The k-Binomial Transforms and the Hankel Transform, Journal of Integer Sequences, Vol. 9 (2006), Article 06.1.1.
Roberto Tauraso, Some Congruences for Central Binomial Sums Involving Fibonacci and Lucas Numbers, JIS 19 (2016) # 16.5.4
Wikipedia, Lucas sequence: Specific names.
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2).
Index entries for linear recurrences with constant coefficients, signature (4,1).

Cf. A000032, A000045, A001077.

Cf. Lucas(k*n): A005248 (k = 2), A056854 (k = 4), A001946 (k = 5), A087215 (k = 6), A087281 (k = 7), A087265 (k = 8), A087287 (k = 9), A089772 (k = 11), A089775 (k = 12).

[Lucas(3*n) : n in [0..100]]; // Vincenzo Librandi, Apr 14 2011

Table[LucasL[3*n], {n,0,100}] (* G. C. Greubel, Nov 07 2018 *)

PARI
```
polsym(x^2-4*x-1,100)
```

a(n)=sum(k=0,n,binomial(n,k)*(fibonacci(n+k-1)+fibonacci(n+k+1))) \\ Paul D. Hanna, Oct 19 2010

[lucas_number2(n,4,-1) for n in range(0, 23)] # Zerinvary Lajos, May 14 2009

G.f.: (2-4*x)/(1-4*x-x^2).
a(n) = 4*a(n-1) +a(n-2) with n>1, a(0)=2, a(1)=4.
a(n) = (2+sqrt(5))^n + (2-sqrt(5))^n.
a(n) = 2*A001077(n).
a(n) = A000032(3*n).
a(n) = Sum_{k=0..n} C(n,k)*Lucas(n+k). - Paul D. Hanna, Oct 19 2010
a(n) = Fibonacci(6*n)/Fibonacci(3*n), n>0. - Gary Detlefs, Dec 26 2010
From Peter Bala, Mar 22 2015: (Start)
a(n) = ( Fibonacci(3*n + 2*k) - F(3*n - 2*k) )/Fibonacci(2*k) for nonzero integer k.
a(n) = ( Fibonacci(3*n + 2*k + 1) + F(3*n - 2*k - 1) )/Fibonacci(2*k + 1) for arbitrary integer k. (End)
a(n) = [x^n] ( (1 + 4*x + sqrt(1 + 8*x + 20*x^2))/2 )^n for n >= 1. - Peter Bala, Jun 23 2015
a(n) = L(n)*(L(n-1)*L(n+1) + 2*(-1)^n). - J. M. Bergot, Feb 05 2016
From Peter Bala Oct 14 2019: (Start)
Sum_{n >= 1} 1/( a(n) + (-1)^(n+1)*20/a(n) ) = 3/16.
Sum_{n >= 1} (-1)^(n+1)/( a(n) + (-1)^(n+1)*20/a(n) ) = 1/16. (End)
a(n) = (15*Fibonacci(n)^2*Lucas(n) + Lucas(n)^3)/4 (Ferns, 1967). - Amiram Eldar, Feb 06 2022
E.g.f.: 2*exp(2*x)*cosh(sqrt(5)*x). - Stefano Spezia, Jan 18 2025

More terms from Erich Friedman

A033887 a(n) = Fibonacci(3*n + 1).

Original entry on oeis.org

1, 3, 13, 55, 233, 987, 4181, 17711, 75025, 317811, 1346269, 5702887, 24157817, 102334155, 433494437, 1836311903, 7778742049, 32951280099, 139583862445, 591286729879, 2504730781961, 10610209857723, 44945570212853, 190392490709135, 806515533049393, 3416454622906707

Offset: 0

N. J. A. Sloane

Binomial transform of A063727, and second binomial transform of (1,1,5,5,25,25,...), which is A074872 with offset 0. - Paul Barry, Jul 16 2003
Equals INVERT transform of A104934: (1, 2, 8, 28, 100, 356, ...) and INVERTi transform of A005054: (1, 4, 20, 100, 500, ...). - Gary W. Adamson, Jul 22 2010
a(n) is the number of compositions of n when there are 3 types of 1 and 4 types of other natural numbers. - Milan Janjic, Aug 13 2010
F(3*n+1) = 3^n*a(n;2/3), where a(n;d), n = 0, 1, ..., d, denote the delta-Fibonacci numbers defined in comments to A000045 (see also the papers by Witula et al.). - Roman Witula, Jul 12 2012
We note that the remark above by Paul Barry can be easily obtained from the following scaling identity for delta-Fibonacci numbers y^n a(n;x/y) = Sum_{k=0..n} binomial(n,k) (y-1)^(n-k) a(k;x) and the fact that a(n;2)=5^floor(n/2). Indeed, for x=y=2 we get 2^n a(n;1) = Sum_{k=0..n} binomial(n,k) a(k;2) and, by A000045: Sum_{k=0..n} binomial(n,k) 2^k a(k;1) = Sum_{k=0..n} binomial(n,k) F(k+1) 2^k = 3^n a(n;2/3) = F(3n+1). - Roman Witula, Jul 12 2012
Except for the first term, this sequence can be generated by Corollary 1 (iv) of Azarian's paper in the references for this sequence. - Mohammad K. Azarian, Jul 02 2015
Number of 1’s in the substitution system {0 -> 110, 1 -> 11100} at step n from initial string "1" (1 -> 11100 -> 111001110011100110110 -> ...). - Ilya Gutkovskiy, Apr 10 2017
The o.g.f. of {F(m*n + 1)}A000045%20and%20L%20=%20A000032.%20-%20_Wolfdieter%20Lang">{n>=0}, for m = 1, 2, ..., is G(m,x) = (1 - F(m-1)*x) / (1 - L(m)*x + (-1)^m*x^2), with F = A000045 and L = A000032. - _Wolfdieter Lang, Feb 06 2023

			a(5) = Fibonacci(3*5 + 1) = Fibonacci(16) = 987. - _Indranil Ghosh_, Feb 04 2017

Indranil Ghosh, Table of n, a(n) for n = 0..1592
Mohammad K. Azarian, Fibonacci Identities as Binomial Sums, International Journal of Contemporary Mathematical Sciences, 7(38) (2012), 1871-1876.
Paul Barry and A. Hennessy, The Euler-Seidel Matrix, Hankel Matrices and Moment Sequences, J. Int. Seq. 13 (2010), #10.8.2, Example 13.
Gary Detlefs and Wolfdieter Lang, Improved Formula for the Multi-Section of the Linear Three-Term Recurrence Sequence, arXiv:2304.12937 [math.CO], 2023.
I. M. Gessel and Ji Li, Compositions and Fibonacci identities, J. Int. Seq. 16 (2013), #13.4.5.
Edyta Hetmaniok, Bożena Piątek, and Roman Wituła, Binomials Transformation Formulae of Scaled Fibonacci Numbers, Open Mathematics, 15(1) (2017), 477-485.
Tanya Khovanova, Recursive Sequences.
Roman Witula, Binomials transformation formulae of scaled Lucas numbers, Demonstratio Mathematica, 46(1) (2013), 15-27.
Roman Witula and Damian Slota, delta-Fibonacci numbers, Appl. Anal. Discr. Math 3 (2009), 310-329, MR2555042.
Index entries for linear recurrences with constant coefficients, signature (4,1).

Cf. A000032, A000045, A104934, A005054, A063727 (inverse binomial transform), A082761 (binomial transform), A001076, A001077.

Cf. A016777, A049310, A049651, A074872, A122070, A167808, A185384.

[Fibonacci(3*n+1): n in [0..100]]; // Vincenzo Librandi, Apr 17 2011

Fibonacci[Range[1,5!,3]] (* Vladimir Joseph Stephan Orlovsky, May 18 2010 *)

a(n)=fibonacci(3*n+1) \\ Charles R Greathouse IV, Feb 03 2014

Vec((1-x)/(1-4*x-x^2) + O(x^100)) \\ Altug Alkan, Dec 10 2015

a(n) = A001076(n) + A001077(n) = A001076(n+1) - A001076(n).
a(n) = 2*A049651(n) + 1.
a(n) = 4*a(n-1) + a(n-2) for n>1, a(0)=1, a(1)=3;
G.f.: (1 - x)/(1 - 4*x - x^2).
a(n) = ((1 + sqrt(5))*(2 + sqrt(5))^n - (1 - sqrt(5))*(2 - sqrt(5))^n)/(2*sqrt(5)).
a(n) = Sum_{k=0..n} Sum_{j=0..n-k} C(n,j)*C(n-j,k)*F(n-j+1). - Paul Barry, May 19 2006
First differences of A001076. - Al Hakanson (hawkuu(AT)gmail.com), May 02 2009
a(n) = A167808(3*n+1). - Reinhard Zumkeller, Nov 12 2009
a(n) = Sum_{k=0..n} C(n,k)*F(n+k+1). - Paul Barry, Apr 19 2010
Let p[1]=3, p[i]=4, (i>1), and A be a Hessenberg matrix of order n defined by: A[i,j] = p[j-i+1] (i <= j), A[i,j]=-1 (i = j+1), and A[i,j] = 0 otherwise. Then, for n >= 1, a(n) = det A. - Milan Janjic, Apr 29 2010
a(n) = Sum_{i=0..n} C(n,n-i)*A063727(i). - Seiichi Kirikami, Mar 06 2012
a(n) = Sum_{k=0..n} A122070(n,k) = Sum_{k=0..n} A185384(n,k). - Philippe Deléham, Mar 13 2012
a(n) = A000045(A016777(n)). - Michel Marcus, Dec 10 2015
a(n) = F(2*n)*L(n+1) + F(n-1)*(-1)^n for n > 0. - J. M. Bergot, Feb 09 2016
a(n) = Sum_{k=0..n} binomial(n,k)*5^floor(k/2)*2^(n-k). - Tony Foster III, Sep 03 2017
2*a(n) = Fibonacci(3*n) + Lucas(3*n). - Bruno Berselli, Oct 13 2017
a(n)^2 is the denominator of continued fraction [4,...,4, 2, 4,...,4], which has n 4's before, and n 4's after, the middle 2. - Greg Dresden and Hexuan Wang, Aug 30 2021
a(n) = i^n*(S(n, -4*i) + i*S(n-1, -4*i)), with i = sqrt(-1), and the Chebyshev S-polynomials (see A049310) with S(n, -1) = 0. From the simplified trisection formula. See the first entry above with A001076. - Gary Detlefs and Wolfdieter Lang, Mar 06 2023
E.g.f.: exp(2*x)*(5*cosh(sqrt(5)*x) + sqrt(5)*sinh(sqrt(5)*x))/5. - Stefano Spezia, May 24 2024

A084057 a(n) = 2a(n-1) + 4a(n-2), a(0)=1, a(1)=1.

Original entry on oeis.org

1, 1, 6, 16, 56, 176, 576, 1856, 6016, 19456, 62976, 203776, 659456, 2134016, 6905856, 22347776, 72318976, 234029056, 757334016, 2450784256, 7930904576, 25664946176, 83053510656, 268766806016, 869747654656, 2814562533376, 9108115685376, 29474481504256

Offset: 0

Paul Barry, May 10 2003

Inverse binomial transform of A001077. Binomial transform of expansion of cosh(sqrt(5)*x) (1,0,5,0,25,...).
The same sequence may be obtained by the following process. Starting a priori with the fraction 1/1, the numerators of fractions built according to the rule: add top and bottom to get the new bottom, add top and 5 times the bottom to get the new top. The limit of the sequence of fractions is sqrt(5). - Cino Hilliard, Sep 25 2005
Numerators of fractions in the approximation of the square root of 5 satisfying: a(n) = (a(n-1)+c)/(a(n-1)+1), with c=5 and a(1)=1. For denominators see A063727. - Mark Dols, Jul 24 2009
Equals right border of triangle A143969. (1, 6, 16, 56, ...) = row sums of triangle A143969 and INVERT transform of (1, 5, 5, 5, ...). - Gary W. Adamson, Sep 06 2008
a(n) is the number of compositions of n when there are 1 type of 1 and 5 types of other natural numbers. - Milan Janjic, Aug 13 2010
From Gary W. Adamson, Jul 30 2016: (Start)
The sequence is case N=1 in an infinite set obtained by taking powers of the 2 X 2 matrix M = [(1,5); (1,N)], then extracting the upper left terms. The infinite set begins:
N=1 (A084057):  1, 6, 16,  56,  176,   576,   1856, ...
N=2 (A108306):  1, 6, 21,  81,  306,  1161,   4401, ...
N=3 (A164549):  1, 6, 26, 116,  516,  2296,  10216, ...
N=4 (A015449):  1, 6, 31, 161,  836,  4341,  22541, ...
N=5 (A000400):  1, 6, 36, 216, 1296,  7776,  46656, ...
N=6 (A049685):  1, 6, 41, 281, 1926, 13201,  90481, ...
N=7 (.......):  1, 6, 46, 356, 2756, 21336, 222712, ...
...
Sequences in the above set can be obtained by taking INVERT transforms of the following:
N=1 INVERT transform of (1, 5,  5,  5,   5,    5, ...
N=2 ..."......"......". (1, 5, 10, 20,  40,   80, ...
N=3 ..."......"......". (1, 5, 15, 45, 135,  405, ...
N=4 ..."......"......". (1, 5, 20, 80, 320, 1280, ...
...
with the pattern (1, 5, N*5, (N^2)*5, (N^3)*5, ...
It appears that the sequence generated from powers (n>0) of the matrix P = [(1,a); (1,b)], (a,b > 0), then extracting the upper left terms, is equal to the INVERT transform of the sequence starting: (1, a, b*a, (b^2)*a, (b^3)*a, ...). (End)

John Derbyshire, Prime Obsession, Joseph Henry Press, April 2004, see p. 16.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,4).

Cf. A046717, A002533, A098158, A143969, A269992.
a(n) = A087131(n)/2.
The following sequences (and others) belong to the same family: A001333, A000129, A026150, A002605, A046717, A015518, A084057, A063727, A002533, A002532, A083098, A083099, A083100, A015519.
Cf. A108306, A164549, A015449, A000400, A049685

I:=[1,1]; [n le 2 select I[n] else 2*Self(n-1)+4*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jul 31 2016

f[n_] := Simplify[((1 + Sqrt[5])^n + (1 - Sqrt[5])^n)/2]; Array[f, 28, 0] (* Or *)
LinearRecurrence[{2, 4}, {1, 1}, 28] (* Robert G. Wilson v, Sep 18 2013 *)
RecurrenceTable[{a[1] == 1, a[2] == 1, a[n] == 2 a[n-1] + 4 a[n-2]}, a, {n, 30}] (* Vincenzo Librandi, Jul 31 2016 *)
Table[2^(n-1) LucasL[n], {n, 0, 20}] (* Vladimir Reshetnikov, Sep 19 2016 *)

lucas(n)=fibonacci(n-1)+fibonacci(n+1)
a(n)=lucas(n)/2*2^n \\ Charles R Greathouse IV, Sep 18 2013

from sage.combinat.sloane_functions import recur_gen2b; it = recur_gen2b(1,1,2,4, lambda n: 0); [next(it) for i in range(1,26)] # Zerinvary Lajos, Jul 09 2008

[lucas_number2(n,2,-4)/2 for n in range(0, 26)] # Zerinvary Lajos, Apr 30 2009

a(n) = ((1+sqrt(5))^n + (1-sqrt(5))^n)/2.
G.f.: (1-x) / (1-2*x-4*x^2).
E.g.f.: exp(x) * cosh(sqrt(5)*x).
a(2n+1) = 2*a(n)*a(n+1) - (-4)^n. - Mario Catalani (mario.catalani(AT)unito.it), Jun 13 2003
a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2*k)*5^k . - Paul Barry, Jul 25 2004
a(n) = Sum_{k=0..n} A098158(n,k)*5^(n-k). - Philippe Deléham, Dec 26 2007
a(n) = 2^(n-1)*A000032(n). - Mark Dols, Jul 24 2009
If p(1)=1, and p(i)=5 for i>1, and if A is the Hessenberg matrix of order n defined by: A(i,j) = p(j-i+1) for i<=j, A(i,j):=-1, (i=j+1), and A(i,j):=0 otherwise, then, for n>=1, a(n)=det A. - Milan Janjic, Apr 29 2010
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(5*k-1)/(x*(5*k+4) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 26 2013
a(n) = A063727(n) - A063272(n-1). - R. J. Mathar, Jun 06 2019
a(n) = 1 + 5*A014335(n). - R. J. Mathar, Jun 06 2019
Sum_{n>=1} 1/a(n) = A269992. - Amiram Eldar, Feb 01 2021

A023039 a(n) = 18*a(n-1) - a(n-2).

Original entry on oeis.org

1, 9, 161, 2889, 51841, 930249, 16692641, 299537289, 5374978561, 96450076809, 1730726404001, 31056625195209, 557288527109761, 10000136862780489, 179445175002939041, 3220013013190122249, 57780789062419261441

Offset: 0

David W. Wilson

The primitive Heronian triangle 3*a(n) +- 2, 4*a(n) has the latter side cut into 2*a(n) +- 3 by the corresponding altitude and has area 10*a(n)*A060645(n). - Lekraj Beedassy, Jun 25 2002
Chebyshev polynomials T(n,x) evaluated at x=9.
{a(n)} gives all (unsigned, integer) solutions of Pell equation a(n)^2 - 80*b(n)^2 = +1 with b(n) = A049660(n), n >= 0.
{a(n)} gives all possible solutions for x in Pell equation x^2 - D*y^2 = 1 for D=5, D=20 and D=80. The corresponding values for y are A060645 (D=5), A207832 (D=20) and A049660 (D=80). - Herbert Kociemba, Jun 05 2022
Also gives solutions to the equation x^2 - 1 = floor(x*r*floor(x/r)) where r=sqrt(5). - Benoit Cloitre, Feb 14 2004
Appears to give all solutions > 1 to the equation: x^2 = ceiling(x*r*floor(x/r)) where r=sqrt(5). - Benoit Cloitre, Feb 24 2004
For all terms x of the sequence, 5*x^2 - 5 is a square, A004292(n)^2.
The a(n) are the x-values in the nonnegative integer solutions of x^2 - 5y^2 = 1, see A060645(n) for the corresponding y-values. - Sture Sjöstedt, Nov 29 2011
Rightmost digits alternate repeatedly: 1 and 9 in fact, a(2) = 18*9 - 1 == 1 (mod 10); a(3) = 18*1 - 9 == 9 (mod 10) therefore a(2n) == 1 (mod 10), a(2n+1) == 9 (mod 10). - Carmine Suriano, Oct 03 2013

			G.f. = 1 + 9*x + 161*x^2 + 2889*x^3 + 51841*x4 + 930249*x^5 + 16692641*x^6 + ...

Seiichi Manyama, Table of n, a(n) for n = 0..750 (terms 0..200 from Vincenzo Librandi)
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Tanya Khovanova, Recursive Sequences
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (18,-1).

Cf. A001077, A115032.
Row 2 of array A188645.
Row 4 of A322790.

I:=[1, 9]; [n le 2 select I[n] else 18*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Feb 13 2012

a := n -> hypergeom([n, -n], [1/2], -4):
seq(simplify(a(n)), n=0..16); # Peter Luschny, Jul 26 2020

LinearRecurrence[{18, -1}, {1, 9}, 50] (* Sture Sjöstedt, Nov 29 2011 *)
CoefficientList[Series[(1-9*x)/(1-18*x+x^2), {x, 0, 50}], x] (* G. C. Greubel, Dec 19 2017 *)

{a(n) = fibonacci(6*n) / 2 + fibonacci(6*n - 1)}; /* Michael Somos, Aug 11 2009 */

x='x+O('x^30); Vec((1-9*x)/(1-18*x+x^2)) \\ G. C. Greubel, Dec 19 2017

a(n) ~ (1/2)*(sqrt(5) + 2)^(2*n). - Joe Keane (jgk(AT)jgk.org), May 15 2002
Limit_{n->infinity} a(n)/a(n-1) = phi^6 = 9 + 4*sqrt(5). - Gregory V. Richardson, Oct 13 2002
a(n) = T(n, 9) = (S(n, 18) - S(n-2, 18))/2, with S(n, x) := U(n, x/2) and T(n, x), resp. U(n, x), are Chebyshev's polynomials of the first, resp. second, kind. See A053120 and A049310. S(-2, x) := -1, S(-1, x) := 0, S(n, 18)=A049660(n+1).
a(n) = sqrt(80*A049660(n)^2 + 1) (cf. Richardson comment).
a(n) = ((9 + 4*sqrt(5))^n + (9 - 4*sqrt(5))^n)/2.
G.f.: (1 - 9*x)/(1 - 18*x + x^2).
a(n) = cosh(2*n*arcsinh(2)). - Herbert Kociemba, Apr 24 2008
a(n) = A001077(2*n). - Michael Somos, Aug 11 2009
From Johannes W. Meijer, Jul 01 2010: (Start)
a(n) = 2*A167808(6*n+1) - A167808(6*n+3).
Limit_{k->infinity} a(n+k)/a(k) = a(n) + A060645(n)*sqrt(5).
Limit_{n->infinity} a(n)/A060645(n) = sqrt(5).
(End)
a(n) = (1/2)*A087215(n) = (1/2)*(sqrt(5) + 2)^(2*n) + (1/2)*(sqrt(5) - 2)^(2*n).
Sum_{n >= 1} 1/( a(n) - 5/a(n) ) = 1/8. Compare with A005248, A002878 and A075796. - Peter Bala, Nov 29 2013
a(n) = 2*A115032(n-1) - 1 = S(n, 18) - 9*S(n-1, 18), with A115032(-1) = 1, and see the above formula with S(n, 18) using its recurrence. - Wolfdieter Lang, Aug 22 2014
a(n) = A128052(3n). - A.H.M. Smeets, Oct 02 2017
a(n) = A049660(n+1) - 9*A049660(n). - R. J. Mathar, May 24 2018
a(n) = hypergeom([n, -n], [1/2], -4). - Peter Luschny, Jul 26 2020
a(n) = L(6*n)/2 for L(n) the Lucas sequence A000032(n). - Greg Dresden, Dec 07 2021
a(n) = cosh(6*n*arccsch(2)). - Peter Luschny, May 25 2022

Chebyshev and Pell comments from Wolfdieter Lang, Nov 08 2002

Sture Sjöstedt's comment corrected and reformulated by Wolfdieter Lang, Aug 24 2014

A049629 a(n) = (F(6n+5) - F(6n+1))/4 = (F(6n+4) + F(6n+2))/4, where F = A000045.

Original entry on oeis.org

1, 19, 341, 6119, 109801, 1970299, 35355581, 634430159, 11384387281, 204284540899, 3665737348901, 65778987739319, 1180356041958841, 21180629767519819, 380070979773397901, 6820097006153642399, 122381675130992165281, 2196050055351705332659, 39406519321199703822581

Offset: 0

Clark Kimberling

x(n) := 2*a(n) and y(n) := A007805(n), n >= 0, give all the positive solutions of the Pell equation x^2 - 5*y^2 = -1.
The Gregory V. Richardson formula follows from this. - Wolfdieter Lang, Jun 20 2013
From Peter Bala, Mar 23 2018: (Start)
Define a binary operation o on the real numbers by x o y = x*sqrt(1 + y^2) + y*sqrt(1 + x^2). The operation o is commutative and associative with identity 0. Then we have
2*a(n) = 2 o 2 o ... o 2 (2*n+1 terms). For example, 2 o 2 = 4*sqrt(5) and 2 o 2 o 2 = 2 o 4*sqrt(5) = 38 = 2*a(1). Cf. A084068.
a(n) = U(2*n+1) where U(n) is the Lehmer sequence [Lehmer, 1930] defined by the recurrence U(n) = sqrt(20)*U(n-1) - U(n-2) with U(0) = 0 and U(1) = 1. The solution to the recurrence is U(n) = (1/4)*( (sqrt(5) + 2)^n - (sqrt(5) - 2)^n ). (End)

			Pell, n=1: (2*19)^2 - 5*17^2 = -1.

G. C. Greubel, Table of n, a(n) for n = 0..795
K. Andersen, L. Carbone, and D. Penta, Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields, Journal of Number Theory and Combinatorics, Vol 2, No. 3 pp 245-278, 2011. See Section 9.
Tanya Khovanova, Recursive Sequences.
D. H. Lehmer, An extended theory of Lucas' functions, Annals of Mathematics, Second Series, Vol. 31, No. 3 (Jul., 1930), pp. 419-448.
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Hyperbola, Forum Geometricorum (2019) Vol. 19, 11-16.
Eric Weisstein's World of Mathematics, Lehmer Number.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
Index entries for sequences related to Chebyshev polynomials..
Index entries for linear recurrences with constant coefficients, signature (18,-1).

Bisection of A001077 divided by 2.
Cf. A000045, A007805, A049310, A049660, A100047, A188378.
Cf. A005013, A033890, A049685, A084608.
Cf. similar sequences of the type (1/k)*sinh((2*n+1)*arcsinh(k)) listed in A097775.

[(Fibonacci(6*n+5) - Fibonacci(6*n+1))/4: n in [0..30]]; // G. C. Greubel, Dec 15 2017

with(numtheory): with(combinat):
seq((fibonacci(6*n+5)-fibonacci(6*n+1))/4,n=0..20); # Muniru A Asiru, Mar 25 2018

a[n_] := Simplify[(2 + Sqrt@5)^(2 n - 1) + (2 - Sqrt@5)^(2 n - 1)]/4; Array[a, 16] (* Robert G. Wilson v, Oct 28 2010 *)

my(x='x+O('x^30)); Vec((1+x)/(1 - 18*x + x^2)) \\ G. C. Greubel, Dec 15 2017

a(n) ~ (1/4)*(sqrt(5) + 2)^(2*n+1). - Joe Keane (jgk(AT)jgk.org), May 15 2002
For all members x of the sequence, 20*x^2 + 5 is a square. Lim_{n -> oo} a(n)/a(n-1) = 9 + 2*sqrt(20) = 9 + 4*sqrt(5). The 20 can be seen to derive from the statement "20*x^2 + 5 is a square". - Gregory V. Richardson, Oct 12 2002
a(n) = (((9 + 4*sqrt(5))^(n+1) - (9 - 4*sqrt(5))^(n+1)) + ((9 + 4*sqrt(5))^n - (9 - 4*sqrt(5))^n)) / (8*sqrt(5)). - Gregory V. Richardson, Oct 12 2002
From R. J. Mathar, Nov 04 2008: (Start)
G.f.: (1+x)/(1 - 18x + x^2).
a(n) = A049660(n) + A049660(n+1). (End)
a(n) = 18*a(n-1) - a(n-2) for n>1; a(0)=1, a(1)=19. - Philippe Deléham, Nov 17 2008
a(n) = S(n,18) + S(n-1,18) with the Chebyshev S-polynomials (A049310). - Wolfdieter Lang, Jun 20 2013
From Peter Bala, Mar 23 2015: (Start)
a(n) = ( Fibonacci(6*n + 6 - 2*k) + Fibonacci(6*n + 2*k) )/( Fibonacci(6 - 2*k) + Fibonacci(2*k) ), for k an arbitrary integer.
a(n) = ( Fibonacci(6*n + 6 - 2*k - 1) - Fibonacci(6*n + 2*k + 1) )/( Fibonacci(6 - 2*k - 1) - Fibonacci(2*k + 1) ), for k an arbitrary integer, k != 1.
The aerated sequence (b(n))n>=1 = [1, 0, 19, 0, 341, 0, 6119, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -16, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for the connection with Chebyshev polynomials. (End)
a(n) = (A188378(n)^3 + (A188378(n)-2)^3) / 8. - Altug Alkan, Jan 24 2016
a(n) = sqrt(5 * Fibonacci(3 + 6*n)^2 - 4)/4. - Gerry Martens, Jul 25 2016
a(n) = Lucas(6*n + 3)/4. - Ehren Metcalfe, Feb 18 2017
From Peter Bala, Mar 23 2018: (Start)
a(n) = 1/4*( (sqrt(5) + 2)^(2*n+1) - (sqrt(5) - 2)^(2*n+1) ).
a(n) = 9*a(n-1) + 2*sqrt(5 + 20*a(n-1)^2).
a(n) = (1/2)*sinh((2*n + 1)*arcsinh(2)). (End)
From Peter Bala, May 09 2025: (Start)
a(n)^2 - 18*a(n)*a(n+1) + a(n+1)^2 = 20.
More generally, for real x, a(n+x)^2 - 18*a(n+x)*a(n+x+1) + a(n+x+1)^2 = 20 with a(n) := (((9 + 4*sqrt(5))^(n+1) - (9 - 4*sqrt(5))^(n+1)) + ((9 + 4*sqrt(5))^n - (9 - 4*sqrt(5))^n)) / (8*sqrt(5)) as given above.
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/20 (telescoping series).
Product_{n >= 1} ((a(n) + 1)/(a(n) - 1)) = sqrt(5)/2 (telescoping product). (End)

A040002 Continued fraction for sqrt(5).

Original entry on oeis.org

2, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4

Offset: 0

N. J. A. Sloane

Decimal expansion of 11/45. - Natan Arie Consigli, Jan 19 2016

			2.236067977499789696409173668... = 2 + 1/(4 + 1/(4 + 1/(4 + 1/(4 + ...)))). - _Harry J. Smith_, Jun 01 2009

John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 186.
James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 276.

Harry J. Smith, Table of n, a(n) for n = 0..20000
G. Xiao, Contfrac
Index entries for continued fractions for constants
Index of eventually constant sequences
Index of divisibility sequences
Index entries for linear recurrences with constant coefficients, signature (1).

Cf. A002163 (decimal expansion), A001077/A001076 (convergents), A248235 (Egyptian fraction).
Cf. Continued fraction for sqrt(a^2+1) = (a, 2a, 2a, 2a....): A040000 (contfrac(sqrt(2)) = (1,2,2,...)), A040002, A040006, A040012, A040020, A040030, A040042, A040056, A040072, A040090, A040110 (contfrac(sqrt(122)) = (11,22,22,...)), A040132, A040156, A040182, A040210, A040240, A040272, A040306, A040342, A040380, A040420 (contfrac(sqrt(442)) = (21,42,42,...)), A040462, A040506, A040552, A040600, A040650, A040702, A040756, A040812, A040870, A040930 (contfrac(sqrt(962)) = (31,62,62,...)).
Essentially the same as A010709.

Digits := 100: convert(evalf(sqrt(N)),confrac,90,'cvgts'):

ContinuedFraction[Sqrt[5],300] (* Vladimir Joseph Stephan Orlovsky, Mar 04 2011 *)
PadRight[{2},120,{4}] (* Harvey P. Dale, Jul 06 2019 *)

{ allocatemem(932245000); default(realprecision, 26000); x=contfrac(sqrt(5)); for (n=0, 20000, write("b040002.txt", n, " ", x[n+1])); } \\ Harry J. Smith, Jun 01 2009

a(0) = 2, a(n) = 4 n>0. - Natan Arie Consigli, Jan 19 2016
From Elmo R. Oliveira, Feb 16 2024: (Start)
G.f.: 2*(1+x)/(1-x).
E.g.f.: 4*exp(x) - 2.
a(n) = 2*A040000(n). (End)

A052924 Expansion of g.f.: (1-x)/(1 - 3*x - x^2).

Original entry on oeis.org

1, 2, 7, 23, 76, 251, 829, 2738, 9043, 29867, 98644, 325799, 1076041, 3553922, 11737807, 38767343, 128039836, 422886851, 1396700389, 4612988018, 15235664443, 50319981347, 166195608484, 548906806799, 1812916028881

Offset: 0

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

Euler encountered this sequence when finding the largest root of z^2 - 3z - 1 = 0. - V. Frederick Rickey (fred-rickey(AT)usma.edu), Aug 20 2003
Let M = a triangle with the Pell series A000129 (1, 2, 5, 12, ...) in each column, with the leftmost column shifted upwards one row. A052924 starting (1, 2, 7, 23, ...) = lim_{n->infinity} M^n, the left-shifted vector considered as a sequence. - Gary W. Adamson, Jul 31 2010
a(n) is the number of compositions of n when there are 2 types of 1 and 3 types of other natural numbers. - Milan Janjic, Aug 13 2010
Equals partial sums of A108300 prefaced with a 1: (1, 1, 5, 16, 53, 175, 578, ...). - Gary W. Adamson, Feb 15 2012

L. Euler, Introductio in analysin infinitorum, 1748, section 338. English translation by John D. Blanton, Introduction to Analysis of the Infinite, 1988, Springer, p. 286.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Sergio Falcón, The k-Fibonacci difference sequences, Chaos, Solitons & Fractals, Volume 87, June 2016, Pages 153-157.
Sergio Falcón and Ángel Plaza, On the Fibonacci k-numbers, Chaos, Solitons & Fractals 2007; 32(5): 1615-24.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 909
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (3,1).

A108300 (first differences), A006190 (partial sums), A355981 (primes).

Cf. A000032, A001077, A078343, A164581, A179237, A180148, A329723.

a:=[1,2];; for n in [3..30] do a[n]:=3*a[n-1]+a[n-2]; od; a; # G. C. Greubel, Jun 09 2019

R:=PowerSeriesRing(Integers(), 30); Coefficients(R!( (1-x)/(1-3*x-x^2) )); // G. C. Greubel, Jun 09 2019

spec:= [S,{S=Sequence(Prod(Sequence(Z),Union(Z,Z,Prod(Z,Z))))}, unlabeled]: seq(combstruct[count](spec,size=n), n=0..30);
seq(coeff(series((1-x)/(1-3*x-x^2), x, n+1), x, n), n = 0..30); # G. C. Greubel, Oct 16 2019

CoefficientList[Series[(1-x)/(1-3*x-x^2), {x,0,30}], x] (* G. C. Greubel, Jun 09 2019 *)

Vec((1-x)/(1-3*x-x^2)+O(x^30)) \\ Charles R Greathouse IV, Nov 20 2011

((1-x)/(1-3*x-x^2)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Jun 09 2019

a(n) = 3*a(n-1) + a(n-2).
a(n) = Sum_{alpha=RootOf(-1+3*x+x^2)} (1/13)*(1+5*alpha)*alpha^(-1-n).
With offset 1: a(1)=1; for n > 1, a(n) = Sum_{i=1..3*n-4} a(ceiling(i/3)). - Benoit Cloitre, Jan 04 2004
Binomial transform of A006130. a(n) = (1/2 - sqrt(13)/26)*(3/2 - sqrt(13)/2)^n + (1/2 + sqrt(13)/26)*(3/2 + sqrt(13)/2)^n. - Paul Barry, Jul 20 2004
From Creighton Dement, Nov 04 2004: (Start)
a(n) = A006190(n+1) - A006190(n);
4*a(n) = 9*A006190(n+1) - A006497(n+1) - 2*A003688(n+1). (End)
Numerators in continued fraction [1, 2, 3, 3, 3, ...], where the latter = 0.69722436226...; the length of an inradius of a right triangle with legs 2 and 3. - Gary W. Adamson, Dec 19 2007
If p[1]=2, p[i]=3, (i>1), and if A is Hessenberg matrix of order n defined by: A[i,j] = p[j-i+1], (i<=j), A[i,j] = -1, (i=j+1), and A[i,j]=0 otherwise. Then, for n >= 1, a(n-1) = det A. - Milan Janjic, Apr 29 2010
a(n) = A006190(n) + A003688(n). - R. J. Mathar, Jul 06 2012
a(n) = Sum_{k=0..n-2} A168561(n-2,k)*3^k + 2 * Sum_{k=0..n-1} A168561(n-1,k)*3^k, n>0. - R. J. Mathar, Feb 14 2024
From Peter Bala, Jul 08 2025: (Start)
The following series telescope:
Sum_{n >= 1} 1/(a(n) + 3*(-1)^(n+1)/a(n)) = 1/2, since 1/(a(n) + 3*(-1)^(n+1)/a(n)) = b(n) - b(n+1), where b(n) = (1/3) * (a(n) + a(n-1)) / (a(n)*a(n-1)).
Sum_{n >= 1} (-1)^(n+1)/(a(n) + 3*(-1)^(n+1)/a(n)) = 1/6, since 1/(a(n) + 3*(-1)^(n+1)/a(n)) = c(n) + c(n+1), where c(n) = (1/3) * (a(n) - a(n-1)) / (a(n)*a(n-1)). (End)

More terms from James Sellers, Jun 06 2000

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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A001076 Denominators of continued fraction convergents to sqrt(5).

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A049660 a(n) = Fibonacci(6*n)/8.

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A014448 Even Lucas numbers: L(3n).

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A084057 a(n) = 2a(n-1) + 4a(n-2), a(0)=1, a(1)=1.

A049629 a(n) = (F(6n+5) - F(6n+1))/4 = (F(6n+4) + F(6n+2))/4, where F = A000045.