A001254 -id:A001254 - cp's OEIS frontend

A106729 Sum of two consecutive squares of Lucas numbers (A001254).

5, 10, 25, 65, 170, 445, 1165, 3050, 7985, 20905, 54730, 143285, 375125, 982090, 2571145, 6731345, 17622890, 46137325, 120789085, 316229930, 827900705, 2167472185, 5674515850, 14856075365, 38893710245, 101825055370, 266581455865

Offset: 0

Lekraj Beedassy, May 14 2005

Positive values of x (or y) satisfying x^2 - 3xy + y^2 + 25 = 0. - Colin Barker, Feb 08 2014
Positive values of x (or y) satisfying x^2 - 7xy + y^2 + 225 = 0. - Colin Barker, Feb 09 2014
Positive values of x (or y) satisfying x^2 - 18xy + y^2 + 1600 = 0. - Colin Barker, Feb 26 2014

Bruno Berselli, Table of n, a(n) for n = 0..300
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (3,-1).

Cf. A000032, A000045, A000204, A001254, A081071, A104457.

[Fibonacci(n-2)^2+Fibonacci(n+3)^2: n in [0..30]]; // Vincenzo Librandi, Jul 09 2011

seq(combinat:-fibonacci(n-2)^2 + combinat:-fibonacci(n+3)^2, n=0..100); # Robert Israel, Nov 23 2014

Table[LucasL[n]^2 + LucasL[n+1]^2, {n, 0, 30}] (* Wesley Ivan Hurt, Nov 23 2014 *)
Total/@Partition[LucasL[Range[0,30]]^2,2,1] (* Harvey P. Dale, Jun 26 2022 *)

a(n) = fibonacci(n-2)^2 + fibonacci(n+3)^2;
vector(30, n, a(n-1)) \\ G. C. Greubel, Dec 17 2017

[fibonacci(n-2)^2 + fibonacci(n+3)^2 for n in (0..30)] # G. C. Greubel, Sep 10 2021

a(n) = Lucas(n)^2 + Lucas(n+1)^2 = 5*(Fibonacci(n)^2 + Fibonacci(n+1)^2) = 5*A001519(n+1).
a(n) = 3*a(n-1) - a(n-2). - T. D. Noe, Dec 11 2006
G.f.: 5*(1-x)/(1-3*x+x^2). - Philippe Deléham, Nov 16 2008
a(n) = Fibonacci(n-2)^2 + Fibonacci(n+3)^2. - Gary Detlefs, Dec 28 2010
a(n) = [1,1; 1,2]^(n-2).{3,4}.{3,4}, for n>=3. - John M. Campbell, Jul 09 2011
a(n) = Lucas(2n) + Lucas(2n+2). - Richard R. Forberg, Nov 23 2014
From Robert Israel, Nov 23 2014: (Start)
a(n) = 5*A000045(2*n+1).
E.g.f.: (5+sqrt(5))/2 * exp((3+sqrt(5))*x/2) + (5-sqrt(5))/2 * exp((3-sqrt(5))*x/2). (End)
From Enrique Navarrete, Mar 24 2025: (Start)
a(n)^2 = 20 + 5*A081071(n).
Limit_{n->oo} a(n+1)/a(n) = (3 + sqrt(5))/2 (see A104457). (End)

Corrected by T. D. Noe, Dec 11 2006

More terms from Bruno Berselli, Jul 17 2011

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

Original entry on oeis.org

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)

A007598 Squared Fibonacci numbers: a(n) = F(n)^2 where F = A000045.

Original entry on oeis.org

0, 1, 1, 4, 9, 25, 64, 169, 441, 1156, 3025, 7921, 20736, 54289, 142129, 372100, 974169, 2550409, 6677056, 17480761, 45765225, 119814916, 313679521, 821223649, 2149991424, 5628750625, 14736260449, 38580030724, 101003831721, 264431464441, 692290561600

Offset: 0

N. J. A. Sloane, Robert G. Wilson v

a(n)*(-1)^(n+1) = (2*(1-T(n,-3/2))/5), n>=0, with Chebyshev's polynomials T(n,x) of the first kind, is the r=-1 member of the r-family of sequences S_r(n) defined in A092184 where more information can be found. - Wolfdieter Lang, Oct 18 2004
From Giorgio Balzarotti, Mar 11 2009: (Start)
Determinant of power series with alternate signs of gamma matrix with determinant 1!.
a(n) = Determinant(A - A^2 + A^3 - A^4 + A^5 - ... - (-1)^n*A^n) where A is the submatrix A(1..2,1..2) of the matrix with factorial determinant.
A = [[1,1,1,1,1,1,...], [1,2,1,2,1,2,...], [1,2,3,1,2,3,...], [1,2,3,4,1,2,...], [1,2,3,4,5,1,...], [1,2,3,4,5,6,...], ...]; note: Determinant A(1..n,1..n) = (n-1)!.
a(n) is even with respect to signs of power of A.
See A158039...A158050 for sequence with matrix 2!, 3!, ... (End)
Equals the INVERT transform of (1, 3, 2, 2, 2, ...). Example: a(7) = 169 = (1, 1, 4, 9, 25, 64) dot (2, 2, 2, 2, 3, 1) = (2 + 2 + 8 + 18 + 75 + 64). - Gary W. Adamson, Apr 27 2009
This is a divisibility sequence.
a(n+1)*(-1)^n, n>=0, is the sequence of the alternating row sums of the Riordan triangle A158454. - Wolfdieter Lang, Dec 18 2010
a(n+1) is the number of tilings of a 2 X 2n rectangle with n tetrominoes of any shape, cf. A230031. - Alois P. Heinz, Nov 29 2013
This is the case P1 = 1, P2 = -6, Q = 1 of the 3 parameter family of 4th-order linear divisibility sequences found by Williams and Guy. - Peter Bala, Mar 31 2014
Differences between successive golden rectangle numbers A001654. - Jonathan Sondow, Nov 05 2015
a(n+1) is the number of 2 X n matrices that can be obtained from a 2 X n matrix by moving each element to an adjacent position, horizontally or vertically. This is because F(n+1) is the number of domino tilings of that matrix, therefore with a checkerboard coloring and two domino tilings we can move the black element of each domino of the first tiling to the white element of the same domino and similarly move the white element of each domino of the second tiling to the black element of the same domino. - Fabio Visonà, May 04 2022
In general, squaring the terms of a second-order linear recurrence with signature (c,d) will result in a third-order linear recurrence with signature (c^2+d,(c^2+d)*d,-d^3). - Gary Detlefs, Jan 05 2023

			G.f. = x + x^2 + 4*x^3 + 9*x^4 + 25*x^5 + 64*x^6 + 169*x^7 + 441*x^8 + ...

Arthur T. Benjamin and Jennifer J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 8.
Ross Honsberger, Mathematical Gems III, M.A.A., 1985, p. 130.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Richard P. Stanley, Enumerative Combinatorics I, Example 4.7.14, p. 251.

Indranil Ghosh, Table of n, a(n) for n = 0..2389 (terms 0..200 from T. D. Noe)
Mohammad K. Azarian, Fibonacci Identities as Binomial Sums, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 38, 2012, pp. 1871-1876.
Mohammad K. Azarian, Fibonacci Identities as Binomial Sums II, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 42, 2012, pp. 2053-2059.
R. C. Alperin, A family of nonlinear recurrences and their linear solutions, Fib. Q., 57:4 (2019), 318-321.
Paul S. Bruckman, Problem B-1023: And a cubic as a sum of two squares, Fibonacci Quarterly, Vol. 45, Number 2; May 2007; p. 186.
Andrej Dujella, A bijective proof of Riordan's theorem on powers of Fibonacci numbers, Discrete Math. 199 (1999), no. 1-3, 217--220. MR1675924 (99k:05016).
Sergio Falcon, Some series of reciprocal k-Fibonacci numbers, Asian Journal of Mathematics and Computer Research, Vol. 11, No. 3 (2016), pp. 184-191; ResearchGate link.
Dominique Foata and Guo-Niu Han, Nombres de Fibonacci et polynômes orthogonaux, in: Marcello Morelli and Marco Tangheroni (eds.), Leonardo Fibonacci : il tempo, le opere, l'ereditá scientifica, Pisa, 23-25 Marzo 1994, Pisa, Pacini Editore, 1994, pp. 179-200.
Svenja Huntemann and Neil A. McKay, Counting Domineering Positions, arXiv:1909.12419 [math.CO], 2019.
Jong Hyun Kim, Hadamard products and tilings, JIS 12 (2009) 09.7.4.
Toufik Mansour, A note on sum of k-th power of Horadam's sequence, arXiv:math/0302015 [math.CO], 2003.
Toufik Mansour, Squaring the terms of an l-th order linear recurrence, arXiv:math/0303138 [math.CO], 2003.
Hilary I. Okagbue, Muminu O. Adamu, Sheila A. Bishop, and Abiodun A. Opanuga, Digit and Iterative Digit Sum of Fibonacci numbers, their identities and powers, International Journal of Applied Engineering Research ISSN 0973-4562 Volume 11, Number 6 (2016) pp 4623-4627.
Pantelimon Stanica, Generating functions, weighted and non-weighted sums of powers of second-order recurrence sequences, arXiv:math/0010149 [math.CO], 2000.
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 linear recurrences with constant coefficients, signature (2,2,-1).
Index entries for two-way infinite sequences.
Index to divisibility sequences.

Cf. A000032, A000045, A001254, A001622, A001654, A047946, A056570, A059260, A061646, A065885, A079291, A080097, A105393.
Bisection of A006498 and A074677. First differences of A001654.
Second row of array A103323.
Half of A175395.

List([0..30], n -> Fibonacci(n)^2); # G. C. Greubel, Dec 10 2018

a007598 = (^ 2) . a000045  -- Reinhard Zumkeller, Sep 01 2013

[Fibonacci(n)^2: n in [0..30]]; // Vincenzo Librandi, Apr 14 2011

with(combinat): seq(fibonacci(n)^2, n=0..27); # Zerinvary Lajos, Sep 21 2007

f[n_] := Fibonacci[n]^2; Array[f, 4!, 0] (* Vladimir Joseph Stephan Orlovsky, Oct 25 2009 *)
LinearRecurrence[{2,2,-1},{0,1,1},41] (* Harvey P. Dale, May 18 2011 *)

PARI
```
{a(n) = fibonacci(n)^2};
```

concat(0, Vec(x*(1-x)/((1+x)*(1-3*x+x^2)) + O(x^30))) \\ Altug Alkan, Nov 06 2015

from sympy import fibonacci
def A007598(n): return fibonacci(n)**2 # Chai Wah Wu, Apr 14 2025

[(fibonacci(n))^2 for n in range(0, 28)]# Zerinvary Lajos, May 15 2009

[fibonacci(n)^2 for n in range(30)] # G. C. Greubel, Dec 10 2018

G.f.: x*(1-x)/((1+x)*(1-3*x+x^2)).
a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3), n > 2. a(0)=0, a(1)=1, a(2)=1.
a(-n) = a(n) for all n in Z.
a(n) = A080097(n-2) + 1.
L.g.f.: 1/5*log((1+3*x+x^2)/(1-6*x+x^2)) = Sum_{n>=0} a(n)/n*x^n; special case of l.g.f. given in A079291. - Joerg Arndt, Apr 13 2011
a(0) = 0, a(1) = 1; a(n) = a(n-1) + Sum(a(n-i)) + k, 0 <= i < n where k = 1 when n is odd, or k = -1 when n is even. E.g., a(2) = 1 = 1 + (1 + 1 + 0) - 1, a(3) = 4 = 1 + (1 + 1 + 0) + 1, a(4) = 9 = 4 + (4 + 1 + 1 + 0) - 1, a(5) = 25 = 9 + (9 + 4 + 1 + 1 + 0) + 1. - Sadrul Habib Chowdhury (adil040(AT)yahoo.com), Mar 02 2004
a(n) = (2*Fibonacci(2*n+1) - Fibonacci(2*n) - 2*(-1)^n)/5. - Ralf Stephan, May 14 2004
a(n) = F(n-1)*F(n+1) - (-1)^n = A059929(n-1) - A033999(n).
Sum_{j=0..2*n} binomial(2*n,j)*a(j) = 5^(n-1)*A005248(n+1) for n >= 1 [P. Stanica]. Sum_{j=0..2*n+1} binomial(2*n+1,j)*a(j) = 5^n*A001519(n+1) [P. Stanica]. - R. J. Mathar, Oct 16 2006
a(n) = (A005248(n) - 2*(-1)^n)/5. - R. J. Mathar, Sep 12 2010
a(n) = (-1)^k*(Fibonacci(n+k)^2-Fibonacci(k)*Fibonacci(2*n+k)), for any k. - Gary Detlefs, Dec 13 2010
a(n) = 3*a(n-1) - a(n-2) + 2*(-1)^(n+1), n > 1. - Gary Detlefs, Dec 20 2010
a(n) = Fibonacci(2*n-2) + a(n-2). - Gary Detlefs, Dec 20 2010
a(n) = (Fibonacci(3*n) - 3*(-1)^n*Fibonacci(n))/(5*Fibonacci(n)), n > 0. - Gary Detlefs, Dec 20 2010
a(n) = (Fibonacci(n)*Fibonacci(n+4) - 3*Fibonacci(n)*Fibonacci(n+1))/2. - Gary Detlefs, Jan 17 2011
a(n) = (((3+sqrt(5))/2)^n + ((3-sqrt(5))/2)^n - 2*(-1)^n)/5; without leading zero we would have a(n) = ((3+sqrt(5))*((3+sqrt(5))/2)^n + (3-sqrt(5))*((3-sqrt(5))/2)^n + 4*(-1)^n)/10. - Tim Monahan, Jul 17 2011
E.g.f.: (exp((phi+1)*x) + exp((2-phi)*x) - 2*exp(-x))/5, with the golden section phi:=(1+sqrt(5))/2. From the Binet-de Moivre formula for F(n). - Wolfdieter Lang, Jan 13 2012
Starting with "1" = triangle A059260 * the Fibonacci sequence as a vector. - Gary W. Adamson, Mar 06 2012
a(0) = 0, a(1) = 1; a(n+1) = (a(n)^(1/2) + a(n-1)^(1/2))^2. - Thomas Ordowski, Jan 06 2013
a(n) + a(n-1) = A001519(n), n > 0. - R. J. Mathar, Mar 19 2014
From Peter Bala, Mar 31 2014: (Start)
a(n) = ( T(n,alpha) - T(n,beta) )/(alpha - beta), where alpha = 3/2 and beta = -1 and T(n,x) denotes the Chebyshev polynomial of the first kind.
a(n) = the bottom left entry of the 2 X 2 matrix T(n, M), where M is the 2 X 2 matrix [0, 3/2; 1, 1/2].
a(n) = U(n-1,i/2)*U(n-1,-i/2), where U(n,x) denotes the Chebyshev polynomial of the second kind.
See the remarks in A100047 for the general connection between Chebyshev polynomials and 4th-order linear divisibility sequences. (End)
a(n) = (F(n+2)*F(n+3) - L(n)*L(n+1))/3 for F = A000045 and L = A000032. - J. M. Bergot, Jun 02 2014
0 = a(n)*(+a(n) - 2*a(n+1) - 2*a(n+2)) + a(n+1)*(+a(n+1) - 2*a(n+2)) + a(n+2)*(+a(n+2)) for all n in Z. - Michael Somos, Jun 03 2014
(F(n)*b(n+2))^2 + (F(n+1)*b(n-1))^2 = F(2*n+1)^3 = A001519(n+1)^3, with b(n) = a(n) + 2*(-1)^n and F(n) = A000045(n) (see Bruckman link). - Michel Marcus, Jan 24 2015
a(n) = 1/4*( a(n-2) - a(n-1) - a(n+1) + a(n+2) ). The same recurrence holds for A001254. - Peter Bala, Aug 18 2015
a(n) = F(n)*F(n+1) - F(n-1)*F(n). - Jonathan Sondow, Nov 05 2015
For n>2, a(n) = F(n-2)*(3*F(n-1) + F(n-3)) + F(2*n-5). Also, for n>2 a(n)=2*F(n-3)*F(n) + F(2*n-3) -(2)*(-1)^n. - J. M. Bergot, Nov 05 2015
a(n) = (F(n+2)^2 + L(n+1)^2) - 2*F(n+2)*L(n+1). - J. M. Bergot, Nov 08 2015
a(n) = F(n+3)^2 - 4*F(n+1)*F(n+2). - J. M. Bergot, Mar 17 2016
a(n) = (F(n-2)*F(n+2) + F(n-1)*F(n+1))/2. - J. M. Bergot, May 25 2017
4*a(n) = L(n+1)*L(n-1) - F(n+2)*F(n-2), where L = A000032. - Bruno Berselli, Sep 27 2017
a(n) = F(n+k)*F(n-k) + (-1)^(n+k)*a(k), for every integer k >= 0. - Federico Provvedi, Dec 10 2018
From Peter Bala, Nov 19 2019: (Start)
Sum_{n >= 3} 1/(a(n) - 1/a(n)) = 4/9.
Sum_{n >= 3} (-1)^n/(a(n) - 1/a(n)) =  (10 - 3*sqrt(5))/18.
Conjecture: Sum_{n >= 1, n != 2*k+1} 1/(a(n) + (-1)^n*a(2*k+1)) = 1/a(4*k+2) for k = 0,1,2,.... (End)
Sum_{n>=1} 1/a(n) = A105393. - Amiram Eldar, Oct 22 2020
Product_{n>=2} (1 + (-1)^n/a(n)) = phi (A001622) (Falcon, 2016, p. 189, eq. (3.1)). - Amiram Eldar, Dec 03 2024

A079291 Squares of Pell numbers.

Original entry on oeis.org

0, 1, 4, 25, 144, 841, 4900, 28561, 166464, 970225, 5654884, 32959081, 192099600, 1119638521, 6525731524, 38034750625, 221682772224, 1292061882721, 7530688524100, 43892069261881, 255821727047184, 1491038293021225

Offset: 0

Ralf Stephan, Feb 08 2003

(-1)^(n+1)*a(n) is the r=-4 member of the r-" of sequences S_r(n), n>=1, defined in A092184 where more information can be found.
Binomial transform of A086346. - Johannes W. Meijer, Aug 01 2010
In general, squaring the terms of a Horadam sequence with signature (c,d) will result in a third-order recurrence with signature (c^2+d, c^2*d+d^2, -d^3). - Gary Detlefs, Nov 11 2021
(Conjectured) For any primitive Pythagorean triple of the form (X, Y, Z=Y+1), it appears that Y or Z will always be (and only be) a square Pell number if X = A001333(n), for n > 1. If n is even, Y is always a square Pell number, and if n is odd, then Z is always a square Pell number. For example: (3, 4, 5), (7, 24, 25), (17, 144, 145), (41, 840, 841), (99, 4900, 4901). - Jules Beauchamp, Feb 02 2022
a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using (1/2,1/2)-fences, black half-squares (1/2 X 1 pieces, always placed so that the shorter sides are horizontal), and white half-squares. A (w,g)-fence is a tile composed of two w X 1 pieces separated by a gap of width g. a(n+1) also equals the number of tilings of an n-board using black (1/4,1/4)-fences, white (1/4,1/4)-fences, and (1/4,3/4)-fences. - Michael A. Allen, Dec 29 2022

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.
Joerg Arndt, Matters Computational (The Fxtbook), sect. 32.1.5, pp. 626-627
Sergio Falcon, Some series of reciprocal k-Fibonacci numbers, Asian Journal of Mathematics and Computer Research, Vol. 11, No. 3 (2016), pp. 184-191; ResearchGate link.
Toufik Mansour, A note on sum of k-th power of Horadam's sequence, arXiv:math/0302015 [math.CO], 2003.
Toufik Mansour, Squaring the terms of an ell-th order linear recurrence, arXiv:math/0303138 [math.CO], 2003.
Pantelimon Stanica, Generating functions, weighted and non-weighted sums for powers of second-order recurrence sequences, arXiv:math/0010149 [math.CO], 2000.
Index entries for linear recurrences with constant coefficients, signature (5,5,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A000129, A001254, A001109, A001541, A001653, A174968.

Cf. A002203, A007598, A084158 (partial sums), A086346, A092184.

I:=[0,1,4]; [n le 3 select I[n] else 5*Self(n-1)+ 5*Self(n-2) - Self(n-3): n in [1..31]]; // Vincenzo Librandi, May 17 2013

with(combinat):seq(fibonacci(i,2)^2, i=0..31); # Zerinvary Lajos, Mar 20 2008

CoefficientList[Series[x(1-x)/((1+x)*(1-6x+x^2)), {x, 0, 30}], x] (* Vincenzo Librandi, May 17 2013 *)
LinearRecurrence[{5,5,-1},{0,1,4},40] (* Harvey P. Dale, Dec 20 2015 *)
Fibonacci[Range[0, 30], 2]^2 (* G. C. Greubel, Sep 17 2021 *)

[lucas_number1(n, 2, -1)^2 for n in (0..30)] # G. C. Greubel, Sep 17 2021

G.f.: x*(1-x)/((1+x)*(1-6*x+x^2)).
a(n) = (r^n + (1/r)^n - 2*(-1)^n)/8, with r = 3 + sqrt(8).
a(n+3) = 5*a(n+2) + 5*a(n+1) - a(n).
L.g.f.: (1/8)*log((1+2*x+x^2)/(1-6*x+x^2)) = Sum_{n>=0} (a(n)/n)*x^n, see p. 627 of the Fxtbook link; special case of the following: let v(0)=0, v(1)=1, and v(n) = u*v(n-1) + v(n-2), then (1/A)*log((1+2*x+x^2)/(1-(2-A)*x+x^2)) = Sum_{n>=0} v(n)^2/n*x^n where A = u^2 + 4. - Joerg Arndt, Apr 08 2011
a(n+1) = Sum_{k=0..n} ( (-1)^(n-k)*A001653(k) ); e.g., 144 = -1 + 5 - 29 + 169; 25 = 1 - 5 + 29. - Charlie Marion, Jul 16 2003
a(n) = A000129(n)^2.
a(n) = (T(n, 3) - (-1)^n)/4 with Chebyshev's polynomials of the first kind evaluated at x=3: T(n, 3) = A001541(n) = ((3 + 2*sqrt(2))^n + (3 - 2*sqrt(2))^n )/2. - Wolfdieter Lang, Oct 18 2004
a(n) is the rightmost term of M^n * [1 0 0] where M is the 3 X 3 matrix [4 4 1 / 2 1 0 / 1 0 0]. a(n+1) = leftmost term. E.g., a(6) = 4900, a(5) = 841 since M^5 * [1 0 0] = [4900 2030 841]. - Gary W. Adamson, Oct 31 2004
a(n) = ( (-1)^(n+1) + A001109(n+1) - 3*A001109(n) )/4. - R. J. Mathar, Nov 16 2007
a(n) = ( (((1 - sqrt(2))^n + (1 + sqrt(2))^n) /2 )^2 + (-1)^(n+1) )/2. - Antonio Pane (apane1(AT)spc.edu), Dec 15 2007
Lim_{k -> infinity} ( a(n+k)/a(k) ) = A001541(n) + 2*A001109(n)*sqrt(2). - Johannes W. Meijer, Aug 01 2010
For n>0, a(2*n) = 6*a(2*n-1) - a(2*n-2) - 2, a(2*n+1) = 6*a(2*n) - a(2*n-1) + 2. - Charlie Marion, Sep 24 2011
a(n) = (1/8)*(A002203(2*n) - 2*(-1)^n). - G. C. Greubel, Sep 17 2021
Conjectured formula for (X, Y, Z) for primitive Pythagorean triple of the form (X, Y, Z=Y+1) is (A001333(n)^2, A079291(n)^2, A079291(n)^2-1) or (A001333(n)^2, A079291(n)^2-1, A079291(n)^2). As a closed formula (X, Y, Z) = (((1-sqrt(2))^n + (1+sqrt(2))^n)/2, (((1-sqrt(2))^n + (1+sqrt(2))^n)^2 - 4)/8, (((1-sqrt(2))^n + (1+sqrt(2))^n)^2 + 4)/8). - Jules Beauchamp, Feb 02 2022
From Michael A. Allen, Dec 29 2022: (Start)
a(n+1) = 6*a(n) - a(n-1) + 2*(-1)^n.
a(n+1) = (1 + (-1)^n)/2 + 4*Sum_{k=1..n} ( k*a(n+1-k) ). (End)
Product_{n>=2} (1 + (-1)^n/a(n)) = (1 + sqrt(2))/2 (A174968) (Falcon, 2016, p. 189, eq. (3.1)). - Amiram Eldar, Dec 03 2024

A203853 a(n) = (1/n) * Sum_{d|n} moebius(n/d) * Lucas(d)^2, where Lucas(n) = A000204(n).

Original entry on oeis.org

1, 4, 5, 10, 24, 50, 120, 270, 640, 1500, 3600, 8610, 20880, 50700, 124024, 304290, 750120, 1854400, 4600200, 11440548, 28527320, 71289000, 178526880, 447910470, 1125750120, 2833885800, 7144449920, 18036373140, 45591631800, 115381697740, 292329067800, 741410800830

Offset: 1

Paul D. Hanna, Jan 07 2012

nonn

Apparently the same as A032170, if n > 2. - R. J. Mathar, Jan 11 2012

			G.f.: F(x) = 1/((1-x-x^2) * (1-3*x^2+x^4)^4 * (1-4*x^3-x^6)^5 * (1-7*x^4+x^8)^10 * (1-11*x^5-x^10)^24 * (1-18*x^6+x^12)^50 * (1-29*x^7-x^14)^120 * ... * (1 - Lucas(n)*x^n + (-1)^n*x^(2*n))^a(n) * ...)
where F(x) = exp( Sum_{n>=1} Lucas(n)^3 * x^n/n ) = g.f. of A203803:
F(x) = 1 + x + 14*x^2 + 35*x^3 + 205*x^4 + 744*x^5 + 3414*x^6 + ...
where
log(F(x)) = x + 3^3*x^2/2 + 4^3*x^3/3 + 7^3*x^4/4 + 11^3*x^5/5 + 18^3*x^6/6 + 29^3*x^7/7 + 47^3*x^8/8 + ... + Lucas(n)^3*x^n/n + ...

Paul D. Hanna, Table of n, a(n) for n = 1..640

Cf. A001254, A203803, A006206, A203854, A203855, A203856, A203857, A203858, A203859, A203800.

a[n_]:= 1/n DivisorSum[n, MoebiusMu[n/#] LucasL[#]^2 &]; Array[a, 30] (* G. C. Greubel, Dec 25 2017 *)

{a(n)=if(n<1, 0, sumdiv(n, d, moebius(n/d)*(fibonacci(d-1)+fibonacci(d+1))^2)/n)}

{Lucas(n)=fibonacci(n-1)+fibonacci(n+1)}
{a(n)=local(F=exp(sum(m=1, n, Lucas(m)^3*x^m/m)+x*O(x^n)));if(n==1,1,polcoeff(F*prod(k=1,n-1,(1 - Lucas(k)*x^k + (-1)^k*x^(2*k) +x*O(x^n))^a(k)),n)/Lucas(n))}

G.f.: 1/Product_{n>=1} (1 - Lucas(n)*x^n + (-1)^n*x^(2*n))^a(n) = exp(Sum_{n>=1} Lucas(n)^3 * x^n/n), which is the g.f. of A203803.

a(n) ~ phi^(2*n) / n, where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Sep 02 2017

A081714 a(n) = F(n)*L(n+1) where F=Fibonacci and L=Lucas numbers.

Original entry on oeis.org

0, 3, 4, 14, 33, 90, 232, 611, 1596, 4182, 10945, 28658, 75024, 196419, 514228, 1346270, 3524577, 9227466, 24157816, 63245987, 165580140, 433494438, 1134903169, 2971215074, 7778742048, 20365011075, 53316291172, 139583862446, 365435296161, 956722026042

Offset: 0

Ralf Stephan, Apr 03 2003

Also convolution of Fibonacci and Lucas numbers.
For n>2, a(n) represents twice the area of the triangle created by the three points (L(n-3), L(n-2)), (L(n-1), L(n)) and (F(n+3), F(n+2)) where L(k)=A000032(k) and F(k)=A000045(k). - J. M. Bergot, May 20 2014
For n>1, a(n) is the remainder when F(n+3)*F(n+4) is divided by F(n+1)*F(n+2). - J. M. Bergot, May 24 2014

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A000045, A000204.

List([0..30], n -> Fibonacci(n)*(Fibonacci(n+2)+Fibonacci(n))); # G. C. Greubel, Jan 07 2019

[Fibonacci(n)*Lucas(n+1): n in [0..30]]; // Vincenzo Librandi, Sep 08 2012

with(combinat): F:=n-> fibonacci(n): L:= n-> F(n+1)+F(n-1):
a:= n-> F(n)*L(n+1): seq(a(n), n=0..30);

Fibonacci[Range[0,50]]*LucasL[Range[0,50]+1] (* Vladimir Joseph Stephan Orlovsky, Mar 17 2011*)

my(x='x+O('x^51));for(n=0,50,print1(polcoeff(serconvol(Ser((1+2*x)/(1-x-x*x)),Ser(x/(1-x-x*x))),n)", "))

a(n)=fibonacci(n)*(fibonacci(n+2)+fibonacci(n))

a(n) = round((-(-1)^n+(2^(-1-n)*((3-sqrt(5))^n*(-1+sqrt(5))+(1+sqrt(5))*(3+sqrt(5))^n))/sqrt(5))) \\ Colin Barker, Sep 28 2016

[fibonacci(n)*(fibonacci(n+2)+fibonacci(n)) for n in (0..30)] # G. C. Greubel, Jan 07 2019

G.f.: x*(3-2*x)/((1+x)*(1-3*x+x^2)).
a(n) = A122367(n) - (-1)^n. - R. J. Mathar, Jul 23 2010
a(n) = (L(n+1)^2 - F(2*n+2))/2 = ( A001254(n+1) - A001906(n+1) )/2. - Gary Detlefs, Nov 28 2010
a(n+1) = - A186679(2*n+1). - Reinhard Zumkeller, Feb 25 2011
a(n) = A035513(1,n-1)*A035513(2,n-1). - R. J. Mathar, Sep 04 2016
a(n)+a(n+1) = A005248(n+1). - R. J. Mathar, Sep 04 2016
a(n) = (-(-1)^n+(2^(-1-n)*((3-sqrt(5))^n*(-1+sqrt(5))+(1+sqrt(5))*(3+sqrt(5))^n)) / sqrt(5)). - Colin Barker, Sep 28 2016

Simpler definition from Michael Somos, Mar 16 2004

A103324 Square array T(n,k) read by antidiagonals: powers of Lucas numbers.

Original entry on oeis.org

2, 4, 1, 8, 1, 3, 16, 1, 9, 4, 32, 1, 27, 16, 7, 64, 1, 81, 64, 49, 11, 128, 1, 243, 256, 343, 121, 18, 256, 1, 729, 1024, 2401, 1331, 324, 29, 512, 1, 2187, 4096, 16807, 14641, 5832, 841, 47, 1024, 1, 6561, 16384, 117649, 161051, 104976, 24389, 2209, 76

Offset: 1

Ralf Stephan, Feb 03 2005

			2,1,3,4,7,11,18,
4,1,9,16,49,121,324,
8,1,27,64,343,1331,5832,
16,1,81,256,2401,14641,104976,
32,1,243,1024,16807,161051,1889568,
64,1,729,4096,117649,1771561,34012224,

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, identity 140.

Rows include A000032, A001254, A075155, A099923, A103325.

T(n, k) = A000032(k)^n, n>=1, k>=0.

T(n, k) = Sum[i_1>=0, Sum[i_2>=0, ... Sum[i_{k-1}>=0, 2^i_1*C(n, i_1)*C(n-i_1, i_2)*C(n-i_2, i_3)*...*C(n-i_{k-2}, i_{k-1}) ] ... ]].

A075155 Cubes of Lucas numbers.

Original entry on oeis.org

8, 1, 27, 64, 343, 1331, 5832, 24389, 103823, 438976, 1860867, 7880599, 33386248, 141420761, 599077107, 2537716544, 10749963743, 45537538411, 192900170952, 817138135549, 3461452853383, 14662949322176, 62113250509227, 263115950765039, 1114577054530568

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Sep 06 2002

G. C. Greubel, Table of n, a(n) for n = 0..1000
Mohammad K. Azarian, Identities Involving Lucas or Fibonacci and Lucas Numbers as Binomial Sums, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 45, 2012, pp. 2221-2227.
Index entries for linear recurrences with constant coefficients, signature (3,6,-3,-1).

Cf. A000032, A061084, A075150, A075151.

Third row of array A103324.

[ Lucas(n)^3 : n in [0..120]]; // Vincenzo Librandi, Apr 14 2011

CoefficientList[Series[(8 - 23*x - 24*x^2 + x^3)/((x^2 + 4*x - 1)*(x^2 - x - 1)), {x,0,50}], x] (* or *) Table[LucasL[n]^3, {n,0,30}] (* or *) LinearRecurrence[{3,6,-3,-1}, {8, 1, 27, 64}, 30] (* G. C. Greubel, Dec 21 2017 *)

a(n)=(fibonacci(n-1)+fibonacci(n+1))^3 \\ Charles R Greathouse IV, Feb 09 2016

from sympy import lucas
def a(n):  return lucas(n)**3
print([a(n) for n in range(25)]) # Michael S. Branicky, Aug 01 2021

a(n) = 3*(-1)^n*L(n) + L(3*n).
a(n) = (-1)^n*A075151(n).
a(n) = A000032(n)^3 = A000032(n) * A001254(n).
a(n) = L(n)*C(n)^2, L(n) = Lucas numbers (A000032), C(n) = reflected Lucas numbers (comment to A061084).
a(n) = 3*a(n-1) + 6*a(n-2) - 3*a(n-3) - a(n-4), n>=4.
G.f.: ( 8-23*x-24*x^2+x^3 )/( (x^2+4*x-1)*(x^2-x-1) ).
a(n) = 2*A001077(n) + 3*A061084(n+1). - R. J. Mathar, Nov 17 2011
a(n) = L(3*n) + (F(n+4) - F(n-4))*(-1)^n, n>3 and F(n)=A000045(n). - J. M. Bergot, Feb 09 2016
a(n) + Sum_{i=0..n+1} a(i) = 19/2 + (5/2)*L(3*n+2). - Greg Dresden, Feb 24 2025

Simpler definition from Ralf Stephan, Nov 01 2004

A099923 Fourth powers of Lucas numbers A000032.

Original entry on oeis.org

16, 1, 81, 256, 2401, 14641, 104976, 707281, 4879681, 33362176, 228886641, 1568239201, 10750371856, 73680216481, 505022001201, 3461445366016, 23725169980801, 162614549665681, 1114577187760656, 7639424429247601

Offset: 0

Ralf Stephan, Nov 01 2004

Arthur T. Benjamin and Jennifer J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 56.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Mohammad K. Azarian, Identities Involving Lucas or Fibonacci and Lucas Numbers as Binomial Sums, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 45, 2012, pp. 2221-2227.
Pridon Davlianidze, Problem B-1270, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 58, No. 2 (2020), p. 179; Four Telescopic Infinite Products, Solution to Problem B-1270 by Jason L. Smith, ibid., Vol. 59, No. 2 (2021), pp. 183-184.
Toufik Mansour, A formula for the generating functions of powers of Horadam's sequence, Australas. J. Combin. 30 (2004) 207-212.
Index entries for linear recurrences with constant coefficients, signature (5,15,-15,-5,1).

Cf. A000032, A000045, A001254, A001622, A075515.

Fourth row of array A103324.

[ Lucas(n)^4 : n in [0..120]]; // Vincenzo Librandi, Apr 14 2011

LucasL[Range[0,20]]^4 (* or *) LinearRecurrence[{5,15,-15,-5,1},{16,1,81,256,2401},21] (* Harvey P. Dale, Jul 04 2014 *)
CoefficientList[Series[(16 - 79 x - 164 x^2 + 76 x^3 + x^4)/((1 - x) (1 + 3*x+x^2)*(1-7*x+x^2)), {x,0,50}], x] (* G. C. Greubel, Dec 21 2017 *)

for(n=0, 30, print1( (fibonacci(n+1) + fibonacci(n-1))^4, ", ")) \\ G. C. Greubel, Dec 21 2017

x='x+O('x^30); Vec((16-79*x-164*x^2+76*x^3+x^4)/((1-x)*(1+3*x+x^2)*(1-7*x+x^2))) \\ G. C. Greubel, Dec 21 2017

a(n) = A000032(n)^4 = A001254(n)^2.
a(n) = L(4*n) + 4*(-1)^n*L(2*n) + 6.
a(n) = L(n-2)*L(n-1)*L(n+1)*L(n+2) + 25, for n >=1.
G.f.: (16-79*x-164*x^2+76*x^3+x^4)/((1-x)*(1+3*x+x^2)*(1-7*x+x^2)). [See Mansour p. 207] - R. J. Mathar, Oct 26 2008
a(0)=16, a(1)=1, a(2)=81, a(3)=256, a(4)=2401, a(n) = 5*a(n-1) + 15*a(n-2) - 15*a(n-3) - 5*a(n-4) + a(n-5). - Harvey P. Dale, Jul 04 2014
Sum_{i=0..n} a(i) = 11 + 6*n + 4*(-1)^n*F(2*n+1) + F(4*n+2), for F = A000045. - Adam Mohamed and Greg Dresden, Jul 02 2024
Product_{n>=2} (1 - 25/a(n)) = phi^5/18, where phi is the golden ratio (A001622) (Davlianidze, 2020). - Amiram Eldar, Dec 04 2024

A200381 Products of Lucas numbers.

Original entry on oeis.org

1, 2, 3, 4, 6, 7, 8, 9, 11, 12, 14, 16, 18, 21, 22, 24, 27, 28, 29, 32, 33, 36, 42, 44, 47, 48, 49, 54, 56, 58, 63, 64, 66, 72, 76, 77, 81, 84, 87, 88, 94, 96, 98, 99, 108, 112, 116, 121, 123, 126, 128, 132, 141, 144, 147, 152, 154, 162, 168, 174, 176, 188

Offset: 1

R. J. Mathar, Nov 17 2011

nonn

Complement of A200995.

Contains A000032, A001254, A075155, A099923, A103325 .. and any products of their members.

			Contains 9=3^2, 14=2*7, 21=3*7, 27=3^3, 42=2*3*7, 141 =3*47 etc.

T. D. Noe, Table of n, a(n) for n = 1..1099
Index entries for Lucas sequences

Cf. A000032, A181156. Complement of A200995.

lim = 11; luc = LucasL[Range[0, lim]]; luc = Delete[luc, 2]; last = luc[[-1]]; t = {1}; Do[t2 = luc[[n]]^Range[Floor[Log[last]/Log[luc[[n]]]]]; s = Select[Union[Flatten[Outer[Times, t, t2]]], # <= last &];  t = Union[t, s], {n, lim}]; t (* T. D. Noe, Nov 17 2011 *)

cp's OEIS Frontend

A106729 Sum of two consecutive squares of Lucas numbers (A001254).

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

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A007598 Squared Fibonacci numbers: a(n) = F(n)^2 where F = A000045.

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A079291 Squares of Pell numbers.

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A203853 a(n) = (1/n) * Sum_{d|n} moebius(n/d) * Lucas(d)^2, where Lucas(n) = A000204(n).

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