A001254 Squares of Lucas numbers.

4, 1, 9, 16, 49, 121, 324, 841, 2209, 5776, 15129, 39601, 103684, 271441, 710649, 1860496, 4870849, 12752041, 33385284, 87403801, 228826129, 599074576, 1568397609, 4106118241, 10749957124, 28143753121, 73681302249, 192900153616, 505019158609, 1322157322201, 3461452808004, 9062201101801, 23725150497409

Offset

0,1

References

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 36, 60.

J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 97.

Thomas Koshy, "Fibonacci and Lucas Numbers and Applications", Wiley, New York, 2001. [Note that Identity 34.7 on page 404 is wrong. - Alonso del Arte, Sep 07 2010]

Links

G. C. Greubel, Table of n, a(n) for n = 0..2375 (terms 0..200 from T. D. Noe)

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.

Tanya Khovanova, Recursive Sequences

T. Mansour, A note on sum of k-th power of Horadam's sequence, arXiv:math/0302015 [math.CO], 2003.

P. 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 (2,2,-1).

Formula

a(n) = A000032(n)^2.

G.f.: ( 4-7*x-x^2 ) / ( (1+x)*(x^2-3*x+1) ). - Len Smiley, Nov 30 2001

From Ralf Stephan, Feb 08 2003: (Start)

a(n) = r^n + (1/r)^n + 2*(-1)^n, with r=(3+sqrt(5))/2.

a(n+3) = 2*a(n+2) + 2*a(n+1) - a(n). (End)

a(n) = L(2*n) + 2*(-1)^n = L(n-1)*L(n+1) + 5(-1)^n.

a(n) = 5*Fibonacci(n)^2 + 4*(-1)^n.

a(n) + a(n+1) = A106729(n). - R. J. Mathar, Nov 17 2011

E.g.f.: 2*exp(-x)*(exp(5*x/2)*cosh(sqrt(5)*x/2)+1). - Wolfdieter Lang, Jan 14 2012

a(n) = 1/4*( a(n-2) - a(n-1) - a(n+1) + a(n+2) ). The same recurrence holds for A007598. - Peter Bala, Aug 18 2015

For n>1, a(n)=(10*F(2*n-1) + 2*L(n-2)*L(n+1))/4 where F(n)=A000045(n), L(n)=A000204(n). - J. M. Bergot, Nov 25 2015

a(n) = (L(n-2)*L(n+2) + L(n-1)*L(n+1))/2 with L(k)=A000032(k). - J. M. Bergot, May 25 2017

From Peter Bala, Nov 13 2019: (Start)

Sum_{n >= 1} 1/a(n) = (1/8)*( theta_3(beta)^4 - 1 ) = A105394, where beta = (3 - sqrt(5))/2 and theta_3(q) = 1 + 2*Sum_{n >= 1} q^(n^2) is a theta function. See Borwein and Borwein, Exercise 7(f), p. 97.

Sum_{n >= 1} 1/(a(n) - 5) = (3 - sqrt(5))/6; Sum_{n >= 1} (-1)^n/(a(n) - 5) = (15 - sqrt(5))/30; Sum_{n >= 1} 1/(a(2*n) - 5) = (5 - sqrt(5))/10.

Sum_{n >= 1} 1/(a(n) - 25/a(n)) = 2/9.

Conjecture: Sum_{n >= 1} 1/(a(n) - 5*(-1)^n*F(2*k+1)^2) = 1/(2*a(2*k+1)) for k = 0,1,2,.... (End)

a(n) = 3*a(n-1) - a(n-2) + 10*(-1)^n. - Greg Dresden, May 18 2020

Maple

with(combinat):seq(5*fibonacci(n)^2+4*(-1)^n, n=0..26)

Mathematica

Table[LucasL[n]^2, {n, 0, 29}] (* Alonso del Arte, Apr 11 2011 *)
LinearRecurrence[{2, 2, -1}, {4, 1, 9}, 33] (* Jean-François Alcover, Jan 07 2019 *)

Prog

(Magma) [ Lucas(n)^2 : n in [0..120]]; // Vincenzo Librandi, Apr 14 2011
(PARI) a(n)=5*fibonacci(n)^2 + 4*(-1)^n \\ Charles R Greathouse IV, Sep 24 2015
(Python)
from sympy import lucas
def a(n):  return lucas(n)**2
print([a(n) for n in range(33)]) # Michael S. Branicky, Apr 01 2021

Crossrefs

Cf. A000032, A000204.

Cf. A007598, A079291.

With alternating signs, cf. A075150.

Bisection of A001638 and A006499. First differences of A005970.

Second row of array A103324.

Sequence in context: A193580 A244761 A075150 * A143763 A278350 A128626

Adjacent sequences: A001251 A001252 A001253 * A001255 A001256 A001257

Keyword

nonn,easy

Author

N. J. A. Sloane

Status

approved