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
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).
Crossrefs
Programs
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Magma
[ Lucas(n)^2 : n in [0..120]]; // Vincenzo Librandi, Apr 14 2011
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Maple
with(combinat):seq(5*fibonacci(n)^2+4*(-1)^n, n=0..26)
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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 *) -
PARI
a(n)=5*fibonacci(n)^2 + 4*(-1)^n \\ Charles R Greathouse IV, Sep 24 2015
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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
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