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%I #96 Sep 08 2022 08:44:28
%S 4,1,9,16,49,121,324,841,2209,5776,15129,39601,103684,271441,710649,
%T 1860496,4870849,12752041,33385284,87403801,228826129,599074576,
%U 1568397609,4106118241,10749957124,28143753121,73681302249,192900153616,505019158609,1322157322201,3461452808004,9062201101801,23725150497409
%N Squares of Lucas numbers.
%D A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 36, 60.
%D J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 97.
%D 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]
%H G. C. Greubel, <a href="/A001254/b001254.txt">Table of n, a(n) for n = 0..2375</a> (terms 0..200 from T. D. Noe)
%H Mohammad K. Azarian, <a href="http://www.m-hikari.com/ijcms/ijcms-2012/45-48-2012/azarianIJCMS45-48-2012.pdf">Identities Involving Lucas or Fibonacci and Lucas Numbers as Binomial Sums</a>, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 45, 2012, pp. 2221-2227.
%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H T. Mansour, <a href="https://arxiv.org/abs/math/0302015">A note on sum of k-th power of Horadam's sequence</a>, arXiv:math/0302015 [math.CO], 2003.
%H P. Stanica, <a href="https://arxiv.org/abs/math/0010149">Generating functions, weighted and non-weighted sums for powers of second-order recurrence sequences</a>, arXiv:math/0010149 [math.CO], 2000.
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (2,2,-1).
%F a(n) = A000032(n)^2.
%F G.f.: ( 4-7*x-x^2 ) / ( (1+x)*(x^2-3*x+1) ). - _Len Smiley_, Nov 30 2001
%F From _Ralf Stephan_, Feb 08 2003: (Start)
%F a(n) = r^n + (1/r)^n + 2*(-1)^n, with r=(3+sqrt(5))/2.
%F a(n+3) = 2*a(n+2) + 2*a(n+1) - a(n). (End)
%F a(n) = L(2*n) + 2*(-1)^n = L(n-1)*L(n+1) + 5(-1)^n.
%F a(n) = 5*Fibonacci(n)^2 + 4*(-1)^n.
%F a(n) + a(n+1) = A106729(n). - _R. J. Mathar_, Nov 17 2011
%F E.g.f.: 2*exp(-x)*(exp(5*x/2)*cosh(sqrt(5)*x/2)+1). - _Wolfdieter Lang_, Jan 14 2012
%F 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
%F 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
%F 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
%F From _Peter Bala_, Nov 13 2019: (Start)
%F 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.
%F 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.
%F Sum_{n >= 1} 1/(a(n) - 25/a(n)) = 2/9.
%F 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)
%F a(n) = 3*a(n-1) - a(n-2) + 10*(-1)^n. - _Greg Dresden_, May 18 2020
%p with(combinat):seq(5*fibonacci(n)^2+4*(-1)^n, n=0..26)
%t Table[LucasL[n]^2, {n, 0, 29}] (* _Alonso del Arte_, Apr 11 2011 *)
%t LinearRecurrence[{2, 2, -1}, {4, 1, 9}, 33] (* _Jean-François Alcover_, Jan 07 2019 *)
%o (Magma) [ Lucas(n)^2 : n in [0..120]]; // _Vincenzo Librandi_, Apr 14 2011
%o (PARI) a(n)=5*fibonacci(n)^2 + 4*(-1)^n \\ _Charles R Greathouse IV_, Sep 24 2015
%o (Python)
%o from sympy import lucas
%o def a(n): return lucas(n)**2
%o print([a(n) for n in range(33)]) # _Michael S. Branicky_, Apr 01 2021
%Y Cf. A000032, A000204.
%Y Cf. A007598, A079291.
%Y With alternating signs, cf. A075150.
%Y Bisection of A001638 and A006499. First differences of A005970.
%Y Second row of array A103324.
%K nonn,easy
%O 0,1
%A _N. J. A. Sloane_
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