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%I A049684 #101 Jul 19 2024 14:27:20
%S A049684 0,1,9,64,441,3025,20736,142129,974169,6677056,45765225,313679521,
%T A049684 2149991424,14736260449,101003831721,692290561600,4745030099481,
%U A049684 32522920134769,222915410843904,1527884955772561,10472279279564025,71778070001175616,491974210728665289
%N A049684 a(n) = Fibonacci(2n)^2.
%C A049684 This is the r=9 member of the r-family of sequences S_r(n) defined in A092184 where more information can be found.
%C A049684 Apparently, this sequence consists of those nonnegative integers k for which x*(x^2-1)*y*(y^2-1) = k*(k^2-1) has a solution in nonnegative integers x, y. If k = a(n),  x = A000045(2*n-1) and y = A000045(2*n+1) are a solution. See A374375 for numbers k*(k^2-1) that can be written as a product of two or more factors of the form x*(x^2-1). - _Pontus von Brömssen_, Jul 14 2024
%D A049684 A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 27.
%D A049684 H. J. H. Tuenter, Fibonacci summation identities arising from Catalan's identity, Fib. Q., 60:4 (2022), 312-319.
%H A049684 Vincenzo Librandi, <a href="/A049684/b049684.txt">Table of n, a(n) for n = 0..500</a>
%H A049684 Marco Abrate, Stefano Barbero, Umberto Cerruti, and Nadir Murru, <a href="https://www.emis.de/journals/INTEGERS/papers/p38/p38.Abstract.html">Polynomial sequences on quadratic curves</a>, Integers, Vol. 15, 2015, #A38.
%H A049684 Mohammad K. Azarian, <a href="http://www.m-hikari.com/ijcms/ijcms-2012/37-40-2012/azarianIJCMS37-40-2012.pdf">Fibonacci Identities as Binomial Sums</a>, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 38, 2012, pp. 1871-1876.
%H A049684 Mohammad K. Azarian, <a href="http://www.m-hikari.com/ijcms/ijcms-2012/41-44-2012/azarianIJCMS41-44-2012.pdf">Fibonacci Identities as Binomial Sums II</a>, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 42, 2012, pp. 2053-2059.
%H A049684 S. Barbero, U. Cerruti, and N. Murru, <a href="http://www.seminariomatematico.polito.it/rendiconti/78-1/BarberoCerrutiMurru.pdf">On polynomial solutions of the Diophantine equation (x + y - 1)^2 = wxy</a>, Rendiconti Sem. Mat. Univ. Pol. Torino (2020) Vol. 78, No. 1, 5-12.
%H A049684 Pridon Davlianidze, <a href="https://www.fq.math.ca/Problems/February2020Elem.pdf">Problem B-1264</a>, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 58, No. 1 (2020), p. 82; <a href="https://www.fq.math.ca/Problems/FQElemProbFeb2021.pdf">It's All About Catalan</a>, Solution to Problem B-1264, ibid., Vol. 59, No. 1 (2021), pp. 87-88.
%H A049684 E. Kilic, Y. T. Ulutas, and N. Omur, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL14/Omur/omur6.html">A Formula for the Generating Functions of Powers of Horadam's Sequence with Two Additional Parameters</a>, J. Int. Seq. 14 (2011) #11.5.6, table 1, k=2.
%H A049684 R. Stephan, <a href="http://www.ark.in-berlin.de/A001110.ps">Boring proof of a nonlinearity</a>
%H A049684 <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials.</a>
%H A049684 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (8,-8,1).
%F A049684 G.f.: (x+x^2) / ((1-x)*(1-7*x+x^2)).
%F A049684 a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3) with n>2, a(0)=0, a(1)=1, a(2)=9.
%F A049684 a(n) = 7*a(n-1) - a(n-2) + 2 = A001906(n)^2.
%F A049684 a(n) = (A000032(4*n)-2)/5. [This is in Koshy's book (reference under A065563) on p. 88, attributed to Lucas 1876.] - _Wolfdieter Lang_, Aug 27 2012
%F A049684 a(n) = 1/5*(-2 + ( (7+sqrt(45))/2 )^n + ( (7-sqrt(45))/2 )^n). - _Ralf Stephan_, Apr 14 2004
%F A049684 a(n) = 2*(T(n, 7/2)-1)/5 with twice the Chebyshev polynomials of the first kind evaluated at x=7/2: 2*T(n, 7/2)= A056854(n). - _Wolfdieter Lang_, Oct 18 2004
%F A049684 a(n) = F(2*n-1)*F(2*n+1)-1. - _Bruno Berselli_, Feb 12 2015
%F A049684 a(n) = Sum_{i=1..n} F(4*i-2) for n>0. - _Bruno Berselli_, Aug 25 2015
%F A049684 From _Peter Bala_, Nov 20 2019: (Start)
%F A049684 Sum_{n >= 1} 1/(a(n) + 1) = (sqrt(5) - 1)/2.
%F A049684 Sum_{n >= 1} 1/(a(n) + 4) = (3*sqrt(5) - 2)/16. More generally, it appears that
%F A049684 Sum_{n >= 1} 1/(a(n) + F(2*k+1)^2) = ((2*k+1)*F(2*k+1)*sqrt(5) - Lucas(2*k+1))/ (2*F(2*k+1)*F(4*k+2)) for k = 0,1,2,....
%F A049684 Sum_{n >= 2} 1/(a(n) - 1) = (8 - 3*sqrt(5))/9. (End)
%F A049684 E.g.f.: (1/5)*(-2*exp(x) + exp((16*x)/(1 + sqrt(5))^4) + exp((1/2)*(7 + 3*sqrt(5))*x)). - _Stefano Spezia_, Nov 23 2019
%F A049684 Product_{n>=2} (1 - 1/a(n)) = phi^2/3, where phi is the golden ratio (A001622) (Davlianidze, 2020). - _Amiram Eldar_, Dec 01 2021
%t A049684 Join[{a=0, b=1}, Table[c=7*b-1*a+2; a=b; b=c, {n, 60}]] (* _Vladimir Joseph Stephan Orlovsky_, Jan 18 2011 *)
%t A049684 Fibonacci[Range[0, 40, 2]]^2 (* _Harvey P. Dale_, Mar 22 2012 *)
%t A049684 Table[Fibonacci[n - 1] Fibonacci[n + 1] - 1, {n, 0, 40, 2}] (* _Bruno Berselli_, Feb 12 2015 *)
%t A049684 LinearRecurrence[{8, -8, 1},{0, 1, 9},21] (* _Ray Chandler_, Sep 23 2015 *)
%o A049684 (PARI) a(n)=fibonacci(2*n)^2
%o A049684 (MuPAD) numlib::fibonacci(2*n)^2 $ n = 0..35; // _Zerinvary Lajos_, May 13 2008
%o A049684 (Sage) [fibonacci(2*n)^2 for n in range(0, 21)] # _Zerinvary Lajos_, May 15 2009
%Y A049684 First differences give A033890.
%Y A049684 First differences of A103434.
%Y A049684 Bisection of A007598 and A064841.
%Y A049684 a(n) = A064170(n+2) - 1 = (1/5) A081070.
%Y A049684 Cf. A000032, A000045, A001622, A001906, A056854, A065563, A092184, A374375.
%K A049684 nonn,nice,easy
%O A049684 0,3
%A A049684 _Clark Kimberling_
%E A049684 Better description and more terms from _Michael Somos_

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