%I M3364 #255 Aug 03 2024 02:30:08

%S 0,1,1,4,9,25,64,169,441,1156,3025,7921,20736,54289,142129,372100,

%T 974169,2550409,6677056,17480761,45765225,119814916,313679521,

%U 821223649,2149991424,5628750625,14736260449,38580030724,101003831721,264431464441,692290561600

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

%C 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

%C From _Giorgio Balzarotti_, Mar 11 2009: (Start)

%C Determinant of power series with alternate signs of gamma matrix with determinant 1!.

%C 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.

%C 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)!.

%C a(n) is even with respect to signs of power of A.

%C See A158039...A158050 for sequence with matrix 2!, 3!, ... (End)

%C 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

%C This is a divisibility sequence.

%C 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

%C 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

%C 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

%C Differences between successive golden rectangle numbers A001654. - _Jonathan Sondow_, Nov 05 2015

%C 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

%C 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

%D R. C. Alperin, A family of nonlinear recurrences and their linear solutions, Fib. Q., 57:4 (2019), 318-321.

%D A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 8.

%D R. Honsberger, Mathematical Gems III, M.A.A., 1985, p. 130.

%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

%D R. P. Stanley, Enumerative Combinatorics I, Example 4.7.14, p. 251.

%H Indranil Ghosh, <a href="/A007598/b007598.txt">Table of n, a(n) for n = 0..2389</a> (terms 0..200 from T. D. Noe)

%H 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 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 Paul S. Bruckman, <a href="http://www.fq.math.ca/Problems/elementary45-2.pdf">Problem B-1023: And a cubic as a sum of two squares</a>, Fibonacci Quarterly, Vol. 45, Number 2; May 2007; p. 186.

%H Andrej Dujella, <a href="http://dx.doi.org/10.1016/S0012-365X(98)00341-0">A bijective proof of Riordan's theorem on powers of Fibonacci numbers</a>, Discrete Math. 199 (1999), no. 1-3, 217--220. MR1675924 (99k:05016).

%H D. Foata and G.-N. Han, <a href="http://www-irma.u-strasbg.fr/~foata/paper/pub71.html">Nombres de Fibonacci et polynomes orthogonaux</a>.

%H Svenja Huntemann and Neil A. McKay, <a href="https://arxiv.org/abs/1909.12419">Counting Domineering Positions</a>, arXiv:1909.12419 [math.CO], 2019.

%H Jong Hyun Kim, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL12/Kim/kim18.html">Hadamard products and tilings</a>, JIS 12 (2009) 09.7.4.

%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 T. Mansour, <a href="https://arxiv.org/abs/math/0303138">Squaring the terms of an l-th order linear recurrence</a>, arXiv:math/0303138 [math.CO], 2003.

%H Hilary I. Okagbue, Muminu O. Adamu, Sheila A. Bishop, and Abiodun A. Opanuga, <a href="https://www.ripublication.com/ijaer16/ijaerv11n6_150.pdf">Digit and Iterative Digit Sum of Fibonacci numbers, their identities and powers</a>, International Journal of Applied Engineering Research ISSN 0973-4562 Volume 11, Number 6 (2016) pp 4623-4627.

%H P. Stanica, <a href="https://arxiv.org/abs/math/0010149">Generating functions, weighted and non-weighted sums of powers of second-order recurrence sequences</a>, arXiv:math/0010149 [math.CO], 2000.

%H H. C. Williams and R. K. Guy, <a href="http://dx.doi.org/10.1142/S1793042111004587">Some fourth-order linear divisibility sequences</a>, Intl. J. Number Theory 7 (5) (2011) 1255-1277.

%H H. C. Williams and R. K. Guy, <a href="http://www.emis.de/journals/INTEGERS/papers/a17self/a17self.Abstract.html">Some Monoapparitic Fourth Order Linear Divisibility Sequences</a> Integers, Volume 12A (2012) The John Selfridge Memorial Volume.

%H <a href="/index/Tu#2wis">Index entries for two-way infinite sequences</a>

%H <a href="/index/Di#divseq">Index to divisibility sequences</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (2,2,-1).

%F G.f.: x*(1-x)/((1+x)*(1-3*x+x^2)).

%F a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3), n > 2. a(0)=0, a(1)=1, a(2)=1.

%F a(-n) = a(n) for all n in Z.

%F a(n) = A080097(n-2) + 1.

%F 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

%F 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

%F a(n) = (2*Fibonacci(2*n+1) - Fibonacci(2*n) - 2*(-1)^n)/5. - _Ralf Stephan_, May 14 2004

%F a(n) = F(n-1)*F(n+1) - (-1)^n = A059929(n-1) - A033999(n).

%F 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

%F a(n) = (A005248(n) - 2*(-1)^n)/5. - _R. J. Mathar_, Sep 12 2010

%F a(n) = (-1)^k*(Fibonacci(n+k)^2-Fibonacci(k)*Fibonacci(2*n+k)), for any k. - _Gary Detlefs_, Dec 13 2010

%F a(n) = 3*a(n-1) - a(n-2) + 2*(-1)^(n+1), n > 1. - _Gary Detlefs_, Dec 20 2010

%F a(n) = Fibonacci(2*n-2) + a(n-2). - _Gary Detlefs_, Dec 20 2010

%F a(n) = (Fibonacci(3*n) - 3*(-1)^n*Fibonacci(n))/(5*Fibonacci(n)), n > 0. - _Gary Detlefs_, Dec 20 2010

%F a(n) = (Fibonacci(n)*Fibonacci(n+4) - 3*Fibonacci(n)*Fibonacci(n+1))/2. - _Gary Detlefs_, Jan 17 2011

%F 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

%F 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

%F Starting with "1" = triangle A059260 * the Fibonacci sequence as a vector. - _Gary W. Adamson_, Mar 06 2012

%F a(0) = 0, a(1) = 1; a(n+1) = (a(n)^(1/2) + a(n-1)^(1/2))^2. - _Thomas Ordowski_, Jan 06 2013

%F a(n) + a(n-1) = A001519(n), n > 0. - _R. J. Mathar_, Mar 19 2014

%F From _Peter Bala_, Mar 31 2014: (Start)

%F 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.

%F 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].

%F a(n) = U(n-1,i/2)*U(n-1,-i/2), where U(n,x) denotes the Chebyshev polynomial of the second kind.

%F See the remarks in A100047 for the general connection between Chebyshev polynomials and 4th-order linear divisibility sequences. (End)

%F 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

%F 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 (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

%F 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

%F a(n) = F(n)*F(n+1) - F(n-1)*F(n). - _Jonathan Sondow_, Nov 05 2015

%F 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

%F a(n) = (F(n+2)^2 + L(n+1)^2) - 2*F(n+2)*L(n+1). - _J. M. Bergot_, Nov 08 2015

%F a(n) = F(n+3)^2 - 4*F(n+1)*F(n+2). - _J. M. Bergot_, Mar 17 2016

%F a(n) = (F(n-2)*F(n+2) + F(n-1)*F(n+1))/2. - _J. M. Bergot_, May 25 2017

%F 4*a(n) = L(n+1)*L(n-1) - F(n+2)*F(n-2), where L = A000032. - _Bruno Berselli_, Sep 27 2017

%F a(n) = F(n+k)*F(n-k) + (-1)^(n+k)*a(k), for every integer k >= 0. - _Federico Provvedi_, Dec 10 2018

%F From _Peter Bala_, Nov 19 2019: (Start)

%F Sum_{n >= 3} 1/(a(n) - 1/a(n)) = 4/9.

%F Sum_{n >= 3} (-1)^n/(a(n) - 1/a(n)) = (10 - 3*sqrt(5))/18.

%F 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)

%F Sum_{n>=1} 1/a(n) = A105393. - _Amiram Eldar_, Oct 22 2020

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

%p with(combinat): seq(fibonacci(n)^2, n=0..27); # _Zerinvary Lajos_, Sep 21 2007

%t f[n_] := Fibonacci[n]^2; Array[f, 4!, 0] (* _Vladimir Joseph Stephan Orlovsky_, Oct 25 2009 *)

%t LinearRecurrence[{2,2,-1},{0,1,1},41] (* _Harvey P. Dale_, May 18 2011 *)

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

%o (PARI) concat(0, Vec(x*(1-x)/((1+x)*(1-3*x+x^2)) + O(x^30))) \\ _Altug Alkan_, Nov 06 2015

%o (Sage) [(fibonacci(n))^2 for n in range(0, 28)]# _Zerinvary Lajos_, May 15 2009

%o (Magma) [Fibonacci(n)^2: n in [0..30]]; // _Vincenzo Librandi_, Apr 14 2011

%o (Haskell)

%o a007598 = (^ 2) . a000045 -- _Reinhard Zumkeller_, Sep 01 2013

%o (Sage) [fibonacci(n)^2 for n in range(30)] # _G. C. Greubel_, Dec 10 2018

%o (GAP) List([0..30], n -> Fibonacci(n)^2); # _G. C. Greubel_, Dec 10 2018

%Y Cf. A000032, A000045, A001254, A001654, A047946, A056570, A059260, A061646, A065885, A079291, A080097, A105393.

%Y Bisection of A006498 and A074677. First differences of A001654.

%Y Second row of array A103323.

%Y Half of A175395.

%K nonn,easy,nice

%O 0,4

%A _N. J. A. Sloane_, _Robert G. Wilson v_