A049684 a(n) = Fibonacci(2n)^2.

0, 1, 9, 64, 441, 3025, 20736, 142129, 974169, 6677056, 45765225, 313679521, 2149991424, 14736260449, 101003831721, 692290561600, 4745030099481, 32522920134769, 222915410843904, 1527884955772561, 10472279279564025, 71778070001175616, 491974210728665289

Offset: 0

Author

Clark Kimberling

Keywords

• nonn
• nice
• easy

Comments

This is the r=9 member of the r-family of sequences S_r(n) defined in A092184 where more information can be found.

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

References

• A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 27.
• H. J. H. Tuenter, Fibonacci summation identities arising from Catalan's identity, Fib. Q., 60:4 (2022), 312-319.

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..500
• Marco Abrate, Stefano Barbero, Umberto Cerruti, and Nadir Murru, Polynomial sequences on quadratic curves, Integers, Vol. 15, 2015, #A38.
• Mohammad K. Azarian, Fibonacci Identities as Binomial Sums, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 38, 2012, pp. 1871-1876.
• Mohammad K. Azarian, Fibonacci Identities as Binomial Sums II, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 42, 2012, pp. 2053-2059.
• S. Barbero, U. Cerruti, and N. Murru, On polynomial solutions of the Diophantine equation (x + y - 1)^2 = wxy, Rendiconti Sem. Mat. Univ. Pol. Torino (2020) Vol. 78, No. 1, 5-12.
• Pridon Davlianidze, Problem B-1264, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 58, No. 1 (2020), p. 82; It's All About Catalan, Solution to Problem B-1264, ibid., Vol. 59, No. 1 (2021), pp. 87-88.
• E. Kilic, Y. T. Ulutas, and N. Omur, A Formula for the Generating Functions of Powers of Horadam's Sequence with Two Additional Parameters, J. Int. Seq. 14 (2011) #11.5.6, table 1, k=2.
• R. Stephan, Boring proof of a nonlinearity
• Index entries for sequences related to Chebyshev polynomials.
• Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Crossrefs

First differences give A033890.

First differences of A103434.

Bisection of A007598 and A064841.

a(n) = A064170(n+2) - 1 = (1/5) A081070.

Cf. A000032, A000045, A001622, A001906, A056854, A065563, A092184, A374375.

Programs

Mathematica

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 *)
Fibonacci[Range[0, 40, 2]]^2 (* Harvey P. Dale, Mar 22 2012 *)
Table[Fibonacci[n - 1] Fibonacci[n + 1] - 1, {n, 0, 40, 2}] (* Bruno Berselli, Feb 12 2015 *)
LinearRecurrence[{8, -8, 1},{0, 1, 9},21] (* Ray Chandler, Sep 23 2015 *)

MuPAD

numlib::fibonacci(2*n)^2 $ n = 0..35; // Zerinvary Lajos, May 13 2008

PARI

a(n)=fibonacci(2*n)^2

Sage

[fibonacci(2*n)^2 for n in range(0, 21)] # Zerinvary Lajos, May 15 2009

Formula

G.f.: (x+x^2) / ((1-x)*(1-7*x+x^2)).

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.

a(n) = 7*a(n-1) - a(n-2) + 2 = A001906(n)^2.

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

a(n) = 1/5*(-2 + ( (7+sqrt(45))/2 )^n + ( (7-sqrt(45))/2 )^n). - Ralf Stephan, Apr 14 2004

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

a(n) = F(2*n-1)*F(2*n+1)-1, see A064170 - Bruno Berselli, Feb 12 2015

a(n) = Sum_{i=1..n} F(4*i-2) for n>0. - Bruno Berselli, Aug 25 2015

From Peter Bala, Nov 20 2019: (Start)

Sum_{n >= 1} 1/(a(n) + 1) = (sqrt(5) - 1)/2.

Sum_{n >= 1} 1/(a(n) + 4) = (3*sqrt(5) - 2)/16. More generally, it appears that

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,....

Sum_{n >= 2} 1/(a(n) - 1) = (8 - 3*sqrt(5))/9. (End)

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

Product_{n>=2} (1 - 1/a(n)) = phi^2/3, where phi is the golden ratio (A001622) (Davlianidze, 2020). - Amiram Eldar, Dec 01 2021

a(n) = A092521(n-1)+A092521(n). - R. J. Mathar, Nov 22 2024

Extensions

Better description and more terms from Michael Somos