A081070 -id:A081070 - cp's OEIS frontend

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

Clark Kimberling

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

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.

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).

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.

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 *)

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

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

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

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

Better description and more terms from Michael Somos

A221364 The simple continued fraction expansion of F(x) := Product_{n >= 0} (1 - x^(4n+3))/(1 - x^(4n+1)) when x = 1/2*(3 - sqrt(5)).

Original entry on oeis.org

1, 1, 1, 5, 1, 16, 1, 45, 1, 121, 1, 320, 1, 841, 1, 2205, 1, 5776, 1, 15125, 1, 39601, 1, 103680, 1, 271441, 1, 710645, 1, 1860496, 1, 4870845, 1, 12752041, 1, 33385280, 1, 87403801, 1, 228826125, 1, 599074576, 1, 1568397605, 1, 4106118241, 1, 10749957120, 1, 28143753121

Offset: 0

Peter Bala, Jan 15 2013

The function F(x) := Product_{n >= 0} (1 - x^(4*n+3))/(1 - x^(4*n+1)) is analytic for |x| < 1. When x is a quadratic irrational of the form x = 1/2*(N - sqrt(N^2 - 4)), N an integer greater than 2, the real number F(x) has a predictable simple continued fraction expansion. The first examples of these expansions, for N = 2, 4, 6 and 8, are due to Hanna. See A174500 through A175503. The present sequence is the case N = 3. See also A221365 (N = 5), A221366 (N = 7), A221369 (N = 9).

If we denote the present sequence by [1, c(1), 1, c(2), 1, c(3), ...] then for k = 1, 2, ..., the simple continued fraction expansion of F((1/2*(3 - sqrt(5)))^k) is given by the sequence [1; c(k), 1, c(2*k), 1, c(3*k), 1, ...]. Examples are given below.

			F(1/2*(3 - sqrt(5))) = 1.53879 34992 88095 08323 ... = 1 + 1/(1 + 1/(1 + 1/(5 + 1/(1 + 1/(16 + 1/(1 + 1/(45 + ...))))))).
F((1/2*(3 - sqrt(5)))^2) = 1.16725 98258 10214 95210 ... = 1 + 1/(5 + 1/(1 + 1/(45 + 1/(1 + 1/(320 + 1/(1 + 1/(2205 + ...))))))).
F((1/2*(3 - sqrt(5)))^3) = 1.05883 42773 67371 19975 ... = 1 + 1/(16 + 1/(1 + 1/(320 + 1/(1 + 1/(5776 + 1/(1 + 1/(103680 + ...))))))).

Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
Index entries for linear recurrences with constant coefficients, signature (0,4,0,-4,0,1).

Cf. A001906, A002878, A004146, A049684, A081070, A081071, A174500 (N = 4), A221365 (N = 5), A221366 (N = 7), A221369 (N = 9).

a(2*n-1) = (1/2*(3 + sqrt(5)))^n + (1/2*(3 - sqrt(5)))^n - 2 = A004146(n); a(2*n) = 1.
a(4*n+1) = A081071(n) = A002878(n)^2;
a(4*n-1) = A081070(n) = 5*A049684(n) = 5*(A001906(n))^2.
a(n) = 4*a(n-2)-4*a(n-4)+a(n-6). G.f.: -(x^4+x^3-3*x^2+x+1) / ((x-1)*(x+1)*(x^2-x-1)*(x^2+x-1)). - Colin Barker, Jan 20 2013

More terms from Michel Marcus, Feb 21 2025

A221366 The simple continued fraction expansion of F(x) := Product_{n >= 0} (1 - x^(4n+3))/(1 - x^(4n+1)) when x = (1/2)(7 - 3sqrt(5)).

Original entry on oeis.org

1, 5, 1, 45, 1, 320, 1, 2205, 1, 15125, 1, 103680, 1, 710645, 1, 4870845, 1, 33385280, 1, 228826125, 1, 1568397605, 1, 10749957120, 1, 73681302245, 1, 505019158605, 1, 3461452808000, 1, 23725150497405, 1

Offset: 0

Peter Bala, Jan 15 2013

The function F(x) := Product_{n >= 0} (1 - x^(4*n+3))/(1 - x^(4*n+1)) is analytic for |x| < 1. When x is a quadratic irrational of the form x = 1/2*(N - sqrt(N^2 - 4)), N an integer greater than 2, the real number F(x) has a predictable simple continued fraction expansion. The first examples of these expansions, for N = 2, 4, 6 and 8, are due to Hanna. See A174500 through A175503. The present sequence is the case N = 7. See also A221364 (N = 3), A221365 (N = 5) and A221367 (N = 9).

If we denote the present sequence by [1, c(1), 1, c(2), 1, c(3), ...] then for k = 1, 2, ..., the simple continued fraction expansion of F((1/2*(7 - sqrt(45)))^k) is given by the sequence [1; c(k), 1, c(2*k), 1, c(3*k), 1, ...]. Examples are given below.

			F(1/2*(7 - sqrt(45))) = 1.16725 98258 10214 95210 ... = 1 + 1/(5 + 1/(1 + 1/(45 + 1/(1 + 1/(320 + 1/(1 + 1/(2205 + ...))))))).
F((1/2*(7 - sqrt(45)))^2) = 1.02173 93445 69104 86504 ... = 1 + 1/(45 + 1/(1 + 1/(2205 + 1/(1 + 1/(103680 + 1/(1 + 1/(4870845 + ...))))))).
F((1/2*(7 - sqrt(45)))^3) = 1.00311 52648 91110 10148 ... = 1 + 1/(320 + 1/(1 + 1/(103680 + 1/(1 + 1/(33385280 + 1/(1 + 1/(10749957120 + ...))))))).

Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
Index entries for linear recurrences with constant coefficients, signature (0,8,0,-8,0,1).

Cf. A004187, A033890, A049682, A081070.

Cf. A174500 (N = 4), A221364 (N = 3), A221365 (N = 5), A221369 (N = 9).

LinearRecurrence[{0,8,0,-8,0,1},{1,5,1,45,1,320},40] (* or *) Riffle[ LinearRecurrence[{8,-8,1},{5,45,320},20],1,{1,-1,2}] (* Harvey P. Dale, Jan 04 2018 *)

a(2*n-1) = (1/2*(7 + sqrt(45)))^n + (1/2*(7 - sqrt(45)))^n - 2 = A081070(n); a(2*n) = 1.
a(4*n-1) = 45*A049682(n) = 45*(A004187(n))^2;
a(4*n+1) = 5*(A033890(n))^2.
a(n) = 8*a(n-2)-8*a(n-4)+a(n-6). G.f.: -(x^4+5*x^3-7*x^2+5*x+1) / ((x-1)*(x+1)*(x^2-3*x+1)*(x^2+3*x+1)). - Colin Barker, Jan 20 2013

cp's OEIS Frontend

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

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A221364 The simple continued fraction expansion of F(x) := Product_{n >= 0} (1 - x^(4n+3))/(1 - x^(4n+1)) when x = 1/2*(3 - sqrt(5)).

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A221366 The simple continued fraction expansion of F(x) := Product_{n >= 0} (1 - x^(4n+3))/(1 - x^(4n+1)) when x = (1/2)(7 - 3sqrt(5)).

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Author

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cp's OEIS Frontend

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

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Mathematica

MuPAD

PARI

Sage

Formula

Extensions

A221364 The simple continued fraction expansion of F(x) := Product_{n >= 0} (1 - x^(4*n+3))/(1 - x^(4*n+1)) when x = 1/2*(3 - sqrt(5)).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Formula

Extensions

A221366 The simple continued fraction expansion of F(x) := Product_{n >= 0} (1 - x^(4*n+3))/(1 - x^(4*n+1)) when x = (1/2)*(7 - 3*sqrt(5)).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A221364 The simple continued fraction expansion of F(x) := Product_{n >= 0} (1 - x^(4n+3))/(1 - x^(4n+1)) when x = 1/2*(3 - sqrt(5)).

A221366 The simple continued fraction expansion of F(x) := Product_{n >= 0} (1 - x^(4n+3))/(1 - x^(4n+1)) when x = (1/2)(7 - 3sqrt(5)).