A081068 a(n) = (Lucas(4*n+2) + 2)/5, or Fibonacci(2*n+1)^2, or A081067(n)/5.

1, 4, 25, 169, 1156, 7921, 54289, 372100, 2550409, 17480761, 119814916, 821223649, 5628750625, 38580030724, 264431464441, 1812440220361, 12422650078084, 85146110326225, 583600122205489, 4000054745112196, 27416783093579881

Offset

0,2

Comments

Indices of 12-gonal numbers which are also squares (A342709). - Bernard Schott, Mar 19 2021

Values of y in solutions of x^2 = 5*y^2 - 4*y in positive integers. See A360467 for how this relates to a problem regarding the subdivision of a square into four triangles of integer area. - Alexander M. Domashenko, Feb 26 2023

And the corresponding x values of x^2 = 5*y^2 - 4*y are in A033890. - Bernard Schott, Feb 26 2023

References

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 19.

Hugh C. Williams, Edouard Lucas and Primality Testing, John Wiley and Sons, 1998, p. 75.

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Mohammad K. Azarian, Fibonacci Identities as Binomial Sums, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 38, 2012, pp. 1871-1876 (See Corollary 1 (vii)).

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.

Derek Jennings, On Sums of Reciprocals of Fibonacci and Lucas Numbers, The Fibonacci Quarterly, Vol. 32, No. 1 (1994), pp. 18-21.

Giovanni Lucca, Integer Sequences and Circle Chains Inside a Circular Segment, Forum Geometricorum, Vol. 18 (2018), 47-55.

Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Formula

a(n) = A001519(n+1)^2 = A122367(n)^2 = A058038(n) + 1.

a(n) = A103433(n+1) - A103433(n).

a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3).

a(n) = Fibonacci(2*n)*Fibonacci(2*n+2) +1. - Gary Detlefs, Apr 01 2012

G.f.: (1-4*x+x^2)/((1-x)*(x^2-7*x+1)). - Colin Barker, Jun 26 2012

Sum_{n>=0} 1/(a(n) + 1) = 1/3*sqrt(5). - Peter Bala, Nov 30 2013

Sum_{n>=0} 1/a(n) = sqrt(5) * Sum_{n>=1} (-1)^(n+1)*n/Fibonacci(2*n) (Jennings, 1994). - Amiram Eldar, Oct 30 2020

Product_{n>=1} (1 + 1/a(n)) = phi^2/2 (A239798), where phi is the golden ratio (A001622) (Davlianidze, 2020). - Amiram Eldar, Dec 01 2021

Maple

luc := proc(n) option remember: if n=0 then RETURN(2) fi: if n=1 then RETURN(1) fi: luc(n-1)+luc(n-2): end: for n from 0 to 40 do printf(`%d, `, (luc(4*n+2)+2)/5) od: # James A. Sellers, Mar 05 2003

Mathematica

CoefficientList[Series[-(1-4*x+x^2)/((x-1)*(x^2-7*x+1)), {x, 0, 40}], x] (* or *) LinearRecurrence[{8, -8, 1}, {1, 4, 25}, 50] (* Vincenzo Librandi, Jun 26 2012 *)
Table[(LucasL[4*n+2] + 2)/5, {n, 0, 30}] (* G. C. Greubel, Dec 17 2017 *)

Prog

(Magma)  I:=[1, 4, 25]; [n le 3 select I[n] else 8*Self(n-1)-8*Self(n-2)+Self(n-3): n in [1..30]]; // Vincenzo Librandi, Jun 26 2012
(PARI) main(size)={ return(concat([1], vector(size, n, fibonacci(2*n+1)^2))) } /* Anders Hellström, Jul 11 2015 */
(Magma) [(Lucas(4*n+2) + 2)/5: n in [0..30]]; // G. C. Greubel, Dec 17 2017
(PARI) for(n=0, 30, print1(fibonacci(2*n+1)^2, ", ")) \\ G. C. Greubel, Dec 17 2017

Crossrefs

Cf. A000045 (Fibonacci numbers), A000032 (Lucas numbers), A081067.

Cf. A001622, A051324, A239798, A342709.

Cf. A033890, A360467.

Sequence in context: A006880 A227693 A175255 * A163072 A278689 A140177

Adjacent sequences: A081065 A081066 A081067 * A081069 A081070 A081071

Keyword

nonn,easy

Author

R. K. Guy, Mar 04 2003

Extensions

More terms from James A. Sellers, Mar 05 2003

Status

approved