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a(n) = (Lucas(4*n+2) + 2)/5, or Fibonacci(2*n+1)^2, or A081067(n)/5.
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%I #80 Dec 16 2023 17:31:44

%S 1,4,25,169,1156,7921,54289,372100,2550409,17480761,119814916,

%T 821223649,5628750625,38580030724,264431464441,1812440220361,

%U 12422650078084,85146110326225,583600122205489,4000054745112196,27416783093579881

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

%C Indices of 12-gonal numbers which are also squares (A342709). - _Bernard Schott_, Mar 19 2021

%C 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

%C And the corresponding x values of x^2 = 5*y^2 - 4*y are in A033890. - _Bernard Schott_, Feb 26 2023

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

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

%H Vincenzo Librandi, <a href="/A081068/b081068.txt">Table of n, a(n) for n = 0..1000</a>

%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 (See Corollary 1 (vii)).

%H 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 Derek Jennings, <a href="https://www.fq.math.ca/Scanned/32-1/jennings.pdf">On Sums of Reciprocals of Fibonacci and Lucas Numbers</a>, The Fibonacci Quarterly, Vol. 32, No. 1 (1994), pp. 18-21.

%H Giovanni Lucca, <a href="http://forumgeom.fau.edu/FG2018volume18/FG201808index.html">Integer Sequences and Circle Chains Inside a Circular Segment</a>, Forum Geometricorum, Vol. 18 (2018), 47-55.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (8,-8,1).

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

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

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

%F a(n) = Fibonacci(2*n)*Fibonacci(2*n+2) +1. - _Gary Detlefs_, Apr 01 2012

%F G.f.: (1-4*x+x^2)/((1-x)*(x^2-7*x+1)). - _Colin Barker_, Jun 26 2012

%F Sum_{n>=0} 1/(a(n) + 1) = 1/3*sqrt(5). - _Peter Bala_, Nov 30 2013

%F 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

%F 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

%p 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

%t 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 *)

%t Table[(LucasL[4*n+2] + 2)/5, {n,0,30}] (* _G. C. Greubel_, Dec 17 2017 *)

%o (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

%o (PARI) main(size)={ return(concat([1],vector(size,n,fibonacci(2*n+1)^2))) } /* _Anders Hellström_, Jul 11 2015 */

%o (Magma) [(Lucas(4*n+2) + 2)/5: n in [0..30]]; // _G. C. Greubel_, Dec 17 2017

%o (PARI) for(n=0,30, print1(fibonacci(2*n+1)^2, ", ")) \\ _G. C. Greubel_, Dec 17 2017

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

%Y Cf. A001622, A051324, A239798, A342709.

%Y Cf. A033890, A360467.

%K nonn,easy

%O 0,2

%A _R. K. Guy_, Mar 04 2003

%E More terms from _James A. Sellers_, Mar 05 2003