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%I #75 Dec 16 2023 15:50:30
%S 0,2,15,104,714,4895,33552,229970,1576239,10803704,74049690,507544127,
%T 3478759200,23843770274,163427632719,1120149658760,7677619978602,
%U 52623190191455,360684711361584,2472169789339634,16944503814015855,116139356908771352,796030994547383610
%N a(n) = (Lucas(4n+1)-1)/5, or Fibonacci(2n)*Fibonacci(2n+1), or A081017(n)/5.
%C Another interpretation of this sequence is: nonnegative integers k such that (k + 1)^2 + (2k)^2 is a perfect square. So apart from a(0) = 0, a(n) + 1 and 2a(n) form the legs of a Pythagorean triple. - Nick Hobson, Jan 13 2007
%C Also solution y of Diophantine equation x^2 + 4*y^2 = k^2 for which x=y+1. - _Carmine Suriano_, Jun 23 2010
%C Also the index of the first of two consecutive heptagonal numbers whose sum is equal to the sum of two consecutive triangular numbers. - _Colin Barker_, Dec 20 2014
%C Nonnegative integers k such that G(x) = k for some rational number x where G(x) = x/(1-x-x^2) is the generating function of the Fibonacci numbers. - _Tom Edgar_, Aug 24 2015
%C The integer solutions of the equation a(b+1) = (a-b)(a-b-1) or, equivalently, binomial(a, b) = binomial(a-1, b+1) are given by (a, b) = (a(n+1), A003482(n)=Fibonacci(2*n) * Fibonacci(2*n+3)) (Lind and Singmaster). - _Tomohiro Yamada_, May 30 2018
%D A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 28.
%D Hugh C. Williams, Edouard Lucas and Primality Testing, John Wiley and Sons, 1998, p. 75.
%H Colin Barker, <a href="/A081018/b081018.txt">Table of n, a(n) for n = 0..1000</a>
%H Pridon Davlianidze, <a href="https://www.fq.math.ca/Problems/ElemProbSolnNov2020.pdf">Problem B-1279</a>, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 58, No. 4 (2020), p. 368; <a href="https://www.fq.math.ca/Problems/ElemProbSolnNov2021.pdf">An Unusual Generalization</a>, Solution to Problem B-1279 by J. N. Senadheera, ibid., Vol. 59, No. 4 (2021), pp. 370-371.
%H Dae S. Hong, <a href="http://www.jstor.org/stable/10.4169/college.math.j.46.2.110">When is the Generating Function of the Fibonacci Numbers an Integer?</a>, The College Mathematics Journal, Vol. 46, No. 2 (March 2015), pp. 110-112
%H D. A. Lind, <a href="https://www.fq.math.ca/Scanned/6-3/lind.pdf">The quadratic field Q(sqrt(5)) and a certain diophantine equation</a>, Fibonacci Quart., Vol. 6, No. 3 (1968), pp. 86-93.
%H David Singmaster, <a href="https://www.fq.math.ca/Scanned/13-4/singmaster.pdf">Repeated binomial coefficients and Fibonacci numbers</a>, Fibonacci Quart., Vol. 13, No. 4 (1973), pp. 295-298.
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (8,-8,1).
%F a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3).
%F a(n) = Fibonacci(3) + Fibonacci(7) + Fibonacci(11) + ... + Fibonacci(4n+3).
%F G.f.: x*(2-x)/((1-x)*(1-7*x+x^2)). - _Colin Barker_, Mar 30 2012
%F E.g.f.: (1/5)^(3/2)*((1+phi^2)*exp(phi^4*x) - (1 + (1/phi^2))*exp(x/phi^4) - sqrt(5)*exp(x)), where 2*phi = 1 + sqrt(5). - _G. C. Greubel_, Aug 24 2015
%F From - _Michael Somos_, Aug 27 2015: (Start)
%F a(n) = -A081016(-1-n) for all n in Z.
%F 0 = a(n) - 7*a(n+1) + a(n+2) - 1 for all n in Z.
%F 0 = a(n)*a(n+2) - a(n+1)^2 + a(n+1) + 2 for all n in Z.
%F 0 = a(n)*(a(n) -7*a(n+1) -1) + a(n+1)*(a(n+1) - 1) - 2 for all n in Z. (End)
%F a(n) = (k(n) + sqrt(k(n)*(4 + 5*k(n))))/2, where k(n) = A049684(n). - _Stefano Spezia_, Mar 11 2021
%F Product_{n>=1} (1 + 1/a(n)) = phi (A001622) (Davlianidze, 2020). - _Amiram Eldar_, Nov 30 2021
%e G.f. = 2*x + 15*x^2 + 104*x^3 + 714*x^4 + 4895*x^5 + 33552*x^6 + ...
%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 25 do printf(`%d,`,(luc(4*n+1)-1)/5) od: # _James A. Sellers_, Mar 03 2003
%t (LucasL[4*Range[0,30]+1]-1)/5 (* or *) LinearRecurrence[{8,-8,1}, {0,2,15}, 30] (* _G. C. Greubel_, Aug 24 2015, modified Jul 14 2019 *)
%o (PARI) concat(0, Vec(x*(2-x)/((1-x)*(1-7*x+x^2)) + O(x^30))) \\ _Colin Barker_, Dec 20 2014
%o (Magma) [(Lucas(4*n+1)-1)/5: n in [0..30]]; // _Vincenzo Librandi_, Aug 24 2015
%o (Sage) [(lucas_number2(4*n+1,1,-1) -1)/5 for n in (0..30)] # _G. C. Greubel_, Jul 14 2019
%o (GAP) List([0..30], n-> (Lucas(1,-1, 4*n+1)[2] -1)/5); # _G. C. Greubel_, Jul 14 2019
%Y Cf. A000045 (Fibonacci numbers), A000032 (Lucas numbers), A081017.
%Y Partial sums of A033891. Bisection of A001654 and A059840.
%Y Equals A089508 + 1.
%Y Cf. A001622, A081007, A081016, A178898.
%K nonn,easy
%O 0,2
%A _R. K. Guy_, Mar 01 2003
%E More terms from _James A. Sellers_, Mar 03 2003
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