A099923 - OEIS

%I #54 Jan 05 2025 19:51:37

%S 16,1,81,256,2401,14641,104976,707281,4879681,33362176,228886641,

%T 1568239201,10750371856,73680216481,505022001201,3461445366016,

%U 23725169980801,162614549665681,1114577187760656,7639424429247601

%N Fourth powers of Lucas numbers A000032.

%D Arthur T. Benjamin and Jennifer J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 56.

%H G. C. Greubel, <a href="/A099923/b099923.txt">Table of n, a(n) for n = 0..1000</a>

%H Mohammad K. Azarian, <a href="http://www.m-hikari.com/ijcms/ijcms-2012/45-48-2012/azarianIJCMS45-48-2012.pdf">Identities Involving Lucas or Fibonacci and Lucas Numbers as Binomial Sums</a>, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 45, 2012, pp. 2221-2227.

%H Pridon Davlianidze, <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Problems/ElemProbMay2020.pdf">Problem B-1270</a>, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 58, No. 2 (2020), p. 179; <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Problems/FQElemProbMay2021.pdf">Four Telescopic Infinite Products</a>, Solution to Problem B-1270 by Jason L. Smith, ibid., Vol. 59, No. 2 (2021), pp. 183-184.

%H Toufik Mansour, <a href="https://ajc.maths.uq.edu.au/pdf/30/ajc_v30_p207.pdf">A formula for the generating functions of powers of Horadam's sequence</a>, Australas. J. Combin. 30 (2004) 207-212.

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,15,-15,-5,1).

%F a(n) = A000032(n)^4 = A001254(n)^2.

%F a(n) = L(4*n) + 4*(-1)^n*L(2*n) + 6.

%F a(n) = L(n-2)*L(n-1)*L(n+1)*L(n+2) + 25, for n >=1.

%F G.f.: (16-79*x-164*x^2+76*x^3+x^4)/((1-x)*(1+3*x+x^2)*(1-7*x+x^2)). [See Mansour p. 207] - _R. J. Mathar_, Oct 26 2008

%F a(0)=16, a(1)=1, a(2)=81, a(3)=256, a(4)=2401, a(n) = 5*a(n-1) + 15*a(n-2) - 15*a(n-3) - 5*a(n-4) + a(n-5). - _Harvey P. Dale_, Jul 04 2014

%F Sum_{i=0..n} a(i) = 11 + 6*n + 4*(-1)^n*F(2*n+1) + F(4*n+2), for F = A000045. - _Adam Mohamed_ and _Greg Dresden_, Jul 02 2024

%F Product_{n>=2} (1 - 25/a(n)) = phi^5/18, where phi is the golden ratio (A001622) (Davlianidze, 2020). - _Amiram Eldar_, Dec 04 2024

%t LucasL[Range[0,20]]^4 (* or *) LinearRecurrence[{5,15,-15,-5,1},{16,1,81,256,2401},21] (* _Harvey P. Dale_, Jul 04 2014 *)

%t CoefficientList[Series[(16 - 79 x - 164 x^2 + 76 x^3 + x^4)/((1 - x) (1 + 3*x+x^2)*(1-7*x+x^2)), {x,0,50}], x] (* _G. C. Greubel_, Dec 21 2017 *)

%o (Magma) [ Lucas(n)^4 : n in [0..120]]; // _Vincenzo Librandi_, Apr 14 2011

%o (PARI) for(n=0, 30, print1( (fibonacci(n+1) + fibonacci(n-1))^4, ", ")) \\ _G. C. Greubel_, Dec 21 2017

%o (PARI) x='x+O('x^30); Vec((16-79*x-164*x^2+76*x^3+x^4)/((1-x)*(1+3*x+x^2)*(1-7*x+x^2))) \\ _G. C. Greubel_, Dec 21 2017

%Y Cf. A000032, A000045, A001254, A001622, A075515.

%Y Fourth row of array A103324.

%K nonn,easy

%O 0,1

%A _Ralf Stephan_, Nov 01 2004