A099923 Fourth powers of Lucas numbers A000032.

16, 1, 81, 256, 2401, 14641, 104976, 707281, 4879681, 33362176, 228886641, 1568239201, 10750371856, 73680216481, 505022001201, 3461445366016, 23725169980801, 162614549665681, 1114577187760656, 7639424429247601

Offset: 0

Author

Ralf Stephan, Nov 01 2004

Keywords

• nonn
• easy

References

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

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• Mohammad K. Azarian, Identities Involving Lucas or Fibonacci and Lucas Numbers as Binomial Sums, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 45, 2012, pp. 2221-2227.
• Pridon Davlianidze, Problem B-1270, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 58, No. 2 (2020), p. 179; Four Telescopic Infinite Products, Solution to Problem B-1270 by Jason L. Smith, ibid., Vol. 59, No. 2 (2021), pp. 183-184.
• Toufik Mansour, A formula for the generating functions of powers of Horadam's sequence, Australas. J. Combin. 30 (2004) 207-212.
• Index entries for linear recurrences with constant coefficients, signature (5,15,-15,-5,1).

Crossrefs

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

Fourth row of array A103324.

Programs

Magma

[ Lucas(n)^4 : n in [0..120]]; // Vincenzo Librandi, Apr 14 2011

Mathematica

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

PARI

for(n=0, 30, print1( (fibonacci(n+1) + fibonacci(n-1))^4, ", ")) \\ G. C. Greubel, Dec 21 2017

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

Formula

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

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

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

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

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

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

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