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A099923
Fourth powers of Lucas numbers A000032.
6
16, 1, 81, 256, 2401, 14641, 104976, 707281, 4879681, 33362176, 228886641, 1568239201, 10750371856, 73680216481, 505022001201, 3461445366016, 23725169980801, 162614549665681, 1114577187760656, 7639424429247601
OFFSET
0,1
REFERENCES
A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 56.
LINKS
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.
Toufik Mansour, A formula for the generating functions of powers of Horadam's sequence, Australas. J. Combin. 30 (2004) 207-212.
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
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 *)
PROG
(Magma) [ Lucas(n)^4 : n in [0..120]]; // Vincenzo Librandi, Apr 14 2011
(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
CROSSREFS
Cf. A075515.
Fourth row of array A103324.
Sequence in context: A329927 A036179 A309132 * A351244 A105671 A145828
KEYWORD
nonn
AUTHOR
Ralf Stephan, Nov 01 2004
STATUS
approved