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A103325
Fifth powers of Lucas numbers.
5
32, 1, 243, 1024, 16807, 161051, 1889568, 20511149, 229345007, 2535525376, 28153056843, 312079600999, 3461619737632, 38387392786601, 425733547012443, 4721411479245824, 52361450147627807, 580696556856146851, 6440026990881070368, 71420978989035821749
OFFSET
0,1
REFERENCES
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.
FORMULA
a(n) = A000032(n)^5 = A000032(n)*A099923(n).
a(n) = L(5*n) + 5*(-1)^n*L(3*n) + 10*L(n), L(n) = A000032, the Lucas numbers.
G.f.: (32-255*x-1045*x^2+960*x^3+235*x^4-x^5)/((1-x-x^2)*(1+4*x-x^2)* (1-11*x-x^2)). [T. Mansour, Australas. J. Comb. 30 (2004), 207] - R. J. Mathar, Oct 26 2008
MATHEMATICA
Table[LucasL[n]^5, {n, 0, 30}] (* or *) CoefficientList[Series[(32 - 255 x - 1045 x^2 + 960 x^3 + 235 x^4 - x^5)/((1-x-x^2)*(1+4*x-x^2)*(1-11*x- x^2)), {x, 0, 50}], x] (* G. C. Greubel, Dec 21 2017 *)
PROG
(Magma) [ Lucas(n)^5 : n in [0..120]]; // Vincenzo Librandi, Apr 14 2011
(PARI) a(n)=(fibonacci(n-1)+fibonacci(n+1))^5 \\ Charles R Greathouse IV, Jun 11 2015
(PARI) x='x+O('x^30); Vec((32-255*x-1045*x^2+960*x^3+235*x^4-x^5)/((1-x-x^2)*(1+4*x-x^2)* (1-11*x-x^2))) \\ G. C. Greubel, Dec 21 2017
CROSSREFS
Fifth row of array A103324.
Sequence in context: A301910 A302154 A221193 * A351245 A373777 A221760
KEYWORD
nonn,easy
AUTHOR
Ralf Stephan, Feb 03 2005
STATUS
approved