A119285 Alternating sum of the fourth powers of the first n Fibonacci numbers.

0, -1, 0, -16, 65, -560, 3536, -25025, 169456, -1166880, 7983745, -54758496, 375223200, -2572072321, 17628580320, -120829829680, 828175410881, -5676410656400, 38906666170736, -266670338968385, 1827785480332240, -12527828615754816, 85867013279034625, -588541268397840576, 4033921854875707200, -27648911743562183425

Offset: 0

Author

Stuart Clary, May 13 2006

Keywords

• sign
• easy

Comments

Natural bilateral extension (brackets mark index 0): ..., -3536, 560, -65, 16, 0, 1, 0, [0], -1, 0, -16, 65, -560, 3536, -25025, ... This is (-A119285)-reversed followed by A119285.

Links

• Kunle Adegoke, Sums of fourth powers of Fibonacci and Lucas numbers, arXiv:1706.00407 [math.NT], 2017.
• Index entries for linear recurrences with constant coefficients, signature (-5,15,15,-5,-1).

Crossrefs

Cf. A005969, A119282, A119283, A119284, A119286, A119287, A128696, A128698.

Programs

Mathematica

a[n_Integer] := If[ n >= 0, Sum[ (-1)^k Fibonacci[k]^4, {k, 1, n} ], Sum[ -(-1)^k Fibonacci[ -k]^4, {k, 1, -n - 1} ] ]
LinearRecurrence[{-5,15,15,-5,-1},{0,-1,0,-16,65},30] (* Harvey P. Dale, Apr 02 2018 *)

Formula

Let F(n) be the Fibonacci number A000045(n) and let L(n) be the Lucas number A000032(n).

a(n) = Sum_{k=1..n} (-1)^k F(k)^4.

Closed form: a(n) = (-1)^n L(4n+2)/75 - (4/25) L(2n+1) + (-1)^n 3/25.

Factored closed form: a(n) = (-1)^n (1/3) F(n-2) F(n) F(n+1) F(n+3).

Recurrence: a(n) + 5 a(n-1) - 15 a(n-2) - 15 a(n-3) + 5 a(n-4) + a(n-5) = 0.

G.f.: A(x) = (-x - 5 x^2 - x^3)/(1 + 5 x - 15 x^2 - 15 x^3 + 5 x^4 + x^5) = -x(1 + 5 x + x^2)/((1 + x)(1 - 3 x + x^2)(1 + 7 x + x^2)).