A203171 Alternating sum of the fourth powers of the first n even-indexed Fibonacci numbers.

0, -1, 80, -4016, 190465, -8960160, 421021536, -19779631105, 929225609456, -43653851217680, 2050801968082945, -96344039926706496, 4526119083346841280, -212631252937414840321, 9989142769386670981520, -469277078911056723578480

Offset: 0

Author

Stuart Clary, Dec 30 2011

Keywords

• sign
• easy

Comments

Natural bilateral extension (brackets mark index 0): ..., 8960160, -190465, 4016, -80, 1, 0, [0], -1, 80, -4016, 190465, -8960160, ... That is, a(-n) = -a(n-1).

Links

• G. C. Greubel, Table of n, a(n) for n = 0..500
• R. S. Melham, Alternating sums of fourth powers of Fibonacci and Lucas numbers, The Fibonacci Quarterly, 38(3):254-259, June-July 2000.
• Index entries for linear recurrences with constant coefficients, signature (-55,-385,-385,-55,-1)

Crossrefs

Cf. A203169, A203170, A203172.

Cf. A156088, A163201.

Programs

Maple

with(combinat): A203171:=n->(-1)^n*(1/21)*fibonacci(2*n)*fibonacci(2*n+2)*(3*fibonacci(2*n+1)^2 - 5): seq(A203171(n), n=0..20); # Wesley Ivan Hurt, Jan 16 2017

Mathematica

a[n_Integer] := (-1)^n (1/525)(3*LucasL[8n+4] - 28*LucasL[4n+2] + 63); Table[a[n], {n, 0, 20}]

PARI

a(n) = sum(k=1, n, (-1)^k*fibonacci(2*k)^4); \\ Michel Marcus, Apr 16 2016

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(2*k)^4.

Closed form: a(n) = (-1)^n (1/525)(3 L(8n+4) - 28 L(4n+2) + 63).

Factored closed form: a(n) = (-1)^n (1/21) F(2n) F(2n+2) (3 F(2n+1)^2 - 5).

Alternate factored closed form: a(n) = (-1)^n (1/21) F(2n) F(2n+2) (3 F(2n) F(2n+2) - 2).

Recurrence: a(n) + 55 a(n-1) + 385 a(n-2) + 385 a(n-3) + 55 a(n-4) + a(n-5) = 0.

G.f.: A(x) = (-x + 25 x^2 - x^3)/(1 + 55 x + 385 x^2 + 385 x^3 + 55 x^4 + x^5) = -x(1 - 25 x + x^2)/((1 + x)(1 + 7 x + x^2)(1 + 47 x + x^2)).