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
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)
Programs
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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
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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
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)).
Comments