A203172 Alternating sum of the fourth powers of the first n odd-indexed Fibonacci numbers.

0, -1, 15, -610, 27951, -1308385, 61433856, -2885861665, 135572548335, -6369013518946, 299207991620175, -14056406104466881, 660351875572408320, -31022481722865482305, 1457396288941918481871, -68466603097469928960610

Offset: 0

Author

Stuart Clary, Dec 30 2011

Keywords

• sign
• easy

Comments

Natural bilateral extension (brackets mark index 0): ..., -1308385, 27951, -610, 15, -1, [0], -1, 15, -610, 27951, -1308385, ... That is, a(-n) = a(n).

Links

• Index entries for linear recurrences with constant coefficients, signature (-54,-330,0,330,54,1)

Crossrefs

Cf. A203169, A203170, A203171.

Cf. A156089, A163202.

Programs

Mathematica

a[n_Integer] := (-1)^n (1/525)(3*LucasL[8n] + 28*LucasL[4n] + 63 - (-1)^n 125); Table[a[n], {n, 0, 20}]
LinearRecurrence[{-54,-330,0,330,54,1},{0,-1,15,-610,27951,-1308385},20] (* Harvey P. Dale, Jun 04 2025 *)

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(2k-1)^4.

Closed form: a(n) = (-1)^n (1/525)(3 L(8n) + 28 L(4n) + 63 - (-1)^n 125).

Alternate closed form: a(n) = (1/21) F(2n)^2 (3 F(2n)^2 + 8) if n is even, a(n) = -(1/21)(3 F(2n)^4 + 8 F(2n)^2 + 10) if n is odd.

Recurrence: a(n) + 54 a(n-1) + 330 a(n-2) - 330 a(n-4) - 54 a(n-5) - a(n-6) = 0.

G.f.: A(x) = -(x + 39 x^2 + 130 x^3 + 39 x^4 + x^5)/(1 + 54 x + 330 x^2 - 330 x^4 - 54 x^5 - x^6) = -x(1 + 39 x + 130 x^2 + 39 x^3 + x^4)/((1 - x)(1 + x)(1 + 7 x + x^2)(1 + 47 x + x^2)).