A119287 Alternating sum of the sixth powers of the first n Fibonacci numbers.

0, -1, 0, -64, 665, -14960, 247184, -4579625, 81186496, -1463617920, 26217022705, -470764268256, 8445336180000, -151560390359569, 2719538168853120, -48800836192146880, 875690649999921929, -15713664197268146000, 281970036429821245616, -5059748557502924705465, 90793493265349521060160, -1629223203785737022267136, 29235223670642547226470625

Offset

0,4

Comments

Natural bilateral extension (brackets mark index 0): ..., 14960, -665, 64, 0, 1, 0, [0], -1, 0, -64, 665, -14960, 247184, ... This is (-A119287)-reversed followed by A119287.

Links

Table of n, a(n) for n=0..22.

Index entries for linear recurrences with constant coefficients, signature (-12,117,156,-520,156,117,-12,-1).

Formula

Let F(n) be the Fibonacci number A000045(n).

a(n) = Sum_{k=1..n} (-1)^k F(k)^6.

a(n) = (-1)^n (1/250) F(6n+3) - (6/125) F(4n+2) + (-1)^n (3/25) F(2n+1) - (2/25)(2 n + 1).

Recurrence: a(n) + 12 a(n-1) - 117 a(n-2) - 156 a(n-3) + 520 a(n-4) - 156 a(n-5) - 117 a(n-6) + 12 a(n-7) + a(n-8) = 0.

G.f.: A(x) = (-x - 12 x^2 + 53 x^3 + 53 x^4 - 12 x^5 - x^6)/(1 + 12 x - 117 x^2 - 156 x^3 + 520 x^4 - 156 x^5 - 117 x^6 + 12 x^7 + x^8) = -x(1 + x)(1 + 11 x - 64 x^2 + 11 x^3 + x^4)/((1 - x)^2 (1 + 3 x + x^2)(1 - 7 x + x^2)(1 + 18 x + x^2)).

Mathematica

a[n_Integer] := If[ n >= 0, Sum[ (-1)^k Fibonacci[k]^6, {k, 1, n} ], Sum[ -(-1)^k Fibonacci[ -k]^6, {k, 1, -n - 1} ] ]
Accumulate[Times@@@Partition[Riffle[Fibonacci[Range[0, 30]]^6, {1, -1}, {2, -1, 2}], 2]] (* Harvey P. Dale, Jul 23 2013 *)

Crossrefs

Cf. A098532, A119282, A119283, A119284, A119285, A119286, A128696, A128698.

Sequence in context: A221509 A283337 A187620 * A318023 A320408 A223963

Adjacent sequences: A119284 A119285 A119286 * A119288 A119289 A119290

Keyword

sign,easy

Author

Stuart Clary, May 13 2006

Status

approved