0,3

Natural bilateral extension (brackets mark index 0): ..., 9800, 539, 27, 0, -1, [-1], 0, 27, 539, 9800, 176175, ... This is A163199-reversed followed by A163199. That is, A163199(-n) = A163199(n-1).

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} F(2k)^3 - 1

a(n) = sum_{k=2..n} F(2k)^3 for n > 0

a(n) = A163198(n) - 1

Closed form: a(n) = F(6n+3)/20 - (3/5) F(2n+1) - 1/2

Closed form: a(n) = (1/4)(F(2n+1)^3 - 3 F(2n+1) - 2)

Factored closed form: a(n) = (1/4) L(n)^2 F(n+1)^2 L(n-1) F(n+2) = A163197(n) if n is even; a(n) = (1/4) F(n)^2 L(n+1)^2 F(n-1) L(n+2) = A163195(n) if n is odd

Recurrence: a(n) - 21 a(n-1) + 56 a(n-2) - 21 a(n-3) + a(n-4) = -8

Recurrence: a(n) - 22 a(n-1) + 77 a(n-2) - 77 a(n-3) + 22 a(n-4) - a(n-5) = 0

G.f.: A(x) = (-1 + 22 x - 50 x^2 + 22 x^3 - x^4)/(1 - 22 x + 77 x^2 - 77 x^3 + 22 x^4 - x^5) = -(1 - 22 x + 50 x^2 - 22 x^3 + x^4)/((1 - x)(1 - 3 x + x^2 )(1 - 18 x + x^2))

a[n_Integer] := If[ n >= 0, Sum[ Fibonacci[2k]^3, {k, 1, n} ] - 1, -Sum[ Fibonacci[-2k]^3, {k, 1, -n - 1} ] - 1 ]

Cf. A163194, A163195, A163196, A163197, A163198

Sequence in context: A016887 A110896 A014928 this_sequence A051561 A163197 A061914

Adjacent sequences: A163196 A163197 A163198 this_sequence A163200 A163201 A163202

sign,easy

Stuart Clary (clary(AT)uakron.edu), Jul 24, 2009