0,3

Natural bilateral extension (brackets mark index 0): ..., 9801, 539, 28, 0, 0, [-1], 1, 27, 540, 9800, 176176, ... This is A163195-reversed followed by A163197. That is, A163197(-n) = A163195(n-1).

Stuart Clary and Paul D. Hemenway, On sums of cubes of Fibonacci numbers, Applications of Fibonacci Numbers, Vol. 5 (St. Andrews, 1992), 123-136, Kluwer Acad. Publ., 1993. See equation (3).

Let F(n) be the Fibonacci number A000045(n) and let L(n) be the Lucas number A000032(n).

a(n) = (1/4) L(n)^2 F(n+1)^2 L(n-1) F(n+2)

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

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

Summation form: a(n) = sum_{k=2..n} F(2k)^3 = A163199(n) if n is even; a(n) = sum_{k=1..n} F(2k)^3 = A163198(n) if n is odd

Recurrence: a(n) - 21 a(n-1) + 56 a(n-2) - 21 a(n-3) + a(n-4) = -(-1)^n 50

Recurrence: a(n) - 20 a(n-1) + 35 a(n-2) + 35 a(n-3) - 20 a(n-4) + a(n-5) = 0

G.f.: A(x) = (-1 + 21 x - 28 x^2)/(1 - 20 x + 35 x^2 + 35 x^3 - 20 x^4 + x^5) = -(1 - 21 x + 28 x^2)/((1 + x)(1 - 3 x + x^2)(1 - 18 x + x^2))

A163195(n) - a(n) = (-1)^n

a[n_Integer] := (1/4) LucasL[n]^2 Fibonacci[n+1]^2 LucasL[n-1] Fibonacci[n+2]

Cf. A163194, A163195, A163196, A163198, A163199

Sequence in context: A014928 A163199 A051561 this_sequence A061914 A076008 A099753

Adjacent sequences: A163194 A163195 A163196 this_sequence A163198 A163199 A163200

sign,easy

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