|
| |
|
|
A163197
|
|
(1/4) L(n)^2 F(n+1)^2 L(n-1) F(n+2) from Fibonacci and Lucas numbers.
|
|
6
| |
|
|
-1, 1, 27, 540, 9800, 176176, 3162159, 56744793, 1018249595, 18271762300, 327873509424, 5883451505856, 105574253853887, 1894453118539345, 33994581881622075, 610008020755286076, 10946149791725643704, 196420688230338021808, 3524626238354441796015, 63246851602149831726825
(list; graph; refs; listen; history; internal format)
|
|
|
|
OFFSET
| 0,3
|
|
|
COMMENTS
| 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).
|
|
|
REFERENCES
| 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).
|
|
|
FORMULA
| 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
|
|
|
MATHEMATICA
| a[n_Integer] := (1/4) LucasL[n]^2 Fibonacci[n+1]^2 LucasL[n-1] Fibonacci[n+2]
|
|
|
CROSSREFS
| Cf. A163194, A163195, A163196, A163198, A163199
Sequence in context: A014928 A163199 A051561 * A061914 A076008 A185891
Adjacent sequences: A163194 A163195 A163196 * A163198 A163199 A163200
|
|
|
KEYWORD
| sign,easy
|
|
|
AUTHOR
| Stuart Clary (clary(AT)uakron.edu), Jul 24, 2009
|
| |
|
|
