0,3

Natural bilateral extension (brackets mark index 0): ..., -37225, 2079, -118, 7, -1, [0], -1, 7, -118, 2079, -37225, ... This is A163202-reversed followed by A163202, without repeating the 0. That is, A163202(-n) = A163202(n). Thus A163202(n) is an even function of n.

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)^3

Closed form: (1/50)(L(6n) + 6 L(2n) - 14) if n is even; -(1/50)(L(6n) + 6 L(2n) + 14) if n is odd

Factored closed form: 1/10) F(n)^2 (L(4 n) + 2 L(2n) + 9) if n is even; -(1/10) F(n)^2 (L(4 n) - 2 L(2n) + 9) if n is odd

Recurrence: a(n) + 21 a(n-1) + 56 a(n-2) + 21 a(n-3) + a(n-4) = -28

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) = (-x - 13 x^2 - 13 x^3 - x^4)/(1 + 20 x + 35 x^2 - 35 x^3 - 20 x^4 - x^5) = -x(1 + x)(1 + 12 x +x^2)/((1 - x)(1 + 3 x + x^2)(1 + 18 x + x^2))

a(-n) = a(n). - Michael Somos Aug 11 2009

-x + 7*x^2 - 118*x^3 + 2079*x^4 - 37225*x^5 + 667744*x^6 - 11981593*x^7 + ... - Michael Somos Aug 11 2009

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

(PARI) {a(n) = ((-1)^n * (fibonacci(6*n) / 2 + fibonacci(6*n - 1) + 3*fibonacci(2*n - 1) + 3*fibonacci(2*n + 1)) - 7) / 25} /* Michael Somos Aug 11 2009 */

Cf. A163198, A163200, A163201, A119284

Sequence in context: A009696 A076283 A097202 this_sequence A057769 A113667 A092612

Adjacent sequences: A163199 A163200 A163201 this_sequence A163203 A163204 A163205

sign,easy

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