A156091 One fourth of the alternating sum of the squares of the first n Fibonacci numbers with index divisible by 3.

0, -1, 15, -274, 4910, -88115, 1581149, -28372580, 509125276, -9135882405, 163936757995, -2941725761526, 52787126949450, -947226559328599, 16997290940965305, -305004010378046920, 5473074895863879224, -98210344115171779145

Offset

0,3

Comments

Natural bilateral extension (brackets mark index 0): ..., 88115, -4910, 274, -15, 1, 0, [0], -1, 15, -274, 4910, -88115, 1581149, ... This is (-A156091)-reversed followed by A156091. That is, A156091(-n) = -A156091(n-1).

Links

Table of n, a(n) for n=0..17.

Index entries for linear recurrences with constant coefficients, signature (-16,34,-16,-1).

Formula

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

a(n) = (1/4) sum_{k=1..n} (-1)^k F(3k)^2.

Closed form: a(n) = (-1)^n F(6n+3)/40 - (2 n + 1)/20.

Recurrence: a(n) + 17 a(n-1) - 17 a(n-2) - a(n-3) = -2.

Recurrence: a(n) + 16 a(n-1) - 34 a(n-2) + 16 a(n-3) + a(n-4) = 0.

G.f.: A(x) = -(x + x^2)/(1 + 16 x - 34 x^2 + 16 x^3 + x^4) = -x(1 + x)/((1 - x)^2 (1 + 18 x + x^2)).

Mathematica

a[n_Integer] := If[ n >= 0, Sum[ (-1)^k (1/4) Fibonacci[3k]^2, {k, 1, n} ], Sum[ -(-1)^k (1/4) Fibonacci[-3k]^2, {k, 1, -n - 1} ] ]

Crossrefs

Cf. A156084, A156085, A156090.

Sequence in context: A013382 A227973 A195615 * A194728 A284077 A339118

Adjacent sequences: A156088 A156089 A156090 * A156092 A156093 A156094

Keyword

sign,easy

Author

Stuart Clary, Feb 04 2009

Status

approved