A156089 Alternating sum of the squares of the first n odd-indexed Fibonacci numbers.

0, -1, 3, -22, 147, -1009, 6912, -47377, 324723, -2225686, 15255075, -104559841, 716663808, -4912086817, 33667943907, -230763520534, 1581676699827, -10840973378257, 74305136947968, -509294985257521, 3490759759854675

Offset: 0

Author

Stuart Clary, Feb 04 2009

Keywords

• sign
• easy

Comments

Natural bilateral extension (brackets mark index 0): ..., 6912, -1009, 147, -22, 3, -1, [0], -1, 3, -22, 147, -1009, 6912, ... This is a(n)-reversed followed by a(n), without repeating the 0. That is, a(-n) = a(n).

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (-7,0,7,1).

Crossrefs

Cf. A000032, A000045, A103434, A103433, A156088.

Programs

Magma

A156089:= func< n | ((-1)^n*(Lucas(4*n)+3) - 5)/15 >; // G. C. Greubel, Jun 12 2025

Mathematica

a[n_]:= If[n >= 0, Sum[(-1)^k Fibonacci[2k-1]^2, {k,n} ], -Sum[-(-1)^k Fibonacci[-2k+1]^2, {k,-n}]];
Join[{0},Accumulate[Times@@@Partition[Riffle[Fibonacci[Range[1,43,2]]^2, {-1,1}],2]]] (* Harvey P. Dale, Aug 18 2011 *)
LinearRecurrence[{-7,0,7,1}, {0,-1,3,-22}, 41] (* G. C. Greubel, Jun 12 2025 *)

SageMath

def A156089(n): return ((-1)^n*(lucas_number2(4*n,1,-1)+3)-5)//15 # G. C. Greubel, Jun 12 2025

Formula

Let F(n) be the n-th Fibonacci number, A000045(n), and L(n) be the n-th Lucas number, A000032(n).

a(n) = Sum_{k=1..n} (-1)^k F(2*k-1)^2.

Closed form: a(n) = (-1)^n * (L(4*n) + 3)/15 - 1/3.

Factored closed form: a(n) = (1/3)*F(2*n)^2 if n is even.

Not-so-factored closed form: a(n) = -(F(2*n)^2 + 2)/3 if n is odd.

Recurrence: a(n) + 7*a(n-1) - 7*a(n-3) - a(n-4) = 0.

G.f.: -x*(1 + 4*x + x^2)/(1 + 7*x - 7*x^3 - x^4) = -x*(1 + 4*x + x^2)/((1 - x)*(1 + x)*(1 + 7*x + x^2)).

E.g.f.: (2/15)*( exp(-7*x/2)*cosh(3*sqrt(5)*x/2) - cosh(x) - 4*sinh(x) ). - G. C. Greubel, Jun 12 2025