A114695 Three consecutive elements of the sequence built from a quadratic form over four consecutive Fibonacci numbers A000045.

2, 2, 4, 104, 143, 169, 4895, 6764, 7921, 229970, 317810, 372100, 10803704, 14930351, 17480761, 507544127, 701408732, 821223649, 23843770274, 32951280098, 38580030724, 1120149658760, 1548008755919, 1812440220361

Offset

0,1

Links

G. C. Greubel, Table of n, a(n) for n = 0..500

Index entries for linear recurrences with constant coefficients, signature (0,0,48,0,0,-48,0,0,1).

Formula

a(3*n) = Fibonacci(4*n+2)*Fibonacci(4*n+3).

a(3*n+1) = Lucas(4*n+1)*Fibonacci(4*n+3).

a(3*n+2) = Fibonacci(4*n+3)*Fibonacci(4*n+3).

From R. J. Mathar, Apr 16 2009: (Start)

a(3*n) = A001654(4*n+2).

a(3*n+1) = A128535(4*n+3).

a(3*n+2) = A007598(4*n+3).

G.f.: (2+2*x+4*x^2+8*x^3+47*x^4-23*x^5-x^6-4*x^7+x^8)/((1-x)*(1+x+x^2)*(1-47*x^3+x^6)).

a(n) = 48*a(n-3) - 48*a(n-6) + a(n-9). (End)

a(n) = F(4*m+3)*( 4*F(4*m+2) - ((n^2 mod 3)*F(4*m+2) + ((n+2)^2 mod 3)*Lucas(4*m+1) + ((n+1)^2 mod 3)*F(4*m+3)) ), where m = floor(n/3) and F = Fibonacci. - G. C. Greubel, May 24 2021

Mathematica

F[n_]:= Fibonacci[n]; Flatten[Table[{F[4*n+2]*F[4*n+3], (F[4*n]+F[4*n+2])*F[4*n+ 3], F[4*n+3]^2}, {n, 0, 12}]] (* modified by G. C. Greubel, May 24 2021 *)
With[{m = Floor[n/3], F = Fibonacci}, Table[F[4*m+3]*(4*F[4*m+2] -(Mod[n^2, 3]*F[4*m +2] +Mod[(n+2)^2, 3]*LucasL[4*m+1] +Mod[(n+1)^2, 3]*F[4*m+3])), {n, 0, 40}]] (* G. C. Greubel, May 24 2021 *)

Prog

(Sage)
f=fibonacci;
def A114695(n): return f(4*(n//3)+3)*( 4*f(4*(n//3)+2) - ((n^2%3)*f(4*(n//3)+2) + ((n+2)^2%3)*(f(4*(n//3)+2) + f(4*(n//3))) + ((n+1)^2%3)*f(4*(n//3)+3) ) )
[A114695(n) for n in (0..40)] # G. C. Greubel, May 24 2021

Crossrefs

Cf. A000032, A000045.

Sequence in context: A270554 A037010 A294184 * A134084 A267346 A264933

Adjacent sequences: A114692 A114693 A114694 * A114696 A114697 A114698

Keyword

nonn,less

Author

Roger L. Bagula, Feb 21 2006

Extensions

Edited by the Associate Editors of the OEIS, Sep 02 2009

Status

approved