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

Author

Roger L. Bagula, Feb 21 2006

Keywords

• nonn
• less

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).

Crossrefs

Cf. A000032, A000045.

Programs

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 *)

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

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

Extensions

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