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A114695
Three consecutive elements of the sequence built from a quadratic form over four consecutive Fibonacci numbers A000045.
6
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
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
Sequence in context: A270554 A037010 A294184 * A134084 A267346 A372238
KEYWORD
nonn,less
AUTHOR
Roger L. Bagula, Feb 21 2006
EXTENSIONS
Edited by the Associate Editors of the OEIS, Sep 02 2009
STATUS
approved