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A088137
Generalized Gaussian Fibonacci integers.
19
0, 1, 2, 1, -4, -11, -10, 13, 56, 73, -22, -263, -460, -131, 1118, 2629, 1904, -4079, -13870, -15503, 10604, 67717, 103622, 4093, -302680, -617639, -327238, 1198441, 3378596, 3161869, -3812050, -17109707, -22783264, 5762593, 79874978, 142462177, 45299420, -336787691
OFFSET
0,3
COMMENTS
The Lucas U(P=2, Q=3) sequence. - R. J. Mathar, Oct 24 2012
Hence for n >= 0, a(n+2)/a(n+1) equals the continued fraction 2 - 3/(2 - 3/(2 - 3/(2 - ... - 3/2))) with n 3's. - Greg Dresden, Oct 06 2019
With different signs, 0, 1, -2, 1, 4, -11, 10, 13, -56, 73, 22, -263, 460, ... also the Lucas U(-2,3) sequence. - R. J. Mathar, Jan 08 2013
From Peter Bala, Apr 01 2018: (Start)
The companion Lucas sequence V(n,2,3) is A087455.
Define a binary operation o on rational numbers by x o y = (x + y)/(1 - 2*x*y). This is a commutative and associative operation with identity 0. Then 1 o 1 o ... o 1 (n terms) = a(n)/A087455(n). Cf. A025172 and A127357. (End)
LINKS
Beata Bajorska-Harapińska, Barbara Smoleń, and Roman Wituła, On Quaternion Equivalents for Quasi-Fibonacci Numbers, Shortly Quaternaccis, Advances in Applied Clifford Algebras (2019) Vol. 29, 54.
Tricia Brown, Alexander Stokolos, and Dmitry Dmitrishin, Convolutions of linear recurrent sequences and Gegenbauer polynomials, ResearchGate (2026). See p. 9.
Ronald Orozco López, Deformed Differential Calculus on Generalized Fibonacci Polynomials, arXiv:2211.04450 [math.CO], 2022.
Mihai Prunescu and Lorenzo Sauras-Altuzarra, On the representation of C-recursive integer sequences by arithmetic terms, arXiv:2405.04083 [math.LO], 2024. See p. 16.
Mihai Prunescu and Joseph M. Shunia, On modular representations of C-recursive integer sequences, arXiv:2502.16928 [math.NT], 2025. See p. 6.
Wikipedia, Lucas sequence
FORMULA
a(n) = 3^(n/2)*sin(n*atan(sqrt(2)))/sqrt(2).
|3*A087455(n) - A087455(n+1)| = 2*a(n+1) or 3*A087455(n) + A087455(n+1) = 2*a(n+1). - Creighton Dement, Aug 02 2004
G.f.: x/(1 - 2*x + 3*x^2).
E.g.f.: exp(x)*sin(sqrt(2)*x)/sqrt(2).
a(n) = 2*a(n-1) - 3*a(n-2) for n > 1, a(0)=0, a(1)=1.
a(n) = ((1 + i*sqrt(2))^n - (1 - i*sqrt(2))^n)/(2*i*sqrt(2)), where i=sqrt(-1).
a(n) = Im((1 + i*sqrt(2))^n/sqrt(2)).
a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2*k+1)(-2)^k.
3^(n+1) = 9*(A087455(n))^2 + 2*(A087455(n+1))^2 - 2*(a(n+2))^2; 3^n = a(n+1)^2 + 3*a(n)^2 - 2*a(n+1)*a(n) for n > 0 - Creighton Dement, Jan 20 2005
G.f.: G(0)*x/(2*(1-x)), where G(k) = 1 + 1/(1 - x*(2*k+1)/(x*(2*k+3) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 25 2013
G.f.: Q(0)*x/2, where Q(k) = 1 + 1/(1 - x*(4*k+2 - 3*x)/( x*(4*k+4 - 3*x) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Sep 06 2013
a(n+1) = Sum_{k=0..n} A123562(n,k). - Philippe Deléham, Nov 23 2013
a(n) = n*hypergeom([(1-n)/2,(2-n)/2],[3/2],-2). - Gerry Martens, Sep 03 2023
MAPLE
A[0]:= 0: A[1]:= 1:
for n from 2 to 100 do A[n]:= 2*A[n-1] - 3*A[n-2] od:
seq(A[n], n=0..100); # Robert Israel, Aug 05 2014
MATHEMATICA
LinearRecurrence[{2, -3}, {0, 1}, 40] (* Harvey P. Dale, Nov 03 2014 *)
PROG
(SageMath) [lucas_number1(n, 2, 3) for n in range(0, 38)] # Zerinvary Lajos, Apr 23 2009
(PARI) x='x+O('x^50); concat([0], Vec(x/(1-2*x+3*x^2))) \\ G. C. Greubel, Oct 22 2018
(Magma) [n le 2 select n-1 else 2*Self(n-1)-3*Self(n-2): n in [1..50]]; // G. C. Greubel, Oct 22 2018
KEYWORD
sign,easy
AUTHOR
Paul Barry, Sep 20 2003
STATUS
approved