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%I #106 Jun 14 2024 09:41:53
%S 0,1,2,1,-4,-11,-10,13,56,73,-22,-263,-460,-131,1118,2629,1904,-4079,
%T -13870,-15503,10604,67717,103622,4093,-302680,-617639,-327238,
%U 1198441,3378596,3161869,-3812050,-17109707,-22783264,5762593,79874978,142462177,45299420,-336787691
%N Generalized Gaussian Fibonacci integers.
%C The Lucas U(P=2, Q=3) sequence. - _R. J. Mathar_, Oct 24 2012
%C 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
%C 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
%C From _Peter Bala_, Apr 01 2018: (Start)
%C The companion Lucas sequence V(n,2,3) is A087455.
%C 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)
%H Vincenzo Librandi, <a href="/A088137/b088137.txt">Table of n, a(n) for n = 0..1000</a>
%H Beata Bajorska-Harapińska, Barbara Smoleń, and Roman Wituła, <a href="https://doi.org/10.1007/s00006-019-0969-9">On Quaternion Equivalents for Quasi-Fibonacci Numbers, Shortly Quaternaccis</a>, Advances in Applied Clifford Algebras (2019) Vol. 29, 54.
%H Ronald Orozco López, <a href="https://arxiv.org/abs/2211.04450">Deformed Differential Calculus on Generalized Fibonacci Polynomials</a>, arXiv:2211.04450 [math.CO], 2022.
%H Mihai Prunescu, <a href="https://arxiv.org/abs/2406.06436">On other two representations of the C-recursive integer sequences by terms in modular arithmetic</a>, arXiv:2406.06436 [math.NT], 2024. See p. 18.
%H Mihai Prunescu and Lorenzo Sauras-Altuzarra, <a href="https://arxiv.org/abs/2405.04083">On the representation of C-recursive integer sequences by arithmetic terms</a>, arXiv:2405.04083 [math.LO], 2024. See p. 16.
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Lucas_sequence">Lucas sequence</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (2,-3).
%H <a href="/index/Lu#Lucas">Index entries for Lucas sequences</a>
%F a(n) = 3^(n/2)*sin(n*atan(sqrt(2)))/sqrt(2).
%F |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
%F G.f.: x/(1 - 2*x + 3*x^2).
%F E.g.f.: exp(x)*sin(sqrt(2)*x)/sqrt(2).
%F a(n) = 2*a(n-1) - 3*a(n-2) for n > 1, a(0)=0, a(1)=1.
%F a(n) = ((1 + i*sqrt(2))^n - (1 - i*sqrt(2))^n)/(2*i*sqrt(2)), where i=sqrt(-1).
%F a(n) = Im((1 + i*sqrt(2))^n/sqrt(2)).
%F a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2*k+1)(-2)^k.
%F 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
%F 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
%F 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
%F a(n+1) = Sum_{k=0..n} A123562(n,k). - _Philippe Deléham_, Nov 23 2013
%F a(n) = n*hypergeom([(1-n)/2,(2-n)/2],[3/2],-2). - _Gerry Martens_, Sep 03 2023
%p A[0]:= 0: A[1]:= 1:
%p for n from 2 to 100 do A[n]:= 2*A[n-1] - 3*A[n-2] od:
%p seq(A[n],n=0..100); # _Robert Israel_, Aug 05 2014
%t LinearRecurrence[{2,-3},{0,1},40] (* _Harvey P. Dale_, Nov 03 2014 *)
%o (Sage) [lucas_number1(n,2,3) for n in range(0, 38)] # _Zerinvary Lajos_, Apr 23 2009
%o (PARI) x='x+O('x^50); concat([0], Vec(x/(1-2*x+3*x^2))) \\ _G. C. Greubel_, Oct 22 2018
%o (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
%Y Cf. A084102, A088138, A045873, A088139.
%Y Cf. A087455, A025172, A123562, A127357.
%K sign,easy
%O 0,3
%A _Paul Barry_, Sep 20 2003
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