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A088137
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Generalized Gaussian Fibonacci integers.
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18
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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
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OFFSET
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0,3
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COMMENTS
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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
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)
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LINKS
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FORMULA
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a(n) = 3^(n/2)*sin(n*atan(sqrt(2)))/sqrt(2).
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.
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) = n*hypergeom([(1-n)/2,(2-n)/2],[3/2],-2). - Gerry Martens, Sep 03 2023
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MAPLE
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A[0]:= 0: A[1]:= 1:
for n from 2 to 100 do A[n]:= 2*A[n-1] - 3*A[n-2] od:
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MATHEMATICA
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LinearRecurrence[{2, -3}, {0, 1}, 40] (* Harvey P. Dale, Nov 03 2014 *)
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PROG
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(Sage) [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
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CROSSREFS
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KEYWORD
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sign,easy
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AUTHOR
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STATUS
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approved
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