A088138 Generalized Gaussian Fibonacci integers.

0, 1, 2, 0, -8, -16, 0, 64, 128, 0, -512, -1024, 0, 4096, 8192, 0, -32768, -65536, 0, 262144, 524288, 0, -2097152, -4194304, 0, 16777216, 33554432, 0, -134217728, -268435456, 0, 1073741824, 2147483648, 0, -8589934592, -17179869184, 0, 68719476736, 137438953472

Offset: 0

Author

Paul Barry, Sep 20 2003

Keywords

• easy
• sign

Comments

The sequence 0,1,-2,0,8,-16,... has g.f. x/(1+2*x-4*x^2), a(n) = 2^n*sin(2n*Pi/3)/sqrt(3) and is the inverse binomial transform of sin(sqrt(3)*x)/sqrt(3): 0,1,-3,0,9,...

a(n+1) is the Hankel transform of A100192. - Paul Barry, Jan 11 2007

a(n+1) is the trinomial transform of A010892: a(n+1) = Sum_{k=0..2n} trinomial(n,k)*A010892(k+1) where trinomial(n, k) = trinomial coefficients (A027907). - Paul Barry, Sep 10 2007

a(n+1) is the Hankel transform of A100067. - Paul Barry, Jun 16 2009

From Paul Curtz, Oct 04 2009: (Start)

1) a(n) = A131577(n)*A128834(n).

2) Binomial transform of 0,1,0,-3,0,9,0,-27, see A000244.

3) Sequence is identical to every 2n-th difference divided by (-3)^n.

4) a(3n) + a(3n+1) + a(3n+2) = (-1)^n*3*A001018(n) for n >= 1.

5) For missing terms in a(n) see A013731 = 4*A001018. (End)

The coefficient of i of Q^n, where Q is the quaternion 1+i+j+k. Due to symmetry, also the coefficients of j and of k. - Stanislav Sykora, Jun 11 2012 [The coefficients of 1 are in A138230. - Wolfdieter Lang, Jan 28 2016]

With different signs, 0, 1, -2, 0, 8, -16, 0, 64, -128, 0, 512, -1024, ... is the Lucas U(-2,4) sequence. - R. J. Mathar, Jan 08 2013

Links

• Robert Israel, Table of n, a(n) for n = 0..3300
• 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.
• Wikipedia, Lucas sequence
• Index entries for linear recurrences with constant coefficients, signature (2,-4).
• Index entries for Lucas sequences

Crossrefs

Cf. A084102, A088137, A045873, A088139, A138230.

Programs

GAP

a:=[0,1];; for n in [3..40] do a[n]:=2*a[n-1]-4*a[n-2]; od; a; # Muniru A Asiru, Oct 23 2018

Magma

I:=[0,1]; [n le 2 select I[n] else 2*Self(n-1) - 4*Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 15 2018

Maple

M:= <<1+I,1+I>|>:
T:= <<-I/2,0>|<0,I/2>>:
seq(LinearAlgebra:-Trace(T.M^n),n=0..100); # Robert Israel, Jan 28 2016

Mathematica

Join[{a=0,b=1},Table[c=2*b-4*a;a=b;b=c,{n,100}]] (* Vladimir Joseph Stephan Orlovsky, Jan 17 2011 *)
LinearRecurrence[{2, -4}, {0, 1}, 40] (* Vincenzo Librandi, Jan 29 2016 *)
Table[2^(n-2)*((-1)^Quotient[n-1,3]+(-1)^Quotient[n,3]), {n,0,40}] (*Federico Provvedi,Apr 24 2022*)

PARI

/* lists powers of any quaternion */
QuaternionToN(a,b,c,d,nmax) = {local (C);C = matrix(nmax+1,4);C[1,1]=1;for(n=2,nmax+1,C[n,1]=a*C[n-1,1]-b*C[n-1,2]-c*C[n-1,3]-d*C[n-1,4];C[n,2]=b*C[n-1,1]+a*C[n-1,2]+d*C[n-1,3]-c*C[n-1,4];C[n,3]=c*C[n-1,1]-d*C[n-1,2]+a*C[n-1,3]+b*C[n-1,4];C[n,4]=d*C[n-1,1]+c*C[n-1,2]-b*C[n-1,3]+a*C[n-1,4];);return (C);} /* Stanislav Sykora, Jun 11 2012 */

PARI

my(x='x+O('x^30)); concat([0], Vec(x/(1-2*x+4*x^2))) \\ G. C. Greubel, Oct 22 2018

PARI

a(n) = 2^(n-1)*polchebyshev(n-1, 2, 1/2); \\ Michel Marcus, May 02 2022

Sage

[lucas_number1(n,2,4) for n in range(0, 39)] # Zerinvary Lajos, Apr 23 2009

Formula

G.f.: x/(1-2*x+4*x^2).

E.g.f.: exp(x)*sin(sqrt(3)*x)/sqrt(3).

a(n) = 2*a(n-1) - 4*a(n-2), a(0)=0, a(1)=1.

a(n) = ((1+i*sqrt(3))^n - (1-i*sqrt(3))^n)/(2*i*sqrt(3)).

a(n) = Im( (1+i*sqrt(3))^n/sqrt(3) ).

a(n) = Sum_{k=0..floor(n/2)} C(n, 2*k+1)*(-3)^k.

From Paul Curtz, Oct 04 2009: (Start)

a(n) = a(n-1) + a(n-2) + 2*a(n-3).

a(n) = 2*a(n-1) - a(n-2) + 2*a(n-3).

a(n) = a(n-1) + 2*a(n-2) - a(n-3) - a(n-4). (End)

E.g.f.: exp(x)*sin(sqrt(3)*x)/sqrt(3) = G(0)*x^2 where G(k)= 1 + (3*k+2)/(2*x - 32*x^5/(16*x^4 - 3*(k+1)*(3*k+2)*(3*k+4)*(3*k+5)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jul 26 2012

G.f.: x/(1-2*x+4*x^2) = 2*x^2*G(0) where G(k)= 1 + 1/(2*x - 32*x^5/(16*x^4 - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jul 27 2012

a(n) = -2^(n-1)*Product_{k=1..n}(1 + 2*cos(k*Pi/n)) for n >= 1. - Peter Luschny, Nov 28 2019

a(n) = 2^(n-1) * U(n-1, 1/2), where U(n, x) is the Chebyshev polynomial of the second kind. - Federico Provvedi, Apr 24 2022