A084102 -id:A084102 - cp's OEIS frontend

A009545 Expansion of e.g.f. sin(x)*exp(x).

0, 1, 2, 2, 0, -4, -8, -8, 0, 16, 32, 32, 0, -64, -128, -128, 0, 256, 512, 512, 0, -1024, -2048, -2048, 0, 4096, 8192, 8192, 0, -16384, -32768, -32768, 0, 65536, 131072, 131072, 0, -262144, -524288, -524288, 0, 1048576, 2097152, 2097152, 0, -4194304, -8388608, -8388608, 0, 16777216, 33554432

Offset: 0

R. H. Hardin

Also first of the two associated sequences a(n) and b(n) built from a(0)=0 and b(0)=1 with the formulas a(n) = a(n-1) + b(n-1) and b(n) = -a(n-1) + b(n-1). The initial terms of the second sequence b(n) are 1, 1, 0, -2, -4, -4, 0, 8, 16, 16, 0, -32, -64, -64, 0, 128, 256, ... The points Mn(a(n)+b(n)*I) of the complex plane are located on the spiral logarithmic rho = 2*(1/2)^(2*theta)/Pi) and on the straight lines drawn from the origin with slopes: infinity, 1/2, 0, -1/2. - Philippe LALLOUET (philip.lallouet(AT)wanadoo.fr), Jun 30 2007
A000225: (1, 3, 7, 15, 31, ...) = 2^n - 1 = INVERT transform of A009545 starting (1, 2, 2, 0, -4, -8, ...). (Cf. comments in A144081). - Gary W. Adamson, Sep 10 2008
Pisano period lengths: 1, 1, 8, 1, 4, 8, 24, 1, 24, 4, 40, 8, 12, 24, 8, 1, 16, 24, 72, 4, ... - R. J. Mathar, Aug 10 2012
The variant 0, 1, -2, 2, 0, -4, 8, -8, 0, 16, -32, 32, 0, -64, (with different signs) is the Lucas U(-2,2) sequence. - R. J. Mathar, Jan 08 2013
(1+i)^n = A146559(n) + a(n)*i where i = sqrt(-1). - Philippe Deléham, Feb 13 2013
This is the Lucas U(2,2) sequence. - Raphie Frank, Nov 28 2015
{A146559, A009545} are the difference analogs of {cos(x),sin(x)} (cf. [Shevelev] link). - Vladimir Shevelev, Jun 08 2017

N. J. A. Sloane, Table of n, a(n) for n = 0..2000, Apr 09 2016 (first 100 terms from T. D. Noe)
Paul Barry, A Catalan Transform and Related Transformations on Integer Sequences, Journal of Integer Sequences, Vol. 8 (2005), Article 05.4.5.
Paul Barry, On Integer-Sequence-Based Constructions of Generalized Pascal Triangles, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.4.
A. F. Horadam, Special properties of the sequence W_n(a,b; p,q), Fib. Quart., 5.5 (1967), 424-434. Case n->n+1, a=0,b=1; p=2, q=-2.
Wolfdieter Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38 (2000) 408-419. Eqs. (38) and (45), lhs, m=2.
Vladimir Shevelev, Combinatorial identities generated by difference analogs of hyperbolic and trigonometric functions of order n, arXiv:1706.01454 [math.CO], 2017.
N. J. A. Sloane, Table of n, (I-1)^n for n = 0..100
Wikipedia, Lucas sequence.
Index entries for linear recurrences with constant coefficients, signature (2,-2).
Index entries for sequences related to Chebyshev polynomials
Index entries for Lucas sequences

Cf. A009116. For minor variants of this sequence see A108520, A084102, A099087.
a(2*n) = A056594(n)*2^n, n >= 1, a(2*n+1) = A057077(n)*2^n.
This is the next term in the sequence A015518, A002605, A000129, A000079, A001477.
Cf. A000225, A144081. - Gary W. Adamson, Sep 10 2008
Cf. A146559.

I:=[0,1,2,2]; [n le 4 select I[n] else -4*Self(n-4): n in [1..60]]; // Vincenzo Librandi, Nov 29 2015

t1 := sum(n*x^n, n=0..100): F := series(t1/(1+x*t1), x, 100): for i from 0 to 50 do printf(`%d, `, coeff(F, x, i)) od: # Zerinvary Lajos, Mar 22 2009
G(x):=exp(x)*sin(x): f[0]:=G(x): for n from 1 to 54 do f[n]:=diff(f[n-1],x) od: x:=0: seq(f[n],n=0..50 ); # Zerinvary Lajos, Apr 05 2009
A009545 := n -> `if`(n<2, n, 2^(n-1)*hypergeom([1-n/2, (1-n)/2], [1-n], 2)):
seq(simplify(A009545(n)), n=0..50); # Peter Luschny, Dec 17 2015

nn=104; Range[0,nn-1]! CoefficientList[Series[Sin[x]Exp[x], {x,0,nn}], x] (* T. D. Noe, May 26 2007 *)
Join[{a=0,b=1},Table[c=2*b-2*a;a=b;b=c,{n,100}]] (* Vladimir Joseph Stephan Orlovsky, Jan 17 2011 *)
f[n_] := (1 + I)^(n - 2) + (1 - I)^(n - 2); Array[f, 51, 0] (* Robert G. Wilson v, May 30 2011 *)
LinearRecurrence[{2,-2},{0,1},110] (* Harvey P. Dale, Oct 13 2011 *)

x='x+O('x^66); Vec(serlaplace(exp(x)*sin(x))) /* Joerg Arndt, Apr 24 2011 */

x='x+O('x^100); concat(0, Vec(x/(1-2*x+2*x^2))) \\ Altug Alkan, Dec 04 2015

def A009545(n): return ((0, 1, 2, 2)[n&3]<<((n>>1)&-2))*(-1 if n&4 else 1) # Chai Wah Wu, Feb 16 2024

[lucas_number1(n,2,2) for n in range(0, 51)] # Zerinvary Lajos, Apr 23 2009

def A146559():
    x, y = 0, -1
    while True:
        yield x
        x, y = x - y, x + y
a = A146559(); [next(a) for i in range(40)]  # Peter Luschny, Jul 11 2013

a(0)=0; a(1)=1; a(2)=2; a(3)=2; a(n) = -4*a(n-4), n>3. - Larry Reeves (larryr(AT)acm.org), Aug 24 2000
Imaginary part of (1+i)^n. - Marc LeBrun
G.f.: x/(1 - 2*x + 2*x^2).
E.g.f.: sin(x)*exp(x).
a(n) = S(n-1, sqrt(2))*(sqrt(2))^(n-1) with S(n, x)= U(n, x/2) Chebyshev's polynomials of the 2nd kind, Cf. A049310, S(-1, x) := 0.
a(n) = ((1+i)^n - (1-i)^n)/(2*i) = 2*a(n-1) - 2*a(n-2) (with a(0)=0 and a(1)=1). - Henry Bottomley, May 10 2001
a(n) = (1+i)^(n-2) + (1-i)^(n-2). - Benoit Cloitre, Oct 28 2002
a(n) = Sum_{k=0..n-1} (-1)^floor(k/2)*binomial(n-1, k). - Benoit Cloitre, Jan 31 2003
a(n) = 2^(n/2)sin(Pi*n/4). - Paul Barry, Sep 17 2003
a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2*k+1)*(-1)^k. - Paul Barry, Sep 20 2003
a(n+1) = Sum_{k=0..n} 2^k*A109466(n,k). - Philippe Deléham, Nov 13 2006
a(n) = 2*((1/2)^(2*theta(n)/Pi))*cos(theta(n)) where theta(4*p+1) = p*Pi + Pi/2, theta(4*p+2) = p*Pi + Pi/4, theta(4*p+3) = p*Pi - Pi/4, theta(4*p+4) = p*Pi - Pi/2, or a(0)=0, a(1)=1, a(2)=2, a(3)=2, and for n>3 a(n)=-4*a(n-4). Same formulas for the second sequence replacing cosines with sines. For example: a(0) = 0, b(0) = 1; a(1) = 0+1 = 1, b(1) = -0+1 = 1; a(2) = 1+1 = 2, b(2) = -1+1 = 0; a(3) = 2+0 = 2, b(3) = -2+0 = -2. - Philippe LALLOUET (philip.lallouet(AT)wanadoo.fr), Jun 30 2007
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3), n > 3, which implies the sequence is identical to its fourth differences. Binomial transform of 0, 1, 0, -1. - Paul Curtz, Dec 21 2007
Logarithm g.f. arctan(x/(1-x)) = Sum_{n>0} a(n)/n*x^n. - Vladimir Kruchinin, Aug 11 2010
a(n) = A046978(n) * A016116(n). - Paul Curtz, Apr 24 2011
E.g.f.: exp(x) * sin(x) = x + x^2/(G(0)-x); G(k) = 2k + 1 + x - x*(2k+1)/(4k+3+x+x^2*(4k+3)/( (2k+2)*(4k+5) - x^2 - x*(2k+2)*(4k+5)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Nov 15 2011
a(n) = Im( (1+i)^n ) where i=sqrt(-1). - Stanislav Sykora, Jun 11 2012
G.f.: x*U(0) where U(k) = 1 + x*(k+3) - x*(k+1)/U(k+1); (continued fraction, 1-step). - Sergei N. Gladkovskii, Oct 10 2012
G.f.: G(0)*x/(2*(1-x)), where G(k) = 1 + 1/(1 - x*(k+1)/(x*(k+2) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 25 2013
G.f.: x + x^2*W(0), where W(k) = 1 + 1/(1 - x*(k+1)/( x*(k+2) + 1/W(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 28 2013
G.f.: Q(0)*x/2, where Q(k) = 1 + 1/(1 - x*(4*k+2 - 2*x)/( x*(4*k+4 - 2*x) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Sep 06 2013
a(n) = (A^n - B^n)/(A - B), where A = 1 + i and B = 1 - i; A and B are solutions of x^2 - 2*x + 2 = 0. - Raphie Frank, Nov 28 2015
a(n) = 2^(n-1)*hypergeom([1-n/2, (1-n)/2], [1-n], 2) for n >= 2. - Peter Luschny, Dec 17 2015
a(k+m) = a(k)*A146559(m) + a(m)*A146559(k). - Vladimir Shevelev, Jun 08 2017

Extended with signs by Olivier Gérard, Mar 15 1997
More terms from Larry Reeves (larryr(AT)acm.org), Aug 24 2000
Definition corrected by Joerg Arndt, Apr 24 2011

A108520 Expansion of 1/(1+2x+2x^2).

Original entry on oeis.org

1, -2, 2, 0, -4, 8, -8, 0, 16, -32, 32, 0, -64, 128, -128, 0, 256, -512, 512, 0, -1024, 2048, -2048, 0, 4096, -8192, 8192, 0, -16384, 32768, -32768, 0, 65536, -131072, 131072, 0, -262144, 524288, -524288, 0, 1048576, -2097152, 2097152, 0, -4194304, 8388608, -8388608

Offset: 0

Michael Somos, Jun 07 2005

Yet another variation on A009545.

Pisano period lengths: 1, 1, 8, 1, 4, 8, 24, 1, 24, 4, 40, 8, 12, 24, 8, 1, 16, 24, 72, 4, ... - R. J. Mathar, Aug 10 2012

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Maran van Heesch, The multiplicative complexity of symmetric functions over a field with characteristic p, Thesis, 2014.
Index entries for linear recurrences with constant coefficients, signature (-2,-2).

a(n) = (-1)^n * A099087(n). a(n) = -A084102(n) if n>0.

Cf. A009116, A009545, A016116, A075553, A084102, A099087.

[n le 2 select n*(-1)^(n-1) else -2*(Self(n-1)+Self(n-2)): n in [1..47]];  // Bruno Berselli, Apr 26 2011

A108520 := n -> `if`(n=0, 1, (-2)^n*hypergeom([1/2-n/2, -n/2], [-n], 2)):
seq(simplify(A108520(n)), n=0..46); # Peter Luschny, Dec 17 2015

CoefficientList[Series[1/(1+2x+2x^2), {x,0,50}], x] (* or *) LinearRecurrence[{-2,-2}, {1,-2}, 50] (* Harvey P. Dale, Sep 30 2012 *)
Table[-(-1-I)^(n-1) - (-1+I)^(n-1), {n, 0, 50}] (* Bruno Berselli, Nov 08 2015 *)
Im[(-1+I)^Range[51]] (* G. C. Greubel, Apr 24 2023 *)

a(n)=if(n<0, 0, polcoeff(1/(1+2*x+2*x^2)+x*O(x^n),n))

a(n)=if(n<1, n==0, -polsym(2+2*x+x^2,n-1)[n])

vector(66,n,imag((-1+I)^n)) /* Joerg Arndt, May 13 2011 */

[imag((-1+I)^(n+1)) for n in range(51)] # G. C. Greubel, Apr 24 2023

G.f.: 1/(1+2*x+2*x^2).
E.g.f.: exp(-x)*(cos(x) - sin(x)).
a(n) = -2*(a(n-1) + a(n-2)).
a(n) = Sum_{k=0..n} Sum_{j=0..n-k} C(k,j)*C(k,n-j)*(-2)^(n-j). - Paul Barry, Mar 09 2006
a(n) = -4 * a(n-4). - Paul Curtz, Apr 24 2011
a(n) = A016116(n+1) * A075553(n+1). - Paul Curtz, Apr 25 2011
From Bruno Berselli, Apr 26 2011: (Start)
a(n) = -(-1-i)^(n-1) - (-1+i)^(n-1), where i=sqrt(-1).
a(n) = -2*A009116(n-1) for n > 0. (End)
Imaginary part of (-1+i)^n, negated real part is A090132. - Joerg Arndt, May 13 2011
E.g.f.: (cos(x) - sin(x))*exp(-x) = G(0); G(k) = 1 - 2*x/(4*k+1+x*(4*k+1)/(2*(2*k+1) -x -2*(x^2)*(2*k+1)/((x^2) -(2*k+2)*(4*k+3)/G(k+1)))); (continued fraction). - Sergei N. Gladkovskii, Nov 26 2011
G.f.: G(0)/(2*(1+x)), where G(k) = 1 + 1/(1 - x*(k+1)/(x*(k+2) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 20 2013
a(n) = (-2)^n*hypergeom([1/2-n/2, -n/2], [-n], 2) for n >= 1. - Peter Luschny, Dec 17 2015

A088137 Generalized Gaussian Fibonacci integers.

Original entry on oeis.org

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

Paul Barry, Sep 20 2003

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)

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
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.
Ronald Orozco López, Deformed Differential Calculus on Generalized Fibonacci Polynomials, arXiv:2211.04450 [math.CO], 2022.
Mihai Prunescu, On other two representations of the C-recursive integer sequences by terms in modular arithmetic, arXiv:2406.06436 [math.NT], 2024. See p. 18.
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
Index entries for linear recurrences with constant coefficients, signature (2,-3).
Index entries for Lucas sequences

Cf. A084102, A088138, A045873, A088139.

Cf. A087455, A025172, A123562, A127357.

[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

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

LinearRecurrence[{2,-3},{0,1},40] (* Harvey P. Dale, Nov 03 2014 *)

x='x+O('x^50); concat([0], Vec(x/(1-2*x+3*x^2))) \\ G. C. Greubel, Oct 22 2018

[lucas_number1(n,2,3) for n in range(0, 38)] # Zerinvary Lajos, Apr 23 2009

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

A087455 Expansion of (1 - x)/(1 - 2x + 3x^2) in powers of x.

Original entry on oeis.org

1, 1, -1, -5, -7, 1, 23, 43, 17, -95, -241, -197, 329, 1249, 1511, -725, -5983, -9791, -1633, 26107, 57113, 35905, -99529, -306773, -314959, 290401, 1525679, 2180155, -216727, -6973919, -13297657, -5673557, 28545857, 74112385, 62587199, -97162757, -382087111, -472685951

Offset: 0

Simone Severini, Oct 23 2003

Type 2 generalized Gaussian Fibonacci integers.
Binomial transform of A077966. - Philippe Deléham, Dec 02 2008
The real component of Q^n, where Q is the quaternion 1 + 0*i + 1*j + 1*k. - Stanislav Sykora, Jun 11 2012
If entries are multiplied by 2*(-1)^n, which gives 2, -2, -2, 10, -14, -2, 46, -86, 34, 190, -482, 394, ..., we obtain the Lucas V(-2,3) sequence. - R. J. Mathar, Jan 08 2013
The real component of (1 + sqrt(-2))^n. - Giovanni Resta, Apr 01 2014
It is an open question whether or not this sequence satisfies Benford's law [Berger-Hill, 2017; Arno Berger, email, Jan 06 2017]. - N. J. A. Sloane, Feb 08 2017
Given an alternated cubic honeycomb with a planar dissection along a plane from edge to opposite edge of the containing cube. The sequence (1 + sqrt(-2))^n contains a real component representing distance along the edge of the tetrahedron/octahedron and an imaginary component representing the orthogonal distance along the sqrt(2) axis in a tetrahedron/octahedron, this generates a unique cevian (line from the apical vertex to a vertex on the triangular tiling composing the opposite face) in this plane with length (sqrt(3))^n. - Jason Pruski, Sep 04 2017, Jan 08 2018
From Peter Bala, Apr 01 2018: (Start)
This sequence is the Lucas sequence V(n,2,3). The companion Lucas sequence U(n,2,3) is A088137.
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) = A088137(n)/a(n). Cf. A025172 and A127357. (End)

			G.f. = 1 + x - x^2 - 5*x^3 - 7*x^4 + x^5 + 23*x6 + 43*x^7 + 17*x^8 - 95*x^9 + ...

Arno Berger and Theodore P. Hill. An Introduction to Benford's Law. Princeton University Press, 2015.
S. Severini, A note on two integer sequences arising from the 3-dimensional hypercube, Technical Report, Department of Computer Science, University of Bristol, Bristol, UK (October 2003).

Robert Israel, Table of n, a(n) for n = 0..3500
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.
A. Berger and T. P. Hill, What is Benford's Law?, Notices, Amer. Math. Soc., 64:2 (2017), 132-134.
F. Beukers, The multiplicity of binary recurrences, Compositio Mathematica, Tome 40 (1980) no. 2 , p. 251-267. See Theorem 2 p. 259.
M. Mignotte, Propriétés arithmétiques des suites récurrentes, Besançon, 1988-1989, see p. 14. In French.
Wikipedia, Lucas sequence
Index entries for linear recurrences with constant coefficients, signature (2,-3).
Index entries for sequences related to Benford's law

Cf. A025172, A048473, A077966, A084102, A088137, A088138, A098158, A124182, A127357.

[n le 2 select 1 else 2*Self(n-1) -3*Self(n-2): n in [1..41]]; // G. C. Greubel, Jan 03 2024

Digits:=100; a:=n->round(abs(evalf((3^(n/2))*cos(n*arctan(sqrt(2))))));
# alternative:
a:= gfun:-rectoproc({a(n) = 2*a(n-1) - 3*a(n-2),a(0)=1,a(1)=1},a(n),remember):
map(a, [$0..100]); # Robert Israel, Jun 23 2015

CoefficientList[Series[(1-x)/(1-2*x+3*x^2), {x, 0, 40}], x] (* Vaclav Kotesovec, Apr 01 2014 *)
a[ n_] := ChebyshevT[ n, 1/Sqrt[3]] Sqrt[3]^n // Simplify; (* Michael Somos, May 15 2015 *)
LinearRecurrence[{2,-3},{1,1},50] (* Harvey P. Dale, Jul 30 2019 *)

{a(n) = real( (1 + quadgen(-8))^n )}; /* Michael Somos, Jul 26 2006 */

{a(n) = real( subst( poltchebi(n), 'x, quadgen(12) / 3) * quadgen(12)^n)}; /* Michael Somos, Jul 26 2006 */

a(n)=simplify(polchebyshev(n,,quadgen(12)/3)*quadgen(12)^n) \\ Charles R Greathouse IV, Jun 26 2013

[sqrt(3)^n*chebyshev_T(n, 1/sqrt(3)) for n in range(41)] # G. C. Greubel, Jan 03 2024

a(n) = (3^(n/2))*cos(n*arctan(sqrt(2))). - Paul Barry, Oct 23 2003
From Paul Barry, Sep 03 2004: (Start)
a(n) = 2*a(n-1) - 3*a(n-2).
a(n) = (-1)^n*Sum_{m=0..n} binomial(n, m)*Sum_{k=0..n} binomial(m, 2k)2^(m-k).
Binomial transform of 1/(1 + 2*x^2), or (1, 0, -2, 0, 4, 0, -8, 0, 16, ...). (End)
a(n+1) = a(n+2) - 2*A088137(n+1), a(n+1) = A088137(n+2) - A088137(n+1). - Creighton Dement, Oct 28 2004
a(n) = upper left and lower right terms of [1,-2, 1,1]^n. - Gary W. Adamson, Mar 28 2008
a(n) = Sum_{k=0..n} A098158(n,k)*(-2)^(n-k). - Philippe Deléham, Nov 14 2008
a(n) = Sum_{k=0..n} A124182(n,k)*(-3)^(n-k). - Philippe Deléham, Nov 15 2008
G.f.: G(0)/2, 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
a(n) = a(-n) * 3^n for all n in Z. - Michael Somos, Aug 25 2014
E.g.f.: (1/2)*(exp((1 - i*sqrt(2))*x) + exp((1 + i*sqrt(2))*x)), where i is the imaginary unit. - Stefano Spezia, Jul 17 2019

The explicit formula was given by Paul Barry.
Corrected and extended by N. J. A. Sloane, Aug 01 2004
More terms from Creighton Dement, Jul 31 2004

A088138 Generalized Gaussian Fibonacci integers.

Original entry on oeis.org

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

Paul Barry, Sep 20 2003

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

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

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

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

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

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

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*)

/* 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 */

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

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

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

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

A045873 a(n) = A006496(n)/2.

Original entry on oeis.org

0, 1, 2, -1, -12, -19, 22, 139, 168, -359, -1558, -1321, 5148, 16901, 8062, -68381, -177072, -12239, 860882, 1782959, -738492, -10391779, -17091098, 17776699, 121008888, 153134281, -298775878, -1363223161, -1232566932

Offset: 0

N. J. A. Sloane

Partial sums of A006495. - Paul Barry, Mar 16 2006
This is the Lucas U(P=2,Q=5) sequence. - R. J. Mathar, Oct 24 2012
With different signs, 0, 1, -2, -1, 12, -19, -22, 139, -168, -359, 1558, ... we obtain the Lucas U(-2,5) sequence. - R. J. Mathar, Jan 08 2013

Vincenzo Librandi, Table of n, a(n) for n = 0..500
Ronald Orozco López, Deformed Differential Calculus on Generalized Fibonacci Polynomials, arXiv:2211.04450 [math.CO], 2022.
Wikipedia, Lucas sequence
Index entries for linear recurrences with constant coefficients, signature (2,-5).
Index entries for Lucas sequences

Cf. A006495, A006496, A084102, A088136, A088137, A088139, A094423.

a:=[0,1];; for n in [3..30] do a[n]:=2*a[n-1]-5*a[n-2]; od; a; # Muniru A Asiru, Oct 23 2018

I:=[0,1]; [n le 2 select I[n] else 2*Self(n-1) - 5*Self(n-2): n in [1..50]]; // G. C. Greubel, Oct 22 2018

seq(coeff(series(x/(1-2*x+5*x^2),x,n+1), x, n), n = 0 .. 30); # Muniru A Asiru, Oct 23 2018

LinearRecurrence[{2,-5}, {0,1}, 40] (* G. C. Greubel, Jan 11 2024 *)

concat(0,Vec(1/(1-2*x+5*x^2)+O(x^99))) \\ Charles R Greathouse IV, Dec 22 2011

[lucas_number1(n,2,5) for n in range(0, 29)] # Zerinvary Lajos, Apr 23 2009

A045873=BinaryRecurrenceSequence(2,-5,0,1)
[A045873(n) for n in range(41)] # G. C. Greubel, Jan 11 2024

a(n)^2 = A094423(n).
From Paul Barry, Sep 20 2003: (Start)
O.g.f.: x/(1 - 2*x + 5*x^2).
E.g.f.: exp(x)*sin(2*x)/2.
a(n) = 2*a(n-1) - 5*a(n-2), a(0)=0, a(1)=1.
a(n) = ((1 + 2*i)^n - (1 - 2*i)^n)/(4*i), where i=sqrt(-1).
a(n) = Im{(1 + 2*i)^n/2}.
a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2k+1)*(-4)^k. (End)
a(n+1) = Sum_{k=0..n} binomial(k,n-k)*2^k*(-5/2)^(n-k). - Paul Barry, Mar 16 2006
G.f.: 1/(4*x - 1/G(0)) where G(k) = 1 - (k+1)/(1 - x/(x - (k+1)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Dec 06 2012
G.f.: Q(0)*x/2, where Q(k) = 1 + 1/(1 - x*(4*k+2 - 5*x)/( x*(4*k+4 - 5*x) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 30 2013
a(n) = 5^((n-1)/2)*ChebyshevU(n-1, 1/sqrt(5)). - G. C. Greubel, Jan 11 2024

More terms from Paul Barry, Sep 20 2003

A084101 Expansion of (1+x)^2/((1-x)*(1+x^2)).

Original entry on oeis.org

1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1

Offset: 0

Paul Barry, May 15 2003

Partial sums of A084099. Inverse binomial transform of A000749 (without leading zeros).
From Klaus Brockhaus, May 31 2010: (Start)
Periodic sequence: Repeat 1, 3, 3, 1.
Interleaving of A010684 and A176040.
Continued fraction expansion of (7 + 5*sqrt(29))/26.
Decimal expansion of 121/909.
a(n) = A143432(n+3) + 1 = 2*A021913(n+1) + 1 = 2*A133872(n+3) + 1.
a(n) = A165207(n+1) - 1.
First differences of A047538.
Binomial transform of A084102. (End)
From Wolfdieter Lang, Feb 09 2012: (Start)
a(n) = A045572(n+1) (Modd 5) := A203571(A045572(n+1)), n >= 0.
For general Modd n (not to be confused with mod n) see a comment on A203571. The nonnegative members of the five residue classes Modd 5, called [m] for m=0,1,...,4, are shown in the array A090298 if there the last row is taken as class [0] after inclusion of 0.
(End)

			From _Wolfdieter Lang_, Feb 09 2012: (Start)
Modd 5 of nonnegative odd numbers restricted mod 5:
A045572: 1, 3, 7, 9, 11, 13, 17, 19, 21, 23, ...
Modd 5:  1, 3, 3, 1,  1,  3,  3,  1,  1,  3, ...
(End)

Guenther Schrack, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1, -1, 1).

Cf. A084102.

Cf. A010684 (repeat 1, 3), A176040 (repeat 3, 1), A178593 (decimal expansion of (7+5*sqrt(29))/26), A143432 (expansion of (1+x^4)/((1-x)*(1+x^2))), A021913 (repeat 0, 0, 1, 1), A133872 (repeat 1, 1, 0, 0), A165207 (repeat 2, 2, 4, 4), A047538 (congruent to 0, 1, 4 or 7 mod 8), A084099 (expansion of (1+x)^2/(1+x^2)), A000749 (expansion of x^3/((1-x)^4-x^4)). - Klaus Brockhaus, May 31 2010

R:=PowerSeriesRing(Integers(), 100); Coefficients(R!( (1+x)^2/((1-x)*(1+x^2)) )); // G. C. Greubel, Feb 28 2019

CoefficientList[Series[(1+x)^2/((1-x)(1+x^2)),{x,0,110}],x] (* or *) PadRight[{},110,{1,3,3,1}] (* Harvey P. Dale, Nov 21 2012 *)

x='x+O('x^100); Vec((1+x)^2/((1-x)*(1+x^2))) \\ Altug Alkan, Dec 24 2015

((1+x)^2/((1-x)*(1+x^2))).series(x, 100).coefficients(x, sparse=False) # G. C. Greubel, Feb 28 2019

a(n) = binomial(3, n mod 4). - Paul Barry, May 25 2003
From Klaus Brockhaus, May 31 2010: (Start)
a(n) = a(n-4) for n > 3; a(0) = a(3) = 1, a(1) = a(2) = 3.
a(n) = (4 - (1+i)*i^n - (1-i)*(-i)^n)/2 where i = sqrt(-1). (End)
E.g.f.: 2*exp(x) + sin(x) - cos(x). - Arkadiusz Wesolowski, Nov 04 2017
a(n) = 2 - (-1)^(n*(n+1)/2). - Guenther Schrack, Feb 26 2019

A088139 a(n) = 2a(n-1) - 6a(n-2), a(0)=0, a(1)=1.

Original entry on oeis.org

0, 1, 2, -2, -16, -20, 56, 232, 128, -1136, -3040, 736, 19712, 35008, -48256, -306560, -323584, 1192192, 4325888, 1498624, -22958080, -54907904, 27932672, 385312768, 603029504, -1105817600, -5829812224, -5024718848, 24929435648, 80007184384

Offset: 0

Paul Barry, Sep 20 2003

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,-6).

Cf. A045873, A084102, A088136, A088137.

a:=[0,1];; for n in [3..30] do a[n]:=2*a[n-1]-6*a[n-2]; od; a; # Muniru A Asiru, Oct 23 2018

I:=[0,1]; [n le 2 select I[n] else 2*Self(n-1) - 6*Self(n-2): n in [1..30]]; // G. C. Greubel, Oct 22 2018

seq(coeff(series(x/(1-2*x+6*x^2),x,n+1), x, n), n = 0 .. 30); # Muniru A Asiru, Oct 23 2018

Join[{a=0,b=1},Table[c=2*b-6*a;a=b;b=c,{n,100}]] (* Vladimir Joseph Stephan Orlovsky, Jan 17 2011*)
TrigExpand@Table[(6^(n/2) Sin[n ArcTan[Sqrt[5]]])/Sqrt[5], {n, 0, 20}] (* or *)
Table[Sum[(-5)^k Binomial[n, 2 k + 1], {k, 0, n/2}], {n, 0, 20}] (* Vladimir Reshetnikov, Sep 20 2016 *)
LinearRecurrence[{2,-6},{0,1},40] (* Harvey P. Dale, Nov 22 2024 *)

x='x+O('x^30); concat([0], Vec(x/(1-2*x+6*x^2))) \\ G. C. Greubel, Oct 22 2018

[lucas_number1(n,2,6) for n in range(0, 30)] # Zerinvary Lajos, Apr 23 2009

G.f.: x/(1-2*x+6*x^2).
E.g.f.: exp(x)*sin(sqrt(5)*x)/sqrt(5).
a(n) = ((1+i*sqrt(5))^n-(1-i*sqrt(5))^n)/(2*i*sqrt(5)).
a(n) = Im{(1+i*sqrt(5))^n/sqrt(5)}.
a(n) = Sum_{k=0..floor(n/2)} C(n, 2k+1)(-5)^k.
a(n+1) = (-1)^n*Sum_{k, 0<=k<=n} A172250(n,k)*(-2)^k. - Philippe Deléham, Feb 15 2012

cp's OEIS Frontend

A009545 Expansion of e.g.f. sin(x)*exp(x).

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A108520 Expansion of 1/(1+2*x+2*x^2).

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A088137 Generalized Gaussian Fibonacci integers.

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A087455 Expansion of (1 - x)/(1 - 2*x + 3*x^2) in powers of x.

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A088138 Generalized Gaussian Fibonacci integers.

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A108520 Expansion of 1/(1+2x+2x^2).

A087455 Expansion of (1 - x)/(1 - 2x + 3x^2) in powers of x.

A088139 a(n) = 2a(n-1) - 6a(n-2), a(0)=0, a(1)=1.