A127357 -id:A127357 - cp's OEIS frontend

A190958 a(n) = 2a(n-1) - 10a(n-2), with a(0) = 0, a(1) = 1.

0, 1, 2, -6, -32, -4, 312, 664, -1792, -10224, -2528, 97184, 219648, -532544, -3261568, -1197696, 30220288, 72417536, -157367808, -1038910976, -504143872, 9380822016, 23803082752, -46202054656, -330434936832, -198849327104, 2906650714112, 7801794699264

Offset: 0

Vladimir Joseph Stephan Orlovsky, May 24 2011

For the difference equation a(n) = c*a(n-1) - d*a(n-2), with a(0) = 0, a(1) = 1, the solution is a(n) = d^((n-1)/2) * ChebyshevU(n-1, c/(2*sqrt(d))) and has the alternate form a(n) = ( ((c + sqrt(c^2 - 4*d))/2)^n - ((c - sqrt(c^2 - 4*d))/2)^n )/sqrt(c^2 - 4*d). In the case c^2 = 4*d then the solution is a(n) = n*d^((n-1)/2). The generating function is x/(1 - c*x + d^2) and the exponential generating function takes the form (2/sqrt(c^2 - 4*d))*exp(c*x/2)*sinh(sqrt(c^2 - 4*d)*x/2) for c^2 > 4*d, (2/sqrt(4*d - c^2))*exp(c*x/2)*sin(sqrt(4*d - c^2)*x/2) for 4*d > c^2, and x*exp(sqrt(d)*x) if c^2 = 4*d. - G. C. Greubel, Jun 10 2022

Vincenzo Librandi, Table of n, a(n) for n = 0..190
Index entries for linear recurrences with constant coefficients, signature (2,-10).

Sequences of the form a(n) = c*a(n-1) - d*a(n-2), with a(0)=0, a(1)=1:
c/d...1.......2.......3.......4.......5.......6.......7.......8.......9......10
.A128834,A107920,A106852,A106853,A106854,A145934,A145976,A145978,A146078,A146080
.A000027,A009545,A088137,A088138,A045873,A088139,A089181,A090591,A127357,A190958
.A001906,A000225,A057083,A049072,A190959,A190960,A190961,A190962,A190963,A190964
.A001353,A007070,A003462,A001787,A099456,A190965,A168175,A190966,A190967,A190968
.A004254,A107839,A116415,A002450,A030191,A001047,A099450,A190969,A190970,A190971
.A001109,A154244,A138395,A084326,A003463,A030192,A081179,A006516,A027471,A106392
.A004187,A186446,A190972,A190973,A190974,A003464,A030240,A152268,A099459,A016127
.A001090,A190975,A190976,A099156,A190977,A190978,A023000,A057084,A154245,A154235
.A018913,A190979,A190980,A190981,A190982,A190983,A190984,A023001,A057085,A178869
A004189,A190869,A190985,A190986,A190987,A190988,A190989,A190990,A002452,A057086

I:=[0,1]; [n le 2 select I[n] else 2*Self(n-1)-10*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Sep 17 2011

Mathematica
```
LinearRecurrence[{2,-10}, {0,1}, 50]
```

a(n)=([0,1; -10,2]^n*[0;1])[1,1] \\ Charles R Greathouse IV, Apr 08 2016

[lucas_number1(n,2,10) for n in (0..50)] # G. C. Greubel, Jun 10 2022

G.f.: x / ( 1 - 2*x + 10*x^2 ). - R. J. Mathar, Jun 01 2011
E.g.f.: (1/3)*exp(x)*sin(3*x). - Franck Maminirina Ramaharo, Nov 13 2018
a(n) = 10^((n-1)/2) * ChebyshevU(n-1, 1/sqrt(10)). - G. C. Greubel, Jun 10 2022
a(n) = (1/3)*10^(n/2)*sin(n*arctan(3)) = Sum_{k=0..floor(n/2)} (-1)^k*3^(2*k)*binomial(n,2*k+1). - Gerry Martens, Oct 15 2022

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

A207538 Triangle of coefficients of polynomials v(n,x) jointly generated with A207537; see Formula section.

Original entry on oeis.org

1, 2, 4, 1, 8, 4, 16, 12, 1, 32, 32, 6, 64, 80, 24, 1, 128, 192, 80, 8, 256, 448, 240, 40, 1, 512, 1024, 672, 160, 10, 1024, 2304, 1792, 560, 60, 1, 2048, 5120, 4608, 1792, 280, 12, 4096, 11264, 11520, 5376, 1120, 84, 1, 8192, 24576, 28160, 15360

Offset: 1

Clark Kimberling, Feb 18 2012

As triangle T(n,k) with 0<=k<=n and with zeros omitted, it is the triangle given by (2, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 1/2, -1/2, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 04 2012
The numbers in rows of the triangle are along "first layer" skew diagonals pointing top-left in center-justified triangle given in A013609 ((1+2*x)^n) and  along (first layer) skew diagonals pointing top-right in center-justified triangle given in A038207 ((2+x)^n), see links. - Zagros Lalo, Jul 31 2018
If s(n) is the row sum at n, then the ratio s(n)/s(n-1) is approximately 2.414213562373095... (A014176: Decimal expansion of the silver mean, 1+sqrt(2)), when n approaches infinity. - Zagros Lalo, Jul 31 2018

			First seven rows:
1
2
4...1
8...4
16..12..1
32..32..6
64..80..24..1
(2, 0, 0, 0, 0, ...) DELTA (0, 1/2, -1/2, 0, 0, 0, ...) begins:
    1
    2,   0
    4,   1,  0
    8,   4,  0, 0
   16,  12,  1, 0, 0
   32,  32,  6, 0, 0, 0
   64,  80, 24, 1, 0, 0, 0
  128, 192, 80, 8, 0, 0, 0, 0

Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 80-83, 357-358.

Jean-Luc Baril and José Luis Ramírez, Descent distribution on Catalan words avoiding ordered pairs of Relations, arXiv:2302.12741 [math.CO], 2023.
S. Halici, On some Pell polynomials , Acta Universitatis Apulensis, No. 29/2012, pp. 105-112.
Zagros Lalo, First layer skew diagonals in center-justified triangle of coefficients in expansion of (1 + 2x)^n
Zagros Lalo, First layer skew diagonals in center-justified triangle of coefficients in expansion of (2 + x)^n

Cf. A028297, A207537, A133156, A038207, A099089.
Cf. A053117, A097610, A091894.
Cf. A013609, A038207.
Cf. A128099.

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := u[n - 1, x] + (x + 1)*v[n - 1, x]
v[n_, x_] := u[n - 1, x] + v[n - 1, x]
Table[Factor[u[n, x]], {n, 1, z}]
Table[Factor[v[n, x]], {n, 1, z}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]  (* A207537, |A028297| *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]  (* A207538, |A133156| *)
t[0, 0] = 1; t[n_, k_] := t[n, k] = If[n < 0 || k < 0, 0, 2 t[n - 1, k] + t[n - 2, k - 1]]; Table[t[n, k], {n, 0, 15}, {k, 0, Floor[n/2]}] // Flatten (* Zagros Lalo, Jul 31 2018 *)
t[n_, k_] := t[n, k] = 2^(n - 2 k) * (n -  k)!/((n - 2 k)! k!) ; Table[t[n, k], {n, 0, 15}, {k, 0, Floor[n/2]} ]  // Flatten (* Zagros Lalo, Jul 31 2018 *)

u(n,x) = u(n-1,x)+(x+1)*v(n-1,x), v(n,x) = u(n-1,x)+v(n-1,x), where u(1,x) = 1, v(1,x) = 1. Also, A207538 = |A133156|.
From Philippe Deléham, Mar 04 2012: (Start)
With 0<=k<=n:
Mirror image of triangle in A099089.
Skew version of A038207.
Riordan array (1/(1-2*x), x^2/(1-2*x)).
G.f.: 1/(1-2*x-y*x^2).
Sum_{k, 0<=k<=n} T(n,k)*x^k = A190958(n+1), A127357(n), A090591(n), A089181(n+1), A088139(n+1), A045873(n+1), A088138(n+1), A088137(n+1), A099087(n), A000027(n+1), A000079(n), A000129(n+1), A002605(n+1), A015518(n+1), A063727(n), A002532(n+1), A083099(n+1), A015519(n+1), A003683(n+1), A002534(n+1), A083102(n), A015520(n+1), A091914(n) for x = -10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 respectively.
T(n,k) = 2*T(n-1,k) + T(-2,k-1) with T(0,0) = 1, T(1,0) = 2, T(1,1) = 0 and T(n, k) = 0 if k<0 or if k>n. (End)
T(n,k) = A013609(n-k, n-2*k+1). - Johannes W. Meijer, Sep 05 2013
From Tom Copeland, Feb 11 2016: (Start)
A053117 is a reflected, aerated and signed version of this entry. This entry belongs to a family discussed in A097610 with parameters h1 = -2 and h2 = -y.
Shifted o.g.f.: G(x,t) = x / (1 - 2 x - t x^2).
The compositional inverse of G(x,t) is Ginv(x,t) = -[(1 + 2x) - sqrt[(1+2x)^2 + 4t x^2]] / (2tx) = x - 2 x^2 + (4-t) x^3 - (8-6t) x^4 + ..., a shifted o.g.f. for A091894 (mod signs with A091894(0,0) = 0).
(End)

A025172 Let phi = arccos(1/3), the dihedral angle of the regular tetrahedron. Then cos(n*phi) = a(n)/3^n.

Original entry on oeis.org

1, 1, -7, -23, 17, 241, 329, -1511, -5983, 1633, 57113, 99529, -314959, -1525679, -216727, 13297657, 28545857, -62587199, -382087111, -200889431, 3037005137, 7882015153, -11569015927, -94076168231, -84031193119, 678623127841, 2113526993753

Offset: 0

Wouter Meeussen

Used when showing that the regular simplex is not "scisssors-dissectible" to a cube, thus answering Hilbert's third problem.
From Peter Bala, Apr 01 2018: (Start)
This sequence is (1/2) * the Lucas sequence V(n,2,9). The companion Lucas sequence U(n,2,9) is A127357.
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 2 o 2 o ... o 2 (n terms) = 2*A127357(n-1)/A025172(n). Cf. A088137 and A087455. (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
J. L. Dupont, Scissors Congruences, Group Homology and Characteristic Classes, World Scientific, 2001. See p. 4.
Wikipedia, Lucas sequence
Index entries for linear recurrences with constant coefficients, signature (2,-9).

Cf. A088137, A087455, A127357.

f:=proc(n) option remember; if n <= 1 then RETURN(1); fi; 2*f(n-1)-9*f(n-2); end;

Table[ n/2 3^n GegenbauerC[ n, 1/3 ], {n, 24} ]
CoefficientList[Series[(1 - x)/(1 - 2 x + 9 x^2), {x, 0, 30}], x] (* Vincenzo Librandi, Oct 17 2013 *)
LinearRecurrence[{2,-9},{1,1},30] (* Harvey P. Dale, Jan 30 2016 *)

{a(n)= if(n<0, 0, 3^(n-1)* subst(3* poltchebi(abs(n)), x, 1/3))} /* Michael Somos, Mar 14 2007 */

a(0) = 1, a(1) = 1; for n >= 2, a(n) = 2*a(n-1) - 9*a(n-2). - Warut Roonguthai, Oct 11 2005
a(n) = (1/2)*(1-2*i*2^(1/2))^n+(1/2)*(1+2*i*2^(1/2))^n, where i=sqrt(-1). - Vladeta Jovovic, Apr 19 2003
a(n) is the permanent of the matrix M^n, where M = [i, 2; 1, i]. - Simone Severini, Apr 27 2007
a(n) = Product_{i=1..n} (2 - tan((i-1/2)*Pi/(2*n))^2). - Gerry Martens, May 26 2011
G.f.: (1-x)/(1-2*x+9*x^2). - Colin Barker, Jun 21 2012
G.f.: G(0)/2, where G(k)= 1 + 1/(1 - x*(8*k+1)/(x*(8*k+9) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 28 2013
E.g.f.: exp(x)*cos(2*sqrt(2)*x). - Vladimir Reshetnikov, Oct 15 2016
a(n) = A127357(n)-A127357(n-1). - R. J. Mathar, Apr 07 2022

Better description from Vladeta Jovovic, Apr 19 2003

Edited by N. J. A. Sloane, Feb 22 2007. Among other things, I changed the offset and the beginning of the sequence, so some of the formulas may need to be adjusted slightly.

A100193 a(n) = Sum_{k=0..n} binomial(2n,n+k)3^k.

Original entry on oeis.org

1, 5, 27, 146, 787, 4230, 22686, 121476, 649731, 3472382, 18546922, 99023292, 528535726, 2820451964, 15048601308, 80283276936, 428271193827, 2284478396334, 12185310873138, 64993897108236, 346655914156602, 1848916875734004, 9861224376230628, 52594507923308856

Offset: 0

Paul Barry, Nov 08 2004

A transform of 3^n under the mapping g(x)->(1/sqrt(1-4*x))*g(x*c(x)^2), where c(x) is the g.f. of the Catalan numbers A000108. A transform of 4^n under the mapping g(x)->(1/(c(x)*sqrt(1-4*x)))*g(x*c(x)).

Hankel transform is A127357. In general, the Hankel transform of Sum_{k=0..n} C(2n,k)*r^(n-k) is the sequence with g.f. 1/(1-2x+r^2*x^2). - Paul Barry, Jan 11 2007

Seiichi Manyama, Table of n, a(n) for n = 0..1000

Cf. A032443, A100192, A000108, A127357.

Table[Binomial[2*n,n]*Hypergeometric2F1[1,-n,1+n,-3],{n,0,20}] (* Vaclav Kotesovec, Feb 03 2014 *)

G.f.: (sqrt(1-4x)+1)/(sqrt(1-4x)*(4*sqrt(1-4x)-2)).
G.f.: sqrt(1-4x)*(3*sqrt(1-4x)-8x+3)/((1-4x)(6-32x)).
a(n) = Sum_{k=0..n} binomial(2n, n-k)*3^k.
a(n) = (Sum_{k=0..n} binomial(2n, n-k))*(Sum_{j=0..n} binomial(n, j)*(-1)^(n-j)*4^j).
a(n) = Sum_{k=0..n} C(n+k-1,k)*4^(n-k). - Paul Barry, Sep 28 2007
Conjecture: 9*n*a(n) + 6*(11-18*n)*a(n-1) + 16*(26*n-37)*a(n-2) + 256*(5-2*n)*a(n-3) = 0. - R. J. Mathar, Nov 09 2012
a(n) ~ (16/3)^n. - Vaclav Kotesovec, Feb 03 2014
a(n) = [x^n] 1/((1 - x)^n*(1 - 4*x)). - Ilya Gutkovskiy, Oct 12 2017

A025170 Expansion of g.f.: 1/(1 + 2x + 9x^2).

Original entry on oeis.org

1, -2, -5, 28, -11, -230, 559, 952, -6935, 5302, 51811, -151340, -163619, 1689298, -1906025, -11391632, 39937489, 22649710, -404736821, 605626252, 2431378885, -10313394038, -1255621889, 95331790120, -179362983239, -499260144602, 2612787138355, -732232975292

Offset: 0

Wouter Meeussen

Reciprocal Chebyshev polynomial of second kind evaluated at 3 multiplied by (-1)^n.
From Sharon Sela (sharonsela(AT)hotmail.com), Jan 19 2002: (Start)
a(n) is (-1)^n times the determinant of the following tridiagonal n X n matrix:
   [2 3 0 . . . . . . .]
   [3 2 3 0 . . . . . .]
   [0 3 2 3 0 . . . . .]
   [. 0 3 2 3 0 . . . .]
   [. . . . . . . . . .]
   [. . . . . . . . . .]
   [. . . . 0 3 2 3 0 .]
   [. . . . . 0 3 2 3 0]
   [. . . . . . 0 3 2 3]
   [. . . . . . . 0 3 2]
(End)

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (-2,-9).
Index entries for sequences related to Chebyshev polynomials.

Variant is A127357.

Cf. A087455, A088137.

[(-3)^n*Evaluate(ChebyshevU(n+1),1/3): n in [0..50]]; // G. C. Greubel, Jan 02 2024

Table[3^n ChebyshevU[n, -1/3], {n, 0, 24}]

a(n)=if(n<0,0,polcoeff(1/(1+2*x+9*x^2)+x*O(x^n),n))

a(n)=if(n<0, 0, 3^n*subst(poltchebi(n+1)+3*poltchebi(n),'x,-1/3)*3/8) /* Michael Somos, Sep 15 2005 */

a(n)=if(n<0, 0, (-1)^n*matdet(matrix(n,n,i,j, if(abs(i-j)<2, 2+abs(i-j))))) /* Michael Somos, Sep 15 2005 */

[3^n*chebyshev_U(n,-1/3) for n in range(41)] # G. C. Greubel, Jan 02 2024

a(n) = 3^n * ChebyshevU(n, -1/3).
a(n) = ( A088137(n+1) )^2 + ( A087455(n+1)/2 )^2 - ( A087455(n+2)/2 )^2. - Creighton Dement, Aug 20 2004
a(n) = -(2*a(n-1) + 9*a(n-2)) for n>1, with a(0)=1, a(1)=-2. - Philippe Deléham, Sep 19 2009
a(n) = (-2)^n*Product_{k=1..n}(1 + 3*cos(k*Pi/(n+1))). - Peter Luschny, Nov 28 2019
From G. C. Greubel, Jan 02 2024: (Start)
a(n) = (-1)^n * A127357(n).
E.g.f.: (1/4)*exp(-x)*(4*cos(2*sqrt(2)*x) - sqrt(2)*sin(2*sqrt(2)*x)). (End)

cp's OEIS Frontend

A190958 a(n) = 2*a(n-1) - 10*a(n-2), with a(0) = 0, a(1) = 1.

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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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A207538 Triangle of coefficients of polynomials v(n,x) jointly generated with A207537; see Formula section.

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A025172 Let phi = arccos(1/3), the dihedral angle of the regular tetrahedron. Then cos(n*phi) = a(n)/3^n.

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A100193 a(n) = Sum_{k=0..n} binomial(2*n,n+k)*3^k.

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A190958 a(n) = 2a(n-1) - 10a(n-2), with a(0) = 0, a(1) = 1.

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

A100193 a(n) = Sum_{k=0..n} binomial(2n,n+k)3^k.

A025170 Expansion of g.f.: 1/(1 + 2x + 9x^2).