A088139 -id:A088139 - 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

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

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)

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

A138229 Expansion of (1-x)/(1-2x+6x^2).

Original entry on oeis.org

1, 1, -4, -14, -4, 76, 176, -104, -1264, -1904, 3776, 18976, 15296, -83264, -258304, -17024, 1515776, 3133696, -2827264, -24456704, -31949824, 82840576, 357380096, 217716736, -1708847104, -4723994624, 805093376

Offset: 0

Paul Barry, Mar 06 2008

Binomial transform of [1, 0, -5, 0, 25, 0, -125, 0, 625, 0, ...]=: powers of -5 with interpolated zeros. - Philippe Deléham, Dec 02 2008

Michael De Vlieger, Table of n, a(n) for n = 0..2571
Beata Bajorska-Harapińska, Barbara Smoleń, Roman Wituła, On Quaternion Equivalents for Quasi-Fibonacci Numbers, Shortly Quaternaccis, Advances in Applied Clifford Algebras (2019) Vol. 29, 54.
Index entries for linear recurrences with constant coefficients, signature (2,-6).

Cf. A088139.

CoefficientList[Series[(1-x)/(1-2x+6x^2),{x,0,30}],x] (* or *) LinearRecurrence[ {2,-6},{1,1},30] (* Harvey P. Dale, Feb 29 2012 *)
TrigExpand@Table[6^(n/2) Cos[n ArcTan[Sqrt[5]]], {n, 0, 20}] (* or *)
Table[Sum[(-5)^k Binomial[n, 2 k], {k, 0, n/2}], {n, 0, 20}] (* Vladimir Reshetnikov, Sep 20 2016 *)

[lucas_number2(n,2,6)/2 for n in range(0,28)] # Zerinvary Lajos, Jul 08 2008

From Philippe Deléham, Nov 14 2008: (Start)
a(n) = 2*a(n-1) - 6*a(n-2), a(0)=1, a(1)=1.
a(n) = Sum_{k=0..n} A098158(n,k)*(-5)^(n-k). (End)
a(n) = Sum_{k=0..n} A124182(n,k)*(-6)^(n-k). - Philippe Deléham, Nov 15 2008
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(5*k+1)/(x*(5*k+6) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 26 2013
a(n) = real part of the quaternion (1 + i + 2*j)^n. - Peter Bala, Mar 29 2015

A172250 Triangle, read by rows, given by [0,1,-1,0,0,0,0,0,0,0,...] DELTA [1,-1,1,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 0, 1, 0, 1, 0, 0, 0, 2, -1, 0, 0, 1, 1, -1, 0, 0, 0, 3, -2, 0, 0, 0, 0, 1, 3, -4, 1, 0, 0, 0, 0, 4, -2, -2, 1, 0, 0, 0, 0, 1, 6, -9, 3, 0, 0, 0, 0, 0, 0, 5, 0, -9, 6, -1, 0, 0, 0, 0, 0, 1, 10, -15, 3, 3, -1, 0, 0, 0, 0, 0, 0, 6, 5, -24, 18, -4, 0, 0, 0, 0, 0, 0, 0, 1, 15, -20, -6, 18, -8, 1

Offset: 0

Philippe Deléham, Jan 29 2010

			Triangle begins:
  1;
  0,  1;
  0,  1,  0;
  0,  0,  2, -1;
  0,  0,  1,  1, -1;
  0,  0,  0,  3, -2,  0;
  0,  0,  0,  1,  3, -4,  1;
  0,  0,  0,  0,  4, -2, -2,  1; ...

Cf. A101950.

T(n,k) = T(n-1,k-1) + T(n-2,k-1) - T(n-2,k-2), T(0,0)=1, T(n,k) = 0 if k > n or if k < 0.
Sum_{k=0..n} T(n,k)*x^k = (-1)^n*A088139(n+1), A001607(n+1), A000007(n), A000012(n), A099087(n), A190960(n+1) for x = -2, -1, 0, 1, 2, 3 respectively. - Philippe Deléham, Feb 15 2012
G.f.: 1/(1-y*x+y*(y-1)*x^2). - Philippe Deléham, Feb 15 2012

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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A088138 Generalized Gaussian Fibonacci integers.

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

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

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A138229 Expansion of (1-x)/(1-2x+6x^2).

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