A002878 -id:A002878 - cp's OEIS frontend

A025169 a(n) = 2Fibonacci(2n+2).

2, 6, 16, 42, 110, 288, 754, 1974, 5168, 13530, 35422, 92736, 242786, 635622, 1664080, 4356618, 11405774, 29860704, 78176338, 204668310, 535828592, 1402817466, 3672623806, 9615053952, 25172538050, 65902560198, 172535142544

Offset: 0

Wouter Meeussen

The pairs (x, y) = (a(n), a(n+1)) satisfy x^2 + y^2 = 3*x*y + 4. - Michel Lagneau, Feb 01 2014

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Hacène Belbachir, Soumeya Merwa Tebtoub, László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Mark W. Coffey, James L. Hindmarsh, Matthew C. Lettington, John Pryce, On Higher Dimensional Interlacing Fibonacci Sequences, Continued Fractions and Chebyshev Polynomials, arXiv:1502.03085 [math.NT], 2015 (see p. 32).
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (3,-1).

Cf. A000032, A000045, A001906, A002878, A122367.

List([0..30], n-> 2*Fibonacci(2*n+2) ); # G. C. Greubel, Jan 16 2020

a025169 n = a025169_list !! n
a025169_list = 2 : 6 : zipWith (-) (map (* 3) $ tail a025169_list) a025169_list
-- Reinhard Zumkeller, Apr 08 2012

[2*Fibonacci(2*n+2): n in [0..30]]; // Vincenzo Librandi, Jul 11 2011

R:=PowerSeriesRing(Integers(), 30); Coefficients(R!( 2/(1-3*x + x^2) )); // Marius A. Burtea, Jan 16 2020

seq( 2*fibonacci(2*n+2), n=0..30); # G. C. Greubel, Jan 16 2020

Table[2Fibonacci[2n+2], {n,0,30}] (* or *)
CoefficientList[Series[2/(1-3x+x^2), {x,0,30}], x] (* Michael De Vlieger, Mar 09 2016 *)
LinearRecurrence[{3, -1}, {2, 6}, 30] (* Jean-François Alcover, Sep 27 2017 *)

PARI
```
a(n)=2*fibonacci(2*n+2)
```

[2*fibonacci(2*n+2) for n in (0..30)] # G. C. Greubel, Jan 16 2020

G.f.: 2/(1 - 3*x + x^2).
a(n) = 3*a(n-1) - a(n-2).
a(n) = 2*A001906(n+1).
a(n) = A111282(n+2). - Reinhard Zumkeller, Apr 08 2012
a(n) = Fibonacci(2*n+1) + Lucas(2*n+1). - Bruno Berselli, Oct 13 2017

Better description from Michael Somos

A155116 a(n) = 3a(n-1) + 3a(n-2), n>2, a(0)=1, a(1)=2, a(2)=8.

Original entry on oeis.org

1, 2, 8, 30, 114, 432, 1638, 6210, 23544, 89262, 338418, 1283040, 4864374, 18442242, 69919848, 265086270, 1005018354, 3810313872, 14445996678, 54768931650, 207644784984, 787241149902, 2984657804658, 11315696863680, 42901064005014

Offset: 0

Philippe Deléham, Jan 20 2009

From Johannes W. Meijer, Aug 14 2010: (Start)
A berserker sequence, see A180140 and A180147. For the central square 16 A[5] vectors with decimal values between 3 and 384 lead to this sequence. These vectors lead for the corner squares to A123620 and for the side squares to A180142.
This sequence belongs to a family of sequences with GF(x)=(1-(2*k-1)*x-k*x^2)/(1-3*x+(k-4)*x^2). Berserker sequences that are members of this family are A000007 (k=2), A155116 (k=1; this sequence), A000302 (k=0), 6*A179606 (k=-1; with leading 1 added) and 2*A180141 (k=-2; n>=1 and a(0)=1). Some other members of this family are (-2)*A003688 (k=3; with leading 1 added), (-4)*A003946 (k=4; with leading 1 added), (-6)*A002878 (k=5; with leading 1 added) and (-8)*A033484 (k=6; with leading 1 added).
Inverse binomial transform of A101368 (without the first leading 1).
(End)

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,3).

Sequences of the form a(n) = m*(a(n-1) + a(n-2)) with a(0)=1, a(1) = m-1, a(2) = m^2 -1: A155020 (m=2), this sequence (m=3), A155117 (m=4), A155119 (m=5), A155127 (m=6), A155130 (m=7), A155132 (m=8), A155144 (m=9), A155157 (m=10).

m:=3; [1] cat [n le 2 select (m-1)*(m*n-(m-1)) else m*(Self(n-1) + Self(n-2)): n in [1..30]]; // G. C. Greubel, Mar 25 2021

With[{m=3}, LinearRecurrence[{m, m}, {1, m-1, m^2-1}, 30]] (* G. C. Greubel, Mar 25 2021 *)

Vec((1-x-x^2)/(1-3*x-3*x^2)+O(x^99)) \\ Charles R Greathouse IV, Jan 12 2012

m=3; [1]+[-(m-1)*(sqrt(m)*i)^(n-2)*chebyshev_U(n, -sqrt(m)*i/2) for n in (1..30)] # G. C. Greubel, Mar 25 2021

G.f.: (1-x-x^2)/(1-3*x-3*x^2).
a(n) = 2*A125145(n-1), n>=1 .
a(n) = ( (2+4*A)*A^(-n-1) + (2+4*B)*B^(-n-1) )/21 with A=(-3+sqrt(21))/6 and B=(-3-sqrt(21))/6 for n>=1 with a(0)=1. [corrected by Johannes W. Meijer, Aug 12 2010]
Contribution from Johannes W. Meijer, Aug 14 2010: (Start)
a(n) = A123620(n)/2 for n>=1.
(End)
a(n) = (1/3)*[n=0] - 2*(sqrt(3)*i)^(n-2)*ChebyshevU(n, -sqrt(3)*i/2). - G. C. Greubel, Mar 25 2021

A082985 Coefficient table for Chebyshev's U(2n,x) polynomial expanded in decreasing powers of (-4(1-x^2)).

Original entry on oeis.org

1, 1, 3, 1, 5, 5, 1, 7, 14, 7, 1, 9, 27, 30, 9, 1, 11, 44, 77, 55, 11, 1, 13, 65, 156, 182, 91, 13, 1, 15, 90, 275, 450, 378, 140, 15, 1, 17, 119, 442, 935, 1122, 714, 204, 17, 1, 19, 152, 665, 1729, 2717, 2508, 1254, 285, 19

Offset: 0

Gary W. Adamson, May 29 2003

Sum of row #n = A000204(2n+1), i.e., A002878(n).
Row #n has the unsigned coefficients of a polynomial whose roots are 2 sin(2*Pi*k/(2n+1)) [for k=1 to 2n].
The positive roots are the diagonal lengths of a regular (2n+1)-gon, inscribed in the unit circle.
Polynomial of row #n = Sum_{m=0..n} (-1)^m T(n,m) x^(2n-2m).
This is also the unsigned coefficient table of Chebyshev's 2*T(2*n+1,x) polynomials expanded in decreasing odd powers of 2*x. - Wolfdieter Lang, Mar 07 2007
The n-th row are the coefficients of the polynomial S(n) where S(0)=1, S(1)=x+3, and S(n) = (x+2)*S(n-1) - S(n-2) (see Sun link). - Michel Marcus, Mar 07 2016

			Expansion of polynomials:
  x^0;
  x^2  -  3*x^0;
  x^4  -  5*x^2 +  5*x^0;
  x^6  -  7*x^4 + 14*x^2 -  7*x^0;
  x^8  -  9*x^6 + 27*x^4 - 30*x^2 +  9*x^0;
  x^10 - 11*x^8 + 44*x^6 - 77*x^4 + 55*x^2 - 11*x^0; ...
Polynomial #4 has 8 roots: 2*sin(2*Pi*k/9) for k=1 to 8.
Coefficients (with signs removed) are
  1;
  1,  3;
  1,  5,  5;
  1,  7, 14,  7;
  1,  9, 27, 30,  9;
  1, 11, 44, 77, 55, 11;
  ...

J. D'Angelo, Several Complex Variables and the Geometry of Real Hypersurfaces, CRC Press, 1992; see pp. 151,175.
Stephen Eberhart, "Mathematical-Physical Correspondence," Number 37-38, Jan 08, 1982.
Theodore J. Rivlin, Chebyshev polynomials: from approximation theory to algebra and number theory, 2. ed., Wiley, New York, 1990.

Vincenzo Librandi, Rows n = 0..100 of triangle, flattened
K. Dilcher and K. B. Stolarsky, A Pascal-type triangle characterizing twin primes, Amer. Math. Monthly, 112 (2005), 673-681.
Zhi-Hong Sun, Expansions and identities concerning Lucas sequences, Fibonacci Quart. 44 (2006), no. 2, 145-153. See Theorem 3.1

Cf. companion triangle A084534.
Cf. diagonals: A005408, A000330, A005585, A050486, A054333, A057788.
Cf. A054142, A172431.

Flat(List([0..10], n-> List([0..n], k-> Binomial(2*n-k,k)*(2*n+1)/(2*n-2*k+1) ))); # G. C. Greubel, Dec 30 2019

[Binomial(2*n-k,k)*(2*n+1)/(2*n-2*k+1): k in [0..n], n in [0..10]]; // G. C. Greubel, Dec 30 2019

A082985 := proc(n,m)
    binomial(2*n-m,m)*(2*n+1)/(2*n-2*m+1) ;
end proc: # R. J. Mathar, Sep 08 2013

T[n_, m_]:= Binomial[2*n-m, m]*(2*n+1)/(2*n-2*m+1); Table[T[n, m], {n, 0, 9}, {m, 0, n}]//Flatten (* Jean-François Alcover, Oct 08 2013, after R. J. Mathar *)

T(n,k)=binomial(2*n-k,k)*(2*n+1)/(2*n-2*k+1); \\ G. C. Greubel, Dec 30 2019

[[binomial(2*n-k,k)*(2*n+1)/(2*n-2*k+1) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Dec 30 2019

Triangle read by rows: row #n has n+1 terms. T(n,0)=1, T(n,n)=2n+1, T(n,m) = T(n-1,m-1) + Sum_{k=0..m} T(n-1-k, m-k).
T(k, s) = ((2k+1)/(2s+1))*binomial(k+s, 2s), 0 <= s <= k; then transpose the triangle. - Gary W. Adamson, May 29 2003
From Wolfdieter Lang, Mar 07 2007: (Start)
Signed version: a(n,m)=0 if n < m, otherwise a(n,m) = ((-1)^m)*binomial(2*n+1-m,m)*(2*n+1)/(2*n+1-m). From the Rivlin reference, p. 37, eq.(1.92), using the differential eq. for T(2*n+1,x). Also from Waring's formula.
Signed version: a(n,m)=0 if n < m, otherwise a(n,m) = ((-1)^m)*(Sum_{k=0..n-m} binomial(m+k,k)*binomial(2*n+1,2*(k+m)))/2^(2*(n-m)). Proof: De Moivre's formula for cos((2*n+1)*phi) rewritten in terms of odd powers of cos(phi). Cf. Rivlin reference p. 4, eq.(1.10).
Signed version: a(n,m) = A084930(n,n-m)/2^(2*(n-m)) (scaled coefficients of Chebyshev's T(2*n+1,x), decreasing odd powers).
Unsigned version: a(n,m)=0 if n < m, otherwise a(n,m) = binomial(2*n-m,m)*(2*n+1)/(2*(n-m)+1). From the differential eq. for U(2*n,x). (End)
T(n,k) = T(n-1,k) + 2*T(n-1,k-1) - T(n-2,k-2). - Philippe Deléham, Feb 24 2012
Sum_{i>=0} T(n-i,n-2*i) = A003945(n). - Philippe Deléham, Feb 24 2012
Sum_{i>=0} T(n-i, n-2*i)*4^i = 3^n = A000244(n). - Philippe Deléham, Feb 24 2012
From Paul Weisenhorn Nov 25 2019: (Start)
T(r,k) = binomial(2*r-k,k-1) + binomial(2*r-1-k,k-2) with 1 <= r and 1 <= k <= r.
For a given n, one gets r = floor((1+sqrt(8*n))/2), k = n-(r^2-r)/2, a(n) = binomial(2*r-k,k-1) + binomial(2*r-1-k,k-2). (End)

Edited by Anne Donovan (anned3005(AT)aol.com), Jun 11 2003

Re-edited by Don Reble, Nov 12 2005

A160682 The list of the A values in the common solutions to 13k+1 = A^2 and 17k+1 = B^2.

Original entry on oeis.org

1, 14, 209, 3121, 46606, 695969, 10392929, 155197966, 2317576561, 34608450449, 516809180174, 7717529252161, 115246129602241, 1720974414781454, 25699370092119569, 383769576967012081, 5730844284413061646, 85578894689228912609, 1277952576054020627489

Offset: 1

Paul Weisenhorn, May 23 2009

This summarizes the case C=13 of common solutions to C*k+1=A^2, (C+4)*k+1=B^2.
The 2 equations are equivalent to the Pell equation x^2-C*(C+4)*y^2=1,
with x=(C*(C+4)*k+C+2)/2; y=A*B/2 and with smallest values x(1) = (C+2)/2, y(1)=1/2.
Generic recurrences are:
A(j+2)=(C+2)*A(j+1)-A(j) with A(1)=1; A(2)=C+1.
B(j+2)=(C+2)*B(j+1)-B(j) with B(1)=1; B(2)=C+3.
k(j+3)=(C+1)*(C+3)*( k(j+2)-k(j+1) )+k(j) with k(1)=0; k(2)=C+2; k(3)=(C+1)*(C+2)*(C+3).
x(j+2)=(C^2+4*C+2)*x(j+1)-x(j) with x(1)=(C+2)/2; x(2)=(C^2+4*C+1)*(C+2)/2;
Binet-type of solutions of these 2nd order recurrences are:
R=C^2+4*C; S=C*sqrt(R); T=(C+2); U=sqrt(R); V=(C+4)*sqrt(R);
A(j)=((R+S)*(T+U)^(j-1)+(R-S)*(T-U)^(j-1))/(R*2^j);
B(j)=((R+V)*(T+U)^(j-1)+(R-V)*(T-U)^(j-1))/(R*2^j);
x(j)+sqrt(R)*y(j)=((T+U)*(C^2*4*C+2+(C+2)*sqrt(R))^(j-1))/2^j;
k(j)=(((T+U)*(R+2+T*U)^(j-1)+(T-U)*(R+2-T*U)^(j-1))/2^j-T)/R. [Paul Weisenhorn, May 24 2009]
.C -A----- -B----- -k-----
01 A001519 A002878 A058038
02 A001653 A002315 A045899/2
03 A004253 A030221 A160695
04 A001653 A002315 A078522/4
05 A049685 A033890 A161582
06 A070997 A057080 A159683/2
07 A070998 A057081 A161585
08 A072256 A054320 A045502/4
09 A078922 A097783 A161586
10 A077417 A077416 A159681/2
11 A085260 A126816 A161588
12 A001570 A028230 A059989/4
13 A160682 A161591 A161584
14 A157456 A159678 A159679/2
15 A161595 A161599 A161583
16 A007805 A049629 A157459/4
For n>=2, a(n) equals the permanent of the (2n-2)X(2n-2) tridiagonal matrix with sqrt(13)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. [John M. Campbell, Jul 08 2011]
Positive values of x (or y) satisfying x^2 - 15xy + y^2 + 13 = 0. - Colin Barker, Feb 11 2014

Vincenzo Librandi, Table of n, a(n) for n = 1..200
J.-C. Novelli, J.-Y. Thibon, Hopf Algebras of m-permutations,(m+1)-ary trees, and m-parking functions, arXiv preprint arXiv:1403.5962 [math.CO], 2014.
Index entries for the Pell equation
Index entries for linear recurrences with constant coefficients, signature (15,-1).

Cf. similar sequences listed in A238379.

I:=[1,14]; [n le 2 select I[n] else 15*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Feb 12 2014

LinearRecurrence[{15,-1},{1,14},20] (* Harvey P. Dale, Oct 08 2012 *)
CoefficientList[Series[(1 - x)/(1 - 15 x + x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Feb 12 2014 *)

a(n) = round((2^(-1-n)*((15-sqrt(221))^n*(13+sqrt(221))+(-13+sqrt(221))*(15+sqrt(221))^n))/sqrt(221)) \\ Colin Barker, Jul 25 2016

a(n) = 15*a(n-1)-a(n-2).
G.f.: (1-x)*x/(1-15*x+x^2).
a(n) = (2^(-1-n)*((15-sqrt(221))^n*(13+sqrt(221))+(-13+sqrt(221))*(15+sqrt(221))^n))/sqrt(221). - Colin Barker, Jul 25 2016

Edited, extended by R. J. Mathar, Sep 02 2009

First formula corrected by Harvey P. Dale, Oct 08 2012

A075796 Numbers k such that 5*k^2 + 5 is a square.

Original entry on oeis.org

2, 38, 682, 12238, 219602, 3940598, 70711162, 1268860318, 22768774562, 408569081798, 7331474697802, 131557975478638, 2360712083917682, 42361259535039638, 760141959546795802, 13640194012307284798, 244763350261984330562, 4392100110703410665318, 78813038642399407645162

Offset: 1

Gregory V. Richardson, Oct 13 2002

Bisection of A001077; a(n) = A001077(2*n-1). - Greg Dresden, Jun 08 2021
From Peter Bala, Aug 25 2022: (Start)
The aerated sequence (b(n))n>=1 = [2, 0, 38, 0, 682, 0, 1238, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). The sequence (1/2)*(b(n))n>=1 is the case P1 = 0, P2 = -16, Q = -1  of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047. (End)

A. H. Beiler, "The Pellian." Ch. 22 in Recreations in the Theory of Numbers: The Queen of Mathematics Entertains. Dover, New York, New York, pp. 248-268, 1966.
L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. AMS Chelsea Publishing, Providence, Rhode Island, 1999, pp. 341-400.
Peter G. L. Dirichlet, Lectures on Number Theory (History of Mathematics Source Series, V. 16); American Mathematical Society, Providence, Rhode Island, 1999, pp. 139-147.

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Tanya Khovanova, Recursive Sequences
J. J. O'Connor and E. F. Robertson, Pell's Equation
Eric Weisstein's World of Mathematics, Pell Equation.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
Index entries for linear recurrences with constant coefficients, signature (18,-1).

Cf. A000290, A306380, A001077.

I:=[2,38]; [n le 2 select I[n] else 18*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Nov 30 2011

[Lucas(6*n-3)/2: n in [1..20]]; // G. C. Greubel, Feb 13 2019

with(combinat); A075796:=n->fibonacci(6*n+3)+fibonacci(6*n)/2; seq(A075796(n), n=1..50); # Wesley Ivan Hurt, Nov 29 2013

LinearRecurrence[{18, -1}, {2, 38}, 50] (* Sture Sjöstedt, Nov 29 2011; typo fixed by Vincenzo Librandi, Nov 30 2011 *)
LucasL[6*Range[20]-3]/2 (* G. C. Greubel, Feb 13 2019 *)
CoefficientList[Series[2*(1+x)/( 1-18*x+x^2 ), {x,0,20}],x] (* Stefano Spezia, Mar 02 2019 *)

vector(20, n, (fibonacci(6*n-2) + fibonacci(6*n-4))/2) \\ G. C. Greubel, Feb 13 2019

[(fibonacci(6*n-2) + fibonacci(6*n-4))/2 for n in (1..20)] # G. C. Greubel, Feb 13 2019

a(n) = (((9 + 4*sqrt(5))^n - (9 - 4*sqrt(5))^n) + ((9 + 4*sqrt(5))^(n-1) - (9 - 4*sqrt(5))^(n-1)))/(4*sqrt(5)).
a(n) = 18*a(n-1) - a(n-2).
a(n) = 2*A049629(n-1).
Limit_{n->oo} a(n)/a(n-1) = 8*phi + 1 = 9 + 4*sqrt(5).
a(n+1) = 9*a(n) + 4*sqrt(5)*sqrt((a(n)^2+1)). - Richard Choulet, Aug 30 2007
G.f.: 2*x*(1 + x)/(1 - 18*x + x^2). - Richard Choulet, Oct 09 2007
From Johannes W. Meijer, Jul 01 2010: (Start)
a(n) = A000045(6*n+3) + A000045(6*n)/2.
a(n) = 2*A167808(6*n+4) - A167808(6*n+6).
Limit_{k->oo} a(n+k)/a(k) = A023039(n)*A060645(n)*sqrt(5).
(End)
5*A007805(n)^2 - 1 = a(n+1)^2. - Sture Sjöstedt, Nov 29 2011
From Peter Bala, Nov 29 2013: (Start)
a(n) = Lucas(6*n - 3)/2.
Sum_{n >= 1} 1/(a(n) + 5/a(n)) = 1/4. Compare with A002878, A005248, A023039. (End)
Limit_{n->oo} a(n)/A007805(n-1) = sqrt(5). - A.H.M. Smeets, May 29 2017
E.g.f.: (exp((9 - 4*sqrt(5))*x)*(- 5 + 2*sqrt(5) + (5 + 2*sqrt(5))*exp(8*sqrt(5)*x)))/(2*sqrt(5)). - Stefano Spezia, Feb 13 2019
Sum_{n > 0} 1/a(n) = (1/log(9 - 4*sqrt(5)))*(- 17 - 38/sqrt(5))*sqrt(5*(9 - 4*sqrt(5)))*(- 9 + 4*sqrt(5))*(psi_{9 - 4*sqrt(5)}(1/2) - psi_{9 - 4*sqrt(5)}(1/2 - (I*Pi)/log(9 - 4*sqrt(5)))) approximately equal to 0.527868600269500798938265500122302016..., where psi_q(x) is the q-digamma function. - Stefano Spezia, Feb 25 2019
a(n) = sinh((6*n - 3)*arccsch(2)). - Peter Luschny, May 25 2022

A129818 Riordan array (1/(1+x), x/(1+x)^2), inverse array is A039599.

Original entry on oeis.org

1, -1, 1, 1, -3, 1, -1, 6, -5, 1, 1, -10, 15, -7, 1, -1, 15, -35, 28, -9, 1, 1, -21, 70, -84, 45, -11, 1, -1, 28, -126, 210, -165, 66, -13, 1, 1, -36, 210, -462, 495, -286, 91, -15, 1, -1, 45, -330, 924, -1287, 1001, -455, 120, -17, 1, 1, -55, 495, -1716, 3003, -3003, 1820, -680, 153, -19, 1

Offset: 0

Philippe Deléham, Jun 09 2007

This sequence is up to sign the same as A129818. - T. D. Noe, Sep 30 2011
Row sums: A057078. - Philippe Deléham, Jun 11 2007
Subtriangle of the triangle given by (0, -1, 0, -1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 19 2012
This triangle provides the coefficients of powers of x^2 for the even-indexed Chebyshev S polynomials (see A049310): S(2*n,x) = Sum_{k=0..n} T(n,k)*x^(2*k), n >= 0. - Wolfdieter Lang, Dec 17 2012
If L(x^n) := C(n) = A000108(n) (Catalan numbers), then the polynomials P_n(x) := Sum_{k=0..n} T(n,k)*x^k are orthogonal with respect to the inner product given by (f(x),g(x)) := L(f(x)*g(x)). - Michael Somos, Jan 03 2019

			Triangle T(n,k) begins:
  n\k  0   1    2     3     4     5    6    7    8   9 10 ...
   0:  1
   1: -1   1
   2:  1  -3    1
   3: -1   6   -5     1
   4:  1 -10   15    -7     1
   5: -1  15  -35    28    -9     1
   6:  1 -21   70   -84    45   -11    1
   7: -1  28 -126   210  -165    66  -13    1
   8:  1 -36  210  -462   495  -286   91  -15    1
   9: -1  45 -330   924 -1287  1001 -455  120  -17   1
  10:  1 -55  495 -1716  3003 -3003 1820 -680  153 -19  1
  ... Reformatted by _Wolfdieter Lang_, Dec 17 2012
Recurrence from the A-sequence A115141:
15 = T(4,2) = 1*6 + (-2)*(-5) + (-1)*1.
(0, -1, 0, -1, 0, 0, ...) DELTA (1, 0, 1, -1, 0, 0, ...) begins:
  1
  0,  1
  0, -1,   1
  0,  1,  -3,   1
  0, -1,   6,  -5,  1
  0,  1, -10,  15, -7,  1
  0, -1,  15, -35, 28, -9, 1. - _Philippe Deléham_, Mar 19 2012
Row polynomial for n=3 in terms of x^2: S(6,x) = -1 + 6*x^2 -5*x^4 + 1*x^6, with Chebyshev's S polynomial. See a comment above. - _Wolfdieter Lang_, Dec 17 2012
Boas-Buck type recurrence: -35 = T(5,2) = (5/3)*(-1*1 +1*(-5) - 1*15) = -3*7 = -35. - _Wolfdieter Lang_, Jun 03 2020

Vincenzo Librandi, Rows n = 1..101, flattened
Paul Barry, A Catalan Transform and Related Transformations on Integer Sequences, Journal of Integer Sequences, Vol. 8 (2005), Article 05.4.5.
Paul Barry and Aoife Hennessy, The Euler-Seidel Matrix, Hankel Matrices and Moment Sequences, J. Int. Seq. 13 (2010) # 10.8.2, example 15.
Aoife Hennessy, A Study of Riordan Arrays with Applications to Continued Fractions, Orthogonal Polynomials and Lattice Paths, Ph. D. Thesis, Waterford Institute of Technology, Oct. 2011.

Cf. A039599, A085478, A123970, A321620.

# The function RiordanSquare is defined in A321620.
RiordanSquare((1 - sqrt(1 - 4*x))/(2*x), 10):
LinearAlgebra[MatrixInverse](%); # Peter Luschny, Jan 04 2019

max = 10; Flatten[ CoefficientList[#, y] & /@ CoefficientList[ Series[ (1 + x)/(1 + (2 - y)*x + x^2), {x, 0, max}], x]] (* Jean-François Alcover, Sep 29 2011, after Wolfdieter Lang *)

@CachedFunction
def A129818(n,k):
    if n< 0: return 0
    if n==0: return 1 if k == 0 else 0
    h = A129818(n-1,k) if n==1 else 2*A129818(n-1,k)
    return A129818(n-1,k-1) - A129818(n-2,k) - h
for n in (0..9): [A129818(n,k) for k in (0..n)] # Peter Luschny, Nov 20 2012

T(n,k) = (-1)^(n-k)*A085478(n,k) = (-1)^(n-k)*binomial(n+k,2*k).
Sum_{k=0..n} T(n,k)*A000531(k) = n^2, with A000531(0)=0. - Philippe Deléham, Jun 11 2007
Sum_{k=0..n} T(n,k)*x^k = A033999(n), A057078(n), A057077(n), A057079(n), A005408(n), A002878(n), A001834(n), A030221(n), A002315(n), A033890(n), A057080(n), A057081(n), A054320(n), A097783(n), A077416(n), A126866(n), A028230(n+1) for x = 0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16, respectively. - Philippe Deléham, Nov 19 2009
O.g.f.: (1+x)/(1+(2-y)*x+x^2). - Wolfdieter Lang, Dec 15 2010
O.g.f. column k with leading zeros (Riordan array, see NAME): (1/(1+x))*(x/(1+x)^2)^k, k >= 0. - Wolfdieter Lang, Dec 15 2010
From Wolfdieter Lang, Dec 20 2010: (Start)
Recurrences from the Z- and A-sequences for Riordan arrays. See the W. Lang link under A006232 for details and references.
T(n,0) = -1*T(n-1,0), n >= 1, from the o.g.f. -1 for the Z-sequence (trivial result).
T(n,k) = Sum_{j=0..n-k} A(j)*T(n-1,k-1+j), n >= k >= 1, with A(j):= A115141(j) = [1,-2,-1,-2,-5,-14,...], j >= 0 (o.g.f. 1/c(x)^2 with the A000108 (Catalan) o.g.f. c(x)). (End)
T(n,k) = (-1)^n*A123970(n,k). - Philippe Deléham, Feb 18 2012
T(n,k) = -2*T(n-1,k) + T(n-1,k-1) - T(n-2,k), T(0,0) = T(1,1) = 1, T(1,0) = -1, T(n,k) = 0 if k < 0 or if k > n. - Philippe Deléham, Mar 19 2012
A039599(m,n) = Sum_{k=0..n} T(n,k) * C(k+m) where C(n) are the Catalan numbers. - Michael Somos, Jan 03 2019
Equals the matrix inverse of the Riordan square (cf. A321620) of the Catalan numbers. - Peter Luschny, Jan 04 2019
Boas-Buck type recurrence for column k >= 0 (see Aug 10 2017 comment in A046521 with references): T(n,k) = ((1 + 2*k)/(n - k))*Sum_{j = k..n-1} (-1)^(n-j)*T(j,k), with input T(n,n) = 1, and T(n,k) = 0 for n < k. - Wolfdieter Lang, Jun 03 2020

A244419 Coefficient triangle of polynomials related to the Dirichlet kernel. Rising powers. Riordan triangle ((1+z)/(1+z^2), 2*z/(1+z^2)).

Original entry on oeis.org

1, 1, 2, -1, 2, 4, -1, -4, 4, 8, 1, -4, -12, 8, 16, 1, 6, -12, -32, 16, 32, -1, 6, 24, -32, -80, 32, 64, -1, -8, 24, 80, -80, -192, 64, 128, 1, -8, -40, 80, 240, -192, -448, 128, 256, 1, 10, -40, -160, 240, 672, -448, -1024, 256, 512, -1, 10, 60, -160, -560, 672, 1792, -1024, -2304, 512, 1024

Offset: 0

Wolfdieter Lang, Jul 29 2014

This is the row reversed version of A180870. See also A157751 and A228565.
The Dirichlet kernel is D(n,x) = Sum_{k=-n..n} exp(i*k*x) = 1 + 2*Sum_{k=1..n} T(n,x) = S(n, 2*y) + S(n-1, 2*y) = S(2*n, sqrt(2*(1+y))) with y = cos(x), n >= 0, with the Chebyshev polynomials T (A053120) and S (A049310). This triangle T(n, k) gives in row n the coefficients of the polynomial Dir(n,y) = D(n,x=arccos(y)) = Sum_{m=0..n} T(n,m)*y^m. See A180870, especially the Peter Bala comments and formulas.
This is the Riordan triangle ((1+z)/(1+z^2), 2*z/(1+z^2)) due to the o.g.f. for Dir(n,y) given by (1+z)/(1 - 2*y*z + z^2) = G(z)/(1 - y*F(z)) with G(z) = (1+z)/(1+z^2) and F(z) = 2*z/(1+z^2) (see the Peter Bala formula under A180870). For Riordan triangles and references see the W. Lang link 'Sheffer a- and z- sequences' under A006232.
The A- and Z- sequences of this Riordan triangle are (see the mentioned W. Lang link in the preceding comment also for the references): The A-sequence has o.g.f. 1+sqrt(1-x^2) and is given by A(2*k+1) = 0 and A(2*k) [2, -1/2, -1/8, -1/16, -5/128, -7/256, -21/1024, -33/2048, -429/32768, -715/65536, ...], k >= 0. (See A098597 and A046161.)
  The Z-sequence has o.g.f. sqrt((1-x)/(1+x)) and is given by
  [1, -1, 1/2, -1/2, 3/8, -3/8, 5/16, -5/16, 35/128, -35/128, ...]. (See A001790 and A046161.)
The column sequences are A057077, 2*(A004526 with even numbers signed), 4*A008805 (signed), 8*A058187 (signed), 16*A189976 (signed), 32*A189980 (signed) for m = 0, 1, ..., 5.
The row sums give A005408 (from the o.g.f. due to the Riordan property), and the alternating row sums give A033999.
The row polynomials Dir(n, x), n >= 0, give solutions to the diophantine equation (a + 1)*X^2 - (a - 1)*Y^2 = 2 by virtue of the identity (a + 1)*Dir(n, -a)^2 - (a - 1)*Dir(n, a)^2 = 2, which is easily proved inductively using the recurrence Dir(n, a) = (1 + a)*(-1)^(n-1)*Dir(n-1, -a) + a*Dir(n-1, a) given below by Wolfdieter Lang. - Peter Bala, May 08 2025

			The triangle T(n,m) begins:
  n\m  0   1   2    3    4    5    6     7     8    9    10 ...
  0:   1
  1:   1   2
  2:  -1   2   4
  3:  -1  -4   4    8
  4:   1  -4 -12    8   16
  5:   1   6 -12  -32   16   32
  6:  -1   6  24  -32  -80   32   64
  7:  -1  -8  24   80  -80 -192   64   128
  8:   1  -8 -40   80  240 -192 -448   128   256
  9:   1  10 -40 -160  240  672 -448 -1024   256  512
  10: -1  10  60 -160 -560  672 1792 -1024 -2304  512  1024
  ...
Example for A-sequence recurrence: T(3,1) = Sum_{j=0..2} A(j)*T(2,j) = 2*(-1) + 0*2 + (-1/2)*4 = -4. Example for Z-sequence recurrence: T(4,0) = Sum_{j=0..3} Z(j)*T(3,j) = 1*(-1) + (-1)*(-4) + (1/2)*4 + (-1/2)*8 = +1. (For the A- and Z-sequences see a comment above.)
Example for the alternate recurrence: T(4,2) = 2*T(3,1) - T(3,2) = 2*(-4) - 4 = -12. T(4,3) = 0*T(3,2) + T(3,3) = T(3,3) = 8. - _Wolfdieter Lang_, Jul 30 2014

Wikipedia, Dirichlet kernel.

Dir(n, x) : A005408 (x = 1), A002878 (x = 3/2), A001834 (x = 2), A030221 (x = 5/2), A002315 (x = 3), A033890 (x = 7/2), A057080 (x = 4), A057081 (x = 9/2), A054320 (x = 5), A077416 (x = 6), A028230 (x = 7), A159678 (x = 8), A049629 (x = 9), A083043 (x = 10),
(-1)^n * Dir(n, x): A122367 (x = -3/2); A079935 (x = -2), A004253 (x = -5/2), A001653 (x = -3), A049685 (x = -7/2), A070997 (x = -4), A070998 (x = -9/2), A072256(n+1) (x = -5).
Cf. A180870 (row reversed), A157751, A228565, A049310, A053120, A057077, A004526, A008805, A058187, A189976, A189980, A005408, A033999.

T[n_, k_] := T[n, k] = Which[k == 0, (-1)^Quotient[n, 2], (0 <= n && n < k) || (n == -1 && k == 1), 0, True, 2 T[n-1, k-1] - T[n-2, k]];
Table[T[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 28 2019, from Sage *)

def T(n, k):
    if k == 0: return (-1)^(n//2)
    if (0 <= n and n < k) or (n == -1 and k == 1): return 0
    return 2*T(n-1, k-1) - T(n-2, k)
for n in range(11): [T(n,k) for k in (0..n)] # Peter Luschny, Jul 29 2014

T(n, m) = [y^m] Dir(n,y) for n >= m >= 0 and 0 otherwise, with the polynomials Dir(y) defined in a comment above.
T(n, m) = 2^m*(S(n,m) + S(n-1,m)) with the entries S(n,m) of A049310 given there explicitly.
O.g.f. for polynomials Dir(y) see a comment above (Riordan triangle ((1+z)/(1+z^2), 2*z/(1+z^2))).
O.g.f. for column m: ((1 + x)/(1 + x^2))*(2*x/(1 + x^2))^m, m >= 0, (Riordan property).
Recurrence for the polynomials: Dir(n, y) = 2*y*Dir(n-1, y) - Dir(n-2, y), n >= 1, with input D(-1, y) = -1 and D(0, y) = 1.
Triangle three-term recurrence: T(n,m) = 2*T(n-1,m-1) - T(n-2,m) for n >= m >= 1 with T(n,m) = 0 if 0 <= n < m, T(0,0) = 1, T(-1,1) = 0 and T(n,0) = A057077(n) = (-1)^(floor(n/2)).
From Wolfdieter Lang, Jul 30 2014: (Start)
In analogy to A157751 one can derive a recurrence for the row polynomials Dir(n, y) = Sum_{m=0..n} T(n,m)*y^m also using a negative argument but only one recursive step: Dir(n,y) = (1+y)*(-1)^(n-1)*Dir(n-1,-y) + y*Dir(n-1,y), n >= 1, Dir(0,y) = 1 (Dir(-1,y) = -1). See also A180870 from where this formula can be derived by row reversion.
This entails another triangle recurrence T(n,m) = (1 + (-1)^(n-m))*T(n-1,m-1) - (-1)^(n-m)*T(n-1,m), for n >= m >= 1 with T(n,m) = 0 if n < m and T(n,0) = (-1)^floor(n/2). (End)
From Peter Bala, Aug 14 2022: (Start)
The row polynomials Dir(n,x), n >= 0, are related to the Chebyshev polynomials of the first kind T(n,x) by the binomial transform as follows:
(2^n)*(x - 1)^(n+1)*Dir(n,x) = (-1) * Sum_{k = 0..2*n+1} binomial(2*n+1,k)*T(k,-x).
Note that Sum_{k = 0..2*n} binomial(2*n,k)*T(k,x) = (2^n)*(1 + x)^n*T(n,x). (End)
From Peter Bala, May 04 2025: (Start)
For n >= 1, the n-th row polynomial Dir(n, x) = (-1)^n * (U(n, -x) - U(n-1, -x)) = U(2*n, sqrt((1+x)/2)), where U(n, x) denotes the n-th Chebyshev polynomial of the second kind.
For n >= 1 and x < 1, Dir(n, x) = (-1)^n * sqrt(2/(1 - x )) * T(2*n+1, sqrt((1 - x)/2)), where T(n, x) denotes the n-th Chebyshev polynomial of the first kind.
Dir(n, x)^2 - 2*x*Dir(n, x)*Dir(n+1, x) + Dir(n+1, x)^2 = 2*(1 + x).
Dir(n, x) = (-1)^n * R(n, -2*(x+1)), where R(n, x) is the n-th row polynomial of the triangle A085478.
Dir(n, x) = Sum_{k = 0..n} (-1)^(n+k) * binomial(n+k, 2*k) * (2*x + 2)^k. (End)

A054486 Expansion of (1+2x)/(1-3x+x^2).

Original entry on oeis.org

1, 5, 14, 37, 97, 254, 665, 1741, 4558, 11933, 31241, 81790, 214129, 560597, 1467662, 3842389, 10059505, 26336126, 68948873, 180510493, 472582606, 1237237325, 3239129369, 8480150782, 22201322977, 58123818149, 152170131470, 398386576261, 1042989597313

Offset: 0

Barry E. Williams, May 06 2000

Binomial transform of A000285. - R. J. Mathar, Oct 26 2011

			G.f. = 1 + 5*x + 14*x^2 + 37*x^3 + 97*x^4 + 254*x^5 + 665*x^6 + 1741*x^7 + ...

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 122-125, 194-196.

G. C. Greubel, Table of n, a(n) for n = 0..1000
I. Adler, Three Diophantine equations - Part II, Fib. Quart., 7 (1969), pp. 181-193.
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (3,-1).

Cf. A000032, A000045, A000285, A002878, A100545, A104449, A228208.

F:=Fibonacci;; List([0..30], n-> F(2*n+2) +2*F(2*n) ); # G. C. Greubel, Nov 08 2019

R:=PowerSeriesRing(Integers(), 30); Coefficients(R!( (1+2*x)/(1-3*x+x^2)) ); // Marius A. Burtea, Nov 05 2019

a:=[1,5]; [n le 2 select a[n] else 3*Self(n-1)-Self(n-2): n in [1..30]]; // Marius A. Burtea, Nov 05 2019

with(combinat); f:=fibonacci; seq(f(2*n+2)+2*f(2*n), n=0..30); # G. C. Greubel, Nov 08 2019

CoefficientList[Series[(2*z+1)/(z^2-3*z+1), {z, 0, 30}], z] (* Vladimir Joseph Stephan Orlovsky, Jul 15 2011 *)
a[ n_]:= 3 Fibonacci[2n] + Fibonacci[2n+1]; (* Michael Somos, Mar 17 2015 *)
LinearRecurrence[{3,-1},{1,5},40] (* Harvey P. Dale, Apr 24 2019 *)

Vec((1+2*x)/(1-3*x+x^2)+O(x^99)) \\ Charles R Greathouse IV, Jul 15 2011

{a(n) = 3*fibonacci(2*n) + fibonacci(2*n+1)}; /* Michael Somos, Mar 17 2015 */

f=fibonacci; [f(2*n+2) +2*f(2*n) for n in (0..30)] # G. C. Greubel, Nov 08 2019

a(n) = 3*a(n-1) - a(n-2), a(0)=1, a(1)=5.
a(n) = (5*(((3+sqrt(5))/2)^n - ((3-sqrt(5))/2)^n) - (((3+sqrt(5))/2)^(n-1) - ((3-sqrt(5))/2)^(n-1)))/sqrt(5).
a(n) + 7*A001519(n) = A005248(n). - Creighton Dement, Oct 30 2004
a(n) = Lucas(2*n+1) + Fibonacci(2*n) = A002878(n) + A001906(n) = A025169(n-1) + A001906(n+1).
a(n) = (-1)^n*Sum_{k = 0..n} A238731(n,k)*(-6)^k. - Philippe Deléham, Mar 05 2014
0 = -11 + a(n)^2 - 3*a(n)*a(n+1) + a(n+1)^2 for all n in Z. - Michael Somos, Mar 17 2015
a(n) = -2*F(n)^2 + 6*F(n)*F(n+1) + F(n+1)^2 for all n in Z where F = Fibonacci. - Michael Somos, Mar 17 2015
a(n) = 3*F(2*n) + F(2*n+1) for all n in Z where F = Fibonacci. - Michael Somos, Mar 17 2015
a(n) = -A100545(-2-n) for all n in Z. - Michael Somos, Mar 17 2015
a(n) = A000285(2*n) = A228208(2*n+1) = A104449(2*n+1) for all n in Z. - Michael Somos, Mar 17 2015
From Klaus Purath, Nov 05 2019: (Start)
a(n) = (a(n-m) + a(n+m))/Lucas(2*m), m <= n.
a(n) = sum of 2*m+1 consecutive terms starting with a(n-m) divided by Lucas(2*m+1), m <= n.
a(n) = alternating sum of 2*m+1 consecutive terms starting with a(n-m) divided by Fibonacci(2*m+1), m <= n.
a(n) + a(n+1) = sum of 2*m+2 consecutive terms starting with a(n-m) divided by Fibonacci(2*m+2), m <= n.
a(n) + a(n+1) = (a(n-m) + a(n+m+1))/Fibonacci(2*m+1), m <= n.
The following formulas are extended to negative indexes:
a(n) = 3*Fibonacci(2*n+1) - Fibonacci(2*n-3).
a(n) = (Fibonacci(2*n+5) - 3* Fibonacci(2*n-1))/2.
a(n) = (4*Lucas(2*n+2) - Lucas(2*n-4))/5.
a(n) = Fibonacci(2*n+5) - 4*Fibonacci(2*n+1).
a(n) = (5*Fibonacci(2*n+5) - Fibonacci(2*n-7))/12. (End)
E.g.f.: exp(-(1/2)*(-3+sqrt(5))*x)*(-7 + sqrt(5) + (7 + sqrt(5))*exp(sqrt(5)*x))/(2*sqrt(5)). - Stefano Spezia, Nov 19 2019
a(n) = 3*n + 1 + Sum_{k=1..n} k*a(n-k). - Yu Xiao, Jun 20 2020

"a(1)=5", not "a(0)=5" from Dan Nielsen (nielsed(AT)uah.edu), Sep 10 2009

A060923 Bisection of Lucas triangle A060922: even-indexed members of column sequences of A060922 (not counting leading zeros).

Original entry on oeis.org

1, 4, 1, 11, 17, 1, 29, 80, 39, 1, 76, 303, 315, 70, 1, 199, 1039, 1687, 905, 110, 1, 521, 3364, 7470, 6666, 2120, 159, 1, 1364, 10493, 29634, 37580, 20965, 4311, 217, 1, 3571, 31885, 109421, 181074, 148545

Offset: 0

Wolfdieter Lang, Apr 20 2001

			Triangle begins:
  {1};
  {4,1};
  {11,17,1};
  {29,80,39,1};
  ...
pLe(2,x) = 1+11*x-11*x^2+4*x^3.

Row sums give A060926.
Column sequences (without leading zeros) are, for m=0..3: A002878, A060934-A060936.
Companion triangle A060924 (odd part).
Cf. A060922.

a(n, m) = A060922(2*n-m, m).
a(n, m) = ((2*(n-m)+1)*A060924(n-1, m-1) + 2*(4*n-3*m)*a(n-1, m-1) + 4*(2*n-m-1)*A060924(n-2, m-1))/(5*m), m >= n >= 1; a(n, 0)= A002878(n); else 0.
G.f. for column m >= 0: x^m*pLe(m+1, x)/(1-3*x+x^2)^(m+1), where pLe(n, x) := Sum_{m=0..n+floor(n/2)} A061186(n, m)*x^m are the row polynomials of the (signed) staircase A061186.
T(n,k) = 3*T(n-1,k) + 2*T(n-1,k-1) - T(n-2,k) + 2*T(n-2,k-1) - T(n-2,k-2) + 4*T(n-3,k-2), T(0,0) = 1, T(1,0) = 4, T(1,1) = 1, T(2,0) = 11, T(2,1) = 17, T(2,2) = 1, T(n,k) = 0 if k < 0 or if k > n. - Philippe Deléham, Jan 21 2014

A081071 a(n) = Lucas(4n+2)-2 = Lucas(2n+1)^2.

Original entry on oeis.org

1, 16, 121, 841, 5776, 39601, 271441, 1860496, 12752041, 87403801, 599074576, 4106118241, 28143753121, 192900153616, 1322157322201, 9062201101801, 62113250390416, 425730551631121, 2918000611027441, 20000273725560976

Offset: 0

R. K. Guy, Mar 04 2003

Conjecture: a(n) = Fibonacci(4*n+3) + Sum_{k=2..2*n} Fibonacci(2*k). - Alex Ratushnyak, May 06 2012

The above conjecture is true for n >= 1. - Nguyen Tuan Anh, Aug 02 2025

Hugh C. Williams, Edouard Lucas and Primality Testing, John Wiley and Sons, 1998, p. 75.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Pridon Davlianidze, Problem B-1270, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 58, No. 2 (2020), p. 179; Four Telescopic Infinite Products, Solution to Problem B-1270 by Jason L. Smith, ibid., Vol. 59, No. 2 (2021), pp. 183-184.
Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A000032 (Lucas numbers), A000045, A001622, A002878 is Lucas(2n+1), A081069.

I:=[1, 16, 121]; [n le 3 select I[n] else 8*Self(n-1)-8*Self(n-2)+Self(n-3): n in [1..30]]; // Vincenzo Librandi, Jun 26 2012

luc := proc(n) option remember: if n=0 then RETURN(2) fi: if n=1 then RETURN(1) fi: luc(n-1)+luc(n-2): end: for n from 0 to 40 do printf(`%d,`,luc(4*n+2)-2) od: # James Sellers, Mar 05 2003

CoefficientList[Series[-(1+8*x+x^2)/((x-1)*(x^2-7*x+1)),{x,0,40}],x] (* or *) LinearRecurrence[{8,-8,1},{1,16,121},50] (* Vincenzo Librandi, Jun 26 2012 *)
LucasL[4*Range[0,20]+2]-2 (* Harvey P. Dale, Nov 25 2012 *)

x='x+O('x^30); Vec((1+8*x+x^2)/((1-x)*(x^2-7*x+1))) \\ G. C. Greubel, Dec 21 2017

a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3).
G.f.: -(1+8*x+x^2)/((x-1)*(x^2-7*x+1)). - Colin Barker, Jun 26 2012
From Peter Bala, Nov 19 2019: (Start)
Sum_{n >= 1} 1/(a(n) + 5) = (3*sqrt(5) - 5)/30.
Sum_{n >= 1} 1/(a(n) - 5) = (15 - 4*sqrt(5) )/60.
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 5) = 1/12.
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 25/a(n)) = (5 + 2*sqrt(5))/120. (End)
Sum_{n>=0} 1/a(n) = (1/sqrt(5)) * Sum_{n>=1} n/F(2*n), where F(n) is the n-th Fibonacci number (A000045). - Amiram Eldar, Oct 05 2020
Product_{n>=1} (1 - 5/a(n)) = phi^2/4, where phi is the golden ratio (A001622) (Davlianidze, 2020). - Amiram Eldar, Dec 04 2024
From Enrique Navarrete, Mar 24 2025: (Start)
20 + 5*a(n) = A106729(n)^2.
Limit_{n->oo} a(n+1)/a(n) = (7 + 3*sqrt(5))/2. (End)

More terms from James Sellers, Mar 05 2003

cp's OEIS Frontend

A025169 a(n) = 2*Fibonacci(2*n+2).

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A155116 a(n) = 3*a(n-1) + 3*a(n-2), n>2, a(0)=1, a(1)=2, a(2)=8.

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A082985 Coefficient table for Chebyshev's U(2*n,x) polynomial expanded in decreasing powers of (-4*(1-x^2)).

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A160682 The list of the A values in the common solutions to 13*k+1 = A^2 and 17*k+1 = B^2.

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A075796 Numbers k such that 5*k^2 + 5 is a square.

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A129818 Riordan array (1/(1+x), x/(1+x)^2), inverse array is A039599.

A025169 a(n) = 2Fibonacci(2n+2).

A155116 a(n) = 3a(n-1) + 3a(n-2), n>2, a(0)=1, a(1)=2, a(2)=8.

A082985 Coefficient table for Chebyshev's U(2n,x) polynomial expanded in decreasing powers of (-4(1-x^2)).

A160682 The list of the A values in the common solutions to 13k+1 = A^2 and 17k+1 = B^2.

A054486 Expansion of (1+2x)/(1-3x+x^2).

A081071 a(n) = Lucas(4n+2)-2 = Lucas(2n+1)^2.