A079935 -id:A079935 - cp's OEIS frontend

A001353 a(n) = 4*a(n-1) - a(n-2) with a(0) = 0, a(1) = 1.

0, 1, 4, 15, 56, 209, 780, 2911, 10864, 40545, 151316, 564719, 2107560, 7865521, 29354524, 109552575, 408855776, 1525870529, 5694626340, 21252634831, 79315912984, 296011017105, 1104728155436, 4122901604639, 15386878263120, 57424611447841, 214311567528244

Offset: 0

N. J. A. Sloane

3*a(n)^2 + 1 is a square. Moreover, 3*a(n)^2 + 1 = (2*a(n) - a(n-1))^2.
Consecutive terms give nonnegative solutions to x^2 - 4*x*y + y^2 = 1. - Max Alekseyev, Dec 12 2012
Values y solving the Pellian x^2 - 3*y^2 = 1; corresponding x values given by A001075(n). Moreover, we have a(n) = 2*a(n-1) + A001075(n-1). - Lekraj Beedassy, Jul 13 2006
Number of spanning trees in 2 X n grid: by examining what happens at the right-hand end we see that a(n) = 3*a(n-1) + 2*a(n-2) + 2*a(n-3) + ... + 2*a(1) + 1, where the final 1 corresponds to the tree ==...=| !. Solving this we get a(n) = 4*a(n-1) - a(n-2).
Complexity of 2 X n grid.
A016064 also describes triangles whose sides are consecutive integers and in which an inscribed circle has an integer radius. A001353 is exactly and precisely mapped to the integer radii of such inscribed circles, i.e., for each term of A016064, the corresponding term of A001353 gives the radius of the inscribed circle. - Harvey P. Dale, Dec 28 2000
n such that 3*n^2 = floor(sqrt(3)*n*ceiling(sqrt(3)*n)). - Benoit Cloitre, May 10 2003
For n>0, ratios a(n+1)/a(n) may be obtained as convergents of the continued fraction expansion of 2+sqrt(3): either as successive convergents of [4;-4] or as odd convergents of [3;1, 2]. - Lekraj Beedassy, Sep 19 2003
Ways of packing a 3 X (2*n-1) rectangle with dominoes, after attaching an extra square to the end of one of the sides of length 3. With reference to A001835, therefore: a(n) = a(n-1) + A001835(n-1) and A001835(n) = 3*A001835(n-1) + 2*a(n-1). - Joshua Zucker and the Castilleja School Math Club, Oct 28 2003
a(n+1) is a Chebyshev transform of 4^n, where the sequence with g.f. G(x) is sent to the sequence with g.f. (1/(1+x^2))G(x/(1+x^2)). - Paul Barry, Oct 25 2004
This sequence is prime-free, because a(2n) = a(n) * (a(n+1)-a(n-1)) and a(2n+1) = a(n+1)^2 - a(n)^2 = (a(n+1)+a(n)) * (a(n+1)-a(n)). - Jianing Song, Jul 06 2019
Numbers such that there is an m with t(n+m) = 3*t(m), where t(n) are the triangular numbers A000217. For instance, t(35) = 3*t(20) = 630, so 35 - 20 = 15 is in the sequence. - Floor van Lamoen, Oct 13 2005
a(n) = number of distinct matrix products in (A + B + C + D)^n where commutator [A,B] = 0 but neither A nor B commutes with C or D. - Paul D. Hanna and Max Alekseyev, Feb 01 2006
For n > 1, middle side (or long leg) of primitive Pythagorean triangles having an angle nearing Pi/3 with larger values of sides. [Complete triple (X, Y, Z), X < Y < Z, is given by X = A120892(n), Y = a(n), Z = A120893(n), with recurrence relations X(i+1) = 2*{X(i) - (-1)^i} + a(i); Z(i+1) = 2*{Z(i) + a(i)} - (-1)^i.] - Lekraj Beedassy, Jul 13 2006
From Dennis P. Walsh, Oct 04 2006: (Start)
Number of 2 X n simple rectangular mazes. A simple rectangular m X n maze is a graph G with vertex set {0, 1, ..., m} X {0, 1, ..., n} that satisfies the following two properties: (i) G consists of two orthogonal trees; (ii) one tree has a path that sequentially connects (0,0),(0,1), ..., (0,n), (1,n), ...,(m-1,n) and the other tree has a path that sequentially connects (1,0), (2,0), ..., (m,0), (m,1), ..., (m,n). For example, a(2) = 4 because there are four 2 X 2 simple rectangular mazes:
                                          
   |  |  |        |   |       |     |        |   |
   |   |        |   |       |  ||        |   |
(End)
[1, 4, 15, 56, 209, ...] is the Hankel transform of [1, 1, 5, 26, 139, 758, ...](see A005573). - Philippe Deléham, Apr 14 2007
The upper principal convergents to 3^(1/2), beginning with 2/1, 7/4, 26/15, 97/56, comprise a strictly decreasing sequence; numerators=A001075, denominators=A001353. - Clark Kimberling, Aug 27 2008
From Gary W. Adamson, Jun 21 2009: (Start)
A001353 and A001835 = bisection of continued fraction [1, 2, 1, 2, 1, 2, ...], i.e., of [1, 3, 4, 11, 15, 41, ...].
For n>0, a(n) equals the determinant of an (n-1) X (n-1) tridiagonal matrix with ones in the super and subdiagonals and (4, 4, 4, ...) as the main diagonal. [Corrected by Johannes Boot, Sep 04 2011]
A001835 and A001353 = right and next to right borders of triangle A125077. (End)
a(n) is equal to the permanent of the (n-1) X (n-1) Hessenberg matrix with 4's along the main diagonal, i's along the superdiagonal and the subdiagonal (i is the imaginary unit), and 0's everywhere else. - John M. Campbell, Jun 09 2011
2a(n) is the number of n-color compositions of 2n consisting of only even parts; see Guo in references. - Brian Hopkins, Jul 19 2011
Pisano period lengths: 1, 2, 6, 4, 3, 6, 8, 4, 18, 6, 10, 12, 12, 8, 6, 8, 18, 18, 5, 12, ... - R. J. Mathar, Aug 10 2012
From Michel Lagneau, Jul 08 2014: (Start)
a(n) is defined also by the recurrence a(1)=1; for n>1, a(n+1) = 2*a(n) + sqrt(3*a(n)^2 + 1) where a(n) is an integer for every n. This sequence is generalizable by the sequence b(n,m) of parameter m with the initial condition b(1,m) = 1, and for n > 1 b(n+1,m) = m*b(n,m) + sqrt((m^2 - 1)*b(n,m)^2 + 1) for m = 2, 3, 4, ... where b(n,m) is an integer for every n.
The first corresponding sequences are
   b(n,2)  = a(n) = A001353(n);
   b(n,3)  = A001109(n);
   b(n,4)  = A001090(n);
   b(n,5)  = A004189(n);
   b(n,6)  = A004191(n);
   b(n,7)  = A007655(n);
   b(n,8)  = A077412(n);
   b(n,9)  = A049660(n);
   b(n,10) = A075843(n);
   b(n,11) = A077421(n);
   ....................
We obtain a general sequence of polynomials {b(n,x)} = {1, 2*x, 4*x^2 - 1, 8*x^3 - 4*x, 16*x^4 - 12*x^2 + 1, 32*x^5 - 32*x^3 + 6*x, ...} with x = m where each b(n,x) is a Gegenbauer polynomial defined by the recurrence b(n,x)- 2*x*b(n-1,x) + b(n-2,x) = 0, the same relation as the Chebyshev recurrence, but with the initial conditions b(x,0) = 1 and b(x,1) = 2*x instead b(x,0) = 1 and b(x,1) = x for the Chebyshev polynomials. (End)
If a(n) denotes the n-th term of the above sequence and we construct a triangle whose sides are a(n) - 1, a(n) + 1 and sqrt(3a(n)^2 + 1), then, for every n the measure of one of the angles of the triangle so constructed will always be 120 degrees. This result of ours was published in Mathematics Spectrum (2012/2013), Vol. 45, No. 3, pp. 126-128. - K. S. Bhanu and Dr. M. N. Deshpande, Professor (Retd), Department of Statistics, Institute of Science, Nagpur (India).
For n >= 1, a(n) equals the number of 01-avoiding words of length n - 1 on alphabet {0, 1, 2, 3}. - Milan Janjic, Jan 25 2015
For n > 0, 10*a(n) is the number of vertices and roots on level n of the {4, 5} mosaic (see L. Németh Table 1 p. 6). - Michel Marcus, Oct 30 2015
(2 + sqrt(3))^n = A001075(n) + a(n)*sqrt(3), n >= 0; integers in the quadratic number field Q(sqrt(3)). - Wolfdieter Lang, Feb 16 2018
A strong divisibility sequence, that is, gcd(a(n), a(m)) = a(gcd(n, m)) for all positive integers n and m. - Michael Somos, Dec 12 2019
The Cholesky decomposition A = C C* for tridiagonal A with A[i,i] = 4 and A[i+1,i] = A[i,i+1] = -1, as it arises in the discretized 2D Laplace operator (Poisson equation...), has nonzero elements C[i,i] = sqrt(a(i+1)/a(i)) = -1/C[i+1,i], i = 1, 2, 3, ... - M. F. Hasler, Mar 12 2021
The triples (a(n-1), 2a(n), a(n+1)), n=2,3,..., are exactly the triples (a,b,c) of positive integers a < b < c in arithmetic progression such that a*b+1, b*c+1, and c*a+1 are perfect squares. - Bernd Mulansky, Jul 10 2021
From Greg Dresden and Linyun Sheng, Jul 01 2025: (Start)
  a(n) is the number of ways to tile this strip of length n,
  .________________
  |  |  |  |  |  |  |\
  ||__||__||__|_\,
  where the last cell is a right triangle, with three types of tiles: 1 X 1 squares, 1 X 1 small right triangles, and large right triangles (with large side length 2) formed by joining two of those small right triangles along a short leg. As an example, here is one of the a(7)=2911 ways to tile the 1 X 7 strip with these kinds of tiles:
  .________________
  |\   /|\ | /|  | / \
  |\/_|\|/|__|/_\,
(End)

			For example, when n = 3:
  ****
  .***
  .***
can be packed with dominoes in 4 different ways: 3 in which the top row is tiled with two horizontal dominoes and 1 in which the top row has two vertical and one horizontal domino, as shown below, so a(2) = 4.
  ---- ---- ---- ||--
  .||| .--| .|-- .|||
  .||| .--| .|-- .|||
G.f. = x + 4*x^2 + 15*x^3 + 56*x^4 + 209*x^5 + 780*x^6 + 2911*x^7 + 10864*x^8 + ...

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Index entries for linear recurrences with constant coefficients, signature (4,-1)
Index entries for sequences related to Chebyshev polynomials.

Cf. A001075, A001542, A001571, A001834, A001835, A002531, A003500, A005246, A016064, A019679, A079935, A082840.
A bisection of A002530.
Cf. A125077.
A row of A116469.
Chebyshev sequence U(n, m): A000027 (m=1), this sequence (m=2), A001109 (m=3), A001090 (m=4), A004189 (m=5), A004191 (m=6), A007655 (m=7), A077412 (m=8), A049660 (m=9), A075843 (m=10), A077421 (m=11), A077423 (m=12), A097309 (m=13), A097311 (m=14), A097313 (m=15), A029548 (m=16), A029547 (m=17), A144128 (m=18), A078987 (m=19), A097316 (m=33).
Cf. A323182.

a:=[0,1];; for n in [3..30] do a[n]:=4*a[n-1]-a[n-2]; od; a; # Muniru A Asiru, Feb 16 2018

a001353 n = a001353_list !! n
a001353_list =
   0 : 1 : zipWith (-) (map (4 *) $ tail a001353_list) a001353_list
-- Reinhard Zumkeller, Aug 14 2011

I:=[0,1]; [n le 2 select I[n] else 4*Self(n-1)-Self(n-2): n in [1..30]]; // G. C. Greubel, Jun 06 2019

A001353 := proc(n) option remember; if n <= 1 then n else 4*A001353(n-1)-A001353(n-2); fi; end;
A001353:=z/(1-4*z+z**2); # Simon Plouffe in his 1992 dissertation.
seq( simplify(ChebyshevU(n-1, 2)), n=0..20); # G. C. Greubel, Dec 23 2019

a[n_] := (MatrixPower[{{1, 2}, {1, 3}}, n].{{1}, {1}})[[2, 1]]; Table[ a[n], {n, 0, 30}] (* Robert G. Wilson v, Jan 13 2005 *)
Table[GegenbauerC[n-1, 1, 2], {n, 0, 30}] (* Zerinvary Lajos, Jul 14 2009 *)
Table[-((I Sin[n ArcCos[2]])/Sqrt[3]), {n, 0, 30}] // FunctionExpand (* Eric W. Weisstein, Jul 16 2011 *)
Table[Sinh[n ArcCosh[2]]/Sqrt[3], {n, 0, 30}] // FunctionExpand (* Eric W. Weisstein, Jul 16 2011 *)
Table[ChebyshevU[n-1, 2], {n, 0, 30}] (* Eric W. Weisstein, Jul 16 2011 *)
a[0]:=0; a[1]:=1; a[n_]:= a[n]= 4a[n-1] - a[n-2]; Table[a[n], {n, 0, 30}] (* Alonso del Arte, Jul 19 2011 *)
LinearRecurrence[{4, -1}, {0, 1}, 30] (* Sture Sjöstedt, Dec 06 2011 *)
Round@Table[Fibonacci[2n, Sqrt[2]]/Sqrt[2], {n, 0, 30}] (* Vladimir Reshetnikov, Sep 15 2016 *)

M = [ 1, 1, 0; 1, 3, 1; 0, 1, 1]; for(i=0,30,print1(([1,0,0]*M^i)[2],",")) \\ Lambert Klasen (Lambert.Klasen(AT)gmx.net), Jan 25 2005

{a(n) = real( (2 + quadgen(12))^n / quadgen(12) )}; /* Michael Somos, Sep 19 2008 */

{a(n) = polchebyshev(n-1, 2, 2)}; /* Michael Somos, Sep 19 2008 */

concat(0, Vec(x/(1-4*x+x^2) + O(x^30))) \\ Altug Alkan, Oct 30 2015

a001353 = [0, 1]
for n in range(30): a001353.append(4*a001353[-1] - a001353[-2])
print(a001353)  # Gennady Eremin, Feb 05 2022

[lucas_number1(n,4,1) for n in range(30)] # Zerinvary Lajos, Apr 22 2009

[chebyshev_U(n-1,2) for n in (0..20)] # G. C. Greubel, Dec 23 2019

G.f.: x/(1-4*x+x^2).
a(n) = ((2 + sqrt(3))^n - (2 - sqrt(3))^n)/(2*sqrt(3)).
a(n) = sqrt((A001075(n)^2 - 1)/3).
a(n) = 2*a(n-1) + sqrt(3*a(n-1)^2 + 1). - Lekraj Beedassy, Feb 18 2002
Limit_{n->oo} a(n)/a(n-1) = 2 + sqrt(3). - Gregory V. Richardson, Oct 06 2002
Binomial transform of A002605.
E.g.f.: exp(2*x)*sinh(sqrt(3)*x)/sqrt(3).
a(n) = S(n-1, 4) = U(n-1, 2); S(-1, x) := 0, Chebyshev's polynomials of the second kind A049310.
a(n+1) = Sum_{k=0..floor(n/2)} binomial(n-k, k)(-1)^k*4^(n - 2*k). - Paul Barry, Oct 25 2004
a(n) = Sum_{k=0..n-1} binomial(n+k,2*k+1)*2^k. - Paul Barry, Nov 30 2004
a(n) = 3*a(n-1) + 3*a(n-2) - a(n-3), n>=3. - Lekraj Beedassy, Jul 13 2006
a(n) = -A106707(n). - R. J. Mathar, Jul 07 2006
M^n * [1,0] = [A001075(n), A001353(n)], where M = the 2 X 2 matrix [2,3; 1,2]; e.g., a(4) = 56 since M^4 * [1,0] = [97, 56] = [A001075(4), A001353(4)]. - Gary W. Adamson, Dec 27 2006
From Michael Somos, Sep 19 2008: (Start)
Sequence satisfies 1 = f(a(n), a(n+1)) where f(u, v) = u^2 + v^2 - 4*u*v.
a(n) = -a(-n) for all integer n. (End)
Rational recurrence: a(n) = (17*a(n-1)*a(n-2) - 4*(a(n-1)^2 + a(n-2)^2))/a(n-3) for n > 3. - Jaume Oliver Lafont, Dec 05 2009
If p[i] = Fibonacci(2i) and if A is the Hessenberg matrix of order n defined by A[i,j] = p[j-i+1], (i <= j), A[i,j] = -1, (i = j + 1), and A[i,j] = 0 otherwise, then, for n >= 1, a(n) = det A. - Milan Janjic, May 08 2010
From Eric W. Weisstein, Jul 16 2011: (Start)
a(n) = C_{n-1}^{(1)}(2), where C_n^{(m)}(x) is the Gegenbauer polynomial.
a(n) = -i*sin(n*arccos(2))/sqrt(3).
a(n) = sinh(n*arccosh(2))/sqrt(3). (End)
a(n) = b such that Integral_{x=0..Pi/2} (sin(n*x))/(2-cos(x)) dx = c + b*log(2). - Francesco Daddi, Aug 02 2011
a(n) = sqrt(A098301(n)) = sqrt([A055793 / 3]), base 3 analog of A031150. - M. F. Hasler, Jan 16 2012
a(n+1) = Sum_{k=0..n} A101950(n,k)*3^k. - Philippe Deléham, Feb 10 2012
1, 4, 15, 56, 209, ... = INVERT(INVERT(1, 2, 3, 4, 5, ...)). - David Callan, Oct 13 2012
From Peter Bala, Dec 23 2012: (Start)
Product_{n >= 1} (1 + 1/a(n)) = 1 + sqrt(3).
Product_{n >= 2} (1 - 1/a(n)) = 1/4*(1 + sqrt(3)). (End)
a(n+1) = (A001834(n) + A001835(n))/2. a(n+1) + a(n) = A001834(n). a(n+1) - a(n) = A001835(n). - Richard R. Forberg, Sep 04 2013
a(n) = -(-i)^(n+1)*Fibonacci(n, 4*i), i = sqrt(-1). - G. C. Greubel, Jun 06 2019
a(n)^2 - a(m)^2 = a(n+m) * a(n-m), a(n+2)*a(n-2) = 16*a(n+1)*a(n-1) - 15*a(n)^2, a(n+3)*a(n-2) = 15*a(n+2)*a(n-1) - 14*a(n+1)*a(n) for all integer n, m. - Michael Somos, Dec 12 2019
a(n) = 2^n*Sum_{k >= n} binomial(2*k,2*n-1)*(1/3)^(k+1). Cf. A102591. - Peter Bala, Nov 29 2021
a(n) = Sum_{k > 0} (-1)^((k-1)/2)*binomial(2*n, n+k)*(k|12), where (k|12) is the Kronecker symbol. - Greg Dresden, Oct 11 2022
Sum_{k=0..n} a(k) = (a(n+1) - a(n) - 1)/2. - Prabha Sivaramannair, Sep 22 2023
a(2n+1) = A001835(n+1) * A001834(n). - M. Farrokhi D. G., Oct 15 2023
Sum_{n>=1} arctan(1/(4*a(n)^2)) = Pi/12 (A019679) (Ohtskua, 2024). - Amiram Eldar, Aug 29 2024
From Peter Bala, May 21 2025: (Start)
Product_{n >= 1} (1 + 1/a(n))^2 = 2*(2 + sqrt(3)) (telescoping product: (1 + 1/a(2*n-1))^2 * (1 + 1/a(2*n-2))^2 = (4 + 2*A251963(n)/A005246(2*n)^2)/(4 + 2*A251963(n-1)/A005246(2*n-2)^2) ).
Product_{n >= 2} (1 - 1/a(n))^2 = (1/8)*(2 + sqrt(3)).
Product_{n >= 1} ((a(2*n) + 1)/(a(2*n) - 1))^2 = 3 (telescoping product: ((a(2*n) + 1)/(a(2*n) - 1))^2 = (3 - 2/A001835(n+1)^2)/(3 - 2/A001835(n)^2) ).
Product_{n >= 2} ((a(2*n-1) + 1)/(a(2*n-1) - 1))^2 = 4/3.
The o.g.f. A(x) satisfies A(x) + A(-x) + 8*A(x)*A(-x) = 0. The o.g.f. for A007655 equals -A(sqrt(x))*A(-sqrt(x)). (End)

A002605 a(n) = 2*(a(n-1) + a(n-2)), a(0) = 0, a(1) = 1.

Original entry on oeis.org

0, 1, 2, 6, 16, 44, 120, 328, 896, 2448, 6688, 18272, 49920, 136384, 372608, 1017984, 2781184, 7598336, 20759040, 56714752, 154947584, 423324672, 1156544512, 3159738368, 8632565760, 23584608256, 64434348032, 176037912576, 480944521216, 1313964867584

Offset: 0

Colin Mallows

Individually, both this sequence and A028859 are convergents to 1 + sqrt(3). Mutually, both sequences are convergents to 2 + sqrt(3) and 1 + sqrt(3)/2. - Klaus E. Kastberg (kastberg(AT)hotkey.net.au), Nov 04 2001
The number of (s(0), s(1), ..., s(n+1)) such that 0 < s(i) < 6 and |s(i) - s(i-1)| <= 1 for i = 1, 2, ..., n + 1, s(0) = 2, s(n+1) = 3. - Herbert Kociemba, Jun 02 2004
The same sequence may be obtained by the following process. Starting a priori with the fraction 1/1, the denominators of fractions built according to the rule: add top and bottom to get the new bottom, add top and 4 times the bottom to get the new top. The limit of the sequence of fractions is sqrt(4). - Cino Hilliard, Sep 25 2005
The Hankel transform of this sequence is [1, 2, 0, 0, 0, 0, 0, 0, 0, ...]. - Philippe Deléham, Nov 21 2007
[1, 3; 1, 1]^n *[1, 0] = [A026150(n), a(n)]. - Gary W. Adamson, Mar 21 2008
(1 + sqrt(3))^n = A026150(n) + a(n)*sqrt(3). - Gary W. Adamson, Mar 21 2008
a(n+1) is the number of ways to tile a board of length n using red and blue tiles of length one and two. - Geoffrey Critzer, Feb 07 2009
Starting with offset 1 = INVERT transform of the Jacobsthal sequence, A001045: (1, 1, 3, 5, 11, 21, ...). - Gary W. Adamson, May 12 2009
Starting with "1" = INVERTi transform of A007482: (1, 3, 11, 39, 139, ...). - Gary W. Adamson, Aug 06 2010
An elephant sequence, see A175654. For the corner squares four A[5] vectors, with decimal values 85, 277, 337 and 340, lead to this sequence (without the leading 0). For the central square these vectors lead to the companion sequence A026150, without the first leading 1. - Johannes W. Meijer, Aug 15 2010
The sequence 0, 1, -2, 6, -16, 44, -120, 328, -896, ... (with alternating signs) is the Lucas U(-2,-2)-sequence. - R. J. Mathar, Jan 08 2013
a(n+1) counts n-walks (closed) on the graph G(1-vertex;1-loop,1-loop,2-loop,2-loop). - David Neil McGrath, Dec 11 2014
Number of binary strings of length 2*n - 2 in the regular language (00+11+0101+1010)*. - Jeffrey Shallit, Dec 14 2015
For n >= 1, a(n) equals the number of words of length n - 1  over {0, 1, 2, 3} in which 0 and 1 avoid runs of odd lengths. - Milan Janjic, Dec 17 2015
a(n+1) is the number of compositions of n into parts 1 and 2, both of two kinds. - Gregory L. Simay, Sep 20 2017
Number of associative, quasitrivial, and order-preserving binary operations on the n-element set {1, ..., n} that have neutral elements. - J. Devillet, Sep 28 2017
(1 + sqrt(3))^n = A026150(n) + a(n)*sqrt(3), for n >= 0; integers in the real quadratic number field Q(sqrt(3)). - Wolfdieter Lang, Feb 10 2018
Starting with 1, 2, 6, 16, ..., number of permutations of length n>0 avoiding the partially ordered pattern (POP) {1>3, 1>4} of length 4. That is, number of length n permutations having no subsequences of length 4 in which the first element is larger than the third and fourth elements. - Sergey Kitaev, Dec 09 2020
a(n) is the number of tilings of a 2 X n board missing one corner cell, with 1 X 1 and L-shaped tiles (where the L-shaped tiles cover 3 squares). Compare to A127864. - Greg Dresden and Yilin Zhu, Jul 17 2025

John Derbyshire, Prime Obsession, Joseph Henry Press, April 2004, p. 16.

Vincenzo Librandi, Table of n, a(n) for n = 0..500
A. Abdurrahman, CM Method and Expansion of Numbers, arXiv:1909.10889 [math.NT], 2019.
Jean-Luc Baril, Nathanaël Hassler, Sergey Kirgizov, and José L. Ramírez, Grand zigzag knight's paths, arXiv:2402.04851 [math.CO], 2024.
Paul Barry, On the Gap-sum and Gap-product Sequences of Integer Sequences, arXiv:2104.05593 [math.CO], 2021.
Paul Barry, Notes on Riordan arrays and lattice paths, arXiv:2504.09719 [math.CO], 2025. See pp. 8, 29.
Martin Burtscher, Igor Szczyrba, and Rafał Szczyrba, Analytic Representations of the n-anacci Constants and Generalizations Thereof, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.5.
M. Couceiro, J. Devillet, and J.-L. Marichal, Quasitrivial semigroups: characterizations and enumerations, arXiv:1709.09162 [math.RA], 2017.
M. Diepenbroek, M. Maus, and A. Stoll, Pattern Avoidance in Reverse Double Lists, Preprint 2015. See Table 3.
Sergio Falcón, Binomial Transform of the Generalized k-Fibonacci Numbers, Communications in Mathematics and Applications (2019) Vol. 10, No. 3, 643-651.
Alice L. L. Gao and Sergey Kitaev, On partially ordered patterns of length 4 and 5 in permutations, arXiv:1903.08946 [math.CO], 2019.
Alice L. L. Gao and Sergey Kitaev, On partially ordered patterns of length 4 and 5 in permutations, The Electronic Journal of Combinatorics 26(3) (2019), P3.26.
Dale Gerdemann Bird Flock, Youtube video, 2011.
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=q=2.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 476
D. Jhala, G. P. S. Rathore, and K. Sisodiya, Some Properties of k-Jacobsthal Numbers with Arithmetic Indexes, Turkish Journal of Analysis and Number Theory, 2014, Vol. 2, No. 4, 119-124.
Tanya Khovanova, Recursive Sequences
Wolfdieter Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38,5 (2000) 408-419; Eqs. (39), (41) and (45), lhs, m=2.
D. H. Lehmer, On Lucas's test for the primality of Mersenne's numbers, Journal of the London Mathematical Society 1.3 (1935): 162-165. See U_n.
Toufik Mansour and Mark Shattuck, Pattern avoidance in flattened derangements, Disc. Math. Lett. (2025) Vol. 15, 67-74. See p. 74.
Yassine Otmani, The 2-Pascal Triangle and a Related Riordan Array, J. Int. Seq. (2025) Vol. 28, Issue 3, Art. No. 25.3.5. See p. 12.
Alan Prince, Counting parses, Rutgers Optimality Archive, 2010.
Index entries for linear recurrences with constant coefficients, signature (2,2).
Index entries for sequences related to Chebyshev polynomials.
Index entries for Lucas sequences.

First differences are given by A026150.
a(n) = A073387(n, 0), n>=0 (first column of triangle).
Equals (1/3) A083337. First differences of A077846. Pairwise sums of A028860 and abs(A077917).
a(n) = A028860(n)/2 apart from the initial terms.
Row sums of A081577 and row sums of triangle A156710.
The following sequences (and others) belong to the same family: A001333, A000129, A026150, A046717, A015518, A084057, A063727, A002533, A002532, A083098, A083099, A083100, A015519.
Cf. A080953, A052948, A080040, A028859, A030195, A106435, A108898, A125145, A265106, A265107, A265278, A270810, A293005, A293006, A293007.
Cf. A175289 (Pisano periods).
Cf. A002530.
Cf. A127864.

a002605 n = a002605_list !! n
a002605_list =
   0 : 1 : map (* 2) (zipWith (+) a002605_list (tail a002605_list))
-- Reinhard Zumkeller, Oct 15 2011

[Floor(((1 + Sqrt(3))^n - (1 - Sqrt(3))^n)/(2*Sqrt(3))): n in [0..30]]; // Vincenzo Librandi, Aug 18 2011

[n le 2 select n-1 else 2*Self(n-1) + 2*Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 07 2018

a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=2*a[n-1]+2*a[n-2]od: seq(a[n], n=0..33); # Zerinvary Lajos, Dec 15 2008
a := n -> `if`(n<3, n, 2^(n-1)*hypergeom([1-n/2, (1-n)/2], [1-n], -2));
seq(simplify(a(n)), n=0..29); # Peter Luschny, Dec 16 2015

Expand[Table[((1 + Sqrt[3])^n - (1 - Sqrt[3])^n)/(2Sqrt[3]), {n, 0, 30}]] (* Artur Jasinski, Dec 10 2006 *)
a[n_]:=(MatrixPower[{{1,3},{1,1}},n].{{1},{1}})[[2,1]]; Table[a[n],{n,-1,40}] (* Vladimir Joseph Stephan Orlovsky, Feb 19 2010 *)
LinearRecurrence[{2, 2}, {0, 1}, 30] (* Robert G. Wilson v, Apr 13 2013 *)
Round@Table[Fibonacci[n, Sqrt[2]] 2^((n - 1)/2), {n, 0, 20}] (* Vladimir Reshetnikov, Oct 15 2016 *)
nxt[{a_,b_}]:={b,2(a+b)}; NestList[nxt,{0,1},30][[All,1]] (* Harvey P. Dale, Sep 17 2022 *)

Vec(x/(1-2*x-2*x^2)+O(x^99)) \\ Charles R Greathouse IV, Jun 10 2011

A002605(n)=([2,2;1,0]^n)[2,1] \\ M. F. Hasler, Aug 06 2018

[lucas_number1(n,2,-2) for n in range(0, 30)] # Zerinvary Lajos, Apr 22 2009

a = BinaryRecurrenceSequence(2,2)
print([a(n) for n in (0..29)])  # Peter Luschny, Aug 29 2016

a(n) = (-I*sqrt(2))^(n-1)*U(n-1, I/sqrt(2)) where U(n, x) is the Chebyshev U-polynomial. - Wolfdieter Lang
G.f.: x/(1 - 2*x - 2*x^2).
From Paul Barry, Sep 17 2003: (Start)
E.g.f.: x*exp(x)*(sinh(sqrt(3)*x)/sqrt(3) + cosh(sqrt(3)*x)).
a(n) = (1 + sqrt(3))^(n-1)*(1/2 + sqrt(3)/6) + (1 - sqrt(3))^(n-1)*(1/2 - sqrt(3)/6), for n>0.
Binomial transform of 1, 1, 3, 3, 9, 9, ... Binomial transform is A079935. (End)
a(n) = Sum_{k=0..floor(n/2)} binomial(n - k, k)*2^(n - k). - Paul Barry, Jul 13 2004
a(n) = A080040(n) - A028860(n+1). - Creighton Dement, Jan 19 2005
a(n) = Sum_{k=0..n} A112899(n,k). - Philippe Deléham, Nov 21 2007
a(n) = Sum_{k=0..n} A063967(n,k). - Philippe Deléham, Nov 03 2006
a(n) = ((1 + sqrt(3))^n - (1 - sqrt(3))^n)/(2*sqrt(3)).
a(n) = Sum_{k=0..n} binomial(n, 2*k + 1) * 3^k.
Binomial transform of expansion of sinh(sqrt(3)x)/sqrt(3) (0, 1, 0, 3, 0, 9, ...). E.g.f.: exp(x)*sinh(sqrt(3)*x)/sqrt(3). - Paul Barry, May 09 2003
a(n) = (1/3)*Sum_{k=1..5} sin(Pi*k/2)*sin(2*Pi*k/3)*(1 + 2*cos(Pi*k/6))^n, n >= 1. - Herbert Kociemba, Jun 02 2004
a(n+1) = ((3 + sqrt(3))*(1 + sqrt(3))^n + (3 - sqrt(3))*(1 - sqrt(3))^n)/6. - Al Hakanson (hawkuu(AT)gmail.com), Jun 29 2009
Antidiagonals sums of A081577. - J. M. Bergot, Dec 15 2012
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, Aug 30 2013
a(n) = 2^(n - 1)*hypergeom([1 - n/2, (1 - n)/2], [1 - n], -2) for n >= 3. - Peter Luschny, Dec 16 2015
Sum_{k=0..n} a(k)*2^(n-k) = a(n+2)/2 - 2^n. - Greg Dresden, Feb 11 2022
a(n) = 2^floor(n/2) * A002530(n). - Gregory L. Simay, Sep 22 2022
From Peter Bala, May 08 2024: (Start)
G.f.: x/(1 - 2*x - 2*x^2) = Sum_{n >= 0} x^(n+1) *( Product_{k = 1..n} (k + 2*x + 1)/(1 + k*x) )
Also x/(1 - 2*x - 2*x^2) = Sum_{n >= 0} (2*x)^n *( x*Product_{k = 1..n} (m*k + 2 - m + x)/(1 + 2*m*k*x) ) for arbitrary m (both series are telescoping). (End)
a(n) = A127864(n-1) + A127864(n-2). - Greg Dresden and Yilin Zhu, Jul 17 2025

Edited by N. J. A. Sloane, Apr 15 2009

A001835 a(n) = 4*a(n-1) - a(n-2), with a(0) = 1, a(1) = 1.

Original entry on oeis.org

1, 1, 3, 11, 41, 153, 571, 2131, 7953, 29681, 110771, 413403, 1542841, 5757961, 21489003, 80198051, 299303201, 1117014753, 4168755811, 15558008491, 58063278153, 216695104121, 808717138331, 3018173449203, 11263976658481, 42037733184721, 156886956080403, 585510091136891

Offset: 0

N. J. A. Sloane

See A079935 for another version.
Number of ways of packing a 3 X 2*(n-1) rectangle with dominoes. - David Singmaster.
Equivalently, number of perfect matchings of the P_3 X P_{2(n-1)} lattice graph. - Emeric Deutsch, Dec 28 2004
The terms of this sequence are the positive square roots of the indices of the octagonal numbers (A046184) - Nicholas S. Horne (nairon(AT)loa.com), Dec 13 1999
Terms are the solutions to: 3*x^2 - 2 is a square. - Benoit Cloitre, Apr 07 2002
Gives solutions x > 0 of the equation floor(x*r*floor(x/r)) == floor(x/r*floor(x*r)) where r = 1 + sqrt(3). - Benoit Cloitre, Feb 19 2004
a(n) = L(n-1,4), where L is defined as in A108299; see also A001834 for L(n,-4). - Reinhard Zumkeller, Jun 01 2005
Values x + y, where (x, y) solves for x^2 - 3*y^2 = 1, i.e., a(n) = A001075(n) + A001353(n). - Lekraj Beedassy, Jul 21 2006
Number of 01-avoiding words of length n on alphabet {0,1,2,3} which do not end in 0. (E.g., for n = 2 we have 02, 03, 11, 12, 13, 21, 22, 23, 31, 32, 33.) - Tanya Khovanova, Jan 10 2007
sqrt(3) = 2/2 + 2/3 + 2/(3*11) + 2/(11*41) + 2/(41*153) + 2/(153*571) + ... - Gary W. Adamson, Dec 18 2007
The lower principal convergents to 3^(1/2), beginning with 1/1, 5/3, 19/11, 71/41, comprise a strictly increasing sequence; numerators = A001834, denominators = A001835. - Clark Kimberling, Aug 27 2008
From Gary W. Adamson, Jun 21 2009: (Start)
A001835 and A001353 = bisection of denominators of continued fraction [1, 2, 1, 2, 1, 2, ...]; i.e., bisection of A002530.
a(n) = determinant of an n*n tridiagonal matrix with 1's in the super- and subdiagonals and (3, 4, 4, 4, ...) as the main diagonal.
Also, the product of the eigenvalues of such matrices: a(n) = Product_{k=1..(n-1)/2)} (4 + 2*cos(2*k*Pi/n).
(End)
Let M = a triangle with the even-indexed Fibonacci numbers (1, 3, 8, 21, ...) in every column, and the leftmost column shifted up one row. a(n) starting (1, 3, 11, ...) = lim_{n->oo} M^n, the left-shifted vector considered as a sequence. - Gary W. Adamson, Jul 27 2010
a(n+1) is the number of compositions of n when there are 3 types of 1 and 2 types of other natural numbers. - Milan Janjic, Aug 13 2010
For n >= 2, a(n) equals the permanent of the (2*n-2) X (2*n-2) tridiagonal matrix with sqrt(2)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
Primes in the sequence are apparently those in A096147. - R. J. Mathar, May 09 2013
Except for the first term, positive values of x (or y) satisfying x^2 - 4xy + y^2 + 2 = 0. - Colin Barker, Feb 04 2014
Except for the first term, positive values of x (or y) satisfying x^2 - 14xy + y^2 + 32 = 0. - Colin Barker, Feb 10 2014
The (1,1) element of A^n where A = (1, 1, 1; 1, 2, 1; 1, 1, 2). - David Neil McGrath, Jul 23 2014
Yong Hao Ng has shown that for any n, a(n) is coprime with any member of A001834 and with any member of A001075. - René Gy, Feb 25 2018
a(n+1) is the number of spanning trees of the graph T_n, where T_n is a 2 X n grid with an additional vertex v adjacent to (1,1) and (2,1). - Kevin Long, May 04 2018
a(n)/A001353(n) is the resistance of an n-ladder graph whose edges are replaced by one-ohm resistors. The resistance in ohms is measured at two nodes at one end of the ladder. It approaches sqrt(3) - 1 for n -> oo. See A342568, A357113, and A357115 for related information. - Hugo Pfoertner, Sep 17 2022
a(n) is the number of ways to tile a 1 X (n-1) strip with three types of tiles: small isosceles right triangles (with small side length 1), 1 X 1 squares formed by joining two of those right triangles along the hypotenuse, and large isosceles right triangles (with large side length 2) formed by joining two of those right triangles along a short leg. As an example, here is one of the a(6)=571 ways to tile a 1 X 5 strip with these kinds of tiles:
   ____________
  | / \ |\   /|  |
  |/_\|\/_||. - Greg Dresden and Arjun Datta, Jun 30 2023
From Klaus Purath, May 11 2024: (Start)
For any two consecutive terms (a(n), a(n+1)) = (x,y): x^2 - 4xy + y^2 = -2 = A028872(-1). In general, the following applies to all sequences (t) satisfying t(i) = 4t(i-1) - t(i-2) with t(0) = 1 and two consecutive terms (x,y): x^2 - 4xy + y^2 = A028872(t(1)-2). This includes and interprets the Feb 04 2014 comments here and on A001075 by Colin Barker and the Dec 12 2012 comment on A001353 by Max Alekseyev. By analogy to this, for three consecutive terms (x,y,z) y^2 - xz = A028872(t(1)-2). This includes and interprets the Jul 10 2021 comment on A001353 by Bernd Mulansky.
If (t) is a sequence satisfying t(k) = 3t(k-1) + 3t(k-2) - t(k-3) or t(k) = 4t(k-1) - t(k-2) without regard to initial values and including this sequence itself, then a(n) = (t(k+2n+1) + t(k))/(t(k+n+1) + t(k+n)) always applies, as long as t(k+n+1) + t(k+n) != 0 for integer k and n >= 1. (End)
Binomial transform of 1, 0, 2, 4, 12, ... (A028860 without the initial -1) and reverse binomial transform of 1, 2, 6, 24, 108, ... (A094433 without the initial 1). - Klaus Purath, Sep 09 2024

Julio R. Bastida, Quadratic properties of a linearly recurrent sequence. Proceedings of the Tenth Southeastern Conference on Combinatorics, Graph Theory and Computing (Florida Atlantic Univ., Boca Raton, Fla., 1979), pp. 163-166, Congress. Numer., XXIII-XXIV, Utilitas Math., Winnipeg, Man., 1979. MR0561042 (81e:10009).
Leonhard Euler, (E388) Vollstaendige Anleitung zur Algebra, Zweiter Theil, reprinted in: Opera Omnia. Teubner, Leipzig, 1911, Series (1), Vol. 1, p. 375.
F. Faase, On the number of specific spanning subgraphs of the graphs G X P_n, Ars Combin. 49 (1998), 129-154.
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 329.
Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
R. P. Stanley, Enumerative Combinatorics I, p. 292.

T. D. Noe, Table of n, a(n) for n = 0..200
Mudit Aggarwal and Samrith Ram, Generating functions for straight polyomino tilings of narrow rectangles, arXiv:2206.04437 [math.CO], 2022.
R. C. Alperin, A family of nonlinear recurrences and their linear solutions, Fib. Q., 57:4 (2019), 318-321.
Krassimir T. Atanassov and Anthony G. Shannon, On intercalated Fibonacci sequences, Notes on Number Theory and Discrete Mathematics (2020) Vol. 26, No. 3, 218-223.
Steve Butler, Paul Horn, and Eric Tressler, Intersecting Domino Tilings, Fibonacci Quart. 48 (2010), no. 2, 114-120.
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 31, 56.
Niccolò Castronuovo, On the number of fixed points of the map gamma, arXiv:2102.02739 [math.NT], 2021. Mentions this sequence.
A. Consilvio et al., Tilings, ordered partitions, and weird languages, MAA FOCUS, June/July 2012, 16-17.
J. B. Cosgrave and K. Dilcher, A role for generalized Fermat numbers, Math. Comp., to appear 2016; (See paper #10).
J. B. Cosgrave and K. Dilcher, A role for generalized Fermat numbers, Math. Comp. 86 (2017), 899-933.
Leonhard Euler, Vollstaendige Anleitung zur Algebra, Zweiter Teil.
F. Faase, On the number of specific spanning subgraphs of the graphs G X P_n, Preliminary version of paper that appeared in Ars Combin. 49 (1998), 129-154.
F. Faase, Counting Hamiltonian cycles in product graphs.
F. Faase, Results from the counting program.
Alex Fink, Richard K. Guy, and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contributions to Discrete Mathematics, Vol 3, No 2 (2008), pp. 76-114. See Section 13.
Darren B. Glass, Critical groups of graphs with dihedral actions. II, Eur. J. Comb. 61, 25-46 (2017).
H. Hosoya and A. Motoyama, An effective algorithm for obtaining polynomials for dimer statistics. Application of operator technique on the topological index to two- and three-dimensional rectangular and torus lattices, J. Math. Physics 26 (1985) 157-167 (Table V).
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 409.
Tanya Khovanova, Recursive Sequences,
Clark Kimberling, Best lower and upper approximates to irrational numbers, Elemente der Mathematik, 52 (1997) 122-126.
David Klarner and Jordan Pollack, Domino tilings of rectangles with fixed width, Disc. Math. 32 (1980) 45-52.
R. J. Mathar, Paving Rectangular Regions with Rectangular Tiles: Tatami and Non-Tatami Tilings, arXiv:1311.6135 [math.CO], 2013, Table 2.
R. J. Mathar, Tilings of rectangular regions by rectangular tiles: Counts derived from transfer matrices, arXiv:1406.7788 (2014), eq. (4).
Valcho Milchev and Tsvetelina Karamfilova, Domino tiling in grid - new dependence, arXiv:1707.09741 [math.HO], 2017.
Yong Hao Ng, A partition in three classes of the set of all prime numbers?, Math StackExchange.
J.-C. Novelli and J.-Y. Thibon, Hopf Algebras of m-permutations,(m+1)-ary trees, and m-parking functions, arXiv preprint arXiv:1403.5962 [math.CO], 2014.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
Jaime Rangel-Mondragon, Polyominoes and Related Families, The Mathematica Journal, 9:3 (2005), 609-640.
John Riordan, Letter to N. J. A. Sloane, Sep 26 1980 with notes on the 1973 Handbook of Integer Sequences. Note that the sequences are identified by their N-numbers, not their A-numbers.
David Singmaster, Letter to N. J. A. Sloane, Oct 3 1982.
Anitha Srinivasan, The Markoff-Fibonacci Numbers, Fibonacci Quart. 58 (2020), no. 5, 222-228.
Thotsaporn "Aek" Thanatipanonda, Statistics of Domino Tilings on a Rectangular Board, Fibonacci Quart. 57 (2019), no. 5, 145-153. See p. 151.
Herman Tulleken, Polyominoes 2.2: How they fit together, (2019).
F. V. Waugh and M. W. Maxfield, Side-and-diagonal numbers, Math. Mag., 40 (1967), 74-83.
Index entries for sequences related to dominoes.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (4,-1).

Row 3 of array A099390.
Essentially the same as A079935.
First differences of A001353.
Partial sums of A052530.
Pairwise sums of A006253.
Bisection of A002530, A005246 and A048788.
First column of array A103997.
Cf. A001519, A003699, A082841, A101265, A125077, A001353, A001542, A096147 (subsequence of primes).

a:=[1,1];; for n in [3..20] do a[n]:=4*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 23 2019

a001835 n = a001835_list !! n
a001835_list =
   1 : 1 : zipWith (-) (map (4 *) $ tail a001835_list) a001835_list
-- Reinhard Zumkeller, Aug 14 2011

[n le 2 select 1 else 4*Self(n-1)-Self(n-2): n in [1..25]]; // Vincenzo Librandi, Sep 16 2016

f:=n->((3+sqrt(3))^(2*n-1)+(3-sqrt(3))^(2*n-1))/6^n; [seq(simplify(expand(f(n))),n=0..20)]; # N. J. A. Sloane, Nov 10 2009

CoefficientList[Series[(1-3x)/(1-4x+x^2), {x, 0, 24}], x] (* Jean-François Alcover, Jul 25 2011, after g.f. *)
LinearRecurrence[{4,-1},{1,1},30] (* Harvey P. Dale, Jun 08 2013 *)
Table[Round@Fibonacci[2n-1, Sqrt[2]], {n, 0, 20}] (* Vladimir Reshetnikov, Sep 15 2016 *)
Table[(3*ChebyshevT[n, 2] - ChebyshevU[n, 2])/2, {n, 0, 20}] (* G. C. Greubel, Dec 23 2019 *)

{a(n) = real( (2 + quadgen(12))^n * (1 - 1 / quadgen(12)) )} /* Michael Somos, Sep 19 2008 */

{a(n) = subst( (polchebyshev(n) + polchebyshev(n-1)) / 3, x, 2)} /* Michael Somos, Sep 19 2008 */

[lucas_number1(n,4,1)-lucas_number1(n-1,4,1) for n in range(25)] # Zerinvary Lajos, Apr 29 2009

[(3*chebyshev_T(n,2) - chebyshev_U(n,2))/2 for n in (0..20)] # G. C. Greubel, Dec 23 2019

G.f.: (1 - 3*x)/(1 - 4*x + x^2). - Simon Plouffe in his 1992 dissertation
a(1-n) = a(n).
a(n) = ((3 + sqrt(3))^(2*n - 1) + (3 - sqrt(3))^(2*n - 1))/6^n. - Dean Hickerson, Dec 01 2002
a(n) = (8 + a(n-1)*a(n-2))/a(n-3). - Michael Somos, Aug 01 2001
a(n+1) = Sum_{k=0..n} 2^k * binomial(n + k, n - k), n >= 0. - Len Smiley, Dec 09 2001
Limit_{n->oo} a(n)/a(n-1) = 2 + sqrt(3). - Gregory V. Richardson, Oct 10 2002
a(n) = 2*A061278(n-1) + 1 for n > 0. - Bruce Corrigan (scentman(AT)myfamily.com), Nov 04 2002
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n - i, i); then q(n, 2) = a(n+1). - Benoit Cloitre, Nov 10 2002
a(n+1) = Sum_{k=0..n} ((-1)^k)*((2*n+1)/(2*n + 1 - k))*binomial(2*n + 1 - k, k)*6^(n - k) (from standard T(n,x)/x, n >= 1, Chebyshev sum formula). The Smiley and Cloitre sum representation is that of the S(2*n, i*sqrt(2))*(-1)^n Chebyshev polynomial. - Wolfdieter Lang, Nov 29 2002
a(n) = S(n-1, 4) - S(n-2, 4) = T(2*n-1, sqrt(3/2))/sqrt(3/2) = S(2*(n-1), i*sqrt(2))*(-1)^(n - 1), with S(n, x) := U(n, x/2), resp. T(n, x), Chebyshev's polynomials of the second, resp. first, kind. See A049310 and A053120. S(-1, x) = 0, S(-2, x) = -1, S(n, 4) = A001353(n+1), T(-1, x) = x.
a(n+1) = sqrt((A001834(n)^2 + 2)/3), n >= 0 (see Cloitre comment).
Sequence satisfies -2 = f(a(n), a(n+1)) where f(u, v) = u^2 + v^2 - 4*u*v. - Michael Somos, Sep 19 2008
a(n) = (1/6)*(3*(2 - sqrt(3))^n + sqrt(3)*(2 - sqrt(3))^n + 3*(2 + sqrt(3))^n - sqrt(3)*(2 + sqrt(3))^n) (Mathematica's solution to the recurrence relation). - Sarah-Marie Belcastro, Jul 04 2009
If p[1] = 3, p[i] = 2, (i > 1), and if A is Hessenberg matrix of order n defined by: A[i,j] = p[j-i+1], (i <= j), A[i,j] = -1, (i = j+1), and A[i,j] = 0 otherwise. Then, for n >= 1, a(n+1) = det A. - Milan Janjic, Apr 29 2010
a(n) = (a(n-1)^2 + 2)/a(n-2). - Irene Sermon, Oct 28 2013
a(n) = A001353(n+1) - 3*A001353(n). - R. J. Mathar, Oct 30 2015
a(n) = a(n-1) + 2*A001353(n-1). - Kevin Long, May 04 2018
From Franck Maminirina Ramaharo, Nov 11 2018: (Start)
a(n) = (-1)^n*(A125905(n) + 3*A125905(n-1)), n > 0.
E.g.f.: exp^(2*x)*(3*cosh(sqrt(3)*x) - sqrt(3)*sinh(sqrt(3)*x))/3. (End)
From Peter Bala, Feb 12 2024: (Start)
For n in Z, a(n) = A001353(n) + A001353(1-n).
For n, j, k in Z, a(n)*a(n+j+k) - a(n+j)*a(n+k) = 2*A001353(j)*A001353(k). The case j = 1, k = 2 is given above. (End)

A061278 a(n) = 5a(n-1) - 5a(n-2) + a(n-3) with a(1) = 1 and a(k) = 0 if k <= 0.

Original entry on oeis.org

0, 1, 5, 20, 76, 285, 1065, 3976, 14840, 55385, 206701, 771420, 2878980, 10744501, 40099025, 149651600, 558507376, 2084377905, 7779004245, 29031639076, 108347552060, 404358569165, 1509086724601, 5631988329240, 21018866592360, 78443478040201, 292755045568445

Offset: 0

Henry Bottomley, Jun 04 2001

Indices m of triangular numbers T(m) which are one-third of another triangular number: 3*T(m) = T(k); the k's are given by A001571. - Bruce Corrigan (scentman(AT)myfamily.com), Oct 31 2002
On the previous comment: for m=0 this is actually one third of the same triangular number. - Zak Seidov, Apr 07 2011
Also numbers n such that the n-th centered 24-gonal number 12*n*(n+1)+1 is a perfect square A001834(n)^2, where A001834(n) is defined by the recursion: a(0) = 1, a(1) = 5, a(n) = 4*a(n-1) - a(n-2) + 1. - Alexander Adamchuk, Apr 21 2007
Also numbers n such that RootMeanSquare(5,...,6*n-1) is an integer. - Ctibor O. Zizka, Dec 17 2008 (Corrected by Robert K. Moniot, Jul 22 2020)
Also numbers n such that n*(n+1) = Sum_{i=1..x} n+i for some x. (This does not apply to the first term.). - Gil Broussard, Dec 23 2008
From John P. McSorley, May 26 2020: (Start)
Consecutive terms (a(n-1), a(n)) = (u,v) give all points on the hyperbola u^2 - u + v^2 - v - 4*u*v = 0 in quadrant I with both coordinates an integer.
Also related to the block sizes of small multi-set designs. (End)
If a(n) white balls and a(n+1) black balls are mixed in a bag, and a pair of balls is drawn without replacement, the probability that one ball of each color is drawn is exactly 1/3. These are the only integers for which the probability is 1/3. For example, if there are 20 white balls and 76 black balls, the probability of drawing one of each is (20/96)*(76/95) + (76/96)*(20/95) = 1/3. - Elliott Line, May 13 2022

			a(2)=5 and T(5)=15 which is 1/3 of 45=T(9).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
R. C. Alperin, A nonlinear recurrence and its relations to Chebyshev polynomials, Fib. Q., Vol. 58, No. 2 (2020), 140-142.
Niccolò Castronuovo, On the number of fixed points of the map gamma, arXiv:2102.02739 [math.NT], 2021. Mentions this sequence.
Brian Lawrence and Will Sawin, The Shafarevich conjecture for hypersurfaces in abelian varieties, arXiv:2004.09046 [math.NT], 2020.
Ioana-Claudia Lazăr, Lucas sequences in t-uniform simplicial complexes, arXiv:1904.06555 [math.GR], 2019.
L. A. Medina and A. Straub, On multiple and infinite log-concavity, 2013.
S. Morier-Genoud, V. Ovsienko and S. Tabachnikov, 2-frieze patterns and the cluster structure of the space of polygons, Annales de l'institut Fourier, 62 no. 3 (2012), 937-987. - From _N. J. A. Sloane_, Dec 26 2012
Robert Phillips, A triangular number result, 2009.
Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv:2101.00998 [math.NT], 2021.
Vladimir Pletser, Closed Form Equations for Triangular Numbers Multiple of Other Triangular Numbers, arXiv:2102.12392 [math.GM], 2021.
Vladimir Pletser, Triangular Numbers Multiple of Triangular Numbers and Solutions of Pell Equations, arXiv:2102.13494 [math.NT], 2021.
Vladimir Pletser, Congruence Properties of Indices of Triangular Numbers Multiple of Other Triangular Numbers, arXiv:2103.03019 [math.GM], 2021.
Vladimir Pletser, Searching for multiple of triangular numbers being triangular numbers, 2021.
Vladimir Pletser, Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers, 2022.
Eric Weisstein, Centered Polygonal Number.
Index entries for linear recurrences with constant coefficients, signature (5,-5,1).

Cf. A001075, A001353, A001571, A001834, A001835, A079935, A101265. Also cf. A212336 for more sequences with g.f. of the type 1/(1-k*x+k*x^2-x^3).

I:=[0, 1]; [n le 2 select I[n] else 4*Self(n-1) - Self(n-2) + 1: n in [1..30]]; // Vincenzo Librandi, Dec 23 2012

f:= gfun:-rectoproc({a(n) = 5*a(n-1) - 5*a(n-2) + a(n-3),a(1)=1,a(0)=0,a(-1)=0},a(n),remember):
map(f, [$0..50]); # Robert Israel, Jun 05 2015

CoefficientList[Series[x/(1 - 5*x + 5*x^2 - x^3), {x, 0, nn}], x] (* T. D. Noe, Jun 04 2012 *)
LinearRecurrence[{5,-5,1},{0,1,5},30] (* Harvey P. Dale, Dec 23 2012 *)

M = [1, 1, 0; 1, 3, 1; 0, 1, 1]; for(i=1, 30, print1(([1, 0, 0]*M^i)[3], ",")) \\ Lambert Klasen (Lambert.Klasen(AT)gmx.net), Jan 25 2005

a(n) = 4*a(n-1) - a(n-2) + 1.
a(n) = A001075(n) - a(n-1) - 1.
a(n) = (A001835(n+1) - 1)/2 = (A001353(n+1) - A001353(n) - 1)/2.
a(n) = a(n-1) + A001353(n), i.e., partial sum of A001353.
From Bruce Corrigan (scentman(AT)myfamily.com), Oct 31 2002: (Start)
a(n+2) = 4*a(n+1) - a(n) + 1 for a(0)=0, a(1)=1.
G.f.: x/((1 - x)*(1 - 4*x + x^2)).
a(n) = (1/12)*((3 - sqrt(3))*(2 - sqrt(3))^n + (3 + sqrt(3))*(2 + sqrt(3))^n-6). (End)
a(n) = (1/12)*(A003500(n) + A003500(n+1)-6). - Mario Catalani (mario.catalani(AT)unito.it), Apr 11 2003
a(n+1) = Sum_{k=0..n} U(k, 2) = Sum_{k=0..n} S(k, 4), where U(n,x) and S(n,x) are Chebyshev polynomials. - Paul Barry, Nov 14 2003
G.f.: x/(1 - 5*x + 5*x^2 - x^3).
a(n) = A079935(n+1) + A001571(n) for n>0, a(0)=0. - Gerry Martens, Jun 05 2015
a(n)*a(n-2) = a(n-1)*(a(n-1) - 1) for n>1. - Bruno Berselli, Nov 29 2016
From John P. McSorley, May 25 2020: (Start)
a(n)^2 - a(n) + a(n-1)^2 - a(n-1) - 4*a(n)*a(n-1) = 0.
a(n) = A001834(n-1) + a(n-2). (End)
(T(a(n)-1) + T(a(n+1)-1))/T(a(n) + a(n+1) - 1) = 2/3 where T(i) is the i-th triangular number. - Robert K. Moniot, Oct 11 2020
E.g.f.: exp(x)*(exp(x)*(3*cosh(sqrt(3)*x) + sqrt(3)*sinh(sqrt(3)*x)) - 3)/6. - Stefano Spezia, Feb 05 2021
a(n) = A101265(n) - 1. - Jon E. Schoenfield, Jan 01 2022

More terms from Lambert Klasen (Lambert.Klasen(AT)gmx.net), Jan 25 2005

A094954 Array T(k,n) read by antidiagonals. G.f.: x(1-x)/(1-kx+x^2), k>1.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 5, 1, 1, 4, 11, 13, 1, 1, 5, 19, 41, 34, 1, 1, 6, 29, 91, 153, 89, 1, 1, 7, 41, 169, 436, 571, 233, 1, 1, 8, 55, 281, 985, 2089, 2131, 610, 1, 1, 9, 71, 433, 1926, 5741, 10009, 7953, 1597, 1, 1, 10, 89, 631, 3409, 13201, 33461, 47956, 29681

Offset: 1

Ralf Stephan, May 31 2004

Also, values of polynomials with coefficients in A098493 (see Fink et al.). See A098495 for negative k.

Number of dimer tilings of the graph S_{k-1} X P_{2n-2}.

			1,1,1,1,1,1,1,1,1,1,1,1,1,1, ...
1,2,5,13,34,89,233,610,1597, ...
1,3,11,41,153,571,2131,7953, ...
1,4,19,91,436,2089,10009,47956, ...
1,5,29,169,985,5741,33461,195025, ...
1,6,41,281,1926,13201,90481,620166, ...

Alex Fink, Richard K. Guy, and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contributions to Discrete Mathematics, Vol 3, No 2 (2008), pp. 76-114.
Elizabeth Wilmer, A note on Stephan's conjecture 87
Elizabeth Wilmer, A note on Stephan's conjecture 87 [cached copy]

Rows are first differences of rows in array A073134.
Rows 2-14 are A000012, A001519, A079935/A001835, A004253, A001653, A049685, A070997, A070998, A072256, A078922, A077417, A085260, A001570. Other rows: A007805 (k=18), A075839 (k=20), A077420 (k=34), A078988 (k=66).
Columns include A028387. Diagonals include A094955, A094956. Antidiagonal sums are A094957.

max = 14; row[k_] := Rest[ CoefficientList[ Series[ x*(1-x)/(1-k*x+x^2), {x, 0, max}], x]]; t = Table[ row[k], {k, 2, max+1}]; Flatten[ Table[ t[[k-n+1, n]], {k, 1, max}, {n, 1, k}]] (* Jean-François Alcover, Dec 27 2011 *)

PARI
```
T(k,n)=polcoeff(x*(1-x)/(1-k*x+x*x),n)
```

Recurrence: T(k, 1) = 1, T(k, 2) = k-1, T(k, n) = kT(k, n-1) - T(k, n-2).
For n>3, T(k, n) = [k(k-2) + T(k, n-1)T(k, n-2)] / T(k, n-3).
T(k, n+1) = S(n, k) - S(n-1, k) = U(n, k/2) - U(n-1, k/2), with S, U = Chebyshev polynomials of second kind.
T(k+2, n+1) = Sum[i=0..n, k^(n-i) * C(2n-i, i)] (from comments by Benoit Cloitre).

A238379 Expansion of (1 - x)/(1 - 36*x + x^2).

Original entry on oeis.org

1, 35, 1259, 45289, 1629145, 58603931, 2108112371, 75833441425, 2727895778929, 98128414600019, 3529895029821755, 126978092658983161, 4567681440693572041, 164309553772309610315, 5910576254362452399299, 212616435603275976764449

Offset: 0

Bruno Berselli, Feb 25 2014

First bisection of A041611.

Bruno Berselli, Table of n, a(n) for n = 0..100
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (36,-1).

Cf. similar sequences with g.f. (1-x)/(1-k*x+x^2): A122367 (k=3), A079935 (k=4), A004253 (k=5), A001653 (k=6), A049685 (k=7), A070997 (k=8), A070998 (k=9), A138288 (k=10), A078922 (k=11), A077417 (k=12), A085260 (k=13), A001570 (k=14), A160682 (k=15), A157456 (k=16), A161595 (k=17). From 18 to 38, even k only, except k=27 and k=31: A007805 (k=18), A075839 (k=20), A157014 (k=22), A159664 (k=24), A153111 (k=26), A097835 (k=27), A159668 (k=28), A157877 (k=30), A111216 (k=31), A159674 (k=32), A077420 (k=34), this sequence (k=36), A097315 (k=38).

[n le 2 select 35^(n-1) else 36*Self(n-1)-Self(n-2): n in [1..20]];

R:=PowerSeriesRing(Integers(), 20); Coefficients(R!( (1 - x)/(1 - 36*x + x^2))); // Marius A. Burtea, Jan 14 2020

CoefficientList[Series[(1 - x)/(1 - 36 x + x^2), {x, 0, 20}], x] (* or *) LinearRecurrence[{36, -1}, {1, 35}, 20]

a(n)=([0,1; -1,36]^n*[1;35])[1,1] \\ Charles R Greathouse IV, May 10 2016

m = 20; L. = PowerSeriesRing(ZZ, m); f = (1-x)/(1-36*x+x^2)
print(f.coefficients())

G.f.: (1 - x)/(1 - 36*x + x^2).
a(n) = a(-n-1) = 36*a(n-1) - a(n-2).
a(n) = ((19-sqrt(323))/38)*(1+(18+sqrt(323))^(2*n+1))/(18+sqrt(323))^n.
a(n+1) - a(n) = 34*A144128(n+1).
323*a(n+1)^2 - ((a(n+2)-a(n))/2)^2 = 34.
Sum_{n>0} 1/(a(n) - 1/a(n)) = 1/34.
See also Tanya Khovanova in Links field:
a(n) = 35*a(n-1) + 34*Sum_{i=0..n-2} a(i).
a(n+2)*a(n) - a(n+1)^2 = 36-2 = 34 = 34*1,
a(n+3)*a(n) - a(n+1)*a(n+2) = 36*(36-2) = 1224 = 34*36.
Generalizing:
a(n+4)*a(n) - a(n+1)*a(n+3) = 44030 = 34*1295,
a(n+5)*a(n) - a(n+1)*a(n+4) = 1583856 = 34*46584,
a(n+6)*a(n) - a(n+1)*a(n+5) = 56974786 = 34*1675729, etc.,
where 1, 36, 1295, 46584, 1675729, ... is the sequence A144128, which is the second bisection of A041611.
a(n)^2 - 36*a(n)*a(n+1) + a(n+1)^2 + 34 = 0 (see comments by Colin Barker in similar sequences).

A014300 Number of nodes of odd outdegree in all ordered rooted (planar) trees with n edges.

Original entry on oeis.org

1, 2, 7, 24, 86, 314, 1163, 4352, 16414, 62292, 237590, 909960, 3497248, 13480826, 52097267, 201780224, 783051638, 3044061116, 11851853042, 46208337584, 180383564228, 704961896036, 2757926215742, 10799653176704, 42326626862636, 166021623024584, 651683311373788

Offset: 1

Emeric Deutsch

Also total number of blocks of odd size in all Catalan(n) possible noncrossing partitions of [n].
Convolution of the sequence of central binomial coefficients 1,2,6,20,70,... (A000984) and of the sequence of Fine numbers 1,0,1,2,6,18,... (A000957).
Row sums of A119307. - Paul Barry, May 13 2006
Hankel transform is A079935. - Paul Barry, Jul 17 2009
Also for n>=1 the number of unimodal functions f:[n]->[n] with f(i)<>f(i+1). a(3) = 7: [1,2,1], [1,2,3], [1,3,1], [1,3,2], [2,3,1], [2,3,2], [3,2,1]. - Alois P. Heinz, May 23 2013
Also, number of sets of n rational numbers on [0,1) such that if x belongs to the set, the fractional part of 2x also belongs to it. - Jianing Song and Andrew Howroyd, May 18 2018
Let A(i, j) denote the infinite array such that the i-th row of this array is the sequence obtained by applying the partial sum operator i times to the function ((-1)^(n + 1) + 1)/2 for n > 0. Then A(n, n) equals a(n) for all n > 0. - John M. Campbell, Jan 20 2019
The Gauss congruences a(n*p^k) == a(n^p^(k-1)) (mod p^k) hold for prime p >= 3 and positive integers n and k. - Peter Bala, Jan 07 2022

Alois P. Heinz, Table of n, a(n) for n = 1..500
Paul Barry, A Catalan Transform and Related Transformations on Integer Sequences, Journal of Integer Sequences, Vol. 8 (2005), Article 05.4.5.
Hacène Belbachir and Abdelghani Mehdaoui, Diagonal sums in Pascal pyramid (1, 2, r), Les Annales RECITS (2019) Vol. 6, 45-52.
N. Dershowitz and S. Zaks, Ordered trees and non-crossing partitions, Discrete Math., 62 (1986), 215-218.
Emeric Deutsch and L. Shapiro, A survey of the Fine numbers, Discrete Math., 241 (2001), 241-265.
Index entries for sequences related to rooted trees

Cf. A059481, A000957, A000984, A072547, A119259, A119307, A079935, A333564.

[(&+[(-1)^(n-k)*Binomial(n+k-1, k-1): k in [0..n]]): n in [1..30]]; // G. C. Greubel, Feb 19 2019

a:= proc(n) a(n):= `if`(n<3, n, ((12-40*n+21*n^2) *a(n-1)+
       2*(3*n-1)*(2*n-3) *a(n-2))/ (2*(3*n-4)*n))
    end:
seq(a(n), n=1..30);  # Alois P. Heinz, Oct 30 2012

Rest[CoefficientList[Series[2x/(1-4x+(1+2x)Sqrt[1-4x]),{x,0,40}],x]]  (* Harvey P. Dale, Apr 25 2011 *)
a[n_] := Sum[Binomial[2k, n-1], {k, 0, n-1}]; Array[a, 30] (* Jean-François Alcover, Dec 25 2015, after Paul Barry *)

a(n) = n--; sum(k=0, n, binomial(2*k,n)); \\ Michel Marcus, May 18 2018

from itertools import count, islice
def A014300_gen(): # generator of terms
    yield from (1,2)
    a, c = 1, 1
    for n in count(1):
        yield (a:=(3*n+5)*(c:=c*((n<<2)+2)//(n+2))-a>>1)
A014300_list = list(islice(A014300_gen(),20)) # Chai Wah Wu, Apr 26 2023

[sum((-1)^(n-k)*binomial(n+k-1, k-1) for k in (0..n)) for n in (1..30)] # G. C. Greubel, Feb 19 2019

a(n) = (binomial(2*n, n) + A000957(n))/3; [simplified by Alexander Burstein, Nov 24 2023]
a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(n+k-1, k-1). - Vladeta Jovovic, Aug 28 2002
G.f.: 2*z/(1-4*z+(1+2*z)*sqrt(1-4*z)).
a(n) = Sum_{j=0..floor((n-1)/2)} binomial(2*n-2*j-2, n-1).
2*a(n) + a(n-1) = (3*n-1)*Catalan(n-1). - Vladeta Jovovic, Dec 03 2004
a(n) = (-1)^n*Sum_{i=0..n} Sum_{j=n..2*n} (-1)^(i+j)*binomial(j, i). - Benoit Cloitre, Jun 18 2005
a(n) = Sum_{k=0..n} C(2*k,n) [offset 0]. - Paul Barry, May 13 2006
a(n) = Sum_{k=0..n} (-1)^(n-k)*C(n+k-1,k-1). - Paul Barry, Jul 18 2006
From Paul Barry, Jul 17 2009: (Start)
a(n) = Sum_{k=0..n} C(2*n-k,n-k)*(1+(-1)^k)/2.
a(n) = Sum_{k=0..n} C(n+k,k)*(1+(-1)^(n-k))/2. (End)
a(n) is the coefficient of x^(n+1)*y^(n+1) in 1/(1- x^2*y/((1-2*x)*(1-y))). - Ira M. Gessel, Oct 30 2012
a(n) = -binomial(2*n,n-1)*hyper2F1([1,2*n+1],[n+2], 2). - Peter Luschny, Jul 25 2014
a(n) = [x^n] x/((1 - x^2)*(1 - x)^n). - Ilya Gutkovskiy, Oct 25 2017
a(n) ~ 4^n / (3*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 25 2017
D-finite with recurrence: 2*n*a(n) +(-3*n-4)*a(n-1) +2*(-9*n+19)*a(n-2) +4*(-2*n+5)*a(n-3)=0. - R. J. Mathar, Feb 20 2020
a(n) = A333564(n)/2^n. - Peter Bala, Apr 09 2020
a(n) = (1/2)*(binomial(2*n,n) - A072547(n)). - Peter Bala, Mar 28 2023

A299045 Rectangular array: A(n,k) = Sum_{j=0..k} (-1)^floor(j/2)binomial(k-floor((j+1)/2), floor(j/2))(-n)^(k-j), n >= 1, k >= 0, read by antidiagonals.

Original entry on oeis.org

1, 1, 0, 1, -1, -1, 1, -2, 1, 1, 1, -3, 5, -1, 0, 1, -4, 11, -13, 1, -1, 1, -5, 19, -41, 34, -1, 1, 1, -6, 29, -91, 153, -89, 1, 0, 1, -7, 41, -169, 436, -571, 233, -1, -1, 1, -8, 55, -281, 985, -2089, 2131, -610, 1, 1, 1, -9, 71, -433, 1926, -5741, 10009, -7953, 1597, -1, 0

Offset: 1

L. Edson Jeffery, Bob Selcoe and Andrew Hone, Feb 01 2018

This array is used to compute A269252: A269252(n) = least k such that |A(n,k)| is a prime, or -1 if no such k exists.
For detailed theory, see [Hone].
The array can be extended to k<0 with A(n, k) = -A(n, -k-1) for all k in Z. - Michael Somos, Jun 19 2023

			Array begins:
1   0  -1     1     0      -1       1         0        -1           1
1  -1   1    -1     1      -1       1        -1         1          -1
1  -2   5   -13    34     -89     233      -610      1597       -4181
1  -3  11   -41   153    -571    2131     -7953     29681     -110771
1  -4  19   -91   436   -2089   10009    -47956    229771    -1100899
1  -5  29  -169   985   -5741   33461   -195025   1136689    -6625109
1  -6  41  -281  1926  -13201   90481   -620166   4250681   -29134601
1  -7  55  -433  3409  -26839  211303  -1663585  13097377  -103115431
1  -8  71  -631  5608  -49841  442961  -3936808  34988311  -310957991
1  -9  89  -881  8721  -86329  854569  -8459361  83739041  -828931049

Robert Price, Table of n, a(n) for n = 1..5050
Andrew N. W. Hone, et al., On a family of sequences related to Chebyshev polynomials, arXiv:1802.01793 [math.NT], 2018.

Cf. A285992, A299107, A299109, A088165, A117522, A299100, A299101, A113501, A269251, A269252, A269253, A269254, A294099, A298675, A298677, A298878, A299045, A299071.
Cf. A094954 (unsigned version of this array, but missing the first row).
Cf. Rows: A057078, A033999, A099496, A079935 (or A001835), A004253, A001653, A049685, A070997, A070998, A138288 (or A072256), ...
Cf. Columns: A000012, A001477 (A000027), A110331 (A165900), A123972, A192398, ...

(* Array: *)
Grid[Table[LinearRecurrence[{-n, -1}, {1, 1 - n}, 10], {n, 10}]]
(*Array antidiagonals flattened (gives this sequence):*)
A299045[n_, k_] := Sum[(-1)^(Floor[j/2]) Binomial[k - Floor[(j + 1)/2], Floor[j/2]] (-n)^(k - j), {j, 0, k}]; Flatten[Table[A299045[n - k, k], {n, 11}, {k, 0, n - 1}]]

{A(n, k) = sum(j=0, k, (-1)^(j\2)*binomial(k-(j+1)\2, j\2)*(-n)^(k-j))}; /* Michael Somos, Jun 19 2023 */

G.f. for row n: (1 + x)/(1 + n*x + x^2), n >= 1.

A(n, k) = B(-n, k) where B = A294099. - Michael Somos, Jun 19 2023

A101879 a(0) = 1, a(1) = 1, a(2) = 2; for n > 2, a(n) = 5a(n-1) - 5a(n-2) + a(n-3).

Original entry on oeis.org

1, 1, 2, 6, 21, 77, 286, 1066, 3977, 14841, 55386, 206702, 771421, 2878981, 10744502, 40099026, 149651601, 558507377, 2084377906, 7779004246, 29031639077, 108347552061, 404358569166, 1509086724602, 5631988329241, 21018866592361, 78443478040202, 292755045568446

Offset: 0

Lambert Klasen (lambert.klasen(AT)gmx.net) and Gary W. Adamson, Jan 28 2005

Consider the matrix M=[1,1,0; 1,3,1; 0,1,1]; characteristic polynomial of M is x^3 - 5*x^2 + 5*x - 1. Use (M^n)[1,1] to define the recursion a(0) = 1, a(1) = 1, a(2) = 2, for n>2 a(n) = 5*a(n-1) - 5*a(n-2) + a(n-3).
a(n+1)/a(n) converges to 2 + sqrt(3) as n goes to infinity, the largest root of the characteristic polynomial. a(n) = A061278(n) + 1; (M^n)[1,2] = A001353(n); (M^n)[1,3] = A061278(n-1) for n>0; all with the same recursive properties.
Consecutive terms of this sequence and consecutive terms of A032908 provide all positive integer pairs for which K=(a+1)/b+(b+1)/a is an integer. For this sequence K=4. - Andrey Vyshnevyy, Sep 18 2015
The two-page Reid Barton article was sent to me around 2002, but for some reason it was not included in the OEIS at that time. I recently rediscovered it in my files. - N. J. A. Sloane, Sep 08 2018

Seiichi Manyama, Table of n, a(n) for n = 0..1750
Reid Barton, A combinatorial interpretation of the sequence 1, 1, 2, 6, 21, 77, ...
Reid Barton, A combinatorial interpretation of the sequence 1, 1, 2, 6, 21, 77, ..., [Annotated scanned copy]
Index entries for linear recurrences with constant coefficients, signature (5,-5,1).

Cf. A061278, A001353, A001835, A005246, A276122, A276271.

I:=[1,1,2]; [n le 3 select I[n] else 5*Self(n-1)-5*Self(n-2)+Self(n-3): n in [1..30]]; // Vincenzo Librandi, Sep 18 2015

LinearRecurrence[{5, -5, 1}, {1, 1, 2}, 30] (* Vincenzo Librandi, Sep 18 2015 *)
CoefficientList[Series[(1 - 4 x + 2 x^2)/((1 - x) (1 - 4 x + x^2)), {x, 0, 27}], x] (* Michael De Vlieger, Aug 11 2016 *)
a[ n_] := If[ n < 1, a[1 - n], SeriesCoefficient[ (1/(1 - x) + (1 - 3 x)/(1 - 4 x + x^2)) / 2, {x, 0, n}]]; (* Michael Somos, Jul 09 2017 *)

M=[1,1,0; 1,3,1; 0,1,1]; for(i=0,40,print1((M^i)[1,1],","))

{a(n) = if( n<1, a(1-n), polcoeff( (1/(1 - x) + (1 - 3*x)/(1 - 4*x + x^2)) / 2 + x * O(x^n), n))}; /* Michael Somos, Jul 09 2017 */

a(n) = A101265(n), n>0. - R. J. Mathar, Aug 30 2008
a(n) = A079935(n+1) - A001571(n). - Gerry Martens, Jun 05 2015
a(0) = a(1) = 1, for n>1 a(n) = (a(n-1) + a(n-1)^2) / a(n-2). - Seiichi Manyama, Aug 11 2016
From Ilya Gutkovskiy, Aug 11 2016: (Start)
G.f.: (1 - 4*x + 2*x^2)/((1 - x)*(1 - 4*x + x^2)).
a(n) = (6+(3-sqrt(3))*(2+sqrt(3))^n + (2-sqrt(3))^n*(3+sqrt(3)))/12. (End)
a(n) = 4*a(n-1) - a(n-2) - 1. - Seiichi Manyama, Aug 26 2016
From Seiichi Manyama, Sep 03 2016: (Start)
a(n) = (a(n-1) + 1)*(a(n-2) + 1) / a(n-3).
a(n) = A005246(n)*A005246(n+1). (End)
From Michael Somos, Jul 09 2017: (Start)
0 = +a(n)*(+1 +a(n) -4*a(n+1)) +a(n+1)*(+1 +a(n+1)) for all n in Z.
a(n) = a(1 - n) = (1 + A001835(n)) / 2 for all n in Z. (End)

a(26)-a(27) from Vincenzo Librandi, Sep 18 2015

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)

cp's OEIS Frontend

A001353 a(n) = 4*a(n-1) - a(n-2) with a(0) = 0, a(1) = 1.

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A002605 a(n) = 2*(a(n-1) + a(n-2)), a(0) = 0, a(1) = 1.

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A001835 a(n) = 4*a(n-1) - a(n-2), with a(0) = 1, a(1) = 1.

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A061278 a(n) = 5*a(n-1) - 5*a(n-2) + a(n-3) with a(1) = 1 and a(k) = 0 if k <= 0.

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A094954 Array T(k,n) read by antidiagonals. G.f.: x(1-x)/(1-kx+x^2), k>1.

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A061278 a(n) = 5a(n-1) - 5a(n-2) + a(n-3) with a(1) = 1 and a(k) = 0 if k <= 0.

A299045 Rectangular array: A(n,k) = Sum_{j=0..k} (-1)^floor(j/2)binomial(k-floor((j+1)/2), floor(j/2))(-n)^(k-j), n >= 1, k >= 0, read by antidiagonals.

A101879 a(0) = 1, a(1) = 1, a(2) = 2; for n > 2, a(n) = 5a(n-1) - 5a(n-2) + a(n-3).