A001075 -id:A001075 - 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)

A001541 a(0) = 1, a(1) = 3; for n > 1, a(n) = 6*a(n-1) - a(n-2).

Original entry on oeis.org

1, 3, 17, 99, 577, 3363, 19601, 114243, 665857, 3880899, 22619537, 131836323, 768398401, 4478554083, 26102926097, 152139002499, 886731088897, 5168247530883, 30122754096401, 175568277047523, 1023286908188737, 5964153172084899, 34761632124320657

Offset: 0

N. J. A. Sloane

Chebyshev polynomials of the first kind evaluated at 3.
This sequence gives the values of x in solutions of the Diophantine equation x^2 - 8*y^2 = 1, the corresponding values of y are in A001109. For n > 0, the ratios a(n)/A001090(n) may be obtained as convergents to sqrt(8): either successive convergents of [3; -6] or odd convergents of [2; 1, 4]. - Lekraj Beedassy, Sep 09 2003 [edited by Jon E. Schoenfield, May 04 2014]
Also gives solutions to the equation x^2 - 1 = floor(x*r*floor(x/r)) where r = sqrt(8). - Benoit Cloitre, Feb 14 2004
Appears to give all solutions greater than 1 to the equation: x^2 = ceiling(x*r*floor(x/r)) where r = sqrt(2). - Benoit Cloitre, Feb 24 2004
This sequence give numbers n such that (n-1)*(n+1)/2 is a perfect square. Remark: (i-1)*(i+1)/2 = (i^2-1)/2 = -1 = i^2 with i = sqrt(-1) so i is also in the sequence. - Pierre CAMI, Apr 20 2005
a(n) is prime for n = {1, 2, 4, 8}. Prime a(n) are {3, 17, 577, 665857}, which belong to A001601(n). a(2k-1) is divisible by a(1) = 3. a(4k-2) is divisible by a(2) = 17. a(8k-4) is divisible by a(4) = 577. a(16k-8) is divisible by a(8) = 665857. - Alexander Adamchuk, Nov 24 2006
The upper principal convergents to 2^(1/2), beginning with 3/2, 17/12, 99/70, 577/408, comprise a strictly decreasing sequence; essentially, numerators=A001541 and denominators=A001542. - Clark Kimberling, Aug 26 2008
Also index of sequence A082532 for which A082532(n) = 1. - Carmine Suriano, Sep 07 2010
Numbers n such that sigma(n-1) and sigma(n+1) are both odd numbers. - Juri-Stepan Gerasimov, Mar 28 2011
Also, numbers such that floor(a(n)^2/2) is a square: base 2 analog of A031149, A204502, A204514, A204516, A204518, A204520, A004275, A001075. - M. F. Hasler, Jan 15 2012
Numbers such that 2n^2 - 2 is a square. Also integer square roots of the expression 2*n^2 + 1, at values of n given by A001542. Also see A228405 regarding 2n^2 -+ 2^k generally for k >= 0. - Richard R. Forberg, Aug 20 2013
Values of x (or y) in the solutions to x^2 - 6xy + y^2 + 8 = 0. - Colin Barker, Feb 04 2014
Panda and Ray call the numbers in this sequence the Lucas-balancing numbers C_n (see references and links).
Partial sums of X or X+1 of Pythagorean triples (X,X+1,Z). - Peter M. Chema, Feb 03 2017
a(n)/A001542(n) is the closest rational approximation to sqrt(2) with a numerator not larger than a(n), and 2*A001542(n)/a(n) is the closest rational approximation to sqrt(2) with a denominator not larger than a(n). These rational approximations together with those obtained from the sequences A001653 and A002315 give a complete set of closest rational approximations to sqrt(2) with restricted numerator or denominator. a(n)/A001542(n) > sqrt(2) > 2*A001542(n)/a(n). - A.H.M. Smeets, May 28 2017
x = a(n), y = A001542(n) are solutions of the Diophantine equation x^2 - 2y^2 = 1 (Pell equation). x = 2*A001542(n), y = a(n) are solutions of the Diophantine equation x^2 - 2y^2 = -2. Both together give the set of fractional approximations for sqrt(2) obtained from limited fractions obtained from continued fraction representation to sqrt(2). - A.H.M. Smeets, Jun 22 2017
a(n) is the radius of the n-th circle among the sequence of circles generated as follows: Starting with a unit circle centered at the origin, every subsequent circle touches the previous circle as well as the two limbs of hyperbola x^2 - y^2 = 1, and lies in the region y > 0. - Kaushal Agrawal, Nov 10 2018
All of the positive integer solutions of a*b+1=x^2, a*c+1=y^2, b*c+1=z^2, x+z=2*y, 0A001542(n), b=A005319(n), c=A001542(n+1), x=A001541(n), y=A001653(n+1), z=A002315(n) with 0Michael Somos, Jun 26 2022

			99^2 + 99^2 = 140^2 + 2. - _Carmine Suriano_, Jan 05 2015
G.f. = 1 + 3*x + 17*x^2 + 99*x^3 + 577*x^4 + 3363*x^5 + 19601*x^6 + 114243*x^7 + ...

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)
J. W. L. Glaisher, On Eulerian numbers (formulas, residues, end-figures), with the values of the first twenty-seven, Quarterly Journal of Mathematics, vol. 45, 1914, pp. 1-51.
G. K. Panda, Some fascinating properties of balancing numbers, In Proc. of Eleventh Internat. Conference on Fibonacci Numbers and Their Applications, Cong. Numerantium 194 (2009), 185-189.
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).
James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 257-258.
P.-F. Teilhet, Query 2376, L'Intermédiaire des Mathématiciens, 11 (1904), 138-139. - N. J. A. Sloane, Mar 08 2022

T. D. Noe, Table of n, a(n) for n = 0..200
I. Adler, Three Diophantine equations - Part II, Fib. Quart., 7 (1969), 181-193.
Christian Aebi and Grant Cairns, Lattice Equable Parallelograms, arXiv:2006.07566 [math.NT], 2020.
Jean-Paul Allouche, Zeta-regularization of arithmetic sequences, EPJ Web of Conferences (2020) Vol. 244, 01008.
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
H. Brocard, Notes élémentaires sur le problème de Peel, Nouvelle Correspondance Mathématique, 4 (1878), 161-169.
John M. Campbell, An Integral Representation of Kekulé Numbers, and Double Integrals Related to Smarandache Sequences, arXiv preprint arXiv:1105.3399 [math.GM], 2011.
P. Catarino, H. Campos, and P. Vasco, On some identities for balancing and cobalancing numbers, Annales Mathematicae et Informaticae, 45 (2015) pp. 11-24.
Kwang-Wu Chen and Yu-Ren Pan, Greatest Common Divisors of Shifted Horadam Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.5.8.
S. Falcon, Relationships between Some k-Fibonacci Sequences, Applied Mathematics, 2014, 5, 2226-2234.
Morgan Fiebig, aBa Mbirika, and Jürgen Spilker, Period patterns, entry points, and orders in the Lucas sequences: theory and applications, arXiv:2408.14632 [math.NT], 2024. See p. 5.
Robert Frontczak, A Note on Hybrid Convolutions Involving Balancing and Lucas-Balancing Numbers, Applied Mathematical Sciences, Vol. 12, 2018, No. 25, 1201-1208.
Robert Frontczak, Sums of Balancing and Lucas-Balancing Numbers with Binomial Coefficients, International Journal of Mathematical Analysis (2018) Vol. 12, No. 12, 585-594.
Robert Frontczak, Powers of Balancing Polynomials and Some Consequences for Fibonacci Sums, International Journal of Mathematical Analysis (2019) Vol. 13, No. 3, 109-115.
Robert Frontczak, Identities for generalized balancing numbers, Notes on Number Theory and Discrete Mathematics (2019) Vol. 25, No. 2, 169-180.
Robert Frontczak and Taras Goy, Additional close links between balancing and Lucas-balancing polynomials, arXiv:2007.14048 [math.NT], 2020.
Robert Frontczak and Taras Goy, More Fibonacci-Bernoulli relations with and without balancing polynomials, arXiv:2007.14618 [math.NT], 2020.
Robert Frontczak and Taras Goy, Lucas-Euler relations using balancing and Lucas-balancing polynomials, arXiv:2009.09409 [math.NT], 2020.
O. Khadir, K. Liptai, and L. Szalay, On the Shifted Product of Binary Recurrences, J. Int. Seq. 13 (2010), 10.6.1.
J. M. Katri and D. R. Byrkit, Problem E1976, Amer. Math. Monthly, 75 (1968), 683-684.
Tanya Khovanova, Recursive Sequences
D. H. Lehmer, A cotangent analogue of continued fractions, Duke Math. J., 4 (1935), 323-340.
D. H. Lehmer, A cotangent analogue of continued fractions, Duke Math. J., 4 (1935), 323-340. [Annotated scanned copy]
D. H. Lehmer, Lacunary recurrence formulas for the numbers of Bernoulli and Euler, Annals Math., 36 (1935), 637-649.
Dino Lorenzini and Z. Xiang, Integral points on variable separated curves, Preprint 2016.
aBa Mbirika, Janeè Schrader, and Jürgen Spilker, Pell and associated Pell braid sequences as GCDs of sums of k consecutive Pell, balancing, and related numbers, arXiv:2301.05758 [math.NT], 2023. See also J. Int. Seq. (2023) Vol. 26, Art. 23.6.4.
A. Patra, G. K. Panda, and T. Khemaratchatakumthorn, Exact divisibility by powers of the balancing and Lucas-balancing numbers, Fibonacci Quart., 59:1 (2021), 57-64; see C(n).
Robert Phillips, Polynomials of the form 1+4ke+4ke^2, 2008.
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
Prasanta K. Ray, Curious congruences for balancing numbers, Int. J. Contemp. Math. Sci. 7 (2012), 881-889.
Soumeya M. Tebtoub, Hacène Belbachir, and László Németh, Integer sequences and ellipse chains inside a hyperbola, Proceedings of the 1st International Conference on Algebras, Graphs and Ordered Sets (ALGOS 2020), hal-02918958 [math.cs], 17-18.
Bibhu Prasad Tripathy and Bijan Kumar Patel, On balancing and Lucas-balancing numbers expressible as product of two k-Fibonacci numbers, arXiv:2508.12238 [math.NT], 2025.
N. J. Wildberger, Pell's equation without irrational numbers, J. Int. Seq. 13 (2010), 10.4.3, Section 4.
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (6,-1).
Index entries for sequences related to Chebyshev polynomials.

Bisection of A001333. A003499(n) = 2a(n).
Cf. A055997 = numbers n such that n(n-1)/2 is a square.
Row 1 of array A188645.
Cf. A046090, A001109, A053142, A084130, A001601, A056771, A077420, A005319, A082532, A001542.
Cf. A055792 (terms squared), A132592.

a001541 n = a001541_list !! (n-1)
a001541_list =
1 : 3 : zipWith (-) (map (* 6) $ tail a001541_list) a001541_list
-- Reinhard Zumkeller, Oct 06 2011
(Scheme, with memoization-macro definec)
(definec (A001541 n) (cond ((zero? n) 1) ((= 1 n) 3) (else (- (* 6 (A001541 (- n 1))) (A001541 (- n 2))))))
;; Antti Karttunen, Oct 04 2016

[n: n in [1..10000000] |IsSquare(8*(n^2-1))]; // Vincenzo Librandi, Nov 18 2010

a[0]:=1: a[1]:=3: for n from 2 to 26 do a[n]:=6*a[n-1]-a[n-2] od: seq(a[n], n=0..20); # Zerinvary Lajos, Jul 26 2006
A001541:=-(-1+3*z)/(1-6*z+z**2); # Simon Plouffe in his 1992 dissertation

Table[Simplify[(1/2) (3 + 2 Sqrt[2])^n + (1/2) (3 - 2 Sqrt[2])^n], {n, 0, 20}] (* Artur Jasinski, Feb 10 2010 *)
a[ n_] := If[n == 0, 1, With[{m = Abs @ n}, m Sum[4^i Binomial[m + i, 2 i]/(m + i), {i, 0, m}]]]; (* Michael Somos, Jul 11 2011 *)
a[ n_] := ChebyshevT[ n, 3]; (* Michael Somos, Jul 11 2011 *)
LinearRecurrence[{6, -1}, {1, 3}, 50] (* Vladimir Joseph Stephan Orlovsky, Feb 12 2012 *)

{a(n) = real((3 + quadgen(32))^n)}; /* Michael Somos, Apr 07 2003 */

{a(n) = subst( poltchebi( abs(n)), x, 3)}; /* Michael Somos, Apr 07 2003 */

{a(n) = if( n<0, a(-n), polsym(1 - 6*x + x^2, n) [n+1] / 2)}; /* Michael Somos, Apr 07 2003 */

{a(n) = polchebyshev( n, 1, 3)}; /* Michael Somos, Jul 11 2011 */

a(n)=([1,2,2;2,1,2;2,2,3]^n)[3,3] \\ Vim Wenders, Mar 28 2007

G.f.: (1-3*x)/(1-6*x+x^2). - Barry E. Williams and Wolfdieter Lang, May 05 2000
E.g.f.: exp(3*x)*cosh(2*sqrt(2)*x). Binomial transform of A084128. - Paul Barry, May 16 2003
From N. J. A. Sloane, May 16 2003: (Start)
a(n) = sqrt(8*((A001109(n))^2) + 1).
a(n) = T(n, 3), with Chebyshev's T-polynomials A053120. (End)
a(n) = ((3+2*sqrt(2))^n + (3-2*sqrt(2))^n)/2.
a(n) = cosh(2*n*arcsinh(1)). - Herbert Kociemba, Apr 24 2008
a(n) ~ (1/2)*(sqrt(2) + 1)^(2*n). - Joe Keane (jgk(AT)jgk.org), May 15 2002
For all elements x of the sequence, 2*x^2 - 2 is a square. Limit_{n -> infinity} a(n)/a(n-1) = 3 + 2*sqrt(2). - Gregory V. Richardson, Oct 10 2002 [corrected by Peter Pein, Mar 09 2009]
a(n) = 3*A001109(n) - A001109(n-1), n >= 1. - Barry E. Williams and Wolfdieter Lang, May 05 2000
For n >= 1, a(n) = A001652(n) - A001652(n-1). - Charlie Marion, Jul 01 2003
From Paul Barry, Sep 18 2003: (Start)
a(n) = ((-1+sqrt(2))^n + (1+sqrt(2))^n + (1-sqrt(2))^n + (-1-sqrt(2))^n)/4 (with interpolated zeros).
E.g.f.: cosh(x)*cosh(sqrt(2)x) (with interpolated zeros). (End)
For n > 0, a(n)^2 + 1 = 2*A001653(n-1)*A001653(n). - Charlie Marion, Dec 21 2003
a(n)^2 + a(n+1)^2 = 2*(A001653(2*n+1) - A001652(2*n)). - Charlie Marion, Mar 17 2003
a(n) = Sum_{k >= 0} binomial(2*n, 2*k)*2^k = Sum_{k >= 0} A086645(n, k)*2^k. - Philippe Deléham, Feb 29 2004
a(n)*A002315(n+k) = A001652(2*n+k) + A001652(k) + 1; for k > 0, a(n+k)*A002315(n) = A001652(2*n+k) - A001652(k-1). - Charlie Marion, Mar 17 2003
For n > k, a(n)*A001653(k) = A011900(n+k) + A053141(n-k-1). For n <= k, a(n)*A001653(k) = A011900(n+k) + A053141(k-n). - Charlie Marion, Oct 18 2004
A053141(n+1) + A055997(n+1) = a(n+1) + A001109(n+1). - Creighton Dement, Sep 16 2004
a(n+1) - A001542(n+1) = A090390(n+1) - A046729(n) = A001653(n); a(n+1) - 4*A079291(n+1) = (-1)^(n+1). Formula generated by the floretion - .5'i + .5'j - .5i' + .5j' - 'ii' + 'jj' - 2'kk' + 'ij' + .5'ik' + 'ji' + .5'jk' + .5'ki' + .5'kj' + e. - Creighton Dement, Nov 16 2004
a(n) = sqrt( A055997(2*n) ). - Alexander Adamchuk, Nov 24 2006
a(2n) = A056771(n). a(2*n+1) = 3*A077420(n). - Alexander Adamchuk, Feb 01 2007
a(n) = (A000129(n)^2)*4 + (-1)^n. - Vim Wenders, Mar 28 2007
2*a(k)*A001653(n)*A001653(n+k) = A001653(n)^2 + A001653(n+k)^2 + A001542(k)^2. - Charlie Marion, Oct 12 2007
a(n) = A001333(2*n). - Ctibor O. Zizka, Aug 13 2008
A028982(a(n)-1) + 2 = A028982(a(n)+1). - Juri-Stepan Gerasimov, Mar 28 2011
a(n) = 2*A001108(n) + 1. - Paul Weisenhorn, Dec 17 2011
a(n) = sqrt(2*x^2 + 1) with x being A001542(n). - Zak Seidov, Jan 30 2013
a(2n) = 2*a(n)^2 - 1 = a(n)^2 + 2*A001542(n)^2. a(2*n+1) = 1 + 2*A002315(n)^2. - Steven J. Haker, Dec 04 2013
a(n) = 3*a(n-1) + 4*A001542(n-1); e.g., a(4) = 99 = 3*17 + 4*12. - Zak Seidov, Dec 19 2013
a(n) = cos(n * arccos(3)) = cosh(n * log(3 + 2*sqrt(2))). - Daniel Suteu, Jul 28 2016
From Ilya Gutkovskiy, Jul 28 2016: (Start)
Inverse binomial transform of A084130.
Exponential convolution of A000079 and A084058.
Sum_{n>=0} (-1)^n*a(n)/n! = cosh(2*sqrt(2))/exp(3) = 0.4226407909842764637... (End)
a(2*n+1) = 2*a(n)*a(n+1) - 3. - Timothy L. Tiffin, Oct 12 2016
a(n) = a(-n) for all n in Z. - Michael Somos, Jan 20 2017
a(2^n) = A001601(n+1). - A.H.M. Smeets, May 28 2017
a(A298210(n)) = A002350(2*n^2). - A.H.M. Smeets, Jan 25 2018
a(n) = S(n, 6) - 3*S(n-1, 6), for n >= 0, with S(n, 6) = A001109(n+1), (Chebyshev S of A049310). See the first comment and the formula a(n) = T(n, 3). - Wolfdieter Lang, Nov 22 2020
From Peter Bala, Dec 31 2021: (Start)
a(n) = [x^n] (3*x + sqrt(1 + 8*x^2))^n.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) hold for all prime p and positive integers n and k.
O.g.f. A(x) = 1 + x*d/dx(log(B(x))), where B(x) = 1/sqrt(1 - 6*x + x^2) is the o.g.f. of A001850. (End)
From Peter Bala, Aug 17 2022: (Start)
Sum_{n >= 1} 1/(a(n) - 2/a(n)) = 1/2.
Sum_{n >= 1} (-1)^(n+1)/(a(n) + 1/a(n)) = 1/4.
Sum_{n >= 1} 1/(a(n)^2 - 2) = 1/2 - 1/sqrt(8). (End)
From  Peter Bala, Jun 23 2025: (Start)
Product_{n >= 0} (1 + 1/a(2^n)) = sqrt(2).
Product_{n >= 0} (1 - 1/(2*a(2^n))) = (4/7)*sqrt(2). See A002812. (End)

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)

A001834 a(0) = 1, a(1) = 5, a(n) = 4*a(n-1) - a(n-2).

Original entry on oeis.org

1, 5, 19, 71, 265, 989, 3691, 13775, 51409, 191861, 716035, 2672279, 9973081, 37220045, 138907099, 518408351, 1934726305, 7220496869, 26947261171, 100568547815, 375326930089, 1400739172541, 5227629760075, 19509779867759, 72811489710961, 271736178976085

Offset: 0

N. J. A. Sloane

Sequence also gives values of x satisfying 3*y^2 - x^2 = 2, the corresponding y being given by A001835(n+1). Moreover, quadruples(p, q, r, s) satisfying p^2 + q^2 + r^2 = s^2, where p = q and r is either p+1 or p-1, are termed nearly isosceles Pythagorean and are given by p = {x + (-1)^n}/3, r = p-(-1)^n, s = y for n > 1. - Lekraj Beedassy, Jul 19 2002
a(n)= A002531(1+2*n). - Anton Vrba (antonvrba(AT)yahoo.com), Feb 14 2007
361 written in base A001835(n+1) - 1 is the square of a(n). E.g., a(12) = 2672279, A001835(13) - 1 = 1542840. We have 361_(1542840) = 3*1542840 + 6*1542840 + 1 = 2672279^2. - Richard Choulet, Oct 04 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
General recurrence is a(n) = (a(1) - 1)*a(n-1) - a(n-2), a(1) >= 4, lim_{n->infinity} a(n) = x*(k*x + 1)^n, k = (a(1) - 3), x = (1 + sqrt((a(1) + 1)/(a(1) - 3)))/2. Examples in OEIS: a(1) = 4 gives A002878, primes in it A121534. a(1) = 5 gives A001834, primes in it A086386. a(1) = 6 gives A030221, primes in it A299109. a(1) = 7 gives A002315, primes in it A088165. a(1) = 8 gives A033890, primes in it not in OEIS (do there exist any?). a(1) = 9 gives A057080, primes in {71, 34649, 16908641, ...}. a(1) = 10 gives A057081, primes in it {389806471, 192097408520951, ...}. - Ctibor O. Zizka, Sep 02 2008
Inverse binomial transform of A030192. - Philippe Deléham, Nov 19 2009
For positive n, a(n) equals the permanent of the (2*n) X (2*n) tridiagonal matrix with sqrt(6)'s along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011
x-values in the solution to 3x^2 + 6 = y^2 (see A082841 for the y-values). - Sture Sjöstedt, Nov 25 2011
Pisano period lengths: 1, 1, 2, 4, 3, 2, 8, 4, 6, 3, 10, 4, 12, 8, 6, 8, 18, 6, 5, 12, ... - R. J. Mathar, Aug 10 2012
The aerated sequence (b(n))A100047%20for%20a%20connection%20with%20Chebyshev%20polynomials.%20-%20_Peter%20Bala">{n>=1} = [1, 0, 5, 0, 19, 0, 71, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -2, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for a connection with Chebyshev polynomials. - _Peter Bala, Mar 22 2015
Yong Hao Ng has shown that for any n, a(n) is coprime with any member of A001835 and with any member of A001075. - René Gy, Feb 26 2018
From Wolfdieter Lang, Oct 15 2020: (Start)
((-1)^n)*a(n) = X(n) = (-1)^n*(S(n, 4) + S(n-1, 4)) and Y(n) = X(n-1) gives all integer solutions (modulo sign flip between X and Y) of X^2 + Y^2 + 4*X*Y = +6, for n = -oo..+oo, with Chebyshev S polynomials (see A049310), with S(-1, x) = 0, and S(-|n|, x) = - S(|n|-2, x), for |n| >= 2.
This binary indefinite quadratic form of discriminant 12, representing 6, has only this family of proper solutions (modulo sign flip), and no improper ones.
This comment is inspired by a paper by Robert K. Moniot (private communication). See his Oct 04 2020 comment in A027941 related to the case of x^2 + y^2 - 3*x*y = -1 (special Markov solutions). (End)
Floretion Algebra Multiplication Program, FAMP Code: A001834 = (4/3)vesseq[ - .25'i + 1.25'j - .25'k - .25i' + 1.25j' - .25k' + 1.25'ii' + .25'jj' - .75'kk' + .75'ij' + .25'ik' + .75'ji' - .25'jk' + .25'ki' - .25'kj' + .25e], apart from initial term

			G.f. = 1 + 5*x + 19*x^2 + 71*x^3 + 265*x^4 + 989*x^5 + 3691*x^6 + ...

Bastida, Julio R. 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.
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).
P.-F. Teilhet, Reply to Query 2094, L'Intermédiaire des Mathématiciens, 10 (1903), 235-238.

T. D. Noe, Table of n, a(n) for n=0..200
Marco Abrate, Stefano Barbero, Umberto Cerruti, and Nadir Murru, Polynomial sequences on quadratic curves, Integers, Vol. 15, 2015, #A38.
K. Andersen, L. Carbone, and D. Penta, Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields, Journal of Number Theory and Combinatorics, Vol 2, No. 3 pp 245-278, 2011. See Section 9.
J. B. Cosgrave and K. Dilcher, A role for generalized Fermat numbers, Math. Comp. 86 (2017), 899-933; see also Paper #10.
Bruno Deschamps, Sur les bonnes valeurs initiales de la suite de Lucas-Lehmer, Journal of Number Theory, Volume 130, Issue 12, December 2010, Pages 2658-2670.
Leonhard Euler, Vollstaendige Anleitung zur Algebra, Zweiter Teil.
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.
Taras Goy and Mark Shattuck, Determinants of Toeplitz-Hessenberg Matrices with Generalized Leonardo Number Entries, Ann. Math. Silesianae (2023). See p. 17.
Christian Kassel and Christophe Reutenauer, Pairs of intertwined integer sequences, arXiv:2507.15780 [math.NT], 2025. See p. 13.
Tanya Khovanova, Recursive Sequences
Seong Ju Kim, R. Stees, and L. Taalman, Sequences of Spiral Knot Determinants, Journal of Integer Sequences, Vol. 19 (2016), # 16.1.4
Clark Kimberling, Best lower and upper approximates to irrational numbers, Elemente der Mathematik, 52 (1997) 122-126.
Wolfdieter Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38 (2000) 408-419. Eq. (44) rhs, m=6.
Ioana-Claudia Lazăr, Lucas sequences in t-uniform simplicial complexes, arXiv:1904.06555 [math.GR], 2019.
Donatella Merlini and Renzo Sprugnoli, Arithmetic into geometric progressions through Riordan arrays, Discrete Mathematics 340.2 (2017): 160-174.
Yong Hao Ng, A partition in three classes of the set of all prime numbers?, Math StackExchange.
S. Northshield, An Analogue of Stern's Sequence for Z[sqrt(2)], Journal of Integer Sequences, 18 (2015), #15.11.6.
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 and Conjectures, Universite du Quebec a Montreal, 1992.
Ryan Stees, Sequences of Spiral Knot Determinants, Senior Honors Projects, Paper 84, James Madison Univ., May 2016.
F. V. Waugh and M. W. Maxfield, Side-and-diagonal numbers, Math. Mag., 40 (1967), 74-83.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences, Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (4,-1).

A bisection of sequence A002531.
Cf. A001352, A001835, A086386 (prime members).
Cf. A026150.
Cf. A082841, A100047.
Cf. A002531, A001353, A049310.
a(n)^2+1 = A094347(n+1).

a001834 n = a001834_list !! (n-1)
a001834_list = 1 : 5 : zipWith (-) (map (* 4) $ tail a001834_list) a001834_list
-- Reinhard Zumkeller, Jan 23 2012

I:=[1,5]; [n le 2 select I[n] else 4*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Mar 22 2015

f:=n->((1+sqrt(3))^(2*n+1)+(1-sqrt(3))^(2*n+1))/2^(n+1); # N. J. A. Sloane, Nov 10 2009

a[0] = 1; a[1] = 5; a[n_] := a[n] = 4a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 25}] (* Robert G. Wilson v, Apr 24 2004 *)
Table[Expand[((1+Sqrt[3])^(2*n+1)+(1+Sqrt[3])^(2*n+1))/2^(n+1)],{n, 0, 20}] (* Anton Vrba, Feb 14 2007 *)
LinearRecurrence[{4, -1}, {1, 5}, 50] (* Sture Sjöstedt, Nov 27 2011 *)
a[c_, n_] := Module[{},
   p := Length[ContinuedFraction[ Sqrt[ c]][[2]]];
   d := Numerator[Convergents[Sqrt[c], n p]];
   t := Table[d[[1 + i]], {i, 0, Length[d] - 1, p}];
   Return[t];
] (* Complement of A002531 *)
a[3, 20] (* Gerry Martens, Jun 07 2015 *)
Round@Table[LucasL[2n+1, Sqrt[2]]/Sqrt[2], {n, 0, 20}] (* Vladimir Reshetnikov, Sep 15 2016 *)

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

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

[(lucas_number2(n,4,1)-lucas_number2(n-1,4,1))/2 for n in range(1, 27)] # Zerinvary Lajos, Nov 10 2009

a(n) = ((1 + sqrt(3))^(2*n + 1) + (1 - sqrt(3))^(2*n + 1))/2^(n + 1). - N. J. A. Sloane, Nov 10 2009
a(n) = (1/2) * ((1 + sqrt(3))*(2 + sqrt(3))^n + (1 - sqrt(3))*(2 - sqrt(3))^n). - Dean Hickerson, Dec 01 2002
From Mario Catalani, Apr 11 2003: (Start)
With a = 2 + sqrt(3), b = 2 - sqrt(3): a(n) = (1/sqrt(2))(a^(n + 1/2) - b^(n + 1/2)).
a(n) - a(n-1) = A003500(n).
a(n) = sqrt(1 + 12*A061278(n) + 12*A061278(n)^2). (End)
a(n) = ((1 + sqrt(3))^(2*n + 1) + (1 - sqrt(3))^(2*n + 1))/2^(n + 1). - Anton Vrba, Feb 14 2007
G.f.: (1 + x)/((1 - 4*x + x^2)). Simon Plouffe in his 1992 dissertation.
a(n) = S(2*n, sqrt(6)) = S(n, 4) + S(n-1, 4); S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310. S(n, 4) = A001353(n).
For all members x of the sequence, 3*x^2 + 6 is a square. Limit_{n->infinity} a(n)/a(n-1) = 2 + sqrt(3). - Gregory V. Richardson, Oct 10 2002
a(n) = 2*A001571(n) + 1. - 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 (-1)^n*q(n, -6) = a(n). - Benoit Cloitre, Nov 10 2002
a(n) = 2^(-n)*Sum_{k>=0} binomial(2*n + 1, 2*k)*3^k; see A091042. - Philippe Deléham, Mar 01 2004
a(n) = floor(sqrt(3)*A001835(n+1)). - Philippe Deléham, Mar 03 2004
a(n+1) - 2*a(n) = 3*A001835(n+1). Using the known relation A001835(n+1) = sqrt((a(n)^2 + 2)/3) it follows that a(n+1) - 2*a(n) = sqrt(3*(a(n)^2 + 2)). Therefore a(n+1)^2 + a(n)^2 - 4*a(n+1)*a(n) - 6 = 0. - Creighton Dement, Apr 18 2005
a(n) = L(n,-4)*(-1)^n, where L is defined as in A108299; see also A001835 for L(n,+4). - Reinhard Zumkeller, Jun 01 2005
a(n) = Jacobi_P(n, 1/2, -1/2, 2)/Jacobi_P(n, -1/2, 1/2, 1). - Paul Barry, Feb 03 2006
Equals binomial transform of A026150 starting (1, 4, 10, 28, 76, ...) and double binomial transform of (1, 3, 3, 9, 9, 27, 27, 81, 81, ...). - Gary W. Adamson, Nov 30 2007
Sequence satisfies 6 = f(a(n), a(n+1)) where f(u, v) = u^2 + v^2 - 4*u*v. - Michael Somos, Sep 19 2008
a(-1-n) = -a(n). - Michael Somos, Sep 19 2008
From Franck Maminirina Ramaharo, Nov 11 2018: (Start)
a(n) = (-1)^n*(5*A125905(n) + A125905(n+1)).
E.g.f.: exp(2*x)*(cosh(sqrt(3)*x) + sqrt(3)*sinh(sqrt(3)*x)). (End)
a(n) = A061278(n+1) - A061278(n-1) for n>=2. - John P. McSorley, Jun 20 2020
From Peter Bala, May 09 2025: (Start)
a(n) = Dir(n, 2), where Dir(n, x) denotes the n-th row polynomial of the triangle A244419.
a(n) - 2*a(n-1) = 3 * A001835(n) for n >= 1.
For arbitrary x, a(n+x)^2 - 4*a(n+x)*a(n+x+1) + a(n+x+1)^2 = 6 with a(n) := (1/2) * ((1 + sqrt(3))*(2 + sqrt(3))^n + (1 - sqrt(3))*(2 - sqrt(3))^n) as above. The particular case x = 0 is noted above,
a(n+1/2) = sqrt(6) * A001353(n+1).
a(n+3/4) + a(n+1/4) = sqrt(6*sqrt(6) + 12) * A001353(n+1).
a(n+3/4) - a(n+1/4) = sqrt(2*sqrt(6) - 4) * A001075(n+1).
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/6 (telescoping series: for n >= 1, 1/(a(n) - 1/a(n)) = 1/A001352(n) + 1/A001352(n+1)).
Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(3) (telescoping product: Product_{n = 1..k} ((a(n) + 1)/(a(n) - 1))^2 = 3*(1 - 2/A102206(k))). (End)

A002530 a(n) = 4*a(n-2) - a(n-4) for n > 1, a(n) = n for n = 0, 1.

Original entry on oeis.org

0, 1, 1, 3, 4, 11, 15, 41, 56, 153, 209, 571, 780, 2131, 2911, 7953, 10864, 29681, 40545, 110771, 151316, 413403, 564719, 1542841, 2107560, 5757961, 7865521, 21489003, 29354524, 80198051, 109552575, 299303201, 408855776, 1117014753, 1525870529, 4168755811

Offset: 0

N. J. A. Sloane

Denominators of continued fraction convergents to sqrt(3), for n >= 1.
Also denominators of continued fraction convergents to sqrt(3) - 1. See A048788 for numerators. - N. J. A. Sloane, Dec 17 2007. Convergents are 1, 2/3, 3/4, 8/11, 11/15, 30/41, 41/56, 112/153, ...
Consider the mapping f(a/b) = (a + 3*b)/(a + b). Taking a = b = 1 to start with and carrying out this mapping repeatedly on each new (reduced) rational number gives the following sequence 1/1, 2/1, 5/3, 7/4, 19/11, ... converging to 3^(1/2). Sequence contains the denominators. The same mapping for N, i.e., f(a/b) = (a + Nb)/(a + b) gives fractions converging to N^(1/2). - Amarnath Murthy, Mar 22 2003
Sqrt(3) = 2/2 + 2/3 + 2/(3*11) + 2/(11*41) + 2/(41*153) + 2/(153*571), ...; the sum of the first 6 terms of this series = 1.7320490367..., while sqrt(3) = 1.7320508075... - Gary W. Adamson, Dec 15 2007
From Clark Kimberling, Aug 27 2008: (Start)
  Related convergents (numerator/denominator):
  lower principal convergents: A001834/A001835
  upper principal convergents: A001075/A001353
  intermediate convergents: A005320/A001075
  principal and intermediate convergents: A143642/A140827
  lower principal and intermediate convergents: A143643/A005246. (End)
Row sums of triangle A152063 = (1, 3, 4, 11, ...). - Gary W. Adamson, Nov 26 2008
From Alois P. Heinz, Apr 13 2011: (Start)
Also number of domino tilings of the 3 X (n-1) rectangle with upper left corner removed iff n is even.  For n=4 the 4 domino tilings of the 3 X 3 rectangle with upper left corner removed are:
   . ._. . ._. . ._. . ._.
   .|__| .|__| .| | | .|___|
   | |_| | | | | | ||| |_| |
   ||__| |||_| ||__| |_|_|  (End)
This is the sequence of Lehmer numbers u_n(sqrt(R),Q) with the parameters R = 2 and Q = -1. It is a strong divisibility sequence, that is, gcd(a(n),a(m)) = a(gcd(n,m)) for all natural numbers n and m. - Peter Bala, Apr 18 2014
2^(-floor(n/2))*(1 + sqrt(3))^n = A002531(n) + a(n)*sqrt(3); integers in the real quadratic number field Q(sqrt(3)). - Wolfdieter Lang, Feb 11 2018
Let T(n) = 2^(n mod 2), U(n) = a(n), V(n) = A002531(n), x(n) = V(n)/U(n). Then T(n*m) * U(n+m) = U(n)*V(m) + U(m)*V(n), T(n*m) * V(n+m) = 3*U(n)*U(m) + V(m)*V(n), x(n+m) = (3 + x(n)*x(m))/(x(n) + x(m)). - Michael Somos, Nov 29 2022

			Convergents to sqrt(3) are: 1, 2, 5/3, 7/4, 19/11, 26/15, 71/41, 97/56, 265/153, 362/209, 989/571, 1351/780, 3691/2131, ... = A002531/A002530 for n >= 1.
1 + 1/(1 + 1/(2 + 1/(1 + 1/2))) = 19/11 so a(5) = 11.
G.f. = x + x^2 + 3*x^3 + 4*x^4 + 11*x^5 + 15*x^6 + 41*x^7 + ... - _Michael Somos_, Mar 18 2022

Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.
Russell Lyons, A bird's-eye view of uniform spanning trees and forests, in Microsurveys in Discrete Probability, AMS, 1998.
I. Niven and H. S. Zuckerman, An Introduction to the Theory of Numbers. 2nd ed., Wiley, NY, 1966, p. 181.
Murat Sahin and Elif Tan, Conditional (strong) divisibility sequences, Fib. Q., 56 (No. 1, 2018), 18-31.
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).
A. Tarn, Approximations to certain square roots and the series of numbers connected therewith, Mathematical Questions and Solutions from the Educational Times, 1 (1916), 8-12.

Harry J. Smith, Table of n, a(n) for n = 0..2000
Peter Bala, Notes on 2-periodic continued fractions and Lehmer sequences
Paul Barry, Symmetric Third-Order Recurring Sequences, Chebyshev Polynomials, and Riordan Arrays, JIS 12 (2009) 09.8.6.
Mario Catalani, Sequences related to convergents to square root of rationals, arXiv:math/0305270 [math.NT], 2003.
Marcia Edson, Scott Lewis and Omer Yayenie, The k-periodic Fibonacci sequence and an extended Binet's formula, INTEGERS 11 (2011) #A32.
Aviezri S. Fraenkel, Jonathan Levitt, and Michael Shimshoni, Characterization of the set of values f(n)=[n alpha], n=1,2,..., Discrete Math. 2 (1972), no.4, 335-345.
Clark Kimberling, Best lower and upper approximates to irrational numbers, Elemente der Mathematik, 52 (1997) 122-126.
Clark Kimberling, Matrix Transformations of Integer Sequences, J. Integer Seqs., Vol. 6, 2003.
MacTutor, D'Arcy Thompson on Greek irrationals
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
Albert Tarn, Approximations to certain square roots and the series of numbers connected therewith [Annotated scanned copy]
D'Arcy Thompson, Excess and Defect: Or the Little More and the Little Less, Mind, New Series, Vol. 38, No. 149 (Jan., 1929), pp. 43-55 (13 pages). See page 48.
Hein van Winkel, Q-quadrangles inscribed in a circle, 2014. See Table 1. [Reference from Antreas Hatzipolakis, Jul 14 2014]
Russell Walsmith, CL-Chemy Transforms Fibonacci-Type Sequences to Arrays
E. W. Weisstein, MathWorld: Lehmer Number
Index entries for "core" sequences
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (0,4,0,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A002531 (numerators of convergents to sqrt(3)), A048788, A003297.
Bisections: A001353 and A001835.
Cf. A152063.
Cf. A108412, A049310.
Analog for sqrt(m): A000129 (m=2), A001076 (m=5), A041007 (m=6), A041009 (m=7), A041011 (m=8), A005668 (m=10), A041015 (m=11), A041017 (m=12), ..., A042935 (m=999), A042937 (m=1000).

I:=[0,1,1,3]; [n le 4 select I[n] else 4*Self(n-2) - Self(n-4): n in [1..50]]; // G. C. Greubel, Feb 25 2019

a := proc(n) option remember; if n=0 then 0 elif n=1 then 1 elif n=2 then 1 elif n=3 then 3 else 4*a(n-2)-a(n-4) fi end; [ seq(a(i),i=0..50) ];
A002530:=-(-1-z+z**2)/(1-4*z**2+z**4); # conjectured (correctly) by Simon Plouffe in his 1992 dissertation

Join[{0},Table[Denominator[FromContinuedFraction[ContinuedFraction[Sqrt[3],n]]], {n,1,50}]] (* Stefan Steinerberger, Apr 01 2006 *)
Join[{0},Denominator[Convergents[Sqrt[3],50]]] (* or *) LinearRecurrence[ {0,4,0,-1},{0,1,1,3},50] (* Harvey P. Dale, Jan 29 2013 *)
a[ n_] := If[n<0, -(-1)^n, 1] SeriesCoefficient[ x*(1+x-x^2)/(1-4*x^2+x^4), {x, 0, Abs@n}]; (* Michael Somos, Apr 18 2019 *)
a[ n_] := ChebyshevU[n-1, Sqrt[-1/2]]*Sqrt[2]^(Mod[n, 2]-1)/I^(n-1) //Simplify; (* Michael Somos, Nov 29 2022 *)

{a(n) = if( n<0, -(-1)^n * a(-n), contfracpnqn(vector(n, i, 1 + (i>1) * (i%2)))[2, 1])}; /* Michael Somos, Jun 05 2003 */

{ for (n=0, 50, a=contfracpnqn(vector(n, i, 1+(i>1)*(i%2)))[2, 1]; write("b002530.txt", n, " ", a); ); } \\ Harry J. Smith, Jun 01 2009

my(w=quadgen(12)); A002530(n)=real((2+w)^(n\/2)*if(bittest(n,0),1-w/3,w/3));
apply(A002530, [0..30]) \\ M. F. Hasler, Nov 04 2019

from functools import cache
@cache
def a(n): return [0, 1, 1, 3][n] if n < 4 else 4*a(n-2) - a(n-4)
print([a(n) for n in range(36)]) # Michael S. Branicky, Nov 13 2022

(x*(1+x-x^2)/(1-4*x^2+x^4)).series(x, 50).coefficients(x, sparse=False) # G. C. Greubel, Feb 25 2019

G.f.: x*(1 + x - x^2)/(1 - 4*x^2 + x^4).
a(n) = 4*a(n-2) - a(n-4). [Corrected by László Szalay, Feb 21 2014]
a(n) = -(-1)^n * a(-n) for all n in Z, would satisfy the same recurrence relation. - Michael Somos, Jun 05 2003
a(2*n) = a(2*n-1) + a(2*n-2), a(2*n+1) = 2*a(2*n) + a(2*n-1).
From Benoit Cloitre, Dec 15 2002: (Start)
a(2*n) = ((2 + sqrt(3))^n - (2 - sqrt(3))^n)/(2*sqrt(3)).
a(2*n) = A001353(n).
a(2*n-1) = ceiling((1 + 1/sqrt(3))/2*(2 + sqrt(3))^n) = ((3 + sqrt(3))^(2*n - 1) + (3 - sqrt(3))^(2*n - 1))/6^n.
a(2*n-1) = A001835(n). (End)
a(n+1) = Sum_{k=0..floor(n/2)} binomial(n - k, k) * 2^floor((n - 2*k)/2). - Paul Barry, Jul 13 2004
a(n) = Sum_{k=0..floor(n/2)} binomial(floor(n/2) + k, floor((n - 1)/2 - k))*2^k. - Paul Barry, Jun 22 2005
G.f.: (sqrt(6) + sqrt(3))/12*Q(0), where Q(k) = 1 - a/(1 + 1/(b^(2*k) - 1 - b^(2*k)/(c + 2*a*x/(2*x - g*m^(2*k)/(1 + a/(1 - 1/(b^(2*k + 1) + 1 - b^(2*k + 1)/(h - 2*a*x/(2*x + g*m^(2*k + 1)/Q(k + 1)))))))))). - Sergei N. Gladkovskii, Jun 21 2012
a(n) = (alpha^n - beta^n)/(alpha - beta) for n odd, and a(n) = (alpha^n - beta^n)/(alpha^2 - beta^2) for n even, where alpha = 1/2*(sqrt(2) + sqrt(6)) and beta = (1/2)*(sqrt(2) - sqrt(6)). Cf. A108412. - Peter Bala, Apr 18 2014
a(n) = (-sqrt(2)*i)^n*S(n, sqrt(2)*i)*2^(-floor(n/2)) = A002605(n)*2^(-floor(n/2)), n >= 0, with i = sqrt(-1) and S the Chebyshev polynomials (A049310). - Wolfdieter Lang, Feb 10 2018
a(n+1)*a(n+2) - a(n+3)*a(n) = (-1)^n, n >= 0. - Kai Wang, Feb 06 2020
E.g.f.: sinh(sqrt(3/2)*x)*(sinh(x/sqrt(2)) + sqrt(2)*cosh(x/sqrt(2)))/sqrt(3). - Stefano Spezia, Feb 07 2020
a(n) = ((1 + sqrt(3))^n - (1 - sqrt(3))^n)/(2*2^floor(n/2))/sqrt(3) = A002605(n)/2^floor(n/2). - Robert FERREOL, Apr 13 2023

Definition edited by M. F. Hasler, Nov 04 2019

A026150 a(0) = a(1) = 1; a(n+2) = 2a(n+1) + 2a(n).

Original entry on oeis.org

1, 1, 4, 10, 28, 76, 208, 568, 1552, 4240, 11584, 31648, 86464, 236224, 645376, 1763200, 4817152, 13160704, 35955712, 98232832, 268377088, 733219840, 2003193856, 5472827392, 14952042496, 40849739776

Offset: 0

N. J. A. Sloane

a(n+1)/A002605(n) converges to sqrt(3). - Mario Catalani (mario.catalani(AT)unito.it), Apr 22 2003
a(n+1)/a(n) converges to 1 + sqrt(3) = 2.732050807568877293.... - Philippe Deléham, Jul 03 2005
Binomial transform of expansion of cosh(sqrt(3)x) (A000244 with interpolated zeros); inverse binomial transform of A001075. - Philippe Deléham, Jul 04 2005
The same sequence may be obtained by the following process. Starting a priori with the fraction 1/1, the numerators of fractions built according to the rule: add top and bottom to get the new bottom, add top and 3 times the bottom to get the new top. The limit of the sequence of fractions is sqrt(3). - Cino Hilliard, Sep 25 2005
Inverse binomial transform of A001075: (1, 2, 7, 26, 97, 362, ...). - Gary W. Adamson, Nov 23 2007
Starting (1, 4, 10, 28, 76, ...), the sequence is the binomial transform of [1, 3, 3, 9, 9, 27, 27, 81, 81, ...], and inverse binomial transform of A001834: (1, 5, 19, 71, 265, ...). - Gary W. Adamson, Nov 30 2007
[1, 3; 1, 1]^n * [1,0] = [a(n), A002605(n)]. - Gary W. Adamson, Mar 21 2008
(1 + sqrt(3))^n = a(n) + A002605(n)*(sqrt(3)). - Gary W. Adamson, Mar 21 2008
Equals right border of triangle A143908. Also, starting (1, 4, 10, 28, ...) = row sums of triangle A143908 and INVERT transform of (1, 3, 3, 3, ...). - Gary W. Adamson, Sep 06 2008
a(n) is the number of compositions of n when there are 1 type of 1 and 3 types of other natural numbers. - Milan Janjic, Aug 13 2010
An elephant sequence, see A175655. For the central square four A[5] vectors, with decimal values 85, 277, 337 and 340, lead to this sequence (without the first leading 1). For the corner squares these vectors lead to the companion sequence A002605 (without the leading 0). - Johannes W. Meijer, Aug 15 2010
Pisano period lengths: 1, 1, 1, 1, 24, 1, 48, 1, 3, 24, 10, 1, 12, 48, 24, 1,144, 3,180, 24, ... - R. J. Mathar, Aug 10 2012
(1 + sqrt(3))^n = a(n) + A002605(n)*sqrt(3), for n >= 0; integers in the real quadratic number field Q(sqrt(3)). - Wolfdieter Lang, Feb 10 2018
a(n) is also the number of solutions for cyclic three-dimensional stable matching instances with master preference lists of size n (Escamocher and O'Sullivan 2018). - Guillaume Escamocher, Jun 15 2018
Starting from a(1), first differences of A005665. - Ivan N. Ianakiev, Nov 22 2019
Number of 3-permutations of n elements avoiding the patterns 231, 312. See Bonichon and Sun. - Michel Marcus, Aug 19 2022

			G.f. = 1 + x + 4*x^2 + 10*x^3 + 28*x^4 + 76*x^5 + 208*x^6 + 568*x^7 + ...

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

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Martin Burtscher, Igor Szczyrba, Rafał Szczyrba, Analytic Representations of the n-anacci Constants and Generalizations Thereof, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.5.
C. Banderier and D. Merlini, Lattice paths with an infinite set of jumps, FPSAC02, Melbourne, 2002.
C. Bautista-Ramos and C. Guillen-Galvan, Fibonacci numbers of generalized Zykov sums, J. Integer Seq., 15 (2012), Article 12.7.8.
Nicolas Bonichon and Pierre-Jean Morel, Baxter d-permutations and other pattern avoiding classes, arXiv:2202.12677 [math.CO], 2022.
A. Burstein, S. Kitaev and T. Mansour, Independent sets in certain classes of (almost) regular graphs, arXiv:math/0310379 [math.CO], 2003.
Guillaume Escamocher and Barry O'Sullivan, Three-Dimensional Matching Instances Are Rich in Stable Matchings, CPAIOR 2018, pages 182-197.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 1052
Tanya Khovanova, Recursive Sequences
Emanuele Munarini, A generalization of André-Jeannin's symmetric identity, Pure Mathematics and Applications (2018) Vol. 27, No. 1, 98-118.
Nathan Sun, On d-permutations and Pattern Avoidance Classes, arXiv:2208.08506 [math.CO], 2022.
Index entries for linear recurrences with constant coefficients, signature (2,2).
Index entries for sequences related to Chebyshev polynomials.

First differences of A002605.
The following sequences (and others) belong to the same family: A001333, A000129, A026150, A002605, A046717, A015518, A084057, A063727, A002533, A002532, A083098, A083099, A083100, A015519.
Cf. A001075, A001834, A083337, A002605, A143908, A028859, A030195, A106435, A108898, A125145, A053120.

a026150 n = a026150_list !! n
a026150_list = 1 : 1 : map (* 2) (zipWith (+) a026150_list (tail
a026150_list))
-- Reinhard Zumkeller, Oct 15 2011

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

with(combstruct):ZL0:=S=Prod(Sequence(Prod(a, Sequence(b))), a):ZL1:=Prod(begin_blockP, Z, end_blockP):ZL2:=Prod(begin_blockLR, Z, Sequence(Prod(mu_length, Z), card>=1), end_blockLR): ZL3:=Prod(begin_blockRL, Sequence(Prod(mu_length, Z), card>=1), Z, end_blockRL):Q:=subs([a=Union(ZL2,ZL2,ZL2), b=ZL1], ZL0), begin_blockP=Epsilon, end_blockP=Epsilon, begin_blockLR=Epsilon, end_blockLR=Epsilon, begin_blockRL=Epsilon, end_blockRL=Epsilon, mu_length=Epsilon:temp15:=draw([S, {Q}, unlabelled], size=15):seq(count([S, {Q}, unlabelled], size=n)/3, n=2..27); # Zerinvary Lajos, Mar 08 2008

Expand[Table[((1 + Sqrt[3])^n + (1 - Sqrt[3])^n)/(2), {n, 0, 30}]] (* Artur Jasinski, Dec 10 2006 *)
LinearRecurrence[{2, 2}, {1, 1}, 30] (* T. D. Noe, Mar 25 2011 *)
Round@Table[LucasL[n, Sqrt[2]] 2^(n/2 - 1), {n, 0, 20}] (* Vladimir Reshetnikov, Oct 15 2016 *)

a(n) := if n<=1 then 1 else 2*a(n-1)+2*a(n-2);
makelist(a(n),n,0,20); /* Emanuele Munarini, Apr 14 2017 */

{a(n) = if( n<0, 0, real((1 + quadgen(12))^n))};

from sage.combinat.sloane_functions import recur_gen2; it = recur_gen2(1,1,2,2); [next(it) for i in range(30)] # Zerinvary Lajos, Jun 25 2008

[lucas_number2(n,2,-2)/2 for n in range(0, 26)] # Zerinvary Lajos, Apr 30 2009

a(n) = (1/2)*((1 + sqrt(3))^n + (1 - sqrt(3))^n). - Benoit Cloitre, Oct 28 2002
G.f.: (1 - x)/(1 - 2*x - 2*x^2).
a(n) = a(n-1) + A083337(n-1). A083337(n)/a(n) converges to sqrt(3). - Mario Catalani (mario.catalani(AT)unito.it), Apr 29 2003
From Paul Barry, May 15 2003: (Start)
a(n) = Sum_{k=0..floor(n/2)} C(n, 2k)*3^k;
E.g.f.: exp(x)*cosh(sqrt(3)x). (End)
a(n) = Sum_{k=0..n} A098158(n,k)*3^(n - k). - Philippe Deléham, Dec 26 2007
a(n) = upper left and lower right terms of [1, 1; 3, 1]^n. (1 + sqrt(3))^n = a(n) + A083337(n)/(sqrt(3)). - Gary W. Adamson, Mar 12 2008
a(n) = A080040(n)/2. - Philippe Deléham, Nov 19 2008
If p[1] = 1, and p[i] = 3, (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) = det A. - Milan Janjic, Apr 29 2010
a(n) = 2 * A052945(n-1). - Vladimir Joseph Stephan Orlovsky, Mar 24 2011
a(n) = round((1 + sqrt(3))^n/2) for n > 0. - Bruno Berselli, Feb 04 2013
G.f.: G(0)/2, where G(k)= 1 + 1/(1 - x*(3*k - 1)/(x*(3*k + 2) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 25 2013
a(n) = (-sqrt(2)*i)^n*T(n,sqrt(2)*i/2), with i = sqrt(-1) and the Chebyshev T-polynomials (A053120). - Wolfdieter Lang, Feb 10 2018

A003500 a(n) = 4*a(n-1) - a(n-2) with a(0) = 2, a(1) = 4.

Original entry on oeis.org

2, 4, 14, 52, 194, 724, 2702, 10084, 37634, 140452, 524174, 1956244, 7300802, 27246964, 101687054, 379501252, 1416317954, 5285770564, 19726764302, 73621286644, 274758382274, 1025412242452, 3826890587534, 14282150107684, 53301709843202, 198924689265124

Offset: 0

N. J. A. Sloane

a(n) gives values of x satisfying x^2 - 3*y^2 = 4; corresponding y values are given by 2*A001353(n).
If M is any given term of the sequence, then the next one is 2*M + sqrt(3*M^2 - 12). - Lekraj Beedassy, Feb 18 2002
For n > 0, the three numbers a(n) - 1, a(n), and a(n) + 1 form a Fleenor-Heronian triangle, i.e., a Heronian triangle with consecutive sides, whose area A(n) may be obtained from the relation [4*A(n)]^2 = 3([a(2n)]^2 - 4); or A(n) = 3*A001353(2*n)/2 and whose semiperimeter is 3*a[n]/2. The sequence is symmetrical about a[0], i.e., a[-n] = a[n].
For n > 0, a(n) + 2 is the number of dimer tilings of a 2*n X 2 Klein bottle (cf. A103999).
Tsumura shows that, for prime p, a(p) is composite (contrary to a conjecture of Juricevic). - Charles R Greathouse IV, Apr 13 2010
Except for the first term, positive values of x (or y) satisfying x^2 - 4*x*y + y^2 + 12 = 0. - Colin Barker, Feb 04 2014
Except for the first term, positive values of x (or y) satisfying x^2 - 14*x*y + y^2 + 192 = 0. - Colin Barker, Feb 16 2014
A268281(n) - 1 is a member of this sequence iff A268281(n) is prime. - Frank M Jackson, Feb 27 2016
a(n) gives values of x satisfying 3*x^2 - 4*y^2 = 12; corresponding y values are given by A005320. - Sture Sjöstedt, Dec 19 2017
Middle side lengths of almost-equilateral Heronian triangles. - Wesley Ivan Hurt, May 20 2020
For all elements k of the sequence, 3*(k-2)*(k+2) is a square. - Davide Rotondo, Oct 25 2020

B. C. Berndt, Ramanujan's Notebooks Part IV, Springer-Verlag, see p. 82.
J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p.91.
Michael P. Cohen, Generating Heronian Triangles With Consecutive Integer Sides. Journal of Recreational Mathematics, vol. 30 no. 2 1999-2000 p. 123.
L. E. Dickson, History of The Theory of Numbers, Vol. 2 pp. 197;198;200;201. Chelsea NY.
Charles R. Fleenor, Heronian Triangles with Consecutive Integer Sides, Journal of Recreational Mathematics, Volume 28, no. 2 (1996-7) 113-115.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
R. P. Stanley. Enumerative combinatorics. Vol. 2, volume 62 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 1999.
V. D. To, "Finding All Fleenor-Heronian Triangles", Journal of Recreational Mathematics vol. 32 no.4 2003-4 pp. 298-301 Baywood NY.

T. D. Noe, Table of n, a(n) for n=0..200
P. Bala, Some simple continued fraction expansions for an infinite product, Part 1
R. A. Beauregard and E. R. Suryanarayan, The Brahmagupta Triangles, The College Mathematics Journal 29(1) 13-7 1998 MAA.
Hacène Belbachir, Soumeya Merwa Tebtoub and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Daniel Birmajer, Juan B. Gil and Michael D. Weiner, Linear recurrence sequences with indices in arithmetic progression and their sums, arXiv preprint arXiv:1505.06339 [math.NT], 2015.
K. S. Brown's Mathpages, Some Properties of the Lucas Sequence(2, 4, 14, 52, 194, ...)
H. W. Gould, A triangle with integral sides and area, Fib. Quart., 11 (1973), 27-39.
Tanya Khovanova, Recursive Sequences
E. Keith Lloyd, The Standard Deviation of 1, 2, ..., n: Pell's Equation and Rational Triangles, Math. Gaz. vol 81 (1997), 231-243.
S. Northshield, An Analogue of Stern's Sequence for Z[sqrt(2)], Journal of Integer Sequences, 18 (2015), #15.11.6.
Hideyuki Ohtskua, proposer, Problem B-1351, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 62, No. 3 (2024), p. 258.
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
Jeffrey Shallit, An interesting continued fraction, Math. Mag., 48 (1975), 207-211.
Jeffrey Shallit, An interesting continued fraction, Math. Mag., 48 (1975), 207-211. [Annotated scanned copy]
Yu Tsumura, On compositeness of special types of integers, arXiv:1004.1244 [math.NT], 2010.
Eric Weisstein's World of Mathematics, Heronian Triangle
Wikipedia, Heronian triangle
A. V. Zarelua, On Matrix Analogs of Fermat's Little Theorem, Mathematical Notes, vol. 79, no. 6, 2006, pp. 783-796. Translated from Matematicheskie Zametki, vol. 79, no. 6, 2006, pp. 840-855.
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for linear recurrences with constant coefficients, signature (4,-1).

Cf. A001075, A001353, A001835.
Cf. A011945 (areas), A334277 (perimeters).
Cf. this sequence (middle side lengths), A016064 (smallest side lengths), A335025 (largest side lengths).
Cf. A001570, A002530, A005320, A006051, A048788, A174500, A268281.
Cf. A011943, A102344, A019673, A105531.

a003500 n = a003500_list !! n
a003500_list = 2 : 4 : zipWith (-)
   (map (* 4) $ tail a003500_list) a003500_list
-- Reinhard Zumkeller, Dec 17 2011

I:=[2,4]; [n le 2 select I[n] else 4*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 14 2018

A003500 := proc(n) option remember; if n <= 1 then 2*n+2 else 4*procname(n-1)-procname(n-2); fi;
end proc;

a[0]=2; a[1]=4; a[n_]:= a[n]= 4a[n-1] -a[n-2]; Table[a[n], {n, 0, 23}]
LinearRecurrence[{4,-1},{2,4},30] (* Harvey P. Dale, Aug 20 2011 *)
Table[Round@LucasL[2n, Sqrt[2]], {n, 0, 20}] (* Vladimir Reshetnikov, Sep 15 2016 *)

x='x+O('x^99); Vec(-2*(-1+2*x)/(1-4*x+x^2)) \\ Altug Alkan, Apr 04 2016

[lucas_number2(n,4,1) for n in range(0, 24)] # Zerinvary Lajos, May 14 2009

a(n) = ( 2 + sqrt(3) )^n + ( 2 - sqrt(3) )^n.
a(n) = 2*A001075(n).
G.f.: 2*(1 - 2*x)/(1 - 4*x + x^2). Simon Plouffe in his 1992 dissertation.
a(n) = A001835(n) + A001835(n+1).
a(n) = trace of n-th power of the 2 X 2 matrix [1 2 / 1 3]. - Gary W. Adamson, Jun 30 2003 [corrected by Joerg Arndt, Jun 18 2020]
From the addition formula, a(n+m) = a(n)*a(m) - a(m-n), it is easy to derive multiplication formulas, such as: a(2*n) = (a(n))^2 - 2, a(3*n) = (a(n))^3 - 3*(a(n)), a(4*n) = (a(n))^4 - 4*(a(n))^2 + 2, a(5*n) = (a(n))^5 - 5*(a(n))^3 + 5*(a(n)), a(6*n) = (a(n))^6 - 6*(a(n))^4 + 9*(a(n))^2 - 2, etc. The absolute values of the coefficients in the expansions are given by the triangle A034807. - John Blythe Dobson, Nov 04 2007
a(n) = 2*A001353(n+1) - 4*A001353(n). - R. J. Mathar, Nov 16 2007
From Peter Bala, Jan 06 2013: (Start)
Let F(x) = Product_{n=0..infinity} (1 + x^(4*n + 1))/(1 + x^(4*n + 3)). Let alpha = 2 - sqrt(3). This sequence gives the simple continued fraction expansion of 1 + F(alpha) = 2.24561 99455 06551 88869 ... = 2 + 1/(4 + 1/(14 + 1/(52 + ...))). Cf. A174500.
Also F(-alpha) = 0.74544 81786 39692 68884 ... has the continued fraction representation 1 - 1/(4 - 1/(14 - 1/(52 - ...))) and the simple continued fraction expansion 1/(1 + 1/((4 - 2) + 1/(1 + 1/((14 - 2) + 1/(1 + 1/((52 - 2) + 1/(1 + ...))))))).
F(alpha)*F(-alpha) has the simple continued fraction expansion 1/(1 + 1/((4^2 - 4) + 1/(1 + 1/((14^2 - 4) + 1/(1 + 1/((52^2 - 4) + 1/(1 + ...))))))).
(End)
a(2^n) = A003010(n). - John Blythe Dobson, Mar 10 2014
a(n) = [x^n] ( (1 + 4*x + sqrt(1 + 8*x + 12*x^2))/2 )^n for n >= 1. - Peter Bala, Jun 23 2015
E.g.f.: 2*exp(2*x)*cosh(sqrt(3)*x). - Ilya Gutkovskiy, Apr 27 2016
a(n) = Sum_{k=0..floor(n/2)} (-1)^k*n*(n - k - 1)!/(k!*(n - 2*k)!)*4^(n - 2*k) for n >= 1. - Peter Luschny, May 10 2016
From Peter Bala, Oct 15 2019: (Start)
a(n) = trace(M^n), where M is the 2 X 2 matrix [0, 1; -1, 4].
Consequently the Gauss congruences hold: a(n*p^k) = a(n*p^(k-1)) ( mod p^k ) for all prime p and positive integers n and k. See Zarelua and also Stanley (Ch. 5, Ex. 5.2(a) and its solution).
2*Sum_{n >= 1} 1/( a(n) - 6/a(n) ) = 1.
6*Sum_{n >= 1} (-1)^(n+1)/( a(n) + 2/a(n) ) = 1.
8*Sum_{n >= 1} 1/( a(n) + 24/(a(n) - 12/(a(n))) ) = 1.
8*Sum_{n >= 1} (-1)^(n+1)/( a(n) + 8/(a(n) + 4/(a(n))) ) = 1.
Series acceleration formulas for sums of reciprocals:
Sum_{n >= 1} 1/a(n) = 1/2 - 6*Sum_{n >= 1} 1/(a(n)*(a(n)^2 - 6)),
Sum_{n >= 1} 1/a(n) = 1/8 + 24*Sum_{n >= 1} 1/(a(n)*(a(n)^2 + 12)),
Sum_{n >= 1} (-1)^(n+1)/a(n) = 1/6 + 2*Sum_{n >= 1} (-1)^(n+1)/(a(n)*(a(n)^2 + 2)) and
Sum_{n >= 1} (-1)^(n+1)/a(n) = 1/8 + 8*Sum_{n >= 1} (-1)^(n+1)/(a(n)*(a(n)^2 + 12)).
Sum_{n >= 1} 1/a(n) = ( theta_3(2-sqrt(3))^2 - 1 )/4 = 0.34770 07561 66992 06261 .... See Borwein and Borwein, Proposition 3.5 (i), p.91.
Sum_{n >= 1} (-1)^(n+1)/a(n) = ( 1 - theta_3(sqrt(3)-2)^2 )/4. Cf. A003499 and A153415. (End)
a(n) = tan(Pi/12)^n + tan(5*Pi/12)^n. - Greg Dresden, Oct 01 2020
From Wolfdieter Lang, Sep 06 2021: (Start)
a(n) = S(n, 4) - S(n-2, 4) = 2*T(n, 2), for n >= 0, with S and T Chebyshev polynomials, with S(-1, x) = 0 and S(-2, x) = -1. S(n, 4) = A001353(n+1), for n >= -1, and T(n, 2) = A001075(n).
a(2*k) = A067902(k), a(2*k+1) = 4*A001570(k+1), for k >= 0. (End)
a(n) = sqrt(2 + 2*A011943(n+1)) = sqrt(2 + 2*A102344(n+1)), n>0. - Ralf Steiner, Sep 23 2021
Sum_{n>=1} arctan(3/a(n)^2) = Pi/6 - arctan(1/3) = A019673 - A105531 (Ohtskua, 2024). - Amiram Eldar, Aug 29 2024

More terms from James Sellers, May 03 2000

Additional comments from Lekraj Beedassy, Feb 14 2002

The fundamental solution of the Pell equation x^2 - 3*(n*y)^2 = 1 is the smallest solution of x^2 - 3*y^2 = 1 satisfying y == 0 (mod n).

For primes p > 2, 2^p-1 is a Mersenne prime if and only if a(2^p-1) = 2^(p-1). For example, a(7) = 4, a(31) = 16, a(127) = 64, but a(2047) = 495 < 1024. - Jianing Song, Jun 02 2022

Further terms are 3049, 7451, 7487, 18869.

All terms must be primes. A117808 lists the corresponding primes floor(A001075(k)/2). Actually, the "floor" here and in the definition is only needed for the initial term 2, since no other even (thus composite) k can be in the sequence, and A001075(k) is even for odd k.

For n >= 2, a(n) == 7 mod 30.
Terms in this sequence have the form A001075(2^k) [see my third comment on A001075]: a(1) = A001075(2^0) = A001075(1), a(2) = A001075(2^1) = A001075(2), a(3) = A001075(2^2) = A001075(4), and a(4) = A001075(2^4) = A001075(16). Are there more terms and, if so, will a(5) = A001075(2^16) = A001075(65536)?
The following terms are not prime and, thus, not in the sequence: A001075(m), for m = 8, 32, 64, 128, 256, 512, 1024. So, a(5) > 2.3619*10^585.
a(5), if it exists, is at least A001075(2^23) and hence has more than four million decimal digits. - Charles R Greathouse IV, Nov 10 2016

cp's OEIS Frontend

A298211 Smallest n such that A001353(a(n)) == 0 (mod n), i.e., x=A001075(a(n)) and y=A001353(a(n)) is the fundamental solution of the Pell equation x^2 - 3*(n*y)^2 = 1.

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A198196 Numbers k such that [V(4,1,k)/4] = floor(cosh(k log(sqrt(3)+2))/2) = A004526(A001075(k)) is a prime.

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A277434 Primes in A001075.

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

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A001541 a(0) = 1, a(1) = 3; for n > 1, a(n) = 6*a(n-1) - a(n-2).

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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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A298211 Smallest n such that A001353(a(n)) == 0 (mod n), i.e., x=A001075(a(n)) and y=A001353(a(n)) is the fundamental solution of the Pell equation x^2 - 3(ny)^2 = 1.

A026150 a(0) = a(1) = 1; a(n+2) = 2a(n+1) + 2a(n).