A016064 -id:A016064 - 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 + ...

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)
G. Everest, A. van der Poorten, I. Shparlinski and T. Ward, Recurrence Sequences, Amer. Math. Soc., 2003; p. 163.
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.
J. D. E. Konhauser et al., Which Way Did the Bicycle Go?, MAA 1996, p. 104.
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).

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.
Christian Aebi and Grant Cairns, Lattice Equable Parallelograms, arXiv:2006.07566 [math.NT], 2020.
Christian Aebi and Grant Cairns, Equable Parallelograms on the Eisenstein Lattice, arXiv:2401.08827 [math.CO], 2024. See p. 14.
W. K. Alt, Enumeration of Domino Tilings on the Projective Grid Graph, A Thesis Presented to The Division of Mathematics and Natural Sciences, Reed College, May 2013.
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.
Francesca Arici and Jens Kaad, Gysin sequences and SU(2)-symmetries of C*-algebras, arXiv:2012.11186 [math.OA], 2020.
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.
J. Austin and L. Schneider, Generalized Fibonacci sequences in Pythagorean triple preserving sequences, Fib. Q., 58:4 (2020), 340-350.
Paul Barry, Notes on Riordan arrays and lattice paths, arXiv:2504.09719 [math.CO], 2025. See pp. 15, 29.
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.
D. Birmajer, J. B. Gil, and M. D. Weiner, On the Enumeration of Restricted Words over a Finite Alphabet, J. Int. Seq. 19 (2016) # 16.1.3, example 12.
K. S. Bhanu and M. N. Deshpande, Integral triangles with 120° angle Mathematics Spectrum, 45 (3) (2012/2013), 126-128.
Latham Boyle and Paul J. Steinhardt, Self-Similar One-Dimensional Quasilattices, arXiv preprint arXiv:1608.08220 [math-ph], 2016.
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 48, 56.
Fabrizio Canfora, Maxim Kurkov, Luigi Rosa, and Patrizia Vitale, The Gribov problem in Noncommutative QED, arXiv preprint arXiv:1505.06342 [hep-th], 2015.
Niccolò Castronuovo, On the number of fixed points of the map gamma, arXiv:2102.02739 [math.NT], 2021. Mentions this sequence.
Z. Cinkir, Effective Resistances, Kirchhoff index and Admissible Invariants of Ladder Graphs, arXiv preprint arXiv:1503.06353 [math.CO], 2015.
J. B. Cosgrave and K. Dilcher, A role for generalized Fermat numbers, Math. Comp. 86 (2017), 899-933 (see also paper #10).
M. N. Deshpande, One Interesting Family of Diophantine Triplets, International Journal of Mathematical Education In Science and Technology, Vol. 33 (No. 2, Mar-Apr), 2002.
M. N. Deshpande, Hansruedi Widmer and Zachary Franco, Simultaneous Squares from Arithmetic Progressions: 10622, The American Mathematical Monthly Vol. 106, No. 9 (Nov., 1999), 867-868.
Tomislav Doslic, Planar polycyclic graphs and their Tutte polynomials, Journal of Mathematical Chemistry, Volume 51, Issue 6, 2013, pp. 1599-1607.
G. Dresden and Y. Li, Periodic Weighted Sums of Binomial Coefficients, arXiv:2210.04322 [math.NT], 2022.
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242.
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
Felix Flicker, Time quasilattices in dissipative dynamical systems, arXiv:1707.09371 [nlin.CD], 2017. Also SciPost Phys. 5, 001 (2018).
D. Fortin, B-spline Toeplitz Inverse Under Corner Perturbations, International Journal of Pure and Applied Mathematics, Volume 77, No. 1, 2012, 107-118. - From _N. J. A. Sloane_, Oct 22 2012
Dale Gerdemann, Fractal images from (4, -1) recursion, YouTube, Oct 27 2014.
Juan B. Gil and Jessica A. Tomasko, Fibonacci colored compositions and applications, arXiv:2108.06462 [math.CO], 2021.
Andrew Granville and Zhi-Wei Sun, Values of Bernoulli polynomials, Pacific J. Math. 172 (1996), 117-137, at p. 119.
T. N. E. Greville, Table for third-degree spline interpolations with equally spaced arguments, Math. Comp., 24 (1970), 179-183.
Y.-H. Guo, n-Colour even self-inverse compositions, Proc. Indian Acad. Sci. (Math. Sci.), 120 (2010), 27-33.
B. Hopkins, Spotted tilings and n-color compositions, INTEGERS 12B (2012/2013), #A6.
A. F. Horadam, Special properties of the sequence W_n(a,b; p,q), Fib. Quart., 5.5 (1967), 424-434. Case a=0,b=1; p=4, q=-1.
W. D. Hoskins, Table for third-degree spline interpolation using equi-spaced knots, Math. Comp., 25 (1971), 797-801.
Milan Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
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.
Germain Kreweras, Complexite et circuits Euleriens dans les sommes tensorielles de graphes, J. Combin. Theory, B 24 (1978), 202-212.
Pablo Lam-Estrada, Myriam Rosalía Maldonado-Ramírez, José Luis López-Bonilla, Fausto Jarquín-Zárate, The sequences of Fibonacci and Lucas for each real quadratic fields Q(Sqrt(d)), arXiv:1904.13002 [math.NT], 2019.
Wolfdieter Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38,5 (2000) 408-419; Eq.(44), lhs, m=6.
Ioana-Claudia Lazăr, Lucas sequences in t-uniform simplicial complexes, arXiv:1904.06555 [math.GR], 2019.
Hojoo Lee, Problems in elementary number theory Problem I 18.
E. Keith Lloyd, The Standard Deviation of 1, 2,..., n: Pell's Equation and Rational Triangles, Math. Gaz. vol 81 (1997), 231-243.
Dino Lorenzini and Z. Xiang, Integral points on variable separated curves, Preprint 2016.
Valcho Milchev and Tsvetelina Karamfilova, Domino tiling in grid - new dependence, arXiv:1707.09741 [math.HO], 2017.
László Németh, Trees on hyperbolic honeycombs, arXiv:1510.08311 [math.CO], 2015.
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.
Ariel D. Procaccia and Jamie Tucker-Foltz, Compact Redistricting Plans Have Many Spanning Trees, Harvard Univ. (2021).
P. Raff, Analysis of the Number of Spanning Trees of K_2 x P_n. Contains sequence, recurrence, generating function, and more. [From _Paul Raff_, Mar 06 2009]
Franck Ramaharo, The bracket polynomial of the Celtic link shadow CK_4^(2n), arXiv:2508.10410 [math.GT], 2025. See p. 6.
Ryan Stees, Sequences of Spiral Knot Determinants, Senior Honors Projects, Paper 84, James Madison Univ., May 2016.
D. P. Walsh, Counting n x 2 Simple Rectangular Mazes
F. V. Waugh and M. W. Maxfield, Side-and-diagonal numbers, Math. Mag., 40 (1967), 74-83.
Eric Weisstein's World of Mathematics, Ladder Graph
Eric Weisstein's World of Mathematics, Spanning Tree
Jianqiang Zhao, Finite Multiple zeta Values and Finite Euler Sums, arXiv preprint arXiv:1507.04917 [math.NT], 2015.
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)

A001075 a(0) = 1, a(1) = 2, a(n) = 4*a(n-1) - a(n-2).

Original entry on oeis.org

1, 2, 7, 26, 97, 362, 1351, 5042, 18817, 70226, 262087, 978122, 3650401, 13623482, 50843527, 189750626, 708158977, 2642885282, 9863382151, 36810643322, 137379191137, 512706121226, 1913445293767, 7141075053842, 26650854921601, 99462344632562, 371198523608647

Offset: 0

N. J. A. Sloane

Chebyshev's T(n,x) polynomials evaluated at x=2.
x = 2^n - 1 is prime if and only if x divides a(2^(n-2)).
Any k in the sequence is succeeded by 2*k + sqrt{3*(k^2 - 1)}. - Lekraj Beedassy, Jun 28 2002
For all elements x of the sequence, 12*x^2 - 12 is a square. Lim_{n -> infinity} a(n)/a(n-1) = 2 + sqrt(3) = (4 + sqrt(12))/2 which preserves the kinship with the equation "12*x^2 - 12 is a square" where the initial "12" ends up appearing as a square root. - Gregory V. Richardson, Oct 10 2002
This sequence gives the values of x in solutions of the Diophantine equation x^2 - 3*y^2 = 1; the corresponding values of y are in A001353. The solution ratios a(n)/A001353(n) are obtained as convergents of the continued fraction expansion of sqrt(3): either as successive convergents of [2;-4] or as odd convergents of [1;1,2]. - Lekraj Beedassy, Sep 19 2003 [edited by Jon E. Schoenfield, May 04 2014]
a(n) is half the central value in a list of three consecutive integers, the lengths of the sides of a triangle with integer sides and area. - Eugene McDonnell (eemcd(AT)mac.com), Oct 19 2003
a(3+6*k) - 1 and a(3+6*k) + 1 are consecutive odd powerful numbers. See A076445. - T. D. Noe, May 04 2006
The intermediate convergents to 3^(1/2), beginning with 3/2, 12/7, 45/26, 168/97, comprise a strictly increasing sequence; essentially, numerators=A005320, denominators=A001075. - Clark Kimberling, Aug 27 2008
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
a(n+1) is the Hankel transform of A000108(n) + A000984(n) = (n+2)*Catalan(n). - Paul Barry, Aug 11 2009
Also, numbers such that floor(a(n)^2/3) is a square: base 3 analog of A031149, A204502, A204514, A204516, A204518, A204520, A004275, A001541. - M. F. Hasler, Jan 15 2012
Pisano period lengths:  1, 2, 2, 4, 3, 2, 8, 4, 6, 6, 10, 4, 12, 8, 6, 8, 18, 6, 5, 12, ... - R. J. Mathar, Aug 10 2012
Except for the first term, positive values of x (or y) satisfying x^2 - 4*x*y + y^2 + 3 = 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 + 48 = 0. - Colin Barker, Feb 10 2014
A triangle with row sums generating the sequence can be constructed by taking the production matrix M. Take powers of M, extracting the top rows.
M =
  1, 1, 0, 0, 0, 0, ...
  2, 0, 3, 0, 0, 0, ...
  2, 0, 0, 3, 0, 0, ...
  2, 0, 0, 0, 3, 0, ...
  2, 0, 0, 0, 0, 3, ...
  ...
The triangle generated from M is:
   1,
   1,  1,
   3,  1, 3,
  11,  3, 3, 9,
  41, 11, 9, 9, 27,
   ...
The left border is A001835 and row sums are (1, 2, 7, 26, 97, ...). - Gary W. Adamson, Jul 25 2016
Even-indexed terms are odd while odd-indexed terms are even. Indeed, a(2*n) = 2*(a(n))^2 - 1 and a(2*n+1) = 2*a(n)*a(n+1) - 2. - Timothy L. Tiffin, Oct 11 2016
For each n, a(0) divides a(n), a(1) divides a(2n+1), a(2) divides a(4*n+2), a(3) divides a(6*n+3), a(4) divides a(8*n+4), a(5) divides a(10n+5), and so on. Thus, a(k) divides a((2*n+1)*k) for each k > 0 and n >= 0. A proof of this can be found in Bhargava-Kedlaya-Ng's first solution to Problem A2 of the 76th Putnam Mathematical Competition. Links to the exam and its solutions can be found below. - Timothy L. Tiffin, Oct 12 2016
From Timothy L. Tiffin, Oct 21 2016: (Start)
If any term a(n) is a prime number, then its index n will be a power of 2. This is a consequence of the results given in the previous two comments. See A277434 for those prime terms.
a(2n) == 1 (mod 6) and a(2*n+1) == 2 (mod 6). Consequently, each odd prime factor of a(n) will be congruent to 1 modulo 6 and, thus, found in A002476.
a(n) == 1 (mod 10) if n == 0 (mod 6), a(n) == 2 (mod 10) if n == {1,-1} (mod 6), a(n) == 7 (mod 10) if n == {2,-2} (mod 6), and a(n) == 6 (mod 10) if n == 3 (mod 6). So, the rightmost digits of a(n) form a repeating cycle of length 6: 1, 2, 7, 6, 7, 2. (End)
a(A298211(n)) = A002350(3*n^2). - A.H.M. Smeets, Jan 25 2018
(2 + sqrt(3))^n = a(n) + A001353(n)*sqrt(3), n >= 0; integers in the quadratic number field Q(sqrt(3)). - Wolfdieter Lang, Feb 16 2018
Yong Hao Ng has shown that for any n, a(n) is coprime with any member of A001834 and with any member of A001835. - René Gy, Feb 26 2018
Positive numbers k such that 3*(k-1)*(k+1) is a square. - Davide Rotondo, Oct 25 2020
a(n)*a(n+1)-1 = a(2*n+1)/2 = A001570(n) divides both a(n)^6+1 and a(n+1)^6+1. In other words, for k = a(2*n+1)/2, (k+1)^6 has divisors congruent to -1 modulo k (cf. A350916). - Max Alekseyev, Jan 23 2022

			2^6 - 1 = 63 does not divide a(2^4) = 708158977, therefore 63 is composite. 2^5 - 1 = 31 divides a(2^3) = 18817, therefore 31 is prime.
G.f. = 1 + 2*x + 7*x^2 + 26*x^3 + 97*x^4 + 362*x^5 + 1351*x^6 + 5042*x^7 + ...

Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.
Eugene McDonnell, "Heron's Rule and Integer-Area Triangles", Vector 12.3 (January 1996) pp. 133-142.
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.

Indranil Ghosh, Table of n, a(n) for n = 0..1745 (terms 0..200 from T. D. Noe)
Christian Aebi and Grant Cairns, Lattice Equable Parallelograms, arXiv:2006.07566 [math.NT], 2020.
Christian Aebi and Grant Cairns, Less than Equable Triangles on the Eisenstein lattice, arXiv:2312.10866 [math.CO], 2023.
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.
C. Banderier and D. Merlini, Lattice paths with an infinite set of jumps, FPSAC02, Melbourne, 2002.
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 [sic], Nouvelle Correspondance Mathématique, 4 (1878), 337-343.
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 30, 43, 56.
Chris Caldwell, Primality Proving, Arndt's theorem.
J. B. Cosgrave and K. Dilcher, A role for generalized Fermat numbers, Math. Comp., 2016.
G. Dresden and Y. Li, Periodic Weighted Sums of Binomial Coefficients, arXiv:2210.04322 [math.NT], 2022.
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242.
Margherita Maria Ferrari and Norma Zagaglia Salvi, Aperiodic Compositions and Classical Integer Sequences, Journal of Integer Sequences, Vol. 20 (2017), Article 17.8.8.
R. K. Guy, Letter to N. J. A. Sloane concerning A001075, A011943, A094347 [Scanned and annotated letter, included with permission]
Kiran S. Kedlaya, The 76th William Lowell Putnam Mathematical Competition, Dec 05 2015.
Kiran S. Kedlaya, Solutions to the 76th William Lowell Putnam Mathematical Competition, Dec 05 2015.
Tanya Khovanova, Recursive Sequences
Clark Kimberling, Best lower and upper approximates to irrational numbers, Elemente der Mathematik, 52 (1997) 122-126.
Pablo Lam-Estrada, Myriam Rosalía Maldonado-Ramírez, José Luis López-Bonilla, and Fausto Jarquín-Zárate, The sequences of Fibonacci and Lucas for each real quadratic fields Q(Sqrt(d)), arXiv:1904.13002 [math.NT], 2019.
Eugene McDonnell, Heron's Rule and Integer-Area Triangles, At Play With J, 2010.
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?, Mathematics Stack Exchange.
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
F. V. Waugh and M. W. Maxfield, Side-and-diagonal numbers, Math. Mag., 40 (1967), 74-83.
Index entries for sequences related to Chebyshev polynomials.
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (4,-1).

Bisections are A011943 and A094347.

Cf. A001353, A065918, A071954, A001571, A001834, A003500, A016064, A082840, A026150, A277434.

a001075 n = a001075_list !! n
a001075_list =
   1 : 2 : zipWith (-) (map (4 *) $ tail a001075_list) a001075_list
-- Reinhard Zumkeller, Aug 11 2011

I:=[1, 2]; [n le 2 select I[n] else 4*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 19 2017

A001075 := proc(n)
    orthopoly[T](n,2) ;
end proc:
seq(A001075(n),n=0..30) ; # R. J. Mathar, Apr 14 2018

Table[ Ceiling[(1/2)*(2 + Sqrt[3])^n], {n, 0, 24}]
CoefficientList[Series[(1-2*x) / (1-4*x+x^2), {x, 0, 24}], x] (* Jean-François Alcover, Dec 21 2011, after Simon Plouffe *)
LinearRecurrence[{4,-1},{1,2},30] (* Harvey P. Dale, Aug 22 2015 *)
Round@Table[LucasL[2n, Sqrt[2]]/2, {n, 0, 20}] (* Vladimir Reshetnikov, Sep 15 2016 *)
ChebyshevT[Range[0, 20], 2] (* Eric W. Weisstein, May 26 2017 *)
a[ n_] := LucasL[2*n, x]/2 /. x->Sqrt[2]; (* Michael Somos, Sep 05 2022 *)

{a(n) = subst(poltchebi(abs(n)), x, 2)};

{a(n) = real((2 + quadgen(12))^abs(n))};

{a(n) = polsym(1 - 4*x + x^2, abs(n))[1 + abs(n)]/2};

a(n)=polchebyshev(n,1,2) \\ Charles R Greathouse IV, Nov 07 2016

my(x='x+O('x^30)); Vec((1-2*x)/(1-4*x+x^2)) \\ G. C. Greubel, Dec 19 2017

[lucas_number2(n,4,1)/2 for n in range(0, 25)] # Zerinvary Lajos, May 14 2009

def a(n):
    Q = QuadraticField(3, 't')
    u = Q.units()[0]
    return (u^n).lift().coeffs()[0]  # Ralf Stephan, Jun 19 2014

G.f.: (1 - 2*x)/(1 - 4*x + x^2). - Simon Plouffe in his 1992 dissertation
E.g.f.: exp(2*x)*cosh(sqrt(3)*x).
a(n) = 4*a(n-1) - a(n-2) = a(-n).
a(n) = (S(n, 4) - S(n-2, 4))/2 = T(n, 2), with S(n, x) := U(n, x/2), S(-1, x) := 0, S(-2, x) := -1. U, resp. T, are Chebyshev's polynomials of the second, resp. first, kind. S(n-1, 4) = A001353(n), n >= 0. See A049310 and A053120.
a(n) = A001353(n+2) - 2*A001353(n+1).
a(n) = sqrt(1 + 3*A001353(n)) (cf. Richardson comment, Oct 10 2002).
a(n) = 2^(-n)*Sum_{k>=0} binomial(2*n, 2*k)*3^k = 2^(-n)*Sum_{k>=0} A086645(n, k)*3^k. - Philippe Deléham, Mar 01 2004
a(n) = ((2 + sqrt(3))^n + (2 - sqrt(3))^n)/2; a(n) = ceiling((1/2)*(2 + sqrt(3))^(n)).
a(n) = cosh(n * log(2 + sqrt(3))).
a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2*k)*2^(n-2*k)*3^k. - Paul Barry, May 08 2003
a(n+2) = 2*a(n+1) + 3*Sum_{k>=0} a(n-k)*2^k. - Philippe Deléham, Mar 03 2004
a(n) = 2*a(n-1) + 3*A001353(n-1). - Lekraj Beedassy, Jul 21 2006
a(n) = left term of M^n * [1,0] where M = the 2 X 2 matrix [2,3; 1,2]. Right term = A001353(n). Example: a(4) = 97 since M^4 * [1,0] = [A001075(4), A001353(4)] = [97, 56]. - Gary W. Adamson, Dec 27 2006
Binomial transform of A026150: (1, 1, 4, 10, 28, 76, ...). - Gary W. Adamson, Nov 23 2007
First differences of A001571. - N. J. A. Sloane, Nov 03 2009
Sequence satisfies -3 = f(a(n), a(n+1)) where f(u, v) = u^2 + v^2 - 4*u*v. - Michael Somos, Sep 19 2008
a(n) = Sum_{k=0..n} A201730(n,k)*2^k. - Philippe Deléham, Dec 06 2011
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(3*k - 4)/(x*(3*k - 1) - 2/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 28 2013
a(n) = Sum_{k=0..n} A238731(n,k). - Philippe Deléham, Mar 05 2014
a(n) = (-1)^n*(A125905(n) + 2*A125905(n-1)), n > 0. - Franck Maminirina Ramaharo, Nov 11 2018
a(n) = (tan(Pi/12)^n + tan(5*Pi/12)^n)/2. - Greg Dresden, Oct 01 2020
From Peter Bala, Aug 17 2022: (Start)
a(n) = (1/2)^n * [x^n] ( 4*x + sqrt(1 + 12*x^2) )^n.
The g.f. A(x) satisfies A(2*x) = 1 + x*B'(x)/B(x), where B(x) = 1/sqrt(1 - 8*x + 4*x^2) is the g.f. of A069835.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p >= 3 and positive integers n and k.
Sum_{n >= 1} 1/(a(n) - (3/2)/a(n)) = 1.
Sum_{n >= 1} (-1)^(n+1)/(a(n) + (1/2)/a(n)) = 1/3.
Sum_{n >= 1} 1/(a(n)^2 - 3/2) = 1 - 1/sqrt(3). (End)
a(n) = binomial(2*n, n) + 2*Sum_{k > 0} binomial(2*n, n+2*k)*cos(k*Pi/3). - Greg Dresden, Oct 11 2022
2*a(n) + 2^n = 3*Sum_{k=-n..n} (-1)^k*binomial(2*n, n+6*k). - Greg Dresden, Feb 07 2023

More terms from James Sellers, Jul 10 2000

Chebyshev comments from Wolfdieter Lang, Oct 31 2002

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

A103974 Smaller sides (a) in (a,a,a+1)-integer triangle with integer area.

Original entry on oeis.org

1, 5, 65, 901, 12545, 174725, 2433601, 33895685, 472105985, 6575588101, 91586127425, 1275630195845, 17767236614401, 247465682405765, 3446752317066305, 48007066756522501, 668652182274248705

Offset: 1

Zak Seidov, Feb 23 2005

Corresponding areas are: 0, 12, 1848, 351780, 68149872, 13219419708, 2564481115560 (see A104009).
What is the next term? Is the sequence finite? The possible last two digits of "a" are (it may help in searching for more terms): {01, 05, 09, 15, 19, 25, 29, 33, 35, 39, 45, 49, 51, 55, 59, 65, 69, 75, 79, 83, 85, 89, 95, 99}.
Equivalently, positive integers a such that 3/16*a^4 + 1/4*a^3 - 1/8*a^2 - 1/4*a - 1/16 is a square (A000290), a direct result of Heron's formula. Conjecture: lim_{n->oo} a(n+1)/a(n) = 7 + 4*sqrt(3) (= 7 + A010502). - Rick L. Shepherd, Sep 04 2005
Values x^2 + y^2, where the pair (x, y) solves for x^2 - 3y^2=1, i.e., a(n)= (A001075(n))^2 + (A001353(n))^2 = A055793(n) + A098301(n). - Lekraj Beedassy, Jul 13 2006
Floretion Algebra Multiplication Program, FAMP Code: 1lestes[ 3'i - 2'j + 'k + 3i' - 2j' + k' - 4'ii' - 3'jj' + 4'kk' - 'ij' - 'ji' + 3'jk' + 3'kj' + 4e ]

Christian Aebi, and Grant Cairns, Lattice Equable Parallelograms, arXiv:2006.07566 [math.NT], 2020.
Project Euler, Problem 94: Almost Equilateral Triangles.
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Cf. A011922, A007655, A001353, A102341, A103975, A016064, A011945, A010502 (4*sqrt(3)), A000290 (square numbers), A350916.

A:=rsolve({-A(n+3)+15*A(n+2)-15*A(n+1)+A(n), A(0) = 1, A(1) = 5, A(2)=65}, A(n), makeproc); # Mihailovs

f[n_] := Simplify[((2 + Sqrt[3])^(2n) + (2 - Sqrt[3])^(2n) + 1)/3]; Table[ f[n], {n, 0, 16}] (* Or *)
a[1] = 1; a[2] = 5; a[3] = 65; a[n_] := a[n] = 15a[n - 1] - 15a[n - 2] + a[n - 3]; Table[ a[n], {n, 17}] (* Or *)
CoefficientList[ Series[(1 - 10x + 5x^2)/(1 - 15x + 15x^2 - x^3), {x, 0, 16}], x] (* Or *)
Range[0, 16]! CoefficientList[ Simplify[ Series[(E^x + E^((7 + 4Sqrt[3])x) + E^((7 - 4Sqrt[3])x))/3, {x, 0, 16}]], x] (* Robert G. Wilson v, Mar 24 2005 *)

for(a=1,10^6, b=a; c=a+1; s=(a+b+c)/2; if(issquare(s*(s-a)*(s-b)*(s-c)), print1(a,","))) /* Uses Heron's formula */ \\ Rick L. Shepherd, Sep 04 2005

Composite of comments from Alec Mihailovs (alec(AT)mihailovs.com) and David Terr, Mar 07 2005: (Start)
"a(n)^2 = A011922(n)^2 + (4*A007655(n))^2, so that A011922(n) = 1/2 base of triangles, A007655(n) = 1/4 height of triangles (conjectured by Paul Hanna).
Area is (a+1)/4*sqrt((3*a+1)*(a-1)). If a is even, the numerator is odd and the area is not an integer. That means a=2*k-1. In this case, Area=k*sqrt((3*k-1)*(k-1)).
Solving equation (3*k-1)*(k-1)=y^2, we get k=(2+sqrt(1+3*y^2))/3. That means that 1+3*y^2=x^2 with integer x and y. This is a Pell equation, all solutions of which have the form x=((2+sqrt(3))^n+(2-sqrt(3))^n)/2, y=((2+sqrt(3))^n-(2-sqrt(3))^n)/(2*sqrt(3)). Therefore k=(x+2)/3 is an integer only for even n. Then a=2*k-1=(2*x+1)/3 with even n. Q.E.D.
a(n)=(1/3)*((2+sqrt(3))^(2*n-2)+(2-sqrt(3))^(2*n-2)+1).
Recurrence: a(n+3)=15*a(n+2)-15*a(n+1)+a(n), a(0)=1, a(1)=5, a(2)=65.
G.f.: x*(1-10*x+5*x^2)/(1-15*x+15*x^2-x^3).
E.g.f.: 1/3*(exp(x)+exp((7+4*sqrt(3))*x)+exp((7-4*sqrt(3))*x)).
a(n) = 4U(n)^2 + 1, where U(1) = 0, U(2)=1 and U(n+1) = 4U(n) - U(n-1) for n>1. (U(n), V(n)) is the n-th solution to Pell's equation 3U(n)^2 + 1 = V(n)^2. (U(n) is the sequence A001353.)" (End)
a(n+1) = A098301(n+1) + A055793(n+2) - Creighton Dement, Apr 18 2005
a(n) = floor((7+4*sqrt(3))*a(n-1))-4, n>=3. - Rick L. Shepherd, Sep 04 2005
a(n)= [1+14*A007655(n+2)-194*A007655(n+1)]/3. - R. J. Mathar, Nov 16 2007
For n>=3, a(n) = 14*a(n-1) - a(n-2) - 4. It is one of 10 second-order linear recurrence sequences satisfying (a(n)*a(n-1)-1) * (a(n)*a(n+1)-1) = (a(n)+1)^4 and together forming A350916. - Max Alekseyev, Jan 22 2022

More terms from Creighton Dement, Apr 18 2005

Edited by Max Alekseyev, Jan 22 2022

A097933 Primes p that divide 3^((p-1)/2) - 1.

Original entry on oeis.org

11, 13, 23, 37, 47, 59, 61, 71, 73, 83, 97, 107, 109, 131, 157, 167, 179, 181, 191, 193, 227, 229, 239, 241, 251, 263, 277, 311, 313, 337, 347, 349, 359, 373, 383, 397, 409, 419, 421, 431, 433, 443, 457, 467, 479, 491, 503, 541, 563, 577, 587, 599, 601, 613

Offset: 1

Cino Hilliard, Sep 04 2004

nonn

Rational primes that decompose in the field Q[sqrt(3)]. - N. J. A. Sloane, Dec 26 2017
For all primes p > 2 and integers gcd(x, y, p) = 1, x^((p-1)/2) +- y^((p-1)/2) is divisible by p. This is because (x^((p-1)/2) - y^((p-1)/2))(x^((p-1)/2) + y^((p-1)/2)) = x^(p-1) - y^(p-1) is divisible by p according to Fermat's Little Theorem (FLT). This sequence lists p that divides 3^((p-1)/2) - 1^((p-1)/2), and A003630 lists the '+' case.
Apart from initial terms, this and A038874 are the same. - N. J. A. Sloane, May 31 2009
Primes in A091998. - Reinhard Zumkeller, Jan 07 2012
Also, primes congruent to 1 or 11 (mod 12). - Vincenzo Librandi, Mar 23 2013
Conjecture: Let r(n) = (a(n) - 1)/(a(n) + 1) if a(n) mod 4 = 1, (a(n) + 1)/(a(n) - 1) otherwise; then Product_{n>=1} r(n) = (6/5) * (6/7) * (12/11) * (18/19) * ... = 2/sqrt(3). - Dimitris Valianatos, Mar 27 2017
Primes p such that Kronecker(12,p) = +1 (12 is the discriminant of Q[sqrt(3)]), that is, odd primes that have 3 as a quadratic residue. - Jianing Song, Nov 21 2018
Comment from Richard R. Forberg, Feb 07 2023: (Start)
Conjecture: These are the exclusive prime factors of the set of integers d > 1 such that there exist primitive Heronian triangles with sides {b, b+d, b+2d} for one or more integers b.
Also b is always > d. For d=11 the b values begin {15, 17, 65, 75, 267, 305, 1025, ...}. For d=1 (not prime, thus not listed) the b values are given by A016064. (End)

			For p = 5, 3^2 - 1 = 8 <> 3*k for any integer k, so 5 is not in this sequence.
For p = 11, 3^5 - 1 = 242 = 11*22, so 11 is in this sequence.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Index to sequences related to decomposition of primes in quadratic fields

Cf. A003630, A010051, A038874, A091998.

a097933 n = a097933_list !! (n-1)
a097933_list = [x | x <- a091998_list, a010051 x == 1]
-- Reinhard Zumkeller, Jan 07 2012

[p: p in PrimesUpTo(1000) | p mod 24 in [1, 11, 13, 23]]; // Vincenzo Librandi, Mar 23 2013

Select[Prime[Range[300]], MemberQ[{1, 11, 13, 23}, Mod[#, 24]]&] (* Vincenzo Librandi, Mar 23 2013 *)
Select[Prime[Range[2,200]],PowerMod[3,(#-1)/2,#]==1&] (* Harvey P. Dale, Jun 02 2020 *)

/* s = +-1, d=diff */ ptopm1d2(n,x,d,s) = { forprime(p=3,n,p2=(p-1)/2; y=x^p2 + s*(x-d)^p2; if(y%p==0,print1(p","))) }

{a(n)= local(m, c); if(n<1, 0, c=0; m=0; while( cMichael Somos, Aug 28 2006 */

A072221 a(n) = 6*a(n-1) - a(n-2) + 2, with a(0)=1, a(1)=4.

Original entry on oeis.org

1, 4, 25, 148, 865, 5044, 29401, 171364, 998785, 5821348, 33929305, 197754484, 1152597601, 6717831124, 39154389145, 228208503748, 1330096633345, 7752371296324, 45184131144601, 263352415571284, 1534930362283105, 8946229758127348, 52142448186480985

Offset: 0

Lekraj Beedassy, Jul 04 2002

The product of three consecutive triangular numbers with middle term A000217(m) where m is in this sequence is a square.
k is in this sequence iff the triangle with sides 3,k,k+1 has integer area. Equivalently, numbers k such that 2*(k+2)*(k-1) is a square. - James R. Buddenhagen, Oct 19 2008
Triangular numbers that are equal to a square plus one have this sequence as indices. For example, the 25th triangular number is 25*26/2 = 325 = 18^2 + 1. - Tanya Khovanova and Alexey Radul, Aug 08 2009
The triangle with sides 3, a(n), a(n)+1 has area A075848(n) if n>=0. - Michael Somos, Dec 25 2018
Compare with A016064 for integers m with triangles with sides 4, m, m+2 and integer area. - Michael Somos, May 11 2019

			From _Michael Somos_, Dec 25 2018: (Start)
For n=1, the triangle (3, 4, 5) has area 6 = A075848(1).
For n=2, the triangle (3, 25, 26) has area 36 = A075848(2). (End)

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Sylvester Robins, Certain Series of Integral, Rational, Scalene Triangles, The American Mathematical Monthly, Vol. 1, No. 1 (Jan., 1894), pp. 13-14. See Example I.
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Cf. A000217, A001108, A001541, A016064, A075848.

a072221 n = a072221_list !! n
a072221_list = 1 : 4 : (map (+ 2) $
   zipWith (-) (map (* 6) $ tail a072221_list) a072221_list)
-- Reinhard Zumkeller, Apr 27 2012

a[n_] := a[n] = 6a[n - 1] - a[n - 2] + 2; a[0] = 1; a[1] = 4; Table[ a[n], {n, 0, 20}]
LinearRecurrence[{7, -7, 1}, {1, 4, 25}, 25] (* T. D. Noe, Dec 09 2013 *)
a[ n_] := (3 ChebyshevT[ n, 3] - 1) / 2; (* Michael Somos, Dec 25 2018 *)

{a(n) = (3 * polchebyshev( n, 1, 3) - 1) / 2}; /* Michael Somos, Dec 25 2018 */

a(n) = (3*A001541(n) - 1)/2.
a(n) = 3*A001108(n) + 1. - David Scheers, Dec 25 2006
From Franz Vrabec, Aug 21 2006: (Start)
a(n) = -1/2 + (3/4)*((3+sqrt(8))^n + (3-sqrt(8))^n) for n >= 0.
a(n) = floor((3/4)*(3+sqrt(8))^n) for n > 0. (End)
G.f.: (1-3x+4x^2)/((1-x)(1-6x+x^2)). - R. J. Mathar, Sep 09 2008
a(n) = a(-n) for all n in Z. - Michael Somos, Dec 25 2018

Edited by Robert G. Wilson v, Jul 08 2002

A011945 Areas of almost-equilateral Heronian triangles (integral side lengths m-1, m, m+1 and integral area).

Original entry on oeis.org

0, 6, 84, 1170, 16296, 226974, 3161340, 44031786, 613283664, 8541939510, 118973869476, 1657092233154, 23080317394680, 321467351292366, 4477462600698444, 62363009058485850, 868604664218103456, 12098102289994962534, 168504827395711372020, 2346969481249964245746

Offset: 1

E. K. Lloyd

Corresponding m's are in A016064. Corresponding values of lesser side give A016064.

Vincenzo Librandi, Table of n, a(n) for n = 1..890
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.
Eric Weisstein's World of Mathematics, Heronian Triangle
Wikipedia, Heronian triangle
P. Yiu, Heron triangles with consecutive sides, Recreational Mathematics, Chap. 9.3, pp. 80/360. (This is a download of 360 pages.)
Index entries for linear recurrences with constant coefficients, signature (14,-1).

Equals 6 * A007655(n+1).
Cf. this sequence (areas), A334277 (perimeters).
Cf. A003500 (middle side lengths), A016064 (smallest side lengths), A335025 (largest side lengths).
Cf. A102341, A103974, A103975.

CoefficientList[Series[6 x/(1 - 14 x + x^2), {x, 0, 30}], x] (* Vincenzo Librandi, Oct 15 2013 *)
LinearRecurrence[{14,-1},{0,6},20] (* Harvey P. Dale, Jan 24 2015 *)

s(n) = floor((a+1)/4)*sqrt(3*(a+3)*(a-1)), where a = A016064(n). - Zak Seidov, Feb 23 2005
a(n) = 14*a(n-1) - a(n-2); a(1) = 0, a(2) = 6.
G.f.: 6*x^2/(1 - 14*x + x^2). - Philippe Deléham, Nov 17 2008
a(n) = (s/4)*((7 + 4*s)^n - (7 - 4*s)^n), where s = sqrt(3). - Zak Seidov, Apr 02 2014
E.g.f.: 6 - exp(7*x)*(12*cosh(4*sqrt(3)*x) - 7*sqrt(3)*sinh(4*sqrt(3)*x))/2. - Stefano Spezia, Dec 12 2022

Entry revised by N. J. A. Sloane, Feb 03 2007

A103975 Smaller side in (a,a+1,a+1)-integer triangle with integer area.

Original entry on oeis.org

16, 240, 3360, 46816, 652080, 9082320, 126500416, 1761923520, 24540428880, 341804080816, 4760716702560, 66308229755040, 923554499868016, 12863454768397200, 179164812257692800, 2495443916839302016, 34757050023492535440, 484103256412056194160

Offset: 1

Zak Seidov, Feb 23 2005

Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Cf. A102341, A103974, A016064, A011945, A103772.

Corresponding areas are given by A104008.

a[n_] := 1/3 (-4 + (2 - Sqrt[3])^(1 + 2 n) + (2 + Sqrt[3])^(1 + 2 n)); A103975 = Expand[a /@ Range[1, 25]] (* Terentyev Oleg, Nov 12 2009 *)
LinearRecurrence[{15,-15,1},{16,240,3360},30] (* Harvey P. Dale, Apr 25 2012 *)

a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3) - Max Alekseyev, May 31 2007
a(n) = 2*A120892(2*n+1) - Max Alekseyev, May 31 2007
a(n) = (1/3)*((2 - sqrt(3))^(1 + 2*n) + (2 + sqrt(3))^(1 + 2*n) - 4). [Terentyev Oleg, Nov 12 2009]
a(n) = (4/3)*(A001570(n+1)-1).
G.f.: -16*x / ((x-1)*(x^2-14*x+1)). - Colin Barker, Apr 09 2013

More terms from Robert G. Wilson v, Mar 24 2005

More terms from Colin Barker, Apr 09 2013

A103772 Larger of two sides in a (k,k,k-1)-integer-sided triangle with integer area.

Original entry on oeis.org

1, 17, 241, 3361, 46817, 652081, 9082321, 126500417, 1761923521, 24540428881, 341804080817, 4760716702561, 66308229755041, 923554499868017, 12863454768397201, 179164812257692801, 2495443916839302017, 34757050023492535441, 484103256412056194161

Offset: 1

Zak Seidov, Feb 23 2005

Corresponding areas are 0, 120, 25080, 4890480, 949077360, 184120982760, ...
Values of (x^2 + y^2)/2, where the pair (x, y) satisfies x^2 - 3*y^2 = -2, i.e., a(n) = {(A001834(n))^2 + (A001835(n))^2}/2 = {(A001834(n))^2 + A046184(n)}/2. - Lekraj Beedassy, Jul 13 2006
The heights of these triangles are given in A028230. (A028230(n), A045899(n), A103772(n)) forms a primitive Pythagorean triple.
Shortest side of (k,k+2,k+3) triangle such that median to longest side is integral. Sequence of such medians is A028230. - James R. Buddenhagen, Nov 22 2013
Numbers n such that (n+1)*(3n-1) is a square. - James R. Buddenhagen, Nov 22 2013

Colin Barker, Table of n, a(n) for n = 1..850
J. B. Cosgrave, The Gauss-Factorial Motzkin connection (Maple worksheet, change suffix to .mw)
J. B. Cosgrave and K. Dilcher, An Introduction to Gauss Factorials, The American Mathematical Monthly, 118 (Nov. 2011), 812-829.
Project Euler, Problem 94: Almost Equilateral Triangles.
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Cf. A102341, A103974, A016064, A011945, A028230.

I:=[1,17]; [n le 2 select I[n] else 14*Self(n-1)-Self(n-2)+4: n in [1..20]]; // Vincenzo Librandi, Mar 05 2016

a[1] = 1; a[2] = 17; a[3] = 241; a[n_] := a[n] = 15a[n - 1] - 15a[n - 2] + a[n - 3]; Table[ a[n] - 1, {n, 17}] (* Robert G. Wilson v, Mar 24 2005 *)
LinearRecurrence[{15,-15,1},{1,17,241},20] (* Harvey P. Dale, Jan 02 2016 *)
RecurrenceTable[{a[1] == 1, a[2] == 17, a[n] == 14 a[n-1] - a[n-2] + 4}, a, {n, 20}] (* Vincenzo Librandi, Mar 05 2016 *)

Vec(x*(1+x)^2/((1-x)*(1-14*x+x^2)) + O(x^25)) \\ Colin Barker, Mar 05 2016

a(n) = (4*A001570(n+1) - 1)/3, n > 0. - Ralf Stephan, May 20 2007
a(n) = A052530(n-1)*A052530(n) + 1. - Johannes Boot, May 21 2011
G.f.: x*(1+x)^2/((1-x)*(1-14*x+x^2)). - Colin Barker, Apr 09 2012
a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3); a(1)=1, a(2)=17, a(3)=241. - Harvey P. Dale, Jan 02 2016
a(n) = (-1+(7-4*sqrt(3))^n*(2+sqrt(3))-(-2+sqrt(3))*(7+4*sqrt(3))^n)/3. - Colin Barker, Mar 05 2016
a(n) = 14*a(n-1) - a(n-2) + 4. - Vincenzo Librandi, Mar 05 2016
a(n) = A001353(n)^2 + A001353(n-1)^2. - Antonio Alberto Olivares, Apr 06 2020

More terms from Robert G. Wilson v, Mar 24 2005

A334277 Perimeters of almost-equilateral Heronian triangles.

Original entry on oeis.org

12, 42, 156, 582, 2172, 8106, 30252, 112902, 421356, 1572522, 5868732, 21902406, 81740892, 305061162, 1138503756, 4248953862, 15857311692, 59180292906, 220863859932, 824275146822, 3076236727356, 11480671762602, 42846450323052, 159905129529606, 596774067795372, 2227191141651882

Offset: 1

Wesley Ivan Hurt, May 20 2020

			a(1) = 12; there is one Heronian triangle with perimeter 12 whose side lengths are consecutive integers, [3,4,5].
a(2) = 42; there is one Heronian triangle with perimeter 42 whose side lengths are consecutive integers, [13,14,15].

Giovanni Resta, Table of n, a(n) for n = 1..1000
Eric Weisstein's World of Mathematics, Heronian Triangle
Wikipedia, Heronian triangle
Wikipedia, Integer Triangle
Index entries for linear recurrences with constant coefficients, signature (4,-1).

Cf. A001075.
Cf. A011945 (areas), this sequence (perimeters).
Cf. A003500 (middle side lengths), A016064 (smallest side lengths), A335025 (largest side lengths).

Table[Expand[3 ((2 + Sqrt[3])^n + (2 - Sqrt[3])^n)], {n, 40}]

a(n) = 3*A003500(n).
a(n) = 3 * ((2 + sqrt(3))^n + (2 - sqrt(3))^n).
From Alejandro J. Becerra Jr., Jan 29 2021: (Start)
G.f.: -6*x*(x - 2)/(x^2 - 4*x + 1).
a(n) = 4*a(n-1) - a(n-2). (End)
a(n) = 6 * A001075(n). - Joerg Arndt, Jan 29 2021
E.g.f.: 6*(exp(2*x)*cosh(sqrt(3)*x) - 1). - Stefano Spezia, Jan 29 2021

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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A001075 a(0) = 1, a(1) = 2, a(n) = 4*a(n-1) - a(n-2).

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

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A103974 Smaller sides (a) in (a,a,a+1)-integer triangle with integer area.

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A097933 Primes p that divide 3^((p-1)/2) - 1.

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