0,3

3*a(n)^2 + 1 is a perfect square.

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) = 4a(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 (hpd1(AT)is2.nyu.edu), Dec 28 2000

If M is any term of the sequence, the next one is 2M + sqrt(3M^2 + 1). - Lekraj Beedassy (blekraj(AT)yahoo.com), Feb 18 2002

n such that 3*n^2=floor(sqrt(3)*n*ceil(sqrt(3)*n)) Benoit Cloitre (benoit7848c(AT)orange.fr), 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 (blekraj(AT)yahoo.com), Sep 19 2003

Ways of packing a 3 X (2n-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*A011835(n-1) + 2*a(n-1). - Joshua Zucker and the Castilleja School Math Club (joshua_zucker(AT)castilleja.org), 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 (pbarry(AT)wit.ie), Oct 25 2004

This sequence generates many brilliant (A078972) numbers for a(p) with prime p: a(2) = 4 = 2 * 2 a(3) = 15 = 3 * 5 a(5) = 209 = 11 * 19 a(7) = 2911 = 41 * 71 a(19) = 21252634831 = 110771 * 191861 a(37) = 419245718107612602961 = 15558008491 * 26947261171. Is this a prime-free sequence? If not, what is its first prime? - Jonathan Vos Post (jvospost2(AT)yahoo.com), Feb 08 2005

Numbers such that there is an m with t(n+m)=3t(m), where t(n) are the triangular numbers A000217. For instance t(35)=3t(20)=630, so 35-20=15 is in the sequence. - comment by Floor van Lamoen (fvlamoen(AT)hotmail.com), Oct 13 2005

a(n) = number of unique 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 (maxal(AT)cs.ucsd.edu), 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 (blekraj(AT)yahoo.com), Jul 13 2006

Values y solving the Pellian x^2 - 3*y^2 = 1; Corresponding x given by A001075(n). Moreover, we have a(n) = 2*a(n-1) + A001075(n-1). - Lekraj Beedassy (blekraj(AT)yahoo.com), Jul 13 2006

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 satsifies 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 2X2 simple rectangular mazes:

.__.............__ ......__.............__

|..|..|........|__...|.......|.....|........|...__|

|...__|........|...__|.......|..|__|........|...__|. - Dennis P. Walsh (dwalsh(AT)mtsu.edu), Oct 04 2006

[1,4,15,56,209,...] is the Hankel transform of [1,1,5,26,139,758,...](see A005573). - Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Apr 14 2007

S. Plouffe, Approximations de S\'{e}ries G\'{e}n\'{e}ratrices et Quelques Conjectures}, Dissertation, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992.

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.

G. Everest, A. van der Poorten, I. Shparlinski and T. Ward, Recurrence Sequences, Amer. Math. Soc., 2003; p. 163.

E. I. Emerson, Recurrent sequences in the equation DQ^2 = R^2 + N, Fib. Quart., 7 (1969), 231-242.

R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 329.

T. N. E. Greville, Table for third-degree spline interpolations with equally spaced arguments, Math. Comp., 24 (1970), 179-183.

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.

J. D. E. Konhauser et al., Which Way Did the Bicycle Go?, MAA 1996, p. 104.

G. Kreweras, Complexite et circuits Euleriens dans la sommes tensorielles de graphes, J. Combin. Theory, B 24 (1978), 202-212.

W. 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.

F. V. Waugh and M. W. Maxfield, Side-and-diagonal numbers, Math. Mag., 40 (1967), 74-83.

T. D. Noe, Table of n, a(n) for n=0..200

S. Plouffe, Approximations de S\'{e}ries G\'{e}n\'{e}ratrices et Quelques Conjectures}, Dissertation, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992.

S. Plouffe, 1031 Generating Functions and Conjectures, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992.

Tanya Khovanova, Recursive Sequences

Hojoo Lee, Problems in elementary number theory Problem I 18.

D. P. Walsh, Counting n x 2 Simple Rectangular Mazes

Index entries for sequences related to Chebyshev polynomials.

a(n)=[(2+sqrt(3))^n-(2-sqrt(3))^n]/(2*sqrt(3)).

Limit as n-> infinity of a(n)/a(n-1) = 2 + sqrt(3). - Gregory V. Richardson (omomom(AT)hotmail.com), Oct 06 2002

Binomial transform of A080953. E.g.f.: exp(2x)sinh(sqrt(3)x)/sqrt(3).

G.f.: x/(1-4*x+x^2). 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-2k)} - Paul Barry (pbarry(AT)wit.ie), Oct 25 2004

a(n)=sum{k=0..n-1, binomial(n+k, 2k+1)2^k} - Paul Barry (pbarry(AT)wit.ie), Nov 30 2004

a(n)=3*a(n-1)+3*a(n-2)-a(n-3); a(0)=0, a(1)=1, a(2)=4. - Lekraj Beedassy (blekraj(AT)yahoo.com), Jul 13 2006

a(n) = 2*a(n-1)+sqrt[3*a(n-1)^2+1]. a(n) = -A106707(n). - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Jul 07 2006

a(n) = 3*(a(n-1)+a(n-2))-a(n-3), a(n) = 5*(a(n-1)-a(n-2))+a(n-3). - Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Sep 20 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 (qntmpkt(AT)yahoo.com), Dec 27 2006

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.

---- ---- ---- ||--

.||| .--| .|-- .|||

.||| .--| .|-- .|||

A001353 := proc(n) option remember; if n <= 1 then 1+3*n else 4*A001353(n-1)-A001353(n-2); fi; end;

A001353:=1/(1-4*z+z**2); [Conjectured by S. Plouffe in his 1992 dissertation.]

a[n_] := (MatrixPower[{{1, 2}, {1, 3}}, n].{{1}, {1}})[[2, 1]]; Table[ a[n], {n, 0, 23}]] (from Robert G. Wilson v Jan 13 2005)

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

a(n) = sqrt((A001075(n)^2-1)/3).

Cf. A003500, A001835.

Cf. A001571, A001834, A002531, A005246, A016064, A082840.

Cf. A079935.

Cf. A078972.

Cf. A001075.

Adjacent sequences: A001350 A001351 A001352 this_sequence A001354 A001355 A001356

Sequence in context: A060111 A077824 A010905 this_sequence A106707 A125905 A026030

nonn,easy,nice

njas