cp's OEIS Frontend

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:

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(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,

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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:

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(End)

Chebyshev T-sequence with Diophantine property. - Wolfdieter Lang, Nov 29 2002

a(n) = L(n,14), where L is defined as in A108299; see also A028230 for L(n,-14). - Reinhard Zumkeller, Jun 01 2005

Numbers x satisfying x^2 + y^3 = (y+1)^3. Corresponding y given by A001921(n)={A028230(n)-1}/2. - Lekraj Beedassy, Jul 21 2006

Mod[ a(n), 12 ] = 1. (a(n) - 1)/12 = A076139(n) = Triangular numbers that are one-third of another triangular number. (a(n) - 1)/4 = A076140(n) = Triangular numbers T(k) that are three times another triangular number. - Alexander Adamchuk, Apr 06 2007

Also numbers n such that RootMeanSquare(1,3,...,2*n-1) is an integer. - Ctibor O. Zizka, Sep 04 2008

a(n), with n>1, is the length of the cevian of equilateral triangle whose side length is the term b(n) of the sequence A028230. This cevian divides the side (2*x+1) of the triangle in two integer segments x and x+1. - Giacomo Fecondo, Oct 09 2010

For n>=2, a(n) equals the permanent of the (2n-2)X(2n-2) tridiagonal matrix with sqrt(12)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011

Beal's conjecture would imply that set intersection of this sequence with the perfect powers (A001597) equals {1}. In other words, existence of a nontrivial perfect power in this sequence would disprove Beal's conjecture. - Max Alekseyev, Mar 15 2015

Numbers n such that there exists positive x with x^2 + x + 1 = 3n^2. - Jeffrey Shallit, Dec 11 2017

Given by the denominators of the continued fractions [1,(1,2)^i,3,(1,2)^{i-1},1]. - Jeffrey Shallit, Dec 11 2017

A near-isosceles integer-sided triangle with an angle of 2*Pi/3 is a triangle whose sides (a, a+1, c) satisfy Diophantine equation (a+1)^3 - a^3 = c^2. For n >= 2, the largest side c is given by a(n) while smallest and middle sides (a, a+1) = (A001921(n-1), A001922(n-1)) (see Julia link). - Bernard Schott, Nov 20 2022

a(n) corresponds also to one-sixth the area of Fleenor-Heronian triangle with middle side A003500(n). - Lekraj Beedassy, Jul 15 2002

a(n) give all (nontrivial, integer) solutions of Pell equation b(n+1)^2 - 48*a(n+1)^2 = +1 with b(n+1)=A011943(n), n>=0.

For n>=3, a(n) equals the permanent of the (n-2) X (n-2) tridiagonal matrix with 14'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

For n>1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,13}. - Milan Janjic, Jan 25 2015

6*a(n)^2 = 6*S(n-1, 14)^2 is the triangular number Tri((T(n, 7) - 1)/2) with Tri = A000217 and T = A053120. This is instance k = 3 of the general k-identity given in a comment to A001109. - Wolfdieter Lang, Feb 01 2016

Or, squares of the form 3k^2+1.

See A023110, A204503, A204512, A204517, A204519, A055812, A055808 and A055792 for the analog in other bases.

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 ]

Gives solutions k to the Diophantine equation m^2 = k*(k+1)/3. - Anton Lorenz Vrba (anton(AT)a-l-v.net), Jun 28 2005

If x=a(n), y=a(n+1), z=a(n+2) are three consecutive terms, then x^2 - 16*y*x + 14*x*z + 16*y^2 - 16*z*y + z^2 = 144. The formula is symmetric in x and z, so it is also valid for x=a(n+2), y=a(n+1), z=a(n). - Alexander Samokrutov, Jul 02 2015

From Bernard Schott, Apr 09 2021 (Start):

Corresponding solutions m (of first comment) are in A011944.

Equivalently, numbers k such that k/3 and k+1 are both perfect squares. (End)

Simple continued fraction expansion of product {n >= 0} {1 - sqrt(m)*[sqrt(m) - sqrt(m-1)]^(4*n+3)}/{1 - sqrt(m)*[sqrt(m) - sqrt(m-1)]^(4*n+1)} at m = 4. For other cases see A221073 (m = 2), A221074 (m = 3) and A221076 (m = 5).

If we denote the present sequence by [2; 12, 1, c(3), 1, c(4), 1, ...] then for k >= 1 the sequence [1; c(2*k+1), 1, c(2*(2*k+1)), 1, c(3*(2*k+1)), 1, ...] gives the simple continued fraction expansion of product {n >= 0} [1-2*{(2-sqrt(3))^(2*k+1)}^(4*n+3)]/[1 - 2*{(2-sqrt(3))^(2*k+1)}^(4*n+1)]. An example is given below

The scatterplot shows an interesting structure where terms are on descending hatches.

Terms on each hatch are quite well fitted by a polynomial of degree 2.

The parity of the term indices alternates from one hatch to the next and that of two consecutive terms alternates on the same hatch.

For terms on a given hatch, the differences of order 2 quickly become constant and equal to 2.

The fixed points begin 1, 16, 225, 3136, etc. They appear to be all squares and to come from A098301.

Numbers k such that k*(5*k+3) is a perfect square. Apparently a(n) = 323*a(n-1) -323*a(n-2) +a(n-3). - R. J. Mathar, May 18 2013