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)

Named after the Newman-Shanks-Williams reference.

Also numbers k such that A125650(3*k^2) is an odd perfect square. Such numbers 3*k^2 form a bisection of A125651. - Alexander Adamchuk, Nov 30 2006

For positive n, a(n) corresponds to the sum of legs of near-isosceles primitive Pythagorean triangles (with consecutive legs). - Lekraj Beedassy, Feb 06 2007

Also numbers m such that m^2 is a centered 16-gonal number; or a number of the form 8k(k+1)+1, where k = A053141(m) = {0, 2, 14, 84, 492, 2870, ...}. - Alexander Adamchuk, Apr 21 2007

The lower principal convergents to 2^(1/2), beginning with 1/1, 7/5, 41/29, 239/169, comprise a strictly increasing sequence; numerators=A002315 and denominators=A001653. - Clark Kimberling, Aug 27 2008

The upper intermediate convergents to 2^(1/2) beginning with 10/7, 58/41, 338/239, 1970/1393 form a strictly decreasing sequence; essentially, numerators=A075870, denominators=A002315. - Clark Kimberling, Aug 27 2008

General recurrence is a(n) = (a(1)-1)*a(n-1) - a(n-2), a(1) >= 4, lim_{n->oo} 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. a(1)=5 gives A001834. a(1)=6 gives A030221. a(1)=7 gives A002315. a(1)=8 gives A033890. a(1)=9 gives A057080. a(1)=10 gives A057081. - Ctibor O. Zizka, Sep 02 2008

Numbers k such that (ceiling(sqrt(k*k/2)))^2 = (1+k*k)/2. - Ctibor O. Zizka, Nov 09 2009

A001109(n)/a(n) converges to cos^2(Pi/8) = 1/2 + 2^(1/2)/4. - Gary Detlefs, Nov 25 2009

The values 2(a(n)^2+1) are all perfect squares, whose square root is given by A075870. - Neelesh Bodas (neelesh.bodas(AT)gmail.com), Aug 13 2010

a(n) represents all positive integers K for which 2(K^2+1) is a perfect square. - Neelesh Bodas (neelesh.bodas(AT)gmail.com), Aug 13 2010

For positive n, a(n) equals the permanent of the (2n) X (2n) tridiagonal matrix with sqrt(8)'s along the main diagonal, and i's along the superdiagonal and subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011

Integers k such that A000217(k-2) + A000217(k-1) + A000217(k) + A000217(k+1) is a square (cf. A202391). - Max Alekseyev, Dec 19 2011

Integer square roots of floor(k^2/2 - 1) or A047838. - Richard R. Forberg, Aug 01 2013

Remark: x^2 - 2*y^2 = +2*k^2, with positive k, and X^2 - 2*Y^2 = +2 reduce to the present Pell equation a^2 - 2*b^2 = -1 with x = k*X = 2*k*b and y = k*Y = k*a. (After a proposed solution for k = 3 by Alexander Samokrutov.) - Wolfdieter Lang, Aug 21 2015

If p is an odd prime, a((p-1)/2) == 1 (mod p). - Altug Alkan, Mar 17 2016

a(n)^2 + 1 = 2*b(n)^2, with b(n) = A001653(n), is the necessary and sufficient condition for a(n) to be a number k for which the diagonal of a 1 X k rectangle is an integer multiple of the diagonal of a 1 X 1 square. If squares are laid out thus along one diagonal of a horizontal 1 X a(n) rectangle, from the lower left corner to the upper right, the number of squares is b(n), and there will always be a square whose top corner lies exactly within the top edge of the rectangle. Numbering the squares 1 to b(n) from left to right, the number of the one square that has a corner in the top edge of the rectangle is c(n) = (2*b(n) - a(n) + 1)/2, which is A055997(n). The horizontal component of the corner of the square in the edge of the rectangle is also an integer, namely d(n) = a(n) - b(n), which is A001542(n). - David Pasino, Jun 30 2016

(a(n)^2)-th triangular number is a square; a(n)^2 = A008843(n) is a subsequence of A001108. - Jaroslav Krizek, Aug 05 2016

a(n-1)/A001653(n) is the closest rational approximation of sqrt(2) with a numerator not larger than a(n-1). These rational approximations together with those obtained from the sequences A001541 and A001542 give a complete set of closest rational approximations of sqrt(2) with restricted numerator or denominator. a(n-1)/A001653(n) < sqrt(2). - A.H.M. Smeets, May 28 2017

Consider the quadrant of a circle with center (0,0) bounded by the positive x and y axes. Now consider, as the start of a series, the circle contained within this quadrant which kisses both axes and the outer bounding circle. Consider further a succession of circles, each kissing the x-axis, the outer bounding circle, and the previous circle in the series. See Holmes link. The center of the n-th circle in this series is ((A001653(n)*sqrt(2)-1)/a(n-1), (A001653(n)*sqrt(2)-1)/a(n-1)^2), the y-coordinate also being its radius. It follows that a(n-1) is the cotangent of the angle subtended at point (0,0) by the center of the n-th circle in the series with respect to the x-axis. - Graham Holmes, Aug 31 2019

There is a link between the two sequences present at the numerator and at the denominator of the fractions that give the coordinates of the center of the kissing circles. A001653 is the sequence of numbers k such that 2*k^2 - 1 is a square, and here, we have 2*A001653(n)^2 - 1 = a(n-1)^2. - Bernard Schott, Sep 02 2019

Let G be a sequence satisfying G(i) = 2*G(i-1) + G(i-2) for arbitrary integers i and without regard to the initial values of G. Then a(n) = (G(i+4*n+2) - G(i))/(2*G(i+2*n+1)) as long as G(i+2*n+1) != 0. - Klaus Purath, Mar 25 2021

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, 0 < a < b < c are given by a=A001542(n), b=A005319(n), c=A001542(n+1), x=A001541(n), y=A001653(n+1), z=A002315(n) with 0 < n. - Michael Somos, Jun 26 2022

3*a(n-1) is the n-th almost Lucas-cobalancing number of second type (see Tekcan and Erdem). - Stefano Spezia, Nov 26 2022

In Moret-Blanc (1881) on page 259 some solution of m^2 - 2n^2 = -1 are listed. The values of m give this sequence, and the values of n give A001653. - Michael Somos, Oct 25 2023

From Klaus Purath, May 11 2024: (Start)

For any two consecutive terms (a(n), a(n+1)) = (x,y): x^2 - 6xy + y^2 = 8 = A028884(1). In general, the following applies to all sequences (t) satisfying t(i) = 6t(i-1) - t(i-2) with t(0) = 1 and two consecutive terms (x,y): x^2 - 6xy + y^2 = A028884(t(1)-6). This includes and interprets the Feb 04 2014 comment on A001541 by Colin Barker as well as the Mar 17 2021 comment on A054489 by John O. Oladokun and the Sep 28 2008 formula on A038723 by Michael Somos. By analogy to this, for three consecutive terms (x,y,z) y^2 - xz = A028884(t(1)-6) always applies.

If (t) is a sequence satisfying t(k) = 7t(k-1) - 7t(k-2) + t(k-3) or t(k) = 6t(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 >= 0. (End)

T(n,k) is the number of lattice paths from (0,0) to (n,n) with steps E = (1,0) and N = (0,1) which touch but do not cross the line x - y = k and only situated above this line; example: T(3,2) = 5 because we have EENNNE, EENNEN, EENENN, ENEENN, NEEENN. - Philippe Deléham, May 23 2005

The matrix inverse of this triangle is the triangular matrix T(n,k) = (-1)^(n+k)* A085478(n,k). - Philippe Deléham, May 26 2005

Essentially the same as A050155 except with a leading diagonal A000108 (Catalan numbers) 1, 1, 2, 5, 14, 42, 132, 429, .... - Philippe Deléham, May 31 2005

Number of Grand Dyck paths of semilength n and having k downward returns to the x-axis. (A Grand Dyck path of semilength n is a path in the half-plane x>=0, starting at (0,0), ending at (2n,0) and consisting of steps u=(1,1) and d=(1,-1)). Example: T(3,2)=5 because we have u(d)uud(d),uud(d)u(d),u(d)u(d)du,u(d)duu(d) and duu(d)u(d) (the downward returns to the x-axis are shown between parentheses). - Emeric Deutsch, May 06 2006

Riordan array (c(x),x*c(x)^2) where c(x) is the g.f. of A000108; inverse array is (1/(1+x),x/(1+x)^2). - Philippe Deléham, Feb 12 2007

The triangle may also be generated from M^n*[1,0,0,0,0,0,0,0,...], where M is the infinite tridiagonal matrix with all 1's in the super and subdiagonals and [1,2,2,2,2,2,2,...] in the main diagonal. - Philippe Deléham, Feb 26 2007

Inverse binomial matrix applied to A124733. Binomial matrix applied to A089942. - Philippe Deléham, Feb 26 2007

Number of standard tableaux of shape (n+k,n-k). - Philippe Deléham, Mar 22 2007

From Philippe Deléham, Mar 30 2007: (Start)

This triangle belongs to the family of triangles defined by: T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0)=x*T(n-1,0)+T(n-1,1), T(n,k)=T(n-1,k-1)+y*T(n-1,k)+T(n-1,k+1) for k>=1. Other triangles arise by choosing different values for (x,y):

(0,0) -> A053121; (0,1) -> A089942; (0,2) -> A126093; (0,3) -> A126970

(1,0) -> A061554; (1,1) -> A064189; (1,2) -> A039599; (1,3) -> A110877;

(1,4) -> A124576; (2,0) -> A126075; (2,1) -> A038622; (2,2) -> A039598;

(2,3) -> A124733; (2,4) -> A124575; (3,0) -> A126953; (3,1) -> A126954;

(3,2) -> A111418; (3,3) -> A091965; (3,4) -> A124574; (4,3) -> A126791;

(4,4) -> A052179; (4,5) -> A126331; (5,5) -> A125906. (End)

The table U(n,k) = Sum_{j=0..n} T(n,j)*k^j is given in A098474. - Philippe Deléham, Mar 29 2007

Sequence read mod 2 gives A127872. - Philippe Deléham, Apr 12 2007

Number of 2n step walks from (0,0) to (2n,2k) and consisting of step u=(1,1) and d=(1,-1) and the path stays in the nonnegative quadrant. Example: T(3,0)=5 because we have uuuddd, uududd, ududud, uduudd, uuddud; T(3,1)=9 because we have uuuudd, uuuddu, uuudud, ududuu, uuduud, uduudu, uudduu, uduuud, uududu; T(3,2)=5 because we have uuuuud, uuuudu, uuuduu, uuduuu, uduuuu; T(3,3)=1 because we have uuuuuu. - Philippe Deléham, Apr 16 2007, Apr 17 2007, Apr 18 2007

Triangular matrix, read by rows, equal to the matrix inverse of triangle A129818. - Philippe Deléham, Jun 19 2007

Let Sum_{n>=0} a(n)*x^n = (1+x)/(1-mx+x^2) = o.g.f. of A_m, then Sum_{k=0..n} T(n,k)*a(k) = (m+2)^n. Related expansions of A_m are: A099493, A033999, A057078, A057077, A057079, A005408, A002878, A001834, A030221, A002315, A033890, A057080, A057081, A054320, A097783, A077416, A126866, A028230, A161591, for m=-3,-2,-1,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15, respectively. - Philippe Deléham, Nov 16 2009

The Kn11, Kn12, Fi1 and Fi2 triangle sums link the triangle given above with three sequences; see the crossrefs. For the definitions of these triangle sums, see A180662. - Johannes W. Meijer, Apr 20 2011

4^n = (n-th row terms) dot (first n+1 odd integer terms). Example: 4^4 = 256 = (14, 28, 20, 7, 1) dot (1, 3, 5, 7, 9) = (14 + 84 + 100 + 49 + 9) = 256. - Gary W. Adamson, Jun 13 2011

The linear system of n equations with coefficients defined by the first n rows solve for diagonal lengths of regular polygons with N= 2n+1 edges; the constants c^0, c^1, c^2, ... are on the right hand side, where c = 2 + 2*cos(2*Pi/N). Example: take the first 4 rows relating to the 9-gon (nonagon), N = 2*4 + 1; with c = 2 + 2*cos(2*Pi/9) = 3.5320888.... The equations are (1,0,0,0) = 1; (1,1,0,0) = c; (2,3,1,0) = c^2; (5,9,5,1) = c^3. The solutions are 1, 2.53208..., 2.87938..., and 1.87938...; the four distinct diagonal lengths of the 9-gon (nonagon) with edge = 1. (Cf. comment in A089942 which uses the analogous operations but with c = 1 + 2*cos(2*Pi/9).) - Gary W. Adamson, Sep 21 2011

Also called the Lobb numbers, after Andrew Lobb, are a natural generalization of the Catalan numbers, given by L(m,n)=(2m+1)*Binomial(2n,m+n)/(m+n+1), where n >= m >= 0. For m=0, we get the n-th Catalan number. See added reference. - Jayanta Basu, Apr 30 2013

From Wolfdieter Lang, Sep 20 2013: (Start)

T(n, k) = A053121(2*n, 2*k). T(n, k) appears in the formula for the (2*n)-th power of the algebraic number rho(N):= 2*cos(Pi/N) = R(N, 2) in terms of the odd-indexed diagonal/side length ratios R(N, 2*k+1) = S(2*k, rho(N)) in the regular N-gon inscribed in the unit circle (length unit 1). S(n, x) are Chebyshev's S polynomials (see A049310):

rho(N)^(2*n) = Sum_{k=0..n} T(n, k)*R(N, 2*k+1), n >= 0, identical in N > = 1. For a proof see the Sep 21 2013 comment under A053121. Note that this is the unreduced version if R(N, j) with j > delta(N), the degree of the algebraic number rho(N) (see A055034), appears.

For the odd powers of rho(n) see A039598. (End)

Unsigned coefficients of polynomial numerators of Eqn. 2.1 of the Chakravarty and Kodama paper, defining the polynomials of A067311. - Tom Copeland, May 26 2016

The triangle is the Riordan square of the Catalan numbers in the sense of A321620. - Peter Luschny, Feb 14 2023

In any generalized Fibonacci sequence {f(i)}, Sum_{i=0..4n+1} f(i) = a(n)*f(2n+2). - Lekraj Beedassy, Dec 31 2002

The continued fraction expansion for F((2n+1)*(k+1))/F((2n+1)*k), k>=1 is [a(n),a(n),...,a(n)] where there are exactly k elements (F(n) denotes the n-th Fibonacci number). E.g., continued fraction for F(12)/F(9) is [4, 4,4]. - Benoit Cloitre, Apr 10 2003

See A135064 for a possible connection with Galois groups of quintics.

Sequence of all positive integers k such that continued fraction [k,k,k,k,k,k,...] belongs to Q(sqrt(5)). - Thomas Baruchel, Sep 15 2003

All positive integer solutions of Pell equation a(n)^2 - 5*b(n)^2 = -4 together with b(n)=A001519(n), n>=0.

a(n) = L(n,-3)*(-1)^n, where L is defined as in A108299; see also A001519 for L(n,+3).

Inverse binomial transform of A030191. - Philippe Deléham, Oct 04 2005

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. a(1)=5 gives A001834. a(1)=6 gives A030221. a(1)=7 gives A002315. a(1)=8 gives A033890. a(1)=9 gives A057080. a(1)=10 gives A057081. - Ctibor O. Zizka, Sep 02 2008

Let r = (2n+1), then a(n), n>0 = Product_{k=1..floor((r-1)/2)} (1 + sin^2 k*Pi/r); e.g., a(3) = 29 = (3.4450418679...)*(4.801937735...)*(1.753020396...). - Gary W. Adamson, Nov 26 2008

a(n+1) is the Hankel transform of A001700(n)+A001700(n+1). - Paul Barry, Apr 21 2009

a(n) is equal to the permanent of the (2n) X (2n) tridiagonal matrix with sqrt(5)'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

Conjecture: for n > 0, a(n) = sqrt(Fibonacci(4*n+3) + Sum_{k=2..2*n} Fibonacci(2*k)). - Alex Ratushnyak, May 06 2012

Pisano period lengths: 1, 3, 4, 3, 2, 12, 8, 6, 12, 6, 5, 12, 14, 24, 4, 12, 18, 12, 9, 6, ... . - R. J. Mathar, Aug 10 2012

The continued fraction [a(n); a(n), a(n), ...] = phi^(2n+1), where phi is the golden ratio, A001622. - Thomas Ordowski, Jun 05 2013

Solutions (x, y) = (a(n), a(n+1)) satisfying x^2 + y^2 = 3xy + 5. - Michel Lagneau, Feb 01 2014

Conjecture: except for the number 3, a(n) are the numbers such that a(n)^2+2 are Lucas numbers. - Michel Lagneau, Jul 22 2014

Comment on the preceding conjecture: It is clear that all a(n) satisfy a(n)^2 + 2 = L(2*(2*n+1)) due to the identity (17 c) of Vajda, p. 177: L(2*n) + 2*(-1)^n = L(n)^2 (take n -> 2*n+1). - Wolfdieter Lang, Oct 10 2014

Limit_{n->oo} a(n+1)/a(n) = phi^2 = phi + 1 = (3+sqrt(5))/2. - Derek Orr, Jun 18 2015

If d[k] denotes the sequence of k-th differences of this sequence, then d[0](0), d[1](1), d[2](2), d[3](3), ... = A048876, cf. message to SeqFan list by P. Curtz on March 2, 2016. - M. F. Hasler, Mar 03 2016

a(n-1) and a(n) are the least phi-antipalindromic numbers (A178482) with 2*n and 2*n+1 digits in base phi, respectively. - Amiram Eldar, Jul 07 2021

Triangulate (hyperbolic) 2-space such that around every vertex exactly 7 triangles touch. Call any 7 triangles having a common vertex the first layer and let the (n+1)-st layer be all triangles that do not appear in any of the first n layers and have a common vertex with the n-th layer. Then the n-th layer contains 7*a(n-1) triangles. E.g., the first layer (by definition) contains 7 triangles, the second layer (the "ring" of triangles around the first layer) consists of 28 triangles, the third layer (the next "ring") consists of 77 triangles, and so on. - Nicolas Nagel, Aug 13 2022

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

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

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

Matrix inverse of A124645.

Let L(n,x) = Sum_{k=0..n} T(n,k)*x^(n-k) and Pi=3.14...:

L(n,x) = Product_{k=1..n} (x - 2*cos((2*k-1)*Pi/(2*n+1)));

Sum_{k=0..n} T(n,k) = L(n,1) = A010892(n+1);

Sum_{k=0..n} abs(T(n,k)) = A000045(n+2);

abs(T(n,k)) = A065941(n,k), T(n,k) = A065941(n,k)*A087960(k);

T(2*n,k) + T(2*n+1,k+1) = 0 for 0 <= k <= 2*n;

T(n,0) = A000012(n) = 1; T(n,1) = -1 for n > 0;

T(n,2) = -(n-1) for n > 1; T(n,3) = A000027(n)=n for n > 2;

T(n,4) = A000217(n-3) for n > 3; T(n,5) = -A000217(n-4) for n > 4;

T(n,6) = -A000292(n-5) for n > 5; T(n,7) = A000292(n-6) for n > 6;

T(n,n-3) = A058187(n-3)*(-1)^floor(n/2) for n > 2;

T(n,n-2) = A008805(n-2)*(-1)^floor((n+1)/2) for n > 1;

T(n,n-1) = A008619(n-1)*(-1)^floor(n/2) for n > 0;

T(n,n) = L(n,0) = (-1)^floor((n+1)/2);

L(n,1) = A010892(n+1); L(n,-1) = A061347(n+2);

L(n,2) = 1; L(n,-2) = A005408(n)*(-1)^n;

L(n,3) = A001519(n); L(n,-3) = A002878(n)*(-1)^n;

L(n,4) = A001835(n+1); L(n,-4) = A001834(n)*(-1)^n;

L(n,5) = A004253(n); L(n,-5) = A030221(n)*(-1)^n;

L(n,6) = A001653(n); L(n,-6) = A002315(n)*(-1)^n;

L(n,7) = A049685(n); L(n,-7) = A033890(n)*(-1)^n;

L(n,8) = A070997(n); L(n,-8) = A057080(n)*(-1)^n;

L(n,9) = A070998(n); L(n,-9) = A057081(n)*(-1)^n;

L(n,10) = A072256(n+1); L(n,-10) = A054320(n)*(-1)^n;

L(n,11) = A078922(n+1); L(n,-11) = A097783(n)*(-1)^n;

L(n,12) = A077417(n); L(n,-12) = A077416(n)*(-1)^n;

L(n,13) = A085260(n);

L(n,14) = A001570(n); L(n,-14) = A028230(n)*(-1)^n;

L(n,n) = A108366(n); L(n,-n) = A108367(n).

Row n of the matrix inverse (A124645) has g.f.: x^floor(n/2)*(1-x)^(n-floor(n/2)). - Paul D. Hanna, Jun 12 2005

From L. Edson Jeffery, Mar 12 2011: (Start)

Conjecture: Let N=2*n+1, with n > 2. Then T(n,k) (0 <= k <= n) gives the k-th coefficient in the characteristic function p_N(x)=0, of degree n in x, for the n X n tridiagonal unit-primitive matrix G_N (see [Jeffery]) of the form

G_N=A_{N,1}=

(0 1 0 ... 0)

(1 0 1 0 ... 0)

(0 1 0 1 0 ... 0)

...

(0 ... 0 1 0 1)

(0 ... 0 1 1),

with solutions phi_j = 2*cos((2*j-1)*Pi/N), j=1,2,...,n. For example, for n=3,

G_7=A_{7,1}=

(0 1 0)

(1 0 1)

(0 1 1).

We have {T(3,k)}=(1,-1,-2,1), while the characteristic function of G_7 is p(x) = x^3-x^2-2*x+1 = 0, with solutions phi_j = 2*cos((2*j-1)*Pi/7), j=1,2,3. (End)

The triangle sums, see A180662 for their definitions, link A108299 with several sequences, see the crossrefs. - Johannes W. Meijer, Aug 08 2011

The roots to the polynomials are chaotic using iterates of the operation (x^2 - 2), with cycle lengths L and initial seeds returning to the same term or (-1)* the seed. Periodic cycle lengths L are shown in A003558 such that for the polynomial represented by row r, the cycle length L is A003558(r-1). The matrices corresponding to the rows as characteristic polynomials are likewise chaotic [cf. Kappraff et al., 2005] with the same cycle lengths but substituting 2*I for the "2" in (x^2 - 2), where I = the Identity matrix. For example, the roots to x^3 - x^2 - 2x + 1 = 0 are 1.801937..., -1.246979..., and 0.445041... With 1.801937... as the initial seed and using (x^2 - 2), we obtain the 3-period trajectory of 8.801937... -> 1.246979... -> -0.445041... (returning to -1.801937...). We note that A003558(2) = 3. The corresponding matrix M is: [0,1,0; 1,0,1; 0,1,1,]. Using seed M with (x^2 - 2*I), we obtain the 3-period with the cycle completed at (-1)*M. - Gary W. Adamson, Feb 07 2012

[Empirical] a(base,n) = a(base-1,n) + 7^(n-1) for base >= 3n-2; a(base,n) = a(base-1,n) + 7^(n-1)-2 when base = 3n-3.

From Johannes W. Meijer, Aug 01 2010: (Start)

The a(n) represent the number of n-move routes of a fairy chess piece starting in a given side square (m = 2, 4, 6 or 8) on a 3 X 3 chessboard. This fairy chess piece behaves like a king on the eight side and corner squares but on the central square the king goes crazy and turns into a red king, see A179596.

For the side squares the 512 red kings lead to 47 different red king sequences, see the cross-references for some examples.

The sequence above corresponds to four A[5] vectors with the decimal [binary] values 367 [1,0,1,1,0,1,1,1,1], 463 [1,1,1,0,0,1,1,1,1], 487 [1,1,1,1,0,0,1,1,1] and 493 [1,1,1,1,0,1,1,0,1]. These vectors lead for the corner squares to A179596 and for the central square to A179597.

This sequence belongs to a family of sequences with g.f. (1+x)/(1-4*x-k*x^2). Red king sequences that are members of this family are A003947 (k=0), A015448 (k=1), A123347 (k=2), A126473 (k=3; this sequence) and A086347 (k=4). Other members of this family are A000351 (k=5), A001834 (k=-1), A111567 (k=-2), A048473 (k=-3) and A053220 (k=-4)

Inverse binomial transform of A154244. (End)

Equals the INVERT transform of A055099: (1, 4, 14, 50, 178, ...). - Gary W. Adamson, Aug 14 2010

Number of one-sided n-step walks taking steps from {E, W, N, NE, NW}. - Shanzhen Gao, May 10 2011

For n>=1, a(n) equals the numbers of words of length n-1 on alphabet {0,1,2,3,4} containing no subwords 00 and 11. - Milan Janjic, Jan 31 2015

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