cp's OEIS Frontend

In the tiling {5,3,4} of 3-dimensional hyperbolic space, the number of regular dodecahedra with right angles of the n generation which are contained in an eighth of space (intersection of three pairwise perpendicular hyperplanes which are supported by the faces of a dodecahedron at a vertex).

Let beta be the greatest real root of the polynomial which is defined by the above recurrent equation. Consider the representation of positive numbers in the basis beta. Then the language which consists of the maximal representations of positive numbers is neither regular nor context-free (M. Margenstern's theorem, see second reference, above).

Column sums of shifted Pascal-like array:

1..1..10..10..100..100.1000.1000

......-1..-2..-30..-40.-500.-600

................1....3...60..100

.........................-1...-4

-------------------------------- +

1..1...9...8...71...63..559..496

Decimal expansion of ratio n/(n+1) is accumulation of Catalan numbers; (5 +/- sqrt(15)).

G.f. for row polynomials S(n,x) (signed triangle): 1/(1-x*z+z^2). Unsigned triangle |a(n,m)| has Fibonacci polynomials F(n+1,x) as row polynomials with g.f. 1/(1-x*z-z^2). |a(n,m)| triangle has rows of Pascal's triangle A007318 in the even-numbered diagonals (odd-numbered ones have only 0's).

Row sums (unsigned triangle) A000045(n+1) (Fibonacci). Row sums (signed triangle) S(n,1) sequence = periodic(1,1,0,-1,-1,0) = A010892.

Alternating row sums A049347(n) = S(n,-1) = periodic(1,-1,0). - Wolfdieter Lang, Nov 04 2011

S(n,x) is the characteristic polynomial of the adjacency matrix of the n-path. - Michael Somos, Jun 24 2002

S(n,x) is also the matching polynomial of the n-path. - Eric W. Weisstein, Apr 10 2017

|T(n,k)| = number of compositions of n+1 into k+1 odd parts. Example: |T(7,3)| = 10 because we have (1,1,3,3), (1,3,1,3), (1,3,3,1), (3,1,1,3), (3,1,3,1), (3,3,1,1), (1,1,1,5), (1,1,5,1), (1,5,1,1) and (5,1,1,1). - Emeric Deutsch, Apr 09 2005

S(n,x)= R(n,x) + S(n-2,x), n >= 2, S(-1,x)=0, S(0,x)=1, R(n,x):=2*T(n,x/2) = Sum_{m=0..n} A127672(n,m)*x^m (monic integer Chebyshev T-Polynomials). This is the rewritten so-called trace of the transfer matrix formula for the T-polynomials. - Wolfdieter Lang, Dec 02 2010

In a regular N-gon inscribed in a unit circle, the side length is d(N,1) = 2*sin(Pi/N). The length ratio R(N,k):=d(N,k)/d(N,1) for the (k-1)-th diagonal, with k from {2,3,...,floor(N/2)}, N >= 4, equals S(k-1,x) = sin(k*Pi/N)/sin(Pi/N) with x=rho(N):=R(N,2) = 2*cos(Pi/N). Example: N=7 (heptagon), rho=R(7,2), sigma:=R(N,3) = S(2,rho) = rho^2 - 1. Motivated by the quoted paper by P. Steinbach. - Wolfdieter Lang, Dec 02 2010

From Wolfdieter Lang, Jul 12 2011: (Start)

In q- or basic analysis, q-numbers are [n]_q := S(n-1,q+1/q) = (q^n-(1/q)^n)/(q-1/q), with the row polynomials S(n,x), n >= 0.

The zeros of the row polynomials S(n-1,x) are (from those of Chebyshev U-polynomials):

x(n-1;k) = +- t(k,rho(n)), k = 1..ceiling((n-1)/2), n >= 2, with t(n,x) the row polynomials of A127672 and rho(n):= 2*cos(Pi/n). The simple vanishing zero for even n appears here as +0 and -0.

Factorization of the row polynomials S(n-1,x), x >= 1, in terms of the minimal polynomials of cos(2 Pi/2), called Psi(n,x), with coefficients given by A181875/A181876:

S(n-1,x) = (2^(n-1))*Product_{n>=1}(Psi(d,x/2), 2 < d | 2n).

(From the rewritten eq. (3) of the Watkins and Zeitlin reference, given under A181872.) [See the W. Lang ArXiv link, Proposition 9, eq. (62). - Wolfdieter Lang, Apr 14 2018]

(End)

The discriminants of the S(n,x) polynomials are found in A127670. - Wolfdieter Lang, Aug 03 2011

This is an example for a subclass of Riordan convolution arrays (lower triangular matrices) called Bell arrays. See the L. W. Shapiro et al. reference under A007318. If a Riordan array is named (G(z),F(z)) with F(z)=z*Fhat(z), the o.g.f. for the row polynomials is G(z)/(1-x*z*Fhat(z)), and it becomes a Bell array if G(z)=Fhat(z). For the present Bell type triangle G(z)=1/(1+z^2) (see the o.g.f. comment above). This leads to the o.g.f. for the column no. k, k >= 0, x^k/(1+x^2)^(k+1) (see the formula section), the one for the row sums and for the alternating row sums (see comments above). The Riordan (Bell) A- and Z-sequences (defined in a W. Lang link under A006232, with references) have o.g.f.s 1-x*c(x^2) and -x*c(x^2), with the o.g.f. of the Catalan numbers A000108. Together they lead to a recurrence given in the formula section. - Wolfdieter Lang, Nov 04 2011

The determinant of the N x N matrix S(N,[x[1], ..., x[N]]) with elements S(m-1,x[n]), for n, m = 1, 2, ..., N, and for any x[n], is identical with the determinant of V(N,[x[1], ..., x[N]]) with elements x[n]^(m-1) (a Vandermondian, which equals Product_{1 <= i < j<= N} (x[j] - x[i])). This is a special instance of a theorem valid for any N >= 1 and any monic polynomial system p(m,x), m>=0, with p(0,x) = 1. For this theorem see the Vein-Dale reference, p. 59. Thanks to L. Edson Jeffery for an email asking for a proof of the non-singularity of the matrix S(N,[x[1], ...., x[N]]) if and only if the x[j], j = 1..N, are pairwise distinct. - Wolfdieter Lang, Aug 26 2013

These S polynomials also appear in the context of modular forms. The rescaled Hecke operator T*n = n^((1-k)/2)*T_n acting on modular forms of weight k satisfies T*(p^n) = S(n, T*p), for each prime p and positive integer n. See the Koecher-Krieg reference, p. 223. - _Wolfdieter Lang, Jan 22 2016

For a shifted o.g.f. (mod signs), its compositional inverse, and connections to Motzkin and Fibonacci polynomials, non-crossing partitions and other combinatorial structures, see A097610. - Tom Copeland, Jan 23 2016

From M. Sinan Kul, Jan 30 2016; edited by Wolfdieter Lang, Jan 31 2016 and Feb 01 2016: (Start)

Solutions of the Diophantine equation u^2 + v^2 - k*u*v = 1 for integer k given by (u(k,n), v(k,n)) = (S(n,k), S(n-1,k)) because of the Cassini-Simson identity: S(n,x)^2 - S(n+1,x)*S(n-1, x) = 1, after use of the S-recurrence. Note that S(-n, x) = -S(-n-2, x), n >= 1, and the periodicity of some S(n, k) sequences.

Hence another way to obtain the row polynomials would be to take powers of the matrix [x, -1; 1,0]: S(n, x) = (([x, -1; 1, 0])^n)[1,1], n >= 0.

See also a Feb 01 2016 comment on A115139 for a well-known S(n, x) sum formula.

Then we have with the present T triangle

A039834(n) = -i^(n+1)*T(n-1, k) where i is the imaginary unit and n >= 0.

A051286(n) = Sum_{i=0..n} T(n,i)^2 (see the Philippe Deléham, Nov 21 2005 formula),

A181545(n) = Sum_{i=0..n+1} abs(T(n,i)^3),

A181546(n) = Sum_{i=0..n+1} T(n,i)^4,

A181547(n) = Sum_{i=0..n+1} abs(T(n,i)^5).

S(n, 0) = A056594(n), and for k = 1..10 the sequences S(n-1, k) with offset n = 0 are A128834, A001477, A001906, A001353, A004254, A001109, A004187, A001090, A018913, A004189.

(End)

For more on the Diophantine equation presented by Kul, see the Ismail paper. - Tom Copeland, Jan 31 2016

The o.g.f. for the Legendre polynomials L(n,x) is 1 / sqrt(1- 2x*z + z^2), and squaring it gives the o.g.f. of U(n,x), A053117, so Sum_{k=0..n} L(k,x/2) L(n-k,x/2) = S(n,x). This gives S(n,x) = L(n/2,x/2)^2 + 2*Sum_{k=0..n/2-1} L(k,x/2) L(n-k,x/2) for n even and S(n,x) = 2*Sum_{k=0..(n-1)/2} L(k,x/2) L(n-k,x/2) for odd n. For a connection to elliptic curves and modular forms, see A053117. For the normalized Legendre polynomials, see A100258. For other properties and relations to other polynomials, see Allouche et al. - Tom Copeland, Feb 04 2016

LG(x,h1,h2) = -log(1 - h1*x + h2*x^2) = Sum_{n>0} F(n,-h1,h2,0,..,0) x^n/n is a log series generator of the bivariate row polynomials of A127672 with A127672(0,0) = 0 and where F(n,b1,b2,..,bn) are the Faber polynomials of A263916. Exp(LG(x,h1,h2)) = 1 / (1 - h1*x + h2*x^2 ) is the o.g.f. of the bivariate row polynomials of this entry. - Tom Copeland, Feb 15 2016 (Instances of the bivariate o.g.f. for this entry are on pp. 5 and 18 of Sunada. - Tom Copeland, Jan 18 2021)

For distinct odd primes p and q the Legendre symbol can be written as Legendre(q,p) = Product_{k=1..P} S(q-1, 2*cos(2*Pi*k/p)), with P = (p-1)/2. See the Lemmermeyer reference, eq. (8.1) on p. 236. Using the zeros of S(q-1, x) (see above) one has S(q-1, x) = Product_{l=1..Q} (x^2 - (2*cos(Pi*l/q))^2), with Q = (q-1)/2. Thus S(q-1, 2*cos(2*Pi*k/p)) = ((-4)^Q)*Product_{l=1..Q} (sin^2(2*Pi*k/p) - sin^2(Pi*l/q)) = ((-4)^Q)*Product_{m=1..Q} (sin^2(2*Pi*k/p) - sin^2(2*Pi*m/q)). For the proof of the last equality see a W. Lang comment on the triangle A057059 for n = Q and an obvious function f. This leads to Eisenstein's proof of the quadratic reciprocity law Legendre(q,p) = ((-1)^(P*Q)) * Legendre(p,q), See the Lemmermeyer reference, pp. 236-237. - Wolfdieter Lang, Aug 28 2016

For connections to generalized Fibonacci polynomials, compare their generating function on p. 5 of the Amdeberhan et al. link with the o.g.f. given above for the bivariate row polynomials of this entry. - Tom Copeland, Jan 08 2017

The formula for Ramanujan's tau function (see A000594) for prime powers is tau(p^k) = p^(11*k/2)*S(k, p^(-11/2)*tau(p)) for k >= 1, and p = A000040(n), n >= 1. See the Hardy reference, p. 164, eqs. (10.3.4) and (10.3.6) rewritten in terms of S. - Wolfdieter Lang, Jan 27 2017

From Wolfdieter Lang, May 08 2017: (Start)

The number of zeros Z(n) of the S(n, x) polynomials in the open interval (-1,+1) is 2*b(n) for even n >= 0 and 1 + 2*b(n) for odd n >= 1, where b(n) = floor(n/2) - floor((n+1)/3). This b(n) is the number of integers k in the interval (n+1)/3 < k <= floor(n/2). See a comment on the zeros of S(n, x) above, and b(n) = A008615(n-2), n >= 0. The numbers Z(n) have been proposed (with a conjecture related to A008611) by Michel Lagneau, as the number of zeros of Fibonacci polynomials on the imaginary axis (-I,+I), with I=sqrt(-1). They are Z(n) = A008611(n-1), n >= 0, with A008611(-1) = 0. Also Z(n) = A194960(n-4), n >= 0. Proof using the A008611 version. A194960 follows from this.

In general the number of zeros Z(a;n) of S(n, x) for n >= 0 in the open interval (-a,+a) for a from the interval (0,2) (x >= 2 never has zeros, and a=0 is trivial: Z(0;n) = 0) is with b(a;n) = floor(n//2) - floor((n+1)*arccos(a/2)/Pi), as above Z(a;n) = 2*b(a;n) for even n >= 0 and 1 + 2*b(a;n) for odd n >= 1. For the closed interval [-a,+a] Z(0;n) = 1 and for a from (0,1) one uses for Z(a;n) the values b(a;n) = floor(n/2) - ceiling((n+1)*arccos(a/2)/Pi) + 1. (End)

The Riordan row polynomials S(n, x) (Chebyshev S) belong to the Boas-Buck class (see a comment and references in A046521), hence they satisfy the Boas-Buck identity: (E_x - n*1)*S(n, x) = (E_x + 1)*Sum_{p=0..n-1} (1 - (-1)^p)*(-1)^((p+1)/2)*S(n-1-p, x), for n >= 0, where E_x = x*d/dx (Euler operator). For the triangle T(n, k) this entails a recurrence for the sequence of column k, given in the formula section. - Wolfdieter Lang, Aug 11 2017

The e.g.f. E(x,t) := Sum_{n>=0} (t^n/n!)*S(n,x) for the row polynomials is obtained via inverse Laplace transformation from the above given o.g.f. as E(x,t) = ((1/xm)*exp(t/xm) - (1/xp)*exp(t/xp) )/(xp - xm) with xp = (x + sqrt(x^2-4))/2 and xm = (x - sqrt(x^2-4))/2. - Wolfdieter Lang, Nov 08 2017

From Wolfdieter Lang, Apr 12 2018: (Start)

Factorization of row polynomials S(n, x), for n >= 1, in terms of C polynomials (not Chebyshev C) with coefficients given in A187360. This is obtained from the factorization into Psi polynomials (see the Jul 12 2011 comment above) but written in terms of minimal polynomials of 2*cos(2*Pi/n) with coefficients in A232624:

S(2*k, x) = Product_{2 <= d | (2*k+1)} C(d, x)*(-1)^deg(d)*C(d, -x), with deg(d) = A055034(d) the degree of C(d, x).

S(2*k+1, x) = Product_{2 <= d | 2*(k+1)} C(d, x) * Product_{3 <= 2*d + 1 | (k+1)} (-1)^(deg(2*d+1))*C(2*d+1, -x).

Note that (-1)^(deg(2*d+1))*C(2*d+1, -x)*C(2*d+1, x) pairs always appear.

The number of C factors of S(2*k, x), for k >= 0, is 2*(tau(2*k+1) - 1) = 2*(A099774(k+1) - 1) = 2*A095374(k), and for S(2*k+1, x), for k >= 0, it is tau(2*(k+1)) + tau_{odd}(k+1) - 2 = A302707(k), with tau(2*k+1) = A099774(k+1), tau(n) = A000005 and tau(2*(k+1)) = A099777(k+1).

For the reverse problem, the factorization of C polynomials into S polynomials, see A255237. (End)

The S polynomials with general initial conditions S(a,b;n,x) = x*S(a,b;n-1,x) - S(a,b;n-2,x), for n >= 1, with S(a,b;-1,x) = a and S(a,b;0,x) = b are S(a,b;n,x) = b*S(n, x) - a*S(n-1, x), for n >= -1. Recall that S(-2, x) = -1 and S(-1, x) = 0. The o.g.f. is G(a,b;z,x) = (b - a*z)/(1 - x*z + z^2). - Wolfdieter Lang, Oct 18 2019

Also the convolution triangle of A101455. - Peter Luschny, Oct 06 2022

From Wolfdieter Lang, Apr 26 2023: (Start)

Multi-section of S-polynomials: S(m*n+k, x) = S(m+k, x)*S(n-1, R(m, x)) - S(k, x)*S(n-2, R(m, x)), with R(n, x) = S(n, x) - S(n-2, x) (see A127672), S(-2, x) = -1, and S(-1, x) = 0, for n >= 0, m >= 1, and k = 0, 1, ..., m-1.

O.g.f. of {S(m*n+k, y)}_{n>=0}: G(m,k,y,x) = (S(k, y) - (S(k, y)*R(m, y) - S(m+k, y))*x)/(1 - R(m,y)*x + x^2).

See eqs. (40) and (49), with r = x or y and s =-1, of the G. Detlefs and W. Lang link at A034807. (End)

S(n, x) for complex n and complex x: S(n, x) = ((-i/2)/sqrt(1 - (x/2)^2))*(q(x/2)*exp(+n*log(q(x/2))) - (1/q(x/2))*exp(-n*log(q(x/2)))), with q(x) = x + sqrt(1 - x^2)*i. Here log(z) = |z| + Arg(z)*i, with Arg(z) from [-Pi,+Pi) (principal branch). This satisfies the recurrence relation for S because it is derived from the Binet - de Moivre formula for S. Examples: S(n/m, 0) = cos((n/m)*Pi/4), for n >= 0 and m >= 1. S(n*i, 0) = (1/2)*(1 + exp(n*Pi))*exp(-(n/2)*Pi), for n >= 0. S(1+i, 2+i) = 0.6397424847... + 1.0355669490...*i. Thanks to Roberto Alfano for asking a question leading to this formula. - Wolfdieter Lang, Jun 05 2023

Lim_{n->oo} S(n, x)/S(n-1, x) = r(x) = (x - sqrt(x^2 -4))/2, for |x| >= 2. For x = +-2, this limit is +-1. - Wolfdieter Lang, Nov 15 2023

8*a(n)^2 + 1 = 8*A001110(n) + 1 = A055792(n+1) is a perfect square. - Gregory V. Richardson, Oct 05 2002

For n >= 2, A001108(n) gives exactly the positive integers m such that 1,2,...,m has a perfect median. The sequence of associated perfect medians is the present sequence. Let a_1,...,a_m be an (ordered) sequence of real numbers, then a term a_k is a perfect median if Sum_{j=1..k-1} a_j = Sum_{j=k+1..m} a_j. See Puzzle 1 in MSRI Emissary, Fall 2005. - Asher Auel, Jan 12 2006

(a(n), b(n)) where b(n) = A082291(n) are the integer solutions of the equation 2*binomial(b,a) = binomial(b+2,a). - Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de); comment revised by Michael Somos, Apr 07 2003

This sequence gives the values of y in solutions of the Diophantine equation x^2 - 8y^2 = 1. It also gives the values of the product xy where (x,y) satisfies x^2 - 2y^2 = +-1, i.e., a(n) = A001333(n)*A000129(n). a(n) also gives the inradius r of primitive Pythagorean triangles having legs whose lengths are consecutive integers, with corresponding semiperimeter s = a(n+1) = {A001652(n) + A046090(n) + A001653(n)}/2 and area rs = A029549(n) = 6*A029546(n). - Lekraj Beedassy, Apr 23 2003 [edited by Jon E. Schoenfield, May 04 2014]

n such that 8*n^2 = floor(sqrt(8)*n*ceiling(sqrt(8)*n)). - Benoit Cloitre, May 10 2003

For n > 0, ratios a(n+1)/a(n) may be obtained as convergents to continued fraction expansion of 3+sqrt(8): either successive convergents of [6;-6] or odd convergents of [5;1, 4]. - Lekraj Beedassy, Sep 09 2003

a(n+1) + A053141(n) = A001108(n+1). Generating floretion: - 2'i + 2'j - 'k + i' + j' - k' + 2'ii' - 'jj' - 2'kk' + 'ij' + 'ik' + 'ji' + 'jk' - 2'kj' + 2e ("jes" series). - Creighton Dement, Dec 16 2004

Kekulé numbers for certain benzenoids (see the Cyvin-Gutman reference). - Emeric Deutsch, Jun 19 2005

Number of D steps on the line y=x in all Delannoy paths of length n (a Delannoy path of length n is a path from (0,0) to (n,n), consisting of steps E=(1,0), N=(0,1) and D=(1,1)). Example: a(2)=6 because in the 13 (=A001850(2)) Delannoy paths of length 2, namely (DD), (D)NE, (D)EN, NE(D), NENE, NEEN, NDE, NNEE, EN(D), ENNE, ENEN, EDN and EENN, we have altogether six D steps on the line y=x (shown between parentheses). - Emeric Deutsch, Jul 07 2005

Define a T-circle to be a first-quadrant circle with integral radius that is tangent to the x- and y-axes. Such a circle has coordinates equal to its radius. Let C(0) be the T-circle with radius 1. Then for n > 0, define C(n) to be the smallest T-circle that does not intersect C(n-1). C(n) has radius a(n+1). Cf. A001653. - Charlie Marion, Sep 14 2005

Numbers such that there is an m with t(n+m)=2t(m), where t(n) are the triangular numbers A000217. For instance, t(20)=2*t(14)=210, so 6 is in the sequence. - Floor van Lamoen, Oct 13 2005

One half the bisection of the Pell numbers (A000129). - Franklin T. Adams-Watters, Jan 08 2006

Pell trapezoids: for n > 0, a(n) = (A000129(n-1)+A000129(n+1))*A000129(n)/2; see also A084158. - Charlie Marion, Apr 01 2006

Tested for 2 < p < 27: If and only if 2^p - 1 (the Mersenne number M(p)) is prime then M(p) divides a(2^(p-1)). - Kenneth J Ramsey, May 16 2006

If 2^p - 1 is prime then M(p) divides a(2^(p-1)-1). - Kenneth J Ramsey, Jun 08 2006; comment corrected by Robert Israel, Mar 18 2007

If 8*n+5 and 8*n+7 are twin primes then their product divides a(4*n+3). - Kenneth J Ramsey, Jun 08 2006

If p is an odd prime, then if p == 1 or 7 (mod 8), then a((p-1)/2) == 0 (mod p) and a((p+1)/2) == 1 (mod p); if p == 3 or 5 (mod 8), then a((p-1)/2) == 1 (mod p) and a((p+1)/2) == 0 (mod p). Kenneth J Ramsey's comment about twin primes follows from this. - Robert Israel, Mar 18 2007

a(n)*(a(n+b) - a(b-2)) = (a(n+1)+1)*(a(n+b-1) - a(b-1)). This identity also applies to any series a(0) = 0 a(1) = 1 a(n) = b*a(n-1) - a(n-2). - Kenneth J Ramsey, Oct 17 2007

For n < 0, let a(n) = -a(-n). Then (a(n+j) + a(k+j)) * (a(n+b+k+j) - a(b-j-2)) = (a(n+j+1) + a(k+j+1)) * (a(n+b+k+j-1) - a(b-j-1)). - Charlie Marion, Mar 04 2011

Sequence gives y values of the Diophantine equation: 0+1+2+...+x = y^2. If (a,b) and (c,d) are two consecutive solutions of the Diophantine equation: 0+1+2+...+x = y^2 with aMohamed Bouhamida, Aug 29 2009

If (p,q) and (r,s) are two consecutive solutions of the Diophantine equation: 0+1+2+...+x = y^2 with p < r then r = 3*p+4*q+1 and s = 2*p+3*q+1. - Mohamed Bouhamida, Sep 02 2009

a(n)/A002315(n) converges to cos^2(Pi/8) (see A201488). - Gary Detlefs, Nov 25 2009

Binomial transform of A086347. - Johannes W. Meijer, Aug 01 2010

If x=a(n), y=A055997(n+1) and z = x^2+y, then x^4 + y^3 = z^2. - Bruno Berselli, Aug 24 2010

In general, if b(0)=1, b(1)=k and for n > 1, b(n) = 6*b(n-1) - b(n-2), then

for n > 0, b(n) = a(n)*k-a(n-1); e.g.,

for k=2, when b(n) = A038725(n), 2 = 1*2 - 0, 11 = 6*2 - 1, 64 = 35*2 - 6, 373 = 204*2 - 35;

for k=3, when b(n) = A001541(n), 3 = 1*3 - 0, 17 = 6*3 - 1; 99 = 35*3 - 6; 577 = 204*3 - 35;

for k=4, when b(n) = A038723(n), 4 = 1*4 - 0, 23 = 6*4 - 1; 134 = 35*4 - 6; 781 = 204*4 - 35;

for k=5, when b(n) = A001653(n), 5 = 1*5 - 0, 29 = 6*5 - 1; 169 = 35*5 - 6; 985 = 204*5 - 35.

See also A002315, A054488, A038761, A054489, A054490.

- Charlie Marion, Dec 08 2010

See a Wolfdieter Lang comment on A001653 on a sequence of (u,v) values for Pythagorean triples (x,y,z) with x=|u^2-v^2|, y=2*u*v and z=u^2+v^2, with u odd and v even, generated from (u(0)=1,v(0)=2), the triple (3,4,5), by a substitution rule given there. The present a(n) appears there as b(n). The corresponding generated triangles have catheti differing by one length unit. - Wolfdieter Lang, Mar 06 2012

a(n)*a(n+2k) + a(k)^2 and a(n)*a(n+2k+1) + a(k)*a(k+1) are triangular numbers. Generalizes description of sequence. - Charlie Marion, Dec 03 2012

a(n)*a(n+2k) + a(k)^2 is the triangular square A001110(n+k). a(n)*a(n+2k+1) + a(k)*a(k+1) is the triangular oblong A029549(n+k). - Charlie Marion, Dec 05 2012

From Richard R. Forberg, Aug 30 2013: (Start)

The squares of a(n) are the result of applying triangular arithmetic to the squares, using A001333 as the "guide" on what integers to square, as follows:

a(2n)^2 = A001333(2n)^2 * (A001333(2n)^2 - 1)/2;

a(2n+1)^2 = A001333(2n+1)^2 * (A001333(2n+1)^2 + 1)/2. (End)

For n >= 1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,5}. - Milan Janjic, Jan 25 2015

Panda and Rout call these "balancing numbers" and note that the period of the sequence modulo a prime p is the same as that modulo p^2 when p = 13, 31, 1546463. But these are precisely the p in A238736 such that p^2 divides A000129(p - (2/p)), where (2/p) is a Jacobi symbol. In light of the above observation by Franklin T. Adams-Watters that the present sequence is one half the bisection of the Pell numbers, i.e., a(n) = A000129(2*n)/2, it follows immediately that modulo a fixed prime p, or any power thereof, the period of a(n) is half that of A000129(n). - John Blythe Dobson, Mar 06 2015

The triangular number = square number identity Tri((T(n, 3) - 1)/2) = S(n-1, 6)^2 with Tri, T, and S given in A000217, A053120 and A049310, is the special case k = 1 of the k-family of identities Tri((T(n, 2*k+1) - 1)/2) = Tri(k)*S(n-1, 2*(2*k+1))^2, k >= 0, n >= 0, with S(-1, x) = 0. For k=2 see A108741(n) for S(n-1, 10)^2. This identity boils down to the identities S(n-1, 2*x)^2 = (T(2*n, x) - 1)/(2*(x^2-1)) and 2*T(n, x)^2 - 1 = T(2*n, x) with x = 2*k+1. - Wolfdieter Lang, Feb 01 2016

a(2)=6 is perfect. For n=2*k, k > 0, k not equal to 1, a(n) is a multiple of a(2) and since every multiple (beyond 1) of a perfect number is abundant, then a(n) is abundant. sigma(a(4)) = 504 > 408 = 2*a(4). For n=2*k+1, k > 0, a(n) mod 10 = A000012(n), so a(n) is odd. If a(n) is a prime number, it is deficient; otherwise a(n) has one or two distinct prime factors and is therefore deficient again. So for n=2k+1, k > 0, a(n) is deficient. sigma(a(5)) = 1260 < 2378 = 2*a(5). - Muniru A Asiru, Apr 14 2016

Behera & Panda call these the balancing numbers, and A001541 are the balancers. - Michel Marcus, Nov 07 2017

In general, a second-order linear recurrence with constant coefficients having a signature of (c,d) will be duplicated by a third-order recurrence having a signature of (x,c^2-c*x+d,-d*x+c*d). The formulas of Olivares and Bouhamida in the formula section which have signatures of (7,-7,1) and (5,5,-1), respectively, are specific instances of this general rule for x = 7 and x = 5. - Gary Detlefs, Jan 29 2021

Note that 6 is the largest triangular number in the sequence, because it is proved that 8 and 9 are the largest perfect powers which are consecutive (Catalan's conjecture). 0 and 1 are also in the sequence because they are also perfect powers and 0*1/2 = 0^2 and 8*9/2 = (2*3)^2. - Metin Sariyar, Jul 15 2021

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)

Chebyshev polynomials of the first kind evaluated at 3.

This sequence gives the values of x in solutions of the Diophantine equation x^2 - 8*y^2 = 1, the corresponding values of y are in A001109. For n > 0, the ratios a(n)/A001090(n) may be obtained as convergents to sqrt(8): either successive convergents of [3; -6] or odd convergents of [2; 1, 4]. - Lekraj Beedassy, Sep 09 2003 [edited by Jon E. Schoenfield, May 04 2014]

Also gives solutions to the equation x^2 - 1 = floor(x*r*floor(x/r)) where r = sqrt(8). - Benoit Cloitre, Feb 14 2004

Appears to give all solutions greater than 1 to the equation: x^2 = ceiling(x*r*floor(x/r)) where r = sqrt(2). - Benoit Cloitre, Feb 24 2004

This sequence give numbers n such that (n-1)*(n+1)/2 is a perfect square. Remark: (i-1)*(i+1)/2 = (i^2-1)/2 = -1 = i^2 with i = sqrt(-1) so i is also in the sequence. - Pierre CAMI, Apr 20 2005

a(n) is prime for n = {1, 2, 4, 8}. Prime a(n) are {3, 17, 577, 665857}, which belong to A001601(n). a(2k-1) is divisible by a(1) = 3. a(4k-2) is divisible by a(2) = 17. a(8k-4) is divisible by a(4) = 577. a(16k-8) is divisible by a(8) = 665857. - Alexander Adamchuk, Nov 24 2006

The upper principal convergents to 2^(1/2), beginning with 3/2, 17/12, 99/70, 577/408, comprise a strictly decreasing sequence; essentially, numerators=A001541 and denominators=A001542. - Clark Kimberling, Aug 26 2008

Also index of sequence A082532 for which A082532(n) = 1. - Carmine Suriano, Sep 07 2010

Numbers n such that sigma(n-1) and sigma(n+1) are both odd numbers. - Juri-Stepan Gerasimov, Mar 28 2011

Also, numbers such that floor(a(n)^2/2) is a square: base 2 analog of A031149, A204502, A204514, A204516, A204518, A204520, A004275, A001075. - M. F. Hasler, Jan 15 2012

Numbers such that 2n^2 - 2 is a square. Also integer square roots of the expression 2*n^2 + 1, at values of n given by A001542. Also see A228405 regarding 2n^2 -+ 2^k generally for k >= 0. - Richard R. Forberg, Aug 20 2013

Values of x (or y) in the solutions to x^2 - 6xy + y^2 + 8 = 0. - Colin Barker, Feb 04 2014

Panda and Ray call the numbers in this sequence the Lucas-balancing numbers C_n (see references and links).

Partial sums of X or X+1 of Pythagorean triples (X,X+1,Z). - Peter M. Chema, Feb 03 2017

a(n)/A001542(n) is the closest rational approximation to sqrt(2) with a numerator not larger than a(n), and 2*A001542(n)/a(n) is the closest rational approximation to sqrt(2) with a denominator not larger than a(n). These rational approximations together with those obtained from the sequences A001653 and A002315 give a complete set of closest rational approximations to sqrt(2) with restricted numerator or denominator. a(n)/A001542(n) > sqrt(2) > 2*A001542(n)/a(n). - A.H.M. Smeets, May 28 2017

x = a(n), y = A001542(n) are solutions of the Diophantine equation x^2 - 2y^2 = 1 (Pell equation). x = 2*A001542(n), y = a(n) are solutions of the Diophantine equation x^2 - 2y^2 = -2. Both together give the set of fractional approximations for sqrt(2) obtained from limited fractions obtained from continued fraction representation to sqrt(2). - A.H.M. Smeets, Jun 22 2017

a(n) is the radius of the n-th circle among the sequence of circles generated as follows: Starting with a unit circle centered at the origin, every subsequent circle touches the previous circle as well as the two limbs of hyperbola x^2 - y^2 = 1, and lies in the region y > 0. - Kaushal Agrawal, Nov 10 2018

All of the positive integer solutions of a*b+1=x^2, a*c+1=y^2, b*c+1=z^2, x+z=2*y, 0A001542(n), b=A005319(n), c=A001542(n+1), x=A001541(n), y=A001653(n+1), z=A002315(n) with 0Michael Somos, Jun 26 2022

Indices of square numbers which are also generalized pentagonal numbers.

If t(n) denotes the n-th triangular number, t(A105038(n))=a(n)*a(n+1). - Robert Phillips (bobanne(AT)bellsouth.net), May 25 2008

The n-th term is a(n) = ((5+sqrt(24))^n - (5-sqrt(24))^n)/(2*sqrt(24)). - Sture Sjöstedt, May 31 2009

For n >= 2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 10'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

a(n) and b(n) (A001079) are the nonnegative proper solutions of the Pell equation b(n)^2 - 6*(2*a(n))^2 = +1. See the cross reference to A001079 below. - Wolfdieter Lang, Jun 26 2013

For n >= 1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,9}. - Milan Janjic, Jan 25 2015

For n > 1, this also gives the number of (n-1)-decimal-digit numbers which avoid a particular two-digit number with distinct digits. For example, there are a(5) = 9701 4-digit numbers which do not include "39" as a substring; see Wikipedia link. - Charles R Greathouse IV, Jan 14 2016

All possible solutions for y in Pell equation x^2 - 24*y^2 = 1. The values for x are given in A001079. - Herbert Kociemba, Jun 05 2022

Dickson on page 384 gives the Diophantine equation "(20) 24x^2 + 1 = y^2" and later states "... three consecutive sets (x_i, y_i) of solutions of (20) or 2x^2 + 1 = 3y^2 satisfy x_{n+1} = 10x_n - x_{n-1}, y_{n+1} = 10y_n - y_{n-1} with (x_1, y_1) = (0, 1) or (1, 1), (x_2, y_2) = (1, 5) or (11, 9), respectively." The first set of values (x_n, y_n) = (A001079(n-1), a(n-1)). - Michael Somos, Jun 19 2023

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

For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 16'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>=2, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,15}. - Milan Janjic, Jan 23 2015

For the difference equation a(n) = c*a(n-1) - d*a(n-2), with a(0) = 0, a(1) = 1, the solution is a(n) = d^((n-1)/2) * ChebyshevU(n-1, c/(2*sqrt(d))) and has the alternate form a(n) = ( ((c + sqrt(c^2 - 4*d))/2)^n - ((c - sqrt(c^2 - 4*d))/2)^n )/sqrt(c^2 - 4*d). In the case c^2 = 4*d then the solution is a(n) = n*d^((n-1)/2). The generating function is x/(1 - c*x + d^2) and the exponential generating function takes the form (2/sqrt(c^2 - 4*d))*exp(c*x/2)*sinh(sqrt(c^2 - 4*d)*x/2) for c^2 > 4*d, (2/sqrt(4*d - c^2))*exp(c*x/2)*sin(sqrt(4*d - c^2)*x/2) for 4*d > c^2, and x*exp(sqrt(d)*x) if c^2 = 4*d. - G. C. Greubel, Jun 10 2022