cp's OEIS Frontend

Number of n-step non-selfintersecting paths starting at (0,0) with steps of types (1,0), (-1,0) or (0,1) [Stanley].

Number of n steps one-sided prudent walks with east, west and north steps. - Shanzhen Gao, Apr 26 2011

Number of ternary strings of length n-1 with subwords (0,2) and (2,0) not allowed. - Olivier Gérard, Aug 28 2012

Number of symmetric 2n X 2 or (2n-1) X 2 crossword puzzle grids: all white squares are edge connected; at least 1 white square on every edge of grid; 180-degree rotational symmetry. - Erich Friedman

a(n+1) is the number of ways to put molecules on a 2 X n ladder lattice so that the molecules do not touch each other.

In other words, a(n+1) is the number of independent vertex sets and vertex covers in the n-ladder graph P_2 X P_n. - Eric W. Weisstein, Apr 04 2017

Number of (n-1) X 2 binary arrays with a path of adjacent 1's from top row to bottom row, see A359576. - R. H. Hardin, Mar 16 2002

a(2*n+1) with b(2*n+1) := A000129(2*n+1), n >= 0, give all (positive integer) solutions to Pell equation a^2 - 2*b^2 = -1.

a(2*n) with b(2*n) := A000129(2*n), n >= 1, give all (positive integer) solutions to Pell equation a^2 - 2*b^2 = +1 (see Emerson reference).

Bisection: a(2*n) = T(n,3) = A001541(n), n >= 0 and a(2*n+1) = S(2*n,2*sqrt(2)) = A002315(n), n >= 0, with T(n,x), resp. S(n,x), Chebyshev's polynomials of the first, resp. second kind. See A053120, resp. A049310.

Binomial transform of A077957. - Paul Barry, Feb 25 2003

For n > 0, the number of (s(0), s(1), ..., s(n)) such that 0 < s(i) < 4 and |s(i) - s(i-1)| <= 1 for i = 1,2,...,n, s(0) = 2, s(n) = 2. - Herbert Kociemba, Jun 02 2004

For n > 1, a(n) corresponds to the longer side of a near right-angled isosceles triangle, one of the equal sides being A000129(n). - Lekraj Beedassy, Aug 06 2004

Exponents of terms in the series F(x,1), where F is determined by the equation F(x,y) = xy + F(x^2*y,x). - Jonathan Sondow, Dec 18 2004

Number of n-words from the alphabet A={0,1,2} which two neighbors differ by at most 1. - Fung Cheok Yin (cheokyin_restart(AT)yahoo.com.hk), Aug 30 2006

Consider the mapping f(a/b) = (a + 2b)/(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, 3/2, 7/5, 17/12, 41/29, ... converging to 2^(1/2). Sequence contains the numerators. - Amarnath Murthy, Mar 22 2003 [Amended by Paul E. Black (paul.black(AT)nist.gov), Dec 18 2006]

Odd-indexed prime numerators are prime RMS numbers (A140480) and also NSW primes (A088165). - Ctibor O. Zizka, Aug 13 2008

The intermediate convergents to 2^(1/2) begin with 4/3, 10/7, 24/17, 58/41; essentially, numerators=A052542 and denominators here. - Clark Kimberling, Aug 26 2008

Equals right border of triangle A143966. Starting (1, 3, 7, ...) equals INVERT transform of (1, 2, 2, 2, ...) and row sums of triangle A143966. - Gary W. Adamson, Sep 06 2008

Inverse binomial transform of A006012; Hankel transform is := [1, 2, 0, 0, 0, 0, 0, 0, 0, ...]. - Philippe Deléham, Dec 04 2008

From Charlie Marion, Jan 07 2009: (Start)

In general, denominators, a(k,n) and numerators, b(k,n), of continued fraction convergents to sqrt((k+1)/k) may be found as follows:

let a(k,0) = 1, a(k,1) = 2k; for n>0, a(k,2n) = 2*a(k,2n-1) + a(k,2n-2) and a(k,2n+1) = (2k)*a(k,2n) + a(k,2n-1);

let b(k,0) = 1, b(k,1) = 2k+1; for n>0, b(k,2n) = 2*b(k,2n-1) + b(k,2n-2) and b(k,2n+1) = (2k)*b(k,2n) + b(k,2n-1).

For example, the convergents to sqrt(2/1) start 1/1, 3/2, 7/5, 17/12, 41/29.

In general, if a(k,n) and b(k,n) are the denominators and numerators, respectively, of continued fraction convergents to sqrt((k+1)/k) as defined above, then

k*a(k,2n)^2 - a(k,2n-1)*a(k,2n+1) = k = k*a(k,2n-2)*a(k,2n) - a(k,2n-1)^2 and

b(k,2n-1)*b(k,2n+1) - k*b(k,2n)^2 = k+1 = b(k,2n-1)^2 - k*b(k,2n-2)*b(k,2n);

for example, if k=1 and n=3, then b(1,n)=a(n+1) and

1*a(1,6)^2 - a(1,5)*a(1,7) = 1*169^2 - 70*408 = 1;

1*a(1,4)*a(1,6) - a(1,5)^2 = 1*29*169 - 70^2 = 1;

b(1,5)*b(1,7) - 1*b(1,6)^2 = 99*577 - 1*239^2 = 2;

b(1,5)^2 - 1*b(1,4)*b(1,6) = 99^2 - 1*41*239 = 2.

Cf. A000129, A142238-A142239, A153313-A153318.

(End)

This sequence occurs in the lower bound of the order of the set of equivalent resistances of n equal resistors combined in series and in parallel (A048211). - Sameen Ahmed Khan, Jun 28 2010

Let M = a triangle with the Fibonacci series in each column, but the leftmost column is shifted upwards one row. A001333 = lim_{n->infinity} M^n, the left-shifted vector considered as a sequence. - Gary W. Adamson, Jul 27 2010

a(n) is the number of compositions of n when there are 1 type of 1 and 2 types of other natural numbers. - Milan Janjic, Aug 13 2010

Equals the INVERTi transform of A055099. - Gary W. Adamson, Aug 14 2010

From L. Edson Jeffery, Apr 04 2011: (Start)

Let U be the unit-primitive matrix (see [Jeffery])

U = U_(8,2) = (0 0 1 0)

(0 1 0 1)

(1 0 2 0)

(0 2 0 1).

Then a(n) = (1/4)*Trace(U^n). (See also A084130, A006012.)

(End)

For n >= 1, row sums of triangle

m/k.|..0.....1.....2.....3.....4.....5.....6.....7

==================================================

.0..|..1

.1..|..1.....2

.2..|..1.....2.....4

.3..|..1.....4.....4.....8

.4..|..1.....4....12.....8....16

.5..|..1.....6....12....32....16....32

.6..|..1.....6....24....32....80....32....64

.7..|..1.....8....24....80....80...192....64...128

which is the triangle for numbers 2^k*C(m,k) with duplicated diagonals. - Vladimir Shevelev, Apr 12 2012

a(n) is also the number of ways to place k non-attacking wazirs on a 2 X n board, summed over all k >= 0 (a wazir is a leaper [0,1]). - Vaclav Kotesovec, May 08 2012

The sequences a(n) and b(n) := A000129(n) are entries of powers of the special case of the Brahmagupta Matrix - for details see Suryanarayan's paper. Further, as Suryanarayan remark, if we set A = 2*(a(n) + b(n))*b(n), B = a(n)*(a(n) + 2*b(n)), C = a(n)^2 + 2*a(n)*b(n) + 2*b(n)^2 we obtain integral solutions of the Pythagorean relation A^2 + B^2 = C^2, where A and B are consecutive integers. - Roman Witula, Jul 28 2012

Pisano period lengths: 1, 1, 8, 4, 12, 8, 6, 4, 24, 12, 24, 8, 28, 6, 24, 8, 16, 24, 40, 12, .... - R. J. Mathar, Aug 10 2012

This sequence and A000129 give the diagonal numbers described by Theon of Smyrna. - Sture Sjöstedt, Oct 20 2012

a(n) is the top left entry of the n-th power of any of the following six 3 X 3 binary matrices: [1, 1, 1; 1, 1, 1; 1, 0, 0] or [1, 1, 1; 1, 1, 0; 1, 1, 0] or [1, 1, 1; 1, 0, 1; 1, 1, 0] or [1, 1, 1; 1, 1, 0; 1, 0, 1] or [1, 1, 1; 1, 0, 1; 1, 0, 1] or [1, 1, 1; 1, 0, 0; 1, 1, 1]. - R. J. Mathar, Feb 03 2014

If p is prime, a(p) == 1 (mod p) (compare with similar comment for A000032). - Creighton Dement, Oct 11 2005, modified by Davide Colazingari, Jun 26 2016

a(n) = A000129(n) + A000129(n-1), where A000129(n) is the n-th Pell Number; e.g., a(6) = 99 = A000129(6) + A000129(5) = 70 + 29. Hence the sequence of fractions has the form 1 + A000129(n-1)/A000129(n), and the ratio A000129(n-1)/A000129(n)converges to sqrt(2) - 1. - Gregory L. Simay, Nov 30 2018

For n > 0, a(n+1) is the length of tau^n(1) where tau is the morphism: 1 -> 101, 0 -> 1. See Song and Wu. - Michel Marcus, Jul 21 2020

For n > 0, a(n) is the number of nonisomorphic quasitrivial semigroups with n elements, see Devillet, Marichal, Teheux. A292932 is the number of labeled quasitrivial semigroups. - Peter Jipsen, Mar 28 2021

a(n) is the permanent of the n X n tridiagonal matrix defined in A332602. - Stefano Spezia, Apr 12 2022

From Greg Dresden, May 08 2023: (Start)

For n >= 2, 4*a(n) is the number of ways to tile this T-shaped figure of length n-1 with two colors of squares and one color of domino; shown here is the figure of length 5 (corresponding to n=6), and it has 4*a(6) = 396 different tilings.

_

|| _

|||_|||

|_|

(End)

12*a(n) = number of walks of length n in the cyclic Kautz digraph CK(3,4). - Miquel A. Fiol, Feb 15 2024

Consider all Pythagorean triples (X,X+1,Z) ordered by increasing Z; sequence gives Z values.

The defining equation is X^2 + (X+1)^2 = Z^2, which when doubled gives 2Z^2 = (2X+1)^2 + 1. So the sequence gives Z's such that 2Z^2 = odd square + 1 (A069894).

(x,y) = (a(n), a(n+1)) are the solutions with x < y of x/(yz) + y/(xz) + z/(xy)=3 with z=2. - Floor van Lamoen, Nov 29 2001

Consequently the sum n^2*(2n^2 - 1) of the first n odd cubes (A002593) is also a square. - Lekraj Beedassy, Jun 05 2002

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

Also, number of domino tilings in S_5 X P_2n. - Ralf Stephan, Mar 30 2004. Here S_5 is the star graph on 5 vertices with the edges {1,2}, {1,3}, {1,4}, {1,5}.

If x is in the sequence then so is x*(8*x^2-3). - James R. Buddenhagen, Jan 13 2005

In general, Sum_{k=0..n} binomial(2n-k,k)j^(n-k) = (-1)^n*U(2n,i*sqrt(j)/2), i=sqrt(-1). - Paul Barry, Mar 13 2005

a(n) = L(n,6), where L is defined as in A108299; see also A002315 for L(n,-6). - Reinhard Zumkeller, Jun 01 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 largest T-circle that intersects C(n-1). C(n) has radius a(n) and the coordinates of its points of intersection with C(n-1) are A001108(n) and A055997(n). Cf. A001109. - Charlie Marion, Sep 14 2005

Number of 01-avoiding words of length n on alphabet {0,1,2,3,4,5} which do not end in 0. - Tanya Khovanova, Jan 10 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 = {a(n)}. - Clark Kimberling, Aug 26 2008

Apparently Ljunggren shows that 169 is the last square term.

If (p,q) is a solution of the Diophantine equation: X^2 + (X+1)^2 = Y^2 then (p+q) or (p+q+1) are perfect squares. If (p,q) is a solution of the Diophantine equation: X^2 + (X+1)^2 = Y^2 then (p+q) or (p+q)/8 are perfect squares. If (p,q) and (r,s) are two consecutive solutions of the Diophantine equation: X^2 + (X+1)^2 = Y^2 with p < r then s-r = p+q+1. - Mohamed Bouhamida, Aug 29 2009

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

Equals INVERT transform of A005054: (1, 4, 20, 100, 500, 2500, ...) and INVERTi transform of A122074: (1, 6, 40, 268, 1796, ...). - Gary W. Adamson, Jul 22 2010

a(n) is the number of compositions of n when there are 5 types of 1 and 4 types of other natural numbers. - Milan Janjic, Aug 13 2010

The remainder after division of a(n) by a(k) appears to belong to a periodic sequence: 1, 5, ..., a(k-1), 0, a(k)-a(k-1), ..., a(k)-1, a(k)-1, ..., a(k)-a(k-1), 0, a(k-1), ..., 5, 1. See Bouhamida's Sep 01 2009 comment. - Charlie Marion, May 02 2011

Apart from initial 1: subsequence of A198389, see also A198385. - Reinhard Zumkeller, Oct 25 2011

(a(n+1), 2*b(n+1)) and (a(n+2), 2*b(n+1)), n >= 0, with b(n):= A001109(n), give the (u(2*n), v(2*n)) and (u(2*n+1), v(2*n+1)) sequences, respectively, for Pythagorean triples (x,y,z), where x=|u^2-v^2|, y=2*u*v and z=u^2+v^2, with u odd and v even, which are generated from (u(0)=1, v(0)=2) by the substitution rule (u,v) -> (2*v+u,v) if u < v and (u,v) -> (u,2*u+v) if u > v. This leads to primitive triples because gcd(u,v) = 1 is respected. This corresponds to (primitive) Pythagorean triangles with |x-y|=1 (the catheti differ by one length unit). This (u,v) sequence starts with (1,2), (5,2), (5,12), (29,12), (29,70) ... - Wolfdieter Lang, Mar 06 2012

Area of the Fibonacci snowflake of order n. - José Luis Ramírez Ramírez, Dec 13 2012

Area of the 3-generalized Fibonacci snowflake of order n, n >= 3. - José Luis Ramírez Ramírez, Dec 13 2012

For the o.g.f. given by Johannes W. Meijer, Aug 01 2010, in the formula section see a comment under A077445. - Wolfdieter Lang, Jan 18 2013

Positive values of x (or y) satisfying x^2 - 6xy + y^2 + 4 = 0. - Colin Barker, Feb 04 2014

Length of period of the continued fraction expansion of a(n)*sqrt(2) is 1, the corresponding repeating value is A077444(n). - Ralf Stephan, Feb 20 2014

Positive values of x (or y) satisfying x^2 - 34xy + y^2 + 144 = 0. - Colin Barker, Mar 04 2014

The value of the hypotenuse in each triple of the Tree of primitive Pythagorean triples (cf. Wikipedia link) starting with root (3,4,5) and recursively selecting the central branch at each triple node of the tree. - Stuart E Anderson, Feb 05 2015

Positive integers z such that z^2 is a centered square number (A001844). - Colin Barker, Feb 12 2015

The aerated sequence (b(n)) n >= 1 = [1, 0, 5, 0, 29, 0, 169, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -8, Q = 1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for the connection with Chebyshev polynomials. - Peter Bala, Mar 25 2015

A002315(n-1)/a(n) is the closest rational approximation of sqrt(2) with a denominator not larger than a(n). 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. A002315(n-1)/a(n) < sqrt(2). - A.H.M. Smeets, May 28 2017

Equivalently, numbers x such that (x-1)*x/2 + x*(x+1)/2 = y^2 + (y+1)^2. y-values are listed in A001652. Example: for x=29 and y=20, 28*29/2 + 29*30/2 = 20^2 + 21^2. - Bruno Berselli, Mar 19 2018

From Wolfdieter Lang, Jun 13 2018: (Start)

(a(n), a(n+1)), with a(0):= 1, give all proper positive solutions m1 = m1(n) and m2 = m2(n), with m1 < m2 and n >= 0, of the Markoff triple (m, m1, m2) (see A002559) for m = 2, i.e., m1^2 - 6*m1*m2 + m2^2 = -4. Hence the unique Markoff triple with largest value m = 2 is (1, 1, 2) (for general m from A002559 this is the famous uniqueness conjecture).

For X = m2 - m1 and Y = m2 this becomes the reduced indefinite quadratic form representation X^2 + 4*X*Y - 4*Y^2 = -4, with discriminant 32, and the only proper fundamental solution (X(0), Y(0)) = (0, 1). For all nonnegative proper (X(n), Y(n)) solutions see (A005319(n) = a(n+1) - a(n), a(n+1)), for n >= 0. (End)

Each Pell(2*k+1) = a(k+1) number with k >= 3 appears as largest number of an ordered Markoff (Markov) triple [x, y, m] with smallest value x = 2 as [2, Pell(2*k-1), Pell(2*k+1)]. This known result follows also from all positive proper solutions of the Pell equation q^2 - 2*m^2 = -1 which are q = q(k) = A002315(k) and m = m(k) = Pell(2*k+1), for k >= 0. y = y(k) = m(k) - 2*q(k) = Pell(2*k-1), with Pell(-1) = 1. The k = 0 and 1 cases do not satisfy x=2 <= y(k) <= m(k). The numbers 1 and 5 appear also as largest Markoff triple members because they are also Fibonacci numbers, and for these triples x=1. - Wolfdieter Lang, Jul 11 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, 0 < a < b < c are given by a=A001542(n), b=A005319(n), c=A001542(n+1), x=A001541(n), y=a(n+1), z=A002315(n) with 0 < n. - Michael Somos, Jun 26 2022

Two times Chebyshev polynomials of the first kind evaluated at 3.

Also 2(a(2*n)-2) and a(2*n+1)-2 are perfect squares. - Mario Catalani (mario.catalani(AT)unito.it), Mar 31 2003

Chebyshev polynomials of the first kind evaluated at 3, then multiplied by 2. - Michael Somos, Apr 07 2003

Also gives solutions > 2 to the equation x^2 - 3 = floor(x*r*floor(x/r)) where r=sqrt(2). - Benoit Cloitre, Feb 14 2004

Output of Lu and Wu's formula for the number of perfect matchings of an m X n Klein bottle where m and n are both even specializes to this sequence for m=2. - Sarah-Marie Belcastro, Jul 04 2009

It appears that for prime P = 8*n +- 3, that a((P-1)/2) == -6 (mod P) and for all composites C = 8*n +- 3, there is at least one i < (C-1)/2 such that a(i) == -6 (mod P). Only a few of the primes P of the form 8*n +-3, e.g., 29, had such an i less than (P-1)/2. As for primes P = 8*n +- 1, it seems that the sum of the two adjacent terms, a((P-1)/2) and a((P+1)/2), is congruent to 8 (mod P). - Kenneth J Ramsey, Feb 14 2012 and Mar 05 2012

For n >= 1, a(n) is also the curvature of circles (rounded to the nearest integer) successively inscribed toward angle 90 degree of tangent lines, starting with a unit circle. The expansion factor is 5.828427... or 1/(3 - 2*sqrt(2)), which is also 3 + 2*sqrt(2) or A156035. See illustration in links. - Kival Ngaokrajang, Sep 04 2013

Except for the first term, positive values of x (or y) satisfying x^2 - 6*x*y + y^2 + 32 = 0. - Colin Barker, Feb 08 2014

The equation "(k^2 + 4)/2 is a square" is a version of the generalized Pell Equation x^2 - D*y^2 = C where x^2 - 2*y^2 = -4.

Sequence of all positive integers k such that continued fraction [k,k,k,k,k,k,...] belongs to Q(sqrt(2)). - Thomas Baruchel, Sep 15 2003

Equivalently, 2*n^2 + 8 is a square.

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

The continued fraction [a(n);a(n),a(n),...] = (1 + sqrt(2))^(2*n-1). - Thomas Ordowski, Jun 07 2013

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

A floretion-generated sequence relating to NSW numbers and numbers n such that (n^2 - 8)/2 is a square. It is also possible to label this sequence as the "tesfor-transform of the zero-sequence" under the floretion given in the program code, below. This is because the sequence "vesseq" would normally have been A046184 (indices of octagonal numbers which are also a square) using the floretion given. This floretion, however, was purposely "altered" in such a way that the sequence "vesseq" would turn into A000004. As (a(n)) would not have occurred under "natural" circumstances, one could speak of it as the transform of A000004.

Floretion Algebra Multiplication Program FAMP code: - tesforseq[ + 3'i - 2'j + 'k + 3i' - 2j' + k' - 4'ii' - 3'jj' + 4'kk' - 'ij' - 'ji' + 3'jk' + 3'kj' + 4e], Note: vesforseq = A000004, lesforseq = A002315, jesforseq = A077445

From Wolfdieter Lang, Feb 05 2015: (Start)

All positive solutions x = a(n) of the (generalized) Pell equation x^2 - 2*y^2 = +7 based on the fundamental solution (x2,y2) = (5,3) of the second class of (proper) solutions. The corresponding y solutions are given by y(n) = A253811(n).

All other positive solutions come from the first class of (proper) solutions based on the fundamental solution (x1,y1) = (3,1). These are given in A038762 and A038761.

All solutions of this Pell equation are found in A077443(n+1) and A077442(n), for n >= 0. See, e.g., the Nagell reference on how to find all solutions.

(End)

Formally Sum_{n>=0} a(n)*(tan(x))^(2*n) = tan(5*x)/ tan(x).

Convergence holds for x from the two open intervals (1-sqrt(6/5), 1-2/sqrt(5)) and (1+2/sqrt(5), 1+sqrt(6/5)), namely (-0.095445115, 0.105572809) and (1.894427191, 2.095445115) (10 digits).

These intervals follow from the denominator of the o.g.f. G(5,x) = (5 - 10*x + x^2)/(1 - 10*x + 5*x^2).

If one replaces x by (tan(x))^2 in this o.g.f. one obtains the formula for tan(5*x)/tan(x) in terms of (tan(x))^2. This formula is the special (n=5) solution of a general recurrence derivable from the addition theorem for tan(n*x) = tan(x + (n-1)*x), namely, with Q(n,x) := tan(n,x)/tan(x), Q(n,x) = (1 + Q(n-1,x))/(1 - v*Q(n-1,x)), where v = v(x) =(tan(x))^2, and the input is Q(1,x) = 1. Read as function of v the solution for Q(5,x) is just G(5,v) with replaced v=v(x).

See the irregular triangles A034867 and A034839 whose row polynomials N(n,x) and D(n,x), respectively, give for n >= 1 the solution to the recurrences N(n,x) = D(n-1,x) + N(n-1,x), D(n,x) = D(n-1,x) + x*N(n-1,x), with inputs N(1,x) = 1 and D(1,x) = 1. The proof by the Pascal triangle A007318 recurrence is trivial. Therefore, Q(n,x) from the preceding comment is given by Q(n,x) = N(n,-v)/D(n,-v) with v=v(x) = (tan(x))^2.

One also has, with Chebyshev's S polynomials (see A049310) Q(n,x) = tan(n*x)/tan(x) = (S(n,y) + S(n-2,y))/(S(n,y) - S(n-2,y)) = y*S(n-1,y)/(S(n,y) - S(n-2,y)) = 1/(1 - (2/y)*S(n-2,y)/S(n-1,y)), where y = y(x) = 2/sqrt(1 + (tan(x))^2). This derives from sin(n*x)/cos(n*x) in terms of Chebyshev S polynomials with argument 2*cos(x) = y(x). Note that S(n-2,y)/S(n-1,y) has the continued fraction representation 1/(y-1/(y- ... -1/(y )..(n-1)brackets..), i.e. (n-1) y's.

These calculations have been motivated by e-mails from Thomas Olsen.

A floretion-generated sequence relating the squares of the numerators of continued fraction convergents to sqrt(2) to the squares of the denominators of continued fraction convergents to sqrt(2) (Pell numbers).

Floretion Algebra Multiplication Program, FAMP Code:

1dia[J]tesseq[ - .5'j + .5'k - .5j' + .5k' - 2'ii' + 'jj' - 'kk' + .5'ij' + .5'ik' + .5'ji' + 'jk' + .5'ki' + 'kj' + e ]. Identity used: dia[I]tes + dia[J]tes + dia[K]tes = jes + fam + 3tes.