cp's OEIS Frontend

We observe that b(n) = log(a(n))/log(2) = A120738(n). Furthermore c(n+1) = b(n+1)-b(n) = A090739(n+1) and c(n+1)-3 = A007814(n+1) for n>=0. - Johannes W. Meijer, Jul 06 2009

Using WolframAlpha, it appears that 2*a(n) gives the coefficients of Pi in the denominators of the residues of f(z) = 2z choose z at odd negative half integers. E.g., the residues of f(z) at z = -1/2, -3/2, -5/2 are 1/(2*Pi), 1/(16*Pi), and 3/(256*Pi) respectively. - Nicholas Juricic, Mar 31 2022

Numerators of sequence of fractions with e.g.f. 1/((1-x)*(1+x)^(1/2)). The denominators are successive powers of 2.

a(n) is the coefficient of x^n in arctan(sqrt(2*x/(1-x)))/sqrt(2*x*(1-x)) multiplied by (2*n+1)!!.

This sequence is the linking pin between the a(n) formulas of the ED1, ED2, ED3 and ED4 array rows, see A167552, A167565, A167580 and A167591. - Johannes W. Meijer, Nov 23 2009

Also numerator of rational part of Haar measure on Grassmannian space G(n,1).

Also rational part of numerator of Gamma(n/2+1)/Gamma(n/2+1/2) (cf. A036039).

Let x(m) = x(m-2) + 1/x(m-1) for m >= 3, with x(1)=x(2)=1. Then the numerator of x(n+2) equals the denominator of n!!/(n+1)!! for n >= 0, where the double factorials are given by A006882. - Joseph E. Cooper III (easonrevant(AT)gmail.com), Nov 07 2010, as corrected in Cooper (2015).

Numerator of (n-1)/( (n-2)/( .../1)), with an empty fraction taken to be 1. - Flávio V. Fernandes, Jan 31 2025

This is the row reversed version of A180870. See also A157751 and A228565.

The Dirichlet kernel is D(n,x) = Sum_{k=-n..n} exp(i*k*x) = 1 + 2*Sum_{k=1..n} T(n,x) = S(n, 2*y) + S(n-1, 2*y) = S(2*n, sqrt(2*(1+y))) with y = cos(x), n >= 0, with the Chebyshev polynomials T (A053120) and S (A049310). This triangle T(n, k) gives in row n the coefficients of the polynomial Dir(n,y) = D(n,x=arccos(y)) = Sum_{m=0..n} T(n,m)*y^m. See A180870, especially the Peter Bala comments and formulas.

This is the Riordan triangle ((1+z)/(1+z^2), 2*z/(1+z^2)) due to the o.g.f. for Dir(n,y) given by (1+z)/(1 - 2*y*z + z^2) = G(z)/(1 - y*F(z)) with G(z) = (1+z)/(1+z^2) and F(z) = 2*z/(1+z^2) (see the Peter Bala formula under A180870). For Riordan triangles and references see the W. Lang link 'Sheffer a- and z- sequences' under A006232.

The A- and Z- sequences of this Riordan triangle are (see the mentioned W. Lang link in the preceding comment also for the references): The A-sequence has o.g.f. 1+sqrt(1-x^2) and is given by A(2*k+1) = 0 and A(2*k) [2, -1/2, -1/8, -1/16, -5/128, -7/256, -21/1024, -33/2048, -429/32768, -715/65536, ...], k >= 0. (See A098597 and A046161.)

The Z-sequence has o.g.f. sqrt((1-x)/(1+x)) and is given by

[1, -1, 1/2, -1/2, 3/8, -3/8, 5/16, -5/16, 35/128, -35/128, ...]. (See A001790 and A046161.)

The column sequences are A057077, 2*(A004526 with even numbers signed), 4*A008805 (signed), 8*A058187 (signed), 16*A189976 (signed), 32*A189980 (signed) for m = 0, 1, ..., 5.

The row sums give A005408 (from the o.g.f. due to the Riordan property), and the alternating row sums give A033999.

The row polynomials Dir(n, x), n >= 0, give solutions to the diophantine equation (a + 1)*X^2 - (a - 1)*Y^2 = 2 by virtue of the identity (a + 1)*Dir(n, -a)^2 - (a - 1)*Dir(n, a)^2 = 2, which is easily proved inductively using the recurrence Dir(n, a) = (1 + a)*(-1)^(n-1)*Dir(n-1, -a) + a*Dir(n-1, a) given below by Wolfdieter Lang. - Peter Bala, May 08 2025

Triangle read by rows ( see example ). The numerator triangle is A274076.

Comments of A273506 give a definition of the fraction triangle, which determines an arbitrary-precision solution to the simple pendulum equations of motion. For more details see "Plane Pendulum and Beyond by Phase Space Geometry" (Klee, 2016).

Triangle read by rows ( see examples ). The denominators are given in A274078.

The rational triangle A273506 / A273507 gives the coefficients for an exact solution of the plane pendulum's phase space trajectory. Differential time dependence for this solution also adheres to the simple form of a triangular summation: dt = dQ(-1+ sum k^n * (T(n, m)/A274078(n, m)) * cos(Q)^(2(n+m)) ); where the sum runs over n = 1,2,3 ... and m = 1,2,3...n. Expanding powers of cosine ( Cf. A273496 ) it is relatively easy to integrate dt ( cf. A274130 ). One period of motion takes Q through the range [ 0 , -2 pi]. Integrating dt over this domain gives another (Cf. A273506) calculation of the series expansion for Elliptic K ( see examples and Mathematica function dtToEllK ). For more details read "Plane Pendulum and Beyond by Phase Space Geometry" (Klee, 2016).

Triangle read by rows (see example). Comments of A274076 give a definition of the fraction triangle, which determines to arbitrary precision the differential time dependence for the time-independent solution (cf. A273506, A273507) of the plane pendulum's equations of motion. For more details see "Plane Pendulum and Beyond by Phase Space Geometry" (Klee, 2016).

Irregular triangle read by rows ( see examples ). The row length sequence is 2*n = A005843(n), n >= 1.The denominators are given in A274131.

The triangles A274076 and A274078 give the coefficients for the exact, differential time dependence of the plane pendulum's equations of motion. Integrating, we obtain time dependence as a Fourier sine series: t = -( (2/pi)K(k) Q + sum k^n * (T(n,m)/A274131(n,m)) * sin(2 m Q) ); where the sum runs over n = 1,2,3 ... and m = 1,2,3,...,2 n. Combining the phase space trajectory and time dependence, it is possible to express Jacobian elliptic functions {cn,dn} in parametric form. For more details read "Plane Pendulum and Beyond by Phase Space Geometry" (Klee, 2016).

Irregular triangle read by rows (see example). The row length sequence is 2*n = A005843(n), n >= 1.

The numerator triangle is A274130.

Comments of A274130 give a definition of the fraction triangle, which determines to arbitrary precision the time dependence for the time-independent solution (cf. A273506, A273507) of the plane pendulum's equations of motion. For more details see "Plane Pendulum and Beyond by Phase Space Geometry" (Klee, 2016).