cp's OEIS Frontend

Sqrt(62) = 787400 * Sum_{n>=0} (A001790(n)/2^A005187(floor(n/2)) * 10^(-6n-5)) where A001790(n) are numerators in expansion of 1/sqrt(1-x) and the denominators in expansion of 1/sqrt(1-x) are 2^A005187(n). 786400 = 62*12700, see A020819 (expansion of 1/sqrt(62)). - Gerald McGarvey, Jan 01 2005

Continued fraction expansion is 7 followed by {1, 6, 1, 14} repeated. - Harry J. Smith, Jun 07 2009

All even coefficients of u_n are 0. Sum_{k=0..n} T(n,k) = 0. 1/u(n)(1/2) = A230142(n)/A230143(n) is an approximation to Pi: Pi-1/u(10)(1/2) = 0.6883935...*10^(-9), Pi-1/u(50)(1/2) = 0.2993600...*10^(-47).

If n >= 1 it appears a(n-1) is equal to the difference between the denominator and the numerator of the ratio (2n-1)!!/(2n-2)!!. In particular a(n-1) is the difference between the denominator and the numerator of the ratio A001147(2n-2)/A000165(2n-1). See examples. - Anthony Hernandez, Feb 05 2020

It can be seen that this is true, e.g., using A001803(n) = (2n+1)!/(n!^2*2^A000120(n)) and A046161(n) = 4^n/2^A000120(n). - M. F. Hasler, Feb 07 2020

Numerators in the expansion of (1-(1-x)^(1/2))/(1-x)^(3/2). Denominators are A046161. Compare to A001790. - Thomas Curtright, Feb 09 2024

This expansion is due to the Riordan property of the triangle A084930. The inverse of the lower triangular matrix built by A084930 is therefore also a (rational) Riordan triangle, namely ((2 - c(z/4))/(1-z), -1 + c(z/4)) in the standard notation, where c is the o.g.f. of A000108 (Catalan).

For the denominators of this triangle see A244421.

The expansion is x^n = sum(R(n,m)*Todd(m, x), m=0..n), n >= 0, with the rational triangle with entries R(n,m) = a(n, m)/b(n, m) with b(n, m) = A244421(n, m).

If one uses instead the expansion of (4*x)^n one finds the integer triangle A111418: (4*x)^n = sum(A111418(n,k) * Todd(k, x), k=0..n).

The row sums of the rational triangle R(n,m) are identically 1. The alternating row sums have o.g.f. 1/sqrt(1-x) which generates A001790(n)/A046161(n) (see a Michael Somos comment on A046161), namely 1, 1/2, 3/8, 5/16, 35/128, 63/256, 231/1024, 429/2048, ...

From Wolfdieter Lang, Jun 13 2016: (Start)

R(n,m) = a(n, m)/ A244421(n, m) is also the rational triangle for the expansion cos^{2*n+1}(x) = Sum_{m=0..n} R(n, m)*cos((2*m+1)*x), n >= 0, m = 0..n. Compare with the odd numbered rows of A273496. In terms of Chebyshev T-polynomials (A053120) this is the identity x^(2*n+1) = Sum_{m=0..n} R(n, m)*T(2*m+1, x).

S(n,m) = (-1)^m*a(n, m)/ A244421(n, m) is the rational triangle for the expansion sin^{2*n+1}(x) = Sum_{m=0..n} S(n, m)*sin((2*m+1)*x), n >= 0, m = 0..n. In terms of Chebyshev S-polynomials (A049310) this is equivalent to the identity (4 - x^2)*n = Sum_{m=0..n} (-1)^m * binomial(n, n-m)*S(2*m,x), n >= 0.

(End)

Irregular triangle read by rows (see examples). Consider an arbitrary anharmonic oscillator with Hamiltonian energy: H=(1/2)*b^2=(1/2)*(p^2+q^2) + Sum_{i=3} 2*v_i*q^i, and a stable minimum at (p,q)=(0,0). The phase space trajectory can be written in polar phase space coordinates as (q,p) = (R(x)cos(x),R(x)sin(x))=(R(Q)Q,R(Q)P). The present triangle determines a power / Fourier series of R(Q): R(Q) = b * (1 + sum b^n*T(n,m)*f(n,m) ); where the sum runs over n = 1,2,3 ... and m = 1,2,3...A000041(n). The basis functions f(n,m) are constructed from partitions of "n" listed in reverse lexicographic order. Partition n=(z_1+z_2+...z_j) becomes 2*Q^((z_1+2)+(z_2+2)+...(z_j+2))*v_{z_1+2}*v_{z_2+2}*...*v_{z_j+2} (see examples). This sequence transforms into A273506/A273507 by setting v_i=0 for odd i, v_i:=(-1)^(i/2-1)/2/(i!) otherwise, and (1/2)*b^2 = 2*k. For more details read "Plane Pendulum and Beyond by Phase Space Geometry" (Klee, 2016).

The phase space trajectory A276738 has phase space angular velocity A276814 and differential time dependence A276815. We calculate the period K = Int dt over the range [2*Pi, 0], trivial to compute from A276815 using A273496. Then K/(2*Pi) = 1 + sum b^(2n)*T(n,m)*f'(n,m); where the sum runs over n = 1, 2, 3 ... and m = 1, 2, 3, ... A000041(2n), and f'(n,m) = f(2n,m) of A276738 with Q=1/2. Choosing one point from the infinite dimensional coefficient space--v_i=0 for odd i, v_i=(-1)^(i/2-1)/2/(i!) otherwise--setting b^2 = 4*k, and summing over the entire table obtains the EllipticK expansion 2*A038534/A038533. For more details read "Plane Pendulum and Beyond by Phase Space Geometry" (Klee, 2016).

Irregular triangle read by rows (see examples).

Consider an axially symmetric oscillator in two dimensions with polar coordinates ( r, y ). By conservation of angular momentum, replace the cyclic angle coordinate y with dy/dt = 1/r^2. The system becomes one-dimensional in r, with an effective potential including the 1/r^2 term. Assume that the effective potential has a minimum around r0 and apply a linear transform r --> q = r-r0. Radial oscillations around the effective potential minimum follow the exact solution of A276738, A276814, A276815, A276816. Now dy = dx (dy/dt) / (dx/dt) = dx * Sum b^n*T(n,m)*F(n,m), with n=1,2,3.... and m=1,2,3...A000070(n). Basis functions F(n,m) are an ordered union over A276738's f(n,m): F(n,m')={ (1/r0^2)*(Q/r0)^n } & Append_{i=1..n}_{m=1..A000041(n)} (1/2/r0^2)*(Q/r0)^(n - i)*f(i,m), where each successive term f(i,m) is appended such that index m' inherets the ordering of each m index (see examples). Integrating dx over a range of 2 Pi loses all odd rows, as in A276815 / A276816. This sequence is a useful tool in classical and relativistic astronomy (follow links to Wolfram demonstrations).

Multiplicative because A034444 is.

The first 2^20 terms are positive. Is the sequence nonnegative?

Records seem to be A001790, occurring at A000302 (apart from 4).

Also rational part of denominator of Gamma(n/2+1)/Gamma(n/2+1/2) (cf. A004731).

A055786 is the preferred version of this sequence.