cp's OEIS Frontend

Continued fraction expansion of (5 + 3*sqrt(5))/10 is A010686.

The horizontal distance between the accumulation point and the outermost point of a golden spiral inscribed inside a golden rectangle with dimensions phi and 1 along the x and y axes, respectively (the vertical distance is A244847). - Amiram Eldar, May 18 2021

This constant c gives the real part of -2*11*Z = (c + d*i), where Z is the (finite) fixed point of a complex function w (of the loxodromic type) mapping iteratively the vertices of golden triangles, starting with vertices (D_1, D_2, D_3), circumscribed by the unit circle with center at the origin, and D_1 = i, D_2 = (s - phi*i)/2 and D_3 = (-s - phi*i)/2. This function is w(z) = a*z + b, with a = (-1 + phi) * exp(-(3*Pi/5)*i) = -((2 - phi) + s*i)/2 and b = (1 - phi)*i, where s = sqrt(3 - phi) = A182007 (the length of the base (D_2, D_3) of the first triangle, also called s_1).

The imaginary part of -2*11*Z is d = -7 + 10*phi = A385446.

If the fixed point Z = -(0.20594... + 0.41728...*i) is chosen as origin then the loxodromic map is W(z') = a*z' (where z' = z - Z and W(z') = w(z'+Z) - Z).

For details see the linked paper, eqs.(5a,b) for w(z), eq.(6) for Z and eq.(7) for W(z'). (In eq.(5b) the i is missing in the exponent.) The nesting of golden triangles as shown in Fig. 1 of the link leads to the fixed point Z.

The vertices of the nested golden triangles can be connected by a spiral built of circular sections with angle 108 degrees, centered at vertices D_{n+3} and shrinking radii s_n =(- 1 + phi)^(n-1)*s. Note that the curvature of this spiral is not continuous.

The length(Z, D_n) =: rho_n of the spokes of the spiral is (-1 + phi)^(n-1)*rho_1, with sqrt(11)*rho_1 = sqrt(8 + 9*phi) = sqrt(5 + 7*phi)*s = A385447.

For the length ratio rho_1/s see A385448.

The logarithmic spiral connecting the vertices D_{n+1} is given in polar coordinates by rho(Phi) = rho_1 * exp((-(5/(3*Pi)) * log(phi)*Phi), with the vertices obtained in polar coordinates for Phi = Phi_n = (3*Pi/5)*n, namely rho(Phi_n) = rho_{n+1}, for n >= 0. For log(phi) see A002390. Note that the nonnegative x-axis is now along Z, D_1. The angle(Z, D1, D4) =: gamma is given by arctan((18 - 11*phi)/s) = arcsin(rho_4 / 2) = 0.169860704... or 9.7323... degrees. See Fig. 3 of the linked paper.

The author conjectures that this is the minimum volume of an axis-aligned bounding box which includes the shortest minimum-link circuit joining all the vertices of the cube {0,1}^3 (i.e., the closed polygonal chains consisting of exactly 6 edges visiting all the points of the set {(0,0,0),(0,0,1),(0,1,0),(0,1,1),(1,0,0),(1,0,1),(1,1,0),(1,1,1)}).

In detail, such a circuit of 6 links is given by (1/2,1+phi,1/2)-((1-phi)/2,0,(1+phi)/2)-((phi+1)/2,0, (1-phi)/2)-(1/2,1+phi,1/2)-((phi+1)/2,0,(phi+1)/2)-((1-phi)/2,0,(1-phi)/2(1/2,1+phi,1/2), where phi := (1+sqrt(5))/2 (see A001622).

Then, phi*(2*phi + 1) = phi^2*(phi + 1) since phi - 1 = 1/phi.

Interleaving of A010716 and A000012.

Also continued fraction expansion of (5+3*sqrt(5))/2.

Also decimal expansion of 17/33.

Essentially first differences of A047264.

Binomial transform of 5 followed by -A122803 without initial terms 1, -2.

Inverse binomial transform of 5 followed by A007283 without initial term 3.

Second inverse binomial transform of A168607 without initial term 3.

Exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + x + 3*x^2 + 3*x^3 + 6*x^4 + 6*x^5 + ... is the o.g.f. for A008805. - Peter Bala, Mar 13 2015

Solution for z in the system {x = 1/y + 1/z, y = x + 1/z, z = y + 1/x}. The corresponding values of x and y are (5^(1/4) + 5^(-1/4))/2 and 5^(1/4).

The largest aspect ratio of a set of three rectangles which have the property that any two of them can be scaled, rotated, and joined at an edge to obtain a rectangle with the third aspect ratio. The other two aspect ratios are given in the comment above.

x = A090550 = 1 + 3*phi = 5.854101966..., where phi is the golden ratio.