# Integrals, the golden ratio, and the plastic constant

Can we generalize this pattern?

I. Lucas numbers

Given the two roots of $1-x-x^2=0$, then we have the formula for the $n$th Lucas number,

$$L_n = \frac{1}{x_1^n}+\frac{1}{x_2^n} = 2, 1, 3, 4, 7, 11, 18, 29, 47,\dots$$

with $L_0 = 2$. A generating function (up to sign) is

$$\sum_{n=0}^\infty L_{n+1}\, x^n =\color{blue}{\frac{1-2x}{1+x-x^2}}=1-3x+4x^2-7x^3+\dots$$

and an integral, a series, and closed-form,

$$\int_0^1\color{blue}{\frac{1+2x}{1+x-x^2}}\,dx = 4\sum_{n=1}^\infty\frac{(-1)^{n+1}\,n!^2}{n\,(2n)!}= \frac{8\ln (\phi)}{\sqrt5}$$

where $\phi=\frac{1}{x_1}=\frac{1+\sqrt5}{2}\approx1.618$ is the golden ratio.

II. Perrin numbers

Given the three roots of $1-x^2-x^3=0$, then we have the formula for the $n$th Perrin number,

$$R_n = \frac{1}{x_1^n}+\frac{1}{x_2^n}+\frac{1}{x_3^n} = 3, 0, 2, 3, 2, 5, 5, 7, 10, 12, 17,\dots$$

with $R_0 = 3$. A generating function is

$$\sum_{n=0}^\infty R_{n+2}\,x^n =\color{blue}{\frac{2+3x}{1-x^2-x^3}}=2+3x+2x^2+5x^3+5x^4+\dots$$

and an integral, a series, and closed-form (found by Vladimir Reshetnikov in this post),

$$\int_0^1\color{blue}{\frac{2-3x}{1-x^2-x^3}}\,dx = \sum_{n=0}^\infty\frac{n!(2n)!}{(3n+2)!}= 3\,\alpha\,\ln(2\,\alpha+1)-\sqrt{\beta\,}\arccos(\gamma)$$

where $\alpha, \beta, \gamma$ are in terms of $P = \frac{1}{x_1}\approx1.3247$ as the plastic constant.

Q: The next sequence is A050443 using $1-x^3-x^4=0$ and its generating function is easy to extrapolate from the previous two. But what is the integral, series, and the closed-form?