Correlation between two sequences of irrational numbers Let us consider the sequence $x(n+1) = \{b+x(n)\}$ with $x(0) = 0$. Here the brackets represent the fractional part function. Thus $x(n)= \{nb\}$ is related to Beatty sequences. If $b$ is irrational, it is known that the numbers $x(n)$ are uniformly distributed, with a lag-$k$ auto-correlation equal to 
$6(b+k)^2 + 6(1-2\lfloor b+k+1\rfloor)(b+k)+6\lfloor b+k+1\rfloor^2
-6\lfloor b+k+1\rfloor+1$.
This result follows from section 5.4 in this article as well as results published here. Also, if $b_1$ and $b_2$ are two irrational numbers that are linearly independent over the set of rational numbers, then the correlation between the two sequences (one generated with $b_1$ and the other one with $b_2$) is zero.
But what if $b_1 = -1 + \sqrt{5}/2$ and $b_2 = 2/\sqrt{5}$? In that case, I know with absolute certainty (yet with no proof so far) that the correlation between the two sequences is 1/20 = 0.05. How do you prove this result?  
 A: Look at the closed subgroup $G\subset \mathbb T^2$ (using the notation that $\mathbb T=\mathbb R / \mathbb Z$) generated by $(b_1,b_2)$; it
 will be of the form $G=\{ (ax,bx):x\in\mathbb T\} $ for certain  integers $a$ and $b$. (In your case,  $(a,b)=(5,4)$.)  Your orbit $\{(nb_
1,nb_2):n=1,2,\ldots\}$ is uniformly distributed in $G$.  Your correlation is obtained by integrating with respect to the uniform (Haar) measure on $G$,
 namely  by $\int_0^1 (ax\bmod1)(bx\bmod1)dx$.
It might help in general to develop the Fourier series for the functions $ x\mapsto ax\bmod1$ and $bx\mapsto x\bmod1$ on $\mathbb T$.
In your specific case, however, the integral $I= \int_0^1 (5x\bmod1)(4x\bmod1)dx$ is given by the following edge-of-tedious expression
$$I=\int_0^{1/5}(5x)(4x)dx+
\int_{1/5}^{1/4}(5x-1)(4x)dx+
\int_{1/4}^{2/5}(5x-1)(4x-1)dx+
\int_{2/5}^{1/2}(5x-2)(4x-1)dx+
\int_{1/2}^{3/5}(5x-2)(4x-2)dx+
\int_{3/5}^{3/4}(5x-3)(4x-2)dx+
\int_{3/4}^{4/5}(5x-3)(4x-3)dx+
\int_{4/5}^{1}(5x-4)(4x-3)dx.
$$
The correlation coefficient is $\frac{I-1/4}{1/12}$.
