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Given any 4 positive numbers $p_{00}\,,p_{01}=p_{10}\,,p_{11}$,such that the sum of the 4 numbers is 1, now I want to find a sequence in $\{0\,,1\}^\mathbb{N}$ such that this sequence has uniform frequencies and recurrent properties. where $p_{ij}$ is the frequency of block $(ij)$ in the sequence. As we all known, uniform frequencies equivalent to unique ergodicity in dynamical systems, a good example is Sturmian sequence which has uniform frequencies and recurrent properties, but the frequencies are determined by the irrational rotation.

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  • $\begingroup$ Do yo need implicit sequence? I guess, Markov process produces a lot of such sequence: for $p=p_{00}/(p_{00}+p_{01})$, $q=p_{10}/(p_{10}+p_{11})$ define $x_{n+1}=0$ or $x_{n+1}=1$ with probabilities $p$ and $1-p$ if $x_n=0$; $x_{n+1}=0$ or $x_{n+1}=1$ with probabilities $q$ and $1-q$ if $x_n=1$. $\endgroup$ Commented Oct 8, 2014 at 15:41

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As you remark, Sturmian words (encoding horizontal and vertical steps when approaching optimally from below a line of given slope by a discrete path (with steps $(1,0)$ and $(0,1)$) of $\mathbb Z^2$) solve the problem if either $p_{00}$ or $p_{11}$ is zero. In the general case, I have the impression that one can combine two Sturmian words as follows: Suppose $p_{00}\geq p_{11}$. We construct first a Sturmian sequence with the correct relative proportions $p_{00}/(1-p_{11})$ and $p_{01}/(1-p_{11})$ of subwords $00$ and $01$ (or $10$). We now replace the isolated $1'$s by $1^a$ and $1^{a+1}$ using a suitable Sturmian word in order to get the correct amount of $11$'s (the correct relative frequencies should be $p_{01}/(1-p_{00})$ and $p_{11}/(1-p_{00})$, I guess). Since this does not alter the relative proportion of subwords $00$ and $01$, this should do the job and the final word has recurrent properties.

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  • $\begingroup$ Thanks, I will think your idea carefully and will find such kind of sequences $\endgroup$
    – Kan
    Commented Mar 14, 2014 at 8:17
  • $\begingroup$ Maybe we can consider a large digit instead of $\{0\,,1\}$ $\endgroup$
    – Kan
    Commented Mar 19, 2014 at 11:26

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