For positive integers $a>b>1$, define
\[x_n = \frac {a^n-1}{b^n-1}\] 
Find the least $d$ such that for any $a,b$, the sequence $x_n$ does not contain $d$ consecutive prime numbers.

[i]V. Senderov[/i]
1. **Define the sequence and initial claim:**
   For positive integers \(a > b > 1\), define the sequence:
   \[
   x_n = \frac{a^n - 1}{b^n - 1}
   \]
   We claim that the least \(d\) such that the sequence \(x_n\) does not contain \(d\) consecutive prime numbers is \(3\).

2. **Example to show necessity:**
   Consider \( (a, b) = (4, 2) \). This gives:
   \[
   x_n = 2^n + 1
   \]
   Notice that:
   \[
   x_1 = 2^1 + 1 = 3 \quad \text{and} \quad x_2 = 2^2 + 1 = 5
   \]
   Both \(x_1\) and \(x_2\) are primes. We need to show that it is impossible for \(x_n, x_{n+1}, x_{n+2}\) to all be primes.

3. **Claim 01: The sequence is increasing.**
   To prove this, we need to show that \(x_n < x_{n+1}\). This can be verified by direct calculation:
   \[
   x_{n+1} = \frac{a^{n+1} - 1}{b^{n+1} - 1}
   \]
   Since \(a > b\), it follows that \(a^{n+1} - 1 > a^n - 1\) and \(b^{n+1} - 1 > b^n - 1\), thus:
   \[
   \frac{a^{n+1} - 1}{b^{n+1} - 1} > \frac{a^n - 1}{b^n - 1}
   \]
   Hence, \(x_n < x_{n+1}\).

4. **Claim 02: If \(x_n, x_{n+1} \in \mathbb{N}\), then \(b - 1 \mid a - 1\).**
   We have:
   \[
   b^n - 1 \mid a^n - 1 \quad \text{and} \quad b^{n+1} - 1 \mid a^{n+1} - 1
   \]
   Let \(p \mid b - 1\). Then:
   \[
   p \mid b - 1 \implies p \mid b^n - 1 \implies p \mid a^n - 1
   \]
   and:
   \[
   p \mid b - 1 \implies p \mid b^{n+1} - 1 \implies p \mid a^{n+1} - 1
   \]
   Therefore:
   \[
   p \mid a^{\gcd(n, n+1)} - 1 = a - 1
   \]
   By the Lifting The Exponent (LTE) lemma, we have:
   \[
   \nu_p(b - 1) + \nu_p(n) = \nu_p(b^n - 1) \le \nu_p(a^n - 1) = \nu_p(a - 1) + \nu_p(n)
   \]
   for all \(p \mid b - 1\). Hence:
   \[
   \nu_p(b - 1) \le \nu_p(a - 1)
   \]
   for all such primes \(p\), forcing \(b - 1 \mid a - 1\).

5. **Contradiction argument:**
   Let \(k = \frac{a - 1}{b - 1}\) be a fixed constant. Define:
   \[
   a_n = a^{n-1} + a^{n-2} + \cdots + 1
   \]
   and similarly for \(b_n\). Notice that \(a_n > b_n\) for all \(n \ge 2\).

   Suppose \(x_n, x_{n+1}, x_{n+2}\) are all primes. Then:
   \[
   \frac{a_{n+1}}{b_{n+1}} = \frac{x_{n+1}}{k} \quad \text{and} \quad \frac{a_{n+2}}{b_{n+2}} = \frac{x_{n+2}}{k}
   \]
   Since \(a_{n+1} > b_{n+1}\) and \(a_{n+2} > b_{n+2}\), we have:
   \[
   \gcd(x_{n+1}, k) = \gcd(x_{n+2}, k) = 1
   \]
   Thus:
   \[
   k \mid b_{n+1} \mid b^{n+1} - 1 \quad \text{and} \quad k \mid b_{n+2} - b_{n+1} = b^{n+1}
   \]
   This forces \(k = 1\), which implies \(a = b\), a contradiction.

The final answer is \(\boxed{3}\).