Factorizing a handiwork number $n$ in two prime factors I have generated two primes $r_0$ and $s_0$ with same bit size $\le 64$, then make another two primes as follows: choose constant $k$ and define $$r_i = r_0 + \alpha_i, \quad 1 \le i \le k, \quad 0 \le \alpha_i \le 1000$$ and 
$$s_i = s_0 + \beta_i, \quad  1 \le i \le k, \quad 0 \le \beta_i \le 1000$$
I have chosen $\alpha_i$ and $\beta_i$ such that both p and q are prime, 
$p = 2\times r_0\times r_1\times\cdots\times r_k + 1$ and $q = 2\times s_0\times s_1\times\cdots\times s_k + 1$. Can we factor $n = p \times q$ with knowing above facts?
The number you constructed is somewhat vulnerable to Pollard's p-1 method. If $r_0$ and $s_0$ are 64 bit long primes, then these are the largest factors in $p-1$ and $q-1$. The method works when $p-1$ (or $q-1$) is powersmooth for some "small" $B$ (choosing $B$ as $64$ bit number isn't that practical - the expected runtime of the algorithm is $O(\log^2(n)B \log(B))$).

In your question you wrote $r_0$ and $s_0$ have less than $64$ bit. If it is actually much smaller, the $p-1$ method will become viable.

Regarding the numbers in the comment: Those are $6\cdot 64 = 384$ bit primes? That is not an acceptable length for RSA - if that was your intention. According to the latest recommendation on keylength.com, today's fatoring moduli should have $2000$ bits at least, meaning you need primes with $1000$ bit each.