Define the sequences $ a_n, b_n, c_n$ as follows. $ a_0 \equal{} k, b_0 \equal{} 4, c_0 \equal{} 1$.
 If $ a_n$ is even then $ a_{n \plus{} 1} \equal{} \frac {a_n}{2}$, $ b_{n \plus{} 1} \equal{} 2b_n$, $ c_{n \plus{} 1} \equal{} c_n$. 
If $ a_n$ is odd, then $ a_{n \plus{} 1} \equal{} a_n \minus{} \frac {b_n}{2} \minus{} c_n$, $ b_{n \plus{} 1} \equal{} b_n$, $ c_{n \plus{} 1} \equal{} b_n \plus{} c_n$. 
Find the number of positive integers $ k < 1995$ such that some $ a_n \equal{} 0$.
1. **Invariant and Initial Conditions:**
   We start by verifying that \( a_n b_n + \frac{c_n^2}{2} \) is an invariant. This means that for all \( n \), the expression \( a_n b_n + \frac{c_n^2}{2} \) remains constant. We also note that \( c_n \equiv 1 \pmod{4} \) for all \( n \).

   If \( a_n = 0 \) for some \( n \), then let \( c_n = 4u + 1 \). This gives us:
   \[
   4a_0 + \frac{1}{2} = \frac{(4u+1)^2}{2}
   \]
   Simplifying, we get:
   \[
   4a_0 + \frac{1}{2} = \frac{16u^2 + 8u + 1}{2}
   \]
   \[
   4a_0 + \frac{1}{2} = 8u^2 + 4u + \frac{1}{2}
   \]
   \[
   4a_0 = 8u^2 + 4u
   \]
   \[
   a_0 = 2u^2 + u
   \]
   Therefore, \( a_0 = u(2u + 1) \).

2. **Proving \( a_0 = u(2u + 1) \) leads to \( a_n = 0 \):**
   We need to show that all \( a_0 \) of the form \( u(2u + 1) \) will eventually lead to \( a_n = 0 \) for some \( n \).

   Let \( u = 2^{x_1} + 2^{x_2} + \cdots + 2^{x_k} \) where \( x_1 < x_2 < \cdots < x_k \). For \( k = 1 \), i.e., \( u = 2^{x_1} \), it's easy to show that \( a_{x_1 + 1} = 0 \).

   For \( k \geq 2 \), denote:
   \[
   v_i = \sum_{j=i+1}^{k} 2^{x_j - x_i} + 1 \quad \forall 1 \leq i \leq k-1
   \]
   and \( v_k = 1 \).

   By induction, we have:
   \[
   a_{x_i + i} = 2^{x_{i+1} - x_i} v_{i+1} (2^{x_{i+1} + 1} v_{i+1} + \sum_{j=1}^{i} 2^{x_{j+2}} + 1)
   \]
   \[
   b_{x_i + i} = 2^{x_i + 2}
   \]
   \[
   c_{x_i + i} = \sum_{j=1}^{i} 2^{x_{j+2}} + 1 \quad \forall 1 \leq i \leq k-1
   \]

   Therefore:
   \[
   a_{x_{k-1} + k-1} = 2^{x_k - x_{k-1}} v_k (2^{x_k + 1} v_k + \sum_{j=1}^{k-1} 2^{x_{j+2}} + 1)
   \]
   \[
   b_{x_{k-1} + k-1} = 2^{x_{k-1} + 2}
   \]
   \[
   c_{x_{k-1} + k-1} = \sum_{j=1}^{k-1} 2^{x_{j+2}} + 1
   \]

   Simplifying further:
   \[
   a_{x_k + k-1} = 2^{x_k + 1} + \sum_{j=1}^{k-1} 2^{x_{j+2}} + 1
   \]
   \[
   b_{x_k + k-1} = 2^{x_k + 2}
   \]
   \[
   c_{x_k + k-1} = \sum_{j=1}^{k-1} 2^{x_{j+2}} + 1
   \]

   Finally:
   \[
   a_{x_k + k} = 0
   \]

   Thus, we have shown that \( a_0 = u(2u + 1) \) will eventually lead to \( a_n = 0 \) for some \( n \).

The final answer is \( \boxed{u(2u + 1)} \)