For each $k$ , find the least $n$ in terms of $k$ st the following holds:
There exists $n$ real numbers $a_1 , a_2 ,\cdot \cdot \cdot , a_n$ st for each $i$ :
$$0 < a_{i+1} - a_{i} < a_i - a_{i-1}$$
 And , there exists $k$ pairs $(i,j)$ st $a_i - a_j = 1$.
1. **Understanding the Problem:**
   We need to find the smallest \( n \) in terms of \( k \) such that there exist \( n \) real numbers \( a_1, a_2, \ldots, a_n \) satisfying:
   - \( 0 < a_{i+1} - a_i < a_i - a_{i-1} \) for each \( i \).
   - There exist \( k \) pairs \((i, j)\) such that \( a_i - a_j = 1 \).

2. **Analyzing the Condition:**
   The condition \( 0 < a_{i+1} - a_i < a_i - a_{i-1} \) implies that the differences between consecutive terms are strictly decreasing. Let \( b_i = a_{i+1} - a_i \). Then, \( b_{i+1} < b_i \).

3. **Sum of Differences:**
   Each \( a_i \) can be written as a sum of some consecutive terms of the form \( b_i \). If \( t \) is the largest number such that \( b_t + b_{t+1} + \cdots + b_s = 1 \), by the condition, \( s-t \) is maximal among all other consecutive terms with sum 1 since \( t \) was maximal.

4. **Bounding \( n \):**
   Since \( s = (s-t) + t > k-1 + k \), this implies \( n \geq 2k \). This is because we need at least \( 2k \) terms to ensure that there are \( k \) pairs \((i, j)\) such that \( a_i - a_j = 1 \).

5. **Constructing an Example:**
   Let \( a_1 = 1 \) and \( a_2 = 2 \). At the \( i \)-th step, consider another number that has a difference of one with the largest number you have (and larger than the current maximum, of course) and put their mean \( +10^{-i} \) between them as another element of the sequence. This ensures that the sequence grows while maintaining the required differences.

6. **Verification:**
   One can easily check that this construction works and satisfies the given conditions. Therefore, the least \( n \) in terms of \( k \) is \( 2k \).

The final answer is \( \boxed{ 2k } \)