It is known that the only numbers not "consecutive summable" are the powers of $2$.  
This is easy to prove: If $m+m+1+\ldots m+k=2^a$ then $2^a=\frac{(k+1)(2m+k)}{2}$.  
This means that $2^{a+1}=(k+1)(2m+k)$ which is impossible since the differences form the two parentheses is an odd number. (And they should be both powers of two).  

This shows that $ld(A)=0$.