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 $ld(A)=0$.

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).
It is not that difficult to show that every number not of the form $2^a, a\in \mathbb{N}$ can be "consecutive summable".

I think the shortest way to prove both is here.
This shows that $ld(A)=0$.

It is known that the only numbers not "consecutive summable" are the powers of $2$.
This shows that $ld(A)=0$.

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 $ld(A)=0$.