The answer is no. Suppose there exists the desired sequence. Let $(b_n)$ be the sequence which is bijective on $\mathbb{N}$, which is: $b_n=a_{n+1}-a_n$ Suppose: $b_n+b{n+1}$ can take all natural value greater or equal to $M$, exactly once. Then, there must exist the sequence: $(c_n)$ which are pairwise distinct such that: $b_{c_i}+b_{c_i+1}=M+i-1$ forall $i$, Sum up for $i=1,2,...,k$, we have: $\sum_{i=1}^k b_{c_i}+\sum_{i=1}^k b_{c_i+1}=kM+\frac{k(k-1)}{2}$ Since $c_i$, $i=1,2,..,k$ are distinct, implies: $\sum_{i=1}^k b_{c_i}\geq 1+2+...+k=\frac{k(k+1)}{2}$ and, also, $\sum_{i=1}^k b_{c_i}\geq \frac{k(k+1)}{2}$, this means, $kM+\frac{k(k-1)}{2}\geq k(k+1)$ which is impossible for sufficient large $k$