Let us call a positive integer $n\in\mathbb{N}$ consecutively summable if there are positive integers $m, k < n$ such that $$n=\sum_{i=0}^k (m+i).$$For $A\subseteq \mathbb{N}$ we set the lower density of $A$ to be $$\text{ld}(A)=\text{lim inf}_{n\to\infty}\frac{|A\cap\{1,\ldots,n\}|}{n}.$$ If $N$ is the set of positive integers that are not consecutively summable, do we have $\text{ld}(N)=0$?


2 Answers 2


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 $ld(A)=0$.


Let us denote $B=\mathbb N\setminus N$, i.e., the set of consecutive-summable numbers.

There already is an answer precisely characterizing the sets $N$ and $B$. (Which is much better result than just estimating the density.) If we are only interested in the density, another way to get that it is equal to zero is to notice the following.

If we fix a $k\in\mathbb N$, then we get the numbers of the form $$n=km+\sum_{i=0}^k k = km + T_k$$ where $T_k$ is the $k$-th triangular number. So we see that with finitely many exceptions $B$ contains the arithmetic progression $k\mathbb N+a$, where $a=T_k \bmod k$.

That means that $N$ is almost (i.e., with finitely many exceptions) contained in the union of the remaining $(k-1)$ congruence classes modulo $k$.

If we take any coprime $k_1,\ldots,k_n$, then we have that $N$ is almost contained in the union of corresponding congruence classes modulo $k_1\cdots k_n$. (Here we are using the Chinese remainder theorem which gives us the correspondence between the system of remainder modulo $k_1,\dots,k_n$ and a single congruence class modulo $k_1\cdots k_n$.) This gives us an estimate for the upper density $$\operatorname{ud}(N) \le \frac{(k_1-1)\cdot (k_2-1) \cdot (k_n-1)}{k_1\cdot k_2 \cdot k_n} = \prod_{j=1}^n \left(1-\frac1{k_j}\right).$$

In particular, if we take $k_n=p_n$ to be the $n$-th prime number, we get that $$\operatorname{ud}(N) \le \prod_{j=1}^n \left(1-\frac1{p_j}\right)$$ and since this is true for every $n$ we have $$\operatorname{ud}(N) \le \prod_{j=1}^\infty \left(1-\frac1{p_j}\right)=0.$$

You may notice that this argument would work for any set which can be related in a similar way to the union of distinct congruence classes if we have $\sum \frac1{k_j}=\infty$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.