Every natural number is covered by consecutive numbers that sum to a prime power.

Solution 1:

For any odd prime $p$, there are $p$ consecutive integers centred on $p$ that sum to $p^2$.

$2+3+4=3^2$
$3+4+5+6+7=5^2$
$4+5+6+7+8+9+10=7^2$
etc.

Let $p_n$ be the $n$-th prime. Then, using Bertrand's postulate in the form $$p_{n+1}<2p_n$$we know that the above sums for consecutive primes overlap.

Finally, we note that $1+2=3$ to complete the proof.

I don't know if this has been shown before, but the proof seems straightforward.

Solution 2:

While nickgard's answer shows how to solve the problem using sums being squares of increasing primes, this answer shows how to do it using the sums being just odd powers of $3$.

As suggested in joriki's question comment, for any integers $1 \le j \le k$, you have

$$\begin{equation}\begin{aligned} \sum_{i=j}^{k}i & = \sum_{i=1}^{k}i - \sum_{i=1}^{j-1}i \\ & = \frac{k(k+1)}{2} - \frac{(j-1)(j)}{2} \\ & = \frac{k^2 + k - j^2 + j}{2} \\ & = \frac{(k-j)(k+j) + k + j}{2} \\ & = \frac{(k+j)(k-j+1)}{2} \end{aligned}\end{equation}\tag{1}\label{eq1A}$$

Consider the ranges $\left[\frac{3^m + 1}{2},\frac{3^{m+1} - 1}{2}\right]$ for $m = 0, 1, 2, \ldots$. The union of these disjoint subsets cover all positive integers. Thus, for any $n \ge 1$, there's a unique $m$ where $n \in \left[\frac{3^m + 1}{2},\frac{3^{m+1} - 1}{2}\right]$. For that $m$, since $\frac{5\left(3^{m}\right)-1}{2} \gt \frac{3^{m+1} - 1}{2}$, you can have $j = \frac{3^m + 1}{2}$ and $k = \frac{5\left(3^{m}\right)-1}{2}$ with $n \in [j,k]$. Using this in \eqref{eq1A} gives

$$\begin{equation}\begin{aligned} \sum_{i=j}^{k}i & = \frac{(k+j)(k-j+1)}{2} \\ & = \frac{\left(\frac{5\left(3^{m}\right)-1}{2}+\frac{3^m + 1}{2}\right)\left(\frac{5\left(3^{m}\right)-1}{2}-\frac{3^m + 1}{2}+1\right)}{2} \\ & = \frac{\left(\frac{6\left(3^{m}\right)}{2}\right)\left(\frac{4\left(3^{m}\right)}{2}-\frac{2}{2}+1\right)}{2} \\ & = \frac{\left(3\left(3^{m}\right)\right)\left(2\left(3^{m}\right)\right)}{2} \\ & = \left(3^{m+1}\right)\left(3^{m}\right) \\ & = 3^{2m+1} \end{aligned}\end{equation}\tag{2}\label{eq2A}$$

The first few examples for $m = 0, 1$ and $2$ are

$$1 + 2 = 3 = 3^{1} \tag{3}\label{eq3A}$$

$$2 + 3 + 4 + 5 + 6 + 7 = 27 = 3^{3} \tag{4}\label{eq4A}$$

$$5 + 6 + \ldots + 21 + 22 = 243 = 3^{5} \tag{5}\label{eq5A}$$