Every natural number is covered by consecutive numbers that sum to a prime power. Conjecture. For every natural number $n \in \Bbb{N}$, there exists a finite set of consecutive numbers $C\subset \Bbb{N}$ containing $n$ such that $\sum\limits_{c\in C} c$ is a prime power.
A list of the first few numbers in $\Bbb{N}$ has several different covers by such consecutive number sets.
One such is:
 3   7  5 13  8  19  11  25    29   16     37    41          49    53
___ ___ _ ___ _ ____ __ _____ _____ __ __ _____ _____       _____ _____    __
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 
        ___   ___                _____             _____ __       ________ 
        11    17                   31                43              81
59               71           3^5
___    __       _____    _________________ 
 30 31 32 33 34 35 36 37 38 39 40 41 42 43 .....
 _____    _____    _____
  61       67        73

Has this been proved already?
 A: 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
$$1 + 2 = 3 = 3^{1} \tag{3}\label{eq3A}$$
$$2 + 3 + 4 + 5 + 6 + 7 = 27 = 3^{3} \tag{4}\label{eq4A}$$
$$5 + 6 + \ldots + 21 + 22 = 243 = 3^{5} \tag{5}\label{eq5A}$$
A: 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.
