# 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


• Apparently, primes are allowed as well. Not sure, whether it has been proven, but this is very likely to be the case. More interesting is to exclude the primes. – Peter Jan 12 '20 at 10:34
• @Peter: The sum can only be a prime if it has at most two terms. (Since $\sum_{k=m}^nk=(n-m+1)(n+m)/2$, which factorizes unless $n-m+1\le2$.) The density of one- and two-term sums that are primes goes to zero, so the brunt of the conjecture has to be borne by the proper prime powers, anyway. – joriki Jan 12 '20 at 13:02
• @Peter a prime power by definition includes the exponent $1$. Therefore, for sake of simplicity, we include the primes. – StudySmarterNotHarder Jan 12 '20 at 17:32

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.

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{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}\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{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}\tag{2}\label{eq2A}

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

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

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

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