# Bounding a Sum of Prime Powers

Suppose $N$ is a positive integer. Let $p_1,p_2,...,p_n$ denote the primes less than or equal to $\sqrt N$. For each of these primes $p$, denote by $k_p$ the largest integer such that $p^{k_p}\le N$. I.e., $k_p=\bigg\lfloor \frac{\log n}{\log p}\bigg\rfloor$.

Are there infinitely many values of $N$ such that $$p_{n}^{k_{n}-1}+\cdots+p_2^{k_2-1}+p_1^{k_1-1}\le N\,\,?$$

When I check in mathematica, it seems to alternate between strings of Trues and then Falses, with the strings of Falses getting larger in each case. I am unsure if this will be eventually false or continue to alternate between strings of True and False.

I believe the inequality holds when $N$ is of the form $2^l-1$ but I cannot show this.

• If you like prime numbers, why you don't work on number theory books ? plouffe.fr/simon/math/IntrodAnalyticNTApostol.pdf – reuns Aug 17 '17 at 4:01
• $\sum_{i \le j} p_i^{\lfloor \log N/\log p \rfloor} \ge N \sum_{i \le j} \frac{1}{p_i}$ – reuns Aug 17 '17 at 4:21

It took me too long to reply because i still don't have a proof about the infinitude of the following numbers,

But first $m \approx \pi(\sqrt{n}) \geq \frac{2\sqrt{n}}{\ln n}$ because i only will be using the primes $2,3,5$ out of all primes less than $\sqrt{n}$.

Secondly define $T_k(m) = \{\ln_m 2^k\}$ where $\{x \}$ gives the fractional part of $x$.

Now i searched for all $k$ such that $(|T_k(3)| \leq \frac{1}{62} or |1-T_k(3)| \leq \frac{1}{62} ) and(|T_k(5)| \leq \frac{1}{62} or |1-T_k(5)| \leq \frac{1}{62} )$

Thirdly define $f(n) = \frac{2^{\lfloor \ln_2(n) \rfloor-1}+3^{\lfloor \ln_3(n) \rfloor-1}+5^{\lfloor \ln_5(n) \rfloor-1}}{n}$ , so if $f(n)>1$ then $m = \pi(\sqrt(n))- u$ where $u$ is the used primes till $f(n)>1$ .

Then i can find three integers $k_2,k_3,k_5$ such that $2^{k_2} \approx 3^{k_3} \approx 5^{k_5}$, its easy to prove that $1-\frac{1}{31} \leq \frac{p_i^{k_i}}{p_j^{k_j}} \leq 1+\frac{1}{31}$, so minimum we get that its $n\left(\frac{1}{2^{1+1/31}}+\frac{1}{3^{1+1/31}}+\frac{1}{5^{1+1/31}}\right) \approx 1.0005529n$

Now for the first such numbers :

1 - $n_1 = 5^{28}$ => $m_1 \geq 2.708*10^8$ => $f(n_1) \approx 1.02153$.

2 - $n_2 = 3^{983}$ => $m_2 \geq 5.92*10^{231}$ => $f(n_2) \approx 1.02648$.

3 - $n_3 = 5^{699}$ => $m_3 \geq 3.466*10^{241}$ => $f(n_3) \approx 1.01783$.

4 - $n_4 = 2^{5503}$ => $m_4 \geq 1.008*10^{825}$ => $f(n_4) \approx 1.02681$.

5 - $n_5 = 3^{4414}$ => $m_5 \geq 4.187*10^{1049}$ => $f(n_5) \approx 1.01742$.

6 - $n_6 = 3^{4455}$ => $m_6 \geq 2.5*10^{1059}$ => $f(n_6) \approx 1.02721$.

7 - $n_7 = 2^{7126}$ => $m_7 \geq 1.5*10^{1069}$ => $f(n_7) \approx 1.03098$.

8 - $n_8 = 5^{3097}$ => $m_8 \geq 9.08*10^{1078}$ => $f(n_8) \approx 1.02111$.

9 - $n_9 = 5^{3740}$ => $m_9 \geq 3.93*10^{1303}$ => $f(n_9) \approx 1.02916$.

And for all powers $k$ for base $2$ less than $10^5$ : $\{65,1558,1623,5503,6996,7061,7126,7191,8684,8749,12629,14122,14187,14252,15745,15810,15875,17368,17433,21248,21313,22806,22871,22936,23001,24494,24559,28439,29932,29997,30062,31555,31620,31685,33178,37058,37123,37188,38681,38746,38811,40304,44184,44249,45742,45807,45872,47365,47430,51310,51375,52868,52933,52998,54491,54556,54621,56114,58436,59929,59994,60059,61552,61617,61682,63175,63240,67120,67185,68678,68743,68808,70301,70366,74246,75739,75804,75869,77362,77427,77492,78985,81372,82865,82930,82995,84488,84553,84618,86111,89991,90056,91549,91614,91679,93172,93237,93302,94795,97117,97182,98675,98740,98805\}$

Even though a proof for the infinitude of such numbers is hard, see three or more prime raised to a powers become very close to each other

But i think you were looking that some small $m$ could be an upper bound, at least from these data we know that $m \geq 3.93*10^{1303}$ which is a huge number, also numerical experiments suggest that $m >> 3.93*10^{1303}$ and may be there is no upper bound constant, every $m$ you choose no matter how big, i can produce a triplet of power to the primes $2,3,5$ that will yield a bigger $m$.

In conclusion for infinitely many numbers $n$ (conjectured to be true but not proven) the sum over $2,3,5$ we be sufficient to produced a bigger number than $n$.