What Is The Ratio Of Sum Of All Prime under N to sum of all composite under N?

170 Views Asked by At

Ratio Of Sum Of Primes Under N To Sum Of Composite Under N

The picture is of the graph of ratio of sum of all prime numbers under 2,000,000 to sum of all composite numbers under 2,000,000 ,which I created using python and matplotlib.

[ X Axis = Range of Numbers ]

[ Y Axis = Ratio]

It appears That the ratio is approaching 0 as we go to $\infty$.

Can anyone explain what it implies ? What does it say about the distribution of prime numbers ? What is the reason for such behaviours ? Is it okay to infer that for larger values , the curve will infinitely come closer to 0 without ever increasing?

2

There are 2 best solutions below

0
On BEST ANSWER

The number of primes under $N$ is asymptotic to $N/(\log N)$, so the sum of those primes is no more than $N^2/(\log N)$. The sum of all numbers under $N$ is essentially $N^2/2$, so the primes make a negligible contribution to this sum (since $\log N$ well exceeds $2$).

0
On

Let $p_n$ be the largest prime not exceeding $N$ and let $p$ be the sequence of primes and $c$ be the sequence of composites. We have

$$ \sum_{p \le p_n} p = \frac{n^2\log n}{2} + \frac{n^2\log \log n}{2} - \frac{3n^2}{4} + o(n^2) $$

and $p_n \sim n\log n + n\log\log n - n$. Hence

$$ \sum_{c \le p_n}c = \sum_{r \le p_n}r - \sum_{p \le p_n} p \sim \frac{n^2\log^2 n}{2} $$

Hence $$ \frac{\sum_{p \le p_n} p}{\sum_{c \le p_n}c} \sim \frac{1}{\log n} $$