Omega

Number Theory Level 5

$\large s = \sum_{n=1}^\infty \dfrac{\omega(n)}{n^2}$

Let $\omega(n)$ denote the number of distinct prime factors of $n$ . Compute $\lceil 1000s\rceil$ .

Details and Assumptions :

You may use the fact that $\displaystyle P(2) = \sum_{p \text{ prime}} \dfrac1{p^2} \approx 0.452247$ .
You may read up Prime zeta function .

The answer is 744.

1 solution

Aareyan Manzoor
Feb 10, 2016

dirichlet series and dirichlet convolution used here. define $L(n)=\begin{cases} 1\text{if n is prime}\\ 0 \text{otherwise}\end{cases}$ we have $L*1=\omega$ $\sum_{n=1} \dfrac{\omega(n)}{n^s}=\sum_{n=1} \dfrac{L(n)}{n^s}\sum_{n=1} \dfrac{1}{n^s}$ $\sum_{n=1} \dfrac{\omega(n)}{n^s}=P(s)\zeta(s)$ Put s=2 $\sum_{n=1} \dfrac{\omega(n)}{n^2}=P(2)\zeta(2)=\dfrac{\pi^2}{6} P(2)\approx 0.\boxed{743}9$

I have changed the value that you gave for $P(2)$ in the question because it was wrong. The answer for the problem is still correct.

Julian Poon - 5 years, 4 months ago

I copy pasted it from wolframalpha.

Aareyan Manzoor - 5 years, 4 months ago

Easter egg: Using the initial value you gave, you can get 404 for the answer.

Julian Poon - 5 years, 4 months ago

There might be a mistake in copying the number because the number you gave was $0.2452247$

Julian Poon - 5 years, 4 months ago

Omega

The answer is 744.

1 solution

0 pending reports