Relationship between primes and practical numbers This is my first post here. I am a musician, and not a mathematician, but I enjoy doing things to prime numbers and seeing what comes out.
I have defined a sequence which takes the following values for $n$:


*

*-1 if $n$ is prime

*1 if $n$ is a practical number

*0 if $n$ is neither or both


I have then taken a sequence of its partial sums. The first 50 terms are 1,1,0,1,0,1,0,1,1,1,0,1,0,0,0,1,0,1,0,1,1,1,0,1,1,1,1,2,1,2,1,2,2,2,2,3,2,2,2,3,2,3,2,2,2,2,1,2,2,2.
The plot for $n<100000$ looks quite linear:

In order to see quite how linear it was, I then divided each term of the sequence by $n$ and got this plot:

It seems to me like it wants to converge to some value. The arithmetic mean of the last 100 terms is 46.3225. 
I vaguely understand that there are some analogies between practical numbers and primes. I am wondering how difficult it would be to establish if the above sequence does in fact converge, and if so, then to what value. I have tried it with other prime-like sequences, such as ludic numbers and lucky numbers, but the other ones didn't seem as neat... 
Thanks!
 A: The plot isn't actually linear, but it indeed looks like one. Here's why:
Let's introduce two functions, $\pi(x)$ and $p(x)$, that respectively count the number of prime and practical numbers less than $x$.
A famous result in number theory, the prime number theorem, tells us that
$$
\pi(x) \sim \frac{x}{\log(x)} \quad \text{for } x \to +\infty
$$
which essentially means that for $x$ large enough $\pi(x)$ behaves almost exactly like $x/\log(x)$.1 Remarkably, just in 2015 Weingartner showed that
$$
p(x) \sim \frac{cx}{\log(x)} \quad \text{for } x \to +\infty
$$
for some constant $c > 0$.
The sequence of partial sums you defined is then simply the sequence $\{s(n)\}_{n=1}^\infty$ where $s(x) := p(x) - \pi(x)$. It follows that
$$
s(x) \sim (c-1) \frac{x}{\log(x)} \quad \text{for } x \to +\infty.
$$
To conclude, here's what $\frac{x}{\log(x)}$ looks like on the interval $[0,10^4]$:

 1. The almost is important: $\pi(x)$ may never be equal to $x/\log(x)$, but after a while the error will grow quite slower than $x/\log(x)$.

Update: I should probably explicitly say something about your second graph, too, but first we need to give a precise meaning to "$\sim$". Given two functions $f,g \colon D \subseteq \Bbb{R} \to \Bbb{R}$ and $\rho \in \Bbb{R} \cup \{\pm\infty\}$, we say that $f \sim g$ for $x \to \rho$ if
$$
\lim_{x \to \rho} \frac{f(x)}{g(x)} = 1.
$$
Now, what can we say about the limit of the sequence depicted in your second graph? From the previous discussion we have
$$
\begin{align}
\lim_{n \to \infty} \frac{s(n)}{n} &= \lim_{n \to \infty} \frac{p(n)-\pi(n)}{n} \\
&= \lim_{n \to \infty} \frac{p(n)-\pi(n)}{n}\; \frac{\log(n)}{\log(n)} \\
&= \lim_{n \to \infty} \frac{p(n)-\pi(n)}{n/\log(n)}\; \frac{1}{\log(n)} \\
&= \left(\lim_{n \to \infty} \frac{p(n)-\pi(n)}{n/\log(n)}\right) \left(\lim_{n \to \infty}\frac{1}{\log(n)}\right) \\
&= (c-1) \lim_{n \to \infty}\frac{1}{\log(n)} \\
&= 0
\end{align}
$$
So $s(n)/n$ does indeed converge to $0$, but I don't know how fast.
