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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!

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  • $\begingroup$ It's just occurred to me now that the reason the other ones aren't as neat is probably because primes are pretty much odd, and practical numbers are even, unlike ludic and lucky numbers. $\endgroup$ – Matt B Dec 30 '15 at 23:08
  • $\begingroup$ From Wikipedia: In number theory, a practical number or panarithmic number is a positive integer n such that all smaller positive integers can be represented as sums of distinct divisors of n. $\endgroup$ – marty cohen Dec 30 '15 at 23:13
  • $\begingroup$ @MattB please can you add to your question the original sequence of -1,1,0 values? (the one that you add to make the final sequence that you use for the grahs) Did you try to compare that sequence with the Möbius function? en.m.wikipedia.org/wiki/M%C3%B6bius_function for instance, what happens if you add or substract both sequences, etc. e.g. Is your original sequence multiplicative? Imho, your original sequence could be more interesting than the accumulated sequence, you can study the behaviour. Is it kind of pseudorandom like the Möbius function seems to be? You did a nice question! $\endgroup$ – iadvd Dec 31 '15 at 16:13
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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 $[0,10^4]$: plot of x/log(x)

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.

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  • $\begingroup$ Right now I don't have time to carefully examine Weingartner's proof, but from the statement of his theorem it isn't clear if $c$ is effective, i.e. if the proof gives a way to compute it, or not. $\endgroup$ – A.P. Dec 31 '15 at 0:51
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    $\begingroup$ The proof of the asymptotic result for practical numbers does unfortunately not seem to give a way to compute the constant $c$. Numerical computations of $p(x)$ for large $x$ suggest that $c$ is approximately $1.3$. $\endgroup$ – Andreas Weingartner Jan 28 '16 at 5:31
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    $\begingroup$ 2020 update: The constant factor $c$ in the asymptotic for $p(x)$ has now been determined to be $c=1.33607...$. See arxiv.org/abs/1906.07819 $\endgroup$ – Andreas Weingartner Aug 5 at 16:44

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