Let $n\geq 2$ integer. The arithmetic function $$\sigma(n)=\sum_{d\mid n}d$$ is the sum of divisor function, and $\pi(x)$ the prime-counting function.

I know how to solve this puzzle that I am written this morning:

Puzzle 1. Find the solutions over integers $n\geq 2$ for $$2\sigma(\pi(n))=n.\tag{1}$$ Solution (Wrong, see comments). The sequence starts as $2,12,26,504\ldots$ One knows that $2\sigma(\pi(n))>2\pi(n)$, and Chebyshev's bounds about (I say for example the identity (8) from this MathWorld ) in the context of the prime number theorem tells me that as there exists a $N_0$ such that $\forall n>N_0$ one has $2\sigma(\pi(n))>\frac{14}{8}n>n$. Thus the solution of our puzzle can be get using a computer since I know that there exist finitely many solution of $(1)$.$\square$

In the same way, trying do a joke using the condition to be a perfect number $$\sigma(n)=2n,$$ with the intention to combine with particular values of the prime-counting function I've consider this more interesting

Puzzle 2. Find the solutions over integers $n\geq 2$ for $$\sigma(\pi(n))=\pi(2n).\tag{2}$$

Computational fact. The sequence for Puzzle 2 starts as $$3, 5, 9, 45, 46, 47, 48, 761, 1301, 1302,\ldots$$

Question. What is the strategy to solve Puzzle 2? Are there finilety many solutions of $(2)$? Many thanks.

Comment: Finally, I tried find in Internet if such sequences were in the literature, I call these sequence of integers as $\pi$erfect numbers (the first kind, the second kind...). I don't know if it is possible write a more nice puzzle using the prime-counting function, and the notion to be perfect (with a more difficult solution).

  • $\begingroup$ To be able to use a computer to solve puzzle 1, don't you need to know what $N_0$ is? $\endgroup$ – mathworker21 Oct 17 '17 at 12:17
  • $\begingroup$ Many thanks for your attention, that I am saying is that Puzzle 1 isn't interesting to me after I know that Chebyshev implies that there are finitely many solutions (with a computer one can find all solutions) @mathworker21 But you are right, in that the serious way is to bound $N_0$ and after work with a computer. $\endgroup$ – user243301 Oct 17 '17 at 12:20
  • $\begingroup$ Chebyshev deals with an inequality for $\frac{n}{\ln(n)}$ not $n$. What you have said for 1 does not follow immediately from that inequality, as $\pi(n) < n$ by definition, and $2\pi(n) > n$ fails immediately for $n > 8$. $\endgroup$ – Paul LeVan Oct 17 '17 at 20:11
  • $\begingroup$ It was a big mistake, I'm sorry I did a wrong calculation, but I believe that Puzzle 1 has only finitely many solutions. Many thaks for your attention and comment @PaulLeVan $\endgroup$ – user243301 Oct 17 '17 at 21:33

To show that there are only finitely many solutions to $2\sigma( \pi(n))=n$, we can use the fact that $\sigma(n)<2 n \log \log n$ for $n>60$, and Rosser and Shoenfeld's results that $$\pi(n)<1.25506 \frac{n}{\log n} \text{ for } n>1$$ to conclude that $$ 2 \sigma(\pi(n))< \frac{5.02024 n \log \log n}{\log n}.$$ We can show that the right-hand expression is less then $n$ for $n<10^6$, and so there are only finitely many solutions fo $2\sigma(\pi(n))=n$.

Checking up to $10^6$, we find all solutions are $2,12,26, 48, $ and $504$.

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  • $\begingroup$ Many thanks for your answer to get Puzzle 1, my calculation was a disaster. I am going to read and study your answer now. As a present for you here you've a new $\pi$uzzle, I say if you want to study it in your home: Solve over integers $n\geq 2$ the equation $\sigma(\pi(n))=2\pi(\sigma(n))$. $\endgroup$ – user243301 Oct 18 '17 at 11:37

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