$f(x) + f(x/2) + f(x/3) + f(x/4) + ... = x$ and $\lim_{n \to \infty} \frac{f(n)}{\pi(n)} = 1$?

Question

$$f(x) + f(x/2) + f(x/3) + f(x/4) + ... = x$$

$$f(n) < \pi(n+1)$$

$$\lim_{n \to \infty} \frac{f(n)}{\pi(n)} = 1$$

where $\pi(n)$ is the prime counting function.

My mentor more or less wrote that when he was around $12$ yo.

If you use the Riemann zeta function $\zeta(s)$ to prove the PNT and you know the easy $f(x) < \frac{x}{\ln(x)-1}$ (easy to show for large $x$ at least) then with some little extra work you can prove the two statements at least for large $x$ or large $n$.

But how about proving it directly ? What if we did not know that $\pi(n)$ is about $= \frac{x}{\ln(x)-1}$ ? What if we do not use the zeta function nor complex dynamics ?

Is this related to one of those elementary proofs of the prime number theorem ?

How does one even come up with the idea of the equation above ?? What is the logic behind it ?

Do not confuse with the similar looking

$$Li(x) + Li(x^{1/2}) + Li(x^{1/3}) + ...$$

you often see in number theory, this is not the one.

And it is also not this one

$$g(x) + g(x)/2 + g(x)/3 + g(x)/4 + ... = x$$

Because that one gives about $g(n) = \frac{n}{H_n}$ where H_n is the n th harmonic number, what is a weak version of the PNT.

It also does not follow from the 3 theorems of Mertens nor does it resemble anything I ever read.

I am not looking for sharp asymptotics for $f(x)$ here, that is not the question here. (although I will probably start a new topic about that soon)

Any ideas ?

Btw this function is even more mysterious when you think about it :

For example

$$f(x) + f(x/2) + f(x/3) + f(x/4) + ... = x$$

substitute $x$ to $x/2$

$$f(x/2) + f(x/4) + f(x/6) + f(x/8) + f(x/10) + ... = x/2$$

This implies that

$$f(x) + f(x/3) + f(x/5) + f(x/7) + f(x/9) + ... = x/2$$

and thus

$$f(x/2) + f(x/4) + f(x/6) + f(x/8) + f(x/10) + ... = f(x) + f(x/3) + f(x/5) + f(x/7) + f(x/9) + ... $$

and

$$f(f(x) + f(x/3) + f(x/5) + f(x/7) + f(x/9) + ...) = f(f(x/2) + f(x/4) + f(x/6) + f(x/8) + f(x/10) + ...) = f(x/2)$$

and similar identities can be constructed. Imo very much like series multisection, fractals and self-referential things.

But getting series expansions for it that are defined everywhere is hard.

Does $f$ need to be continu ?

Maybe we should relax the equation and write

$$f(n) + f(n/2) + f(n/3) + f(n/4) + ...= n$$

and the divisions get rounded or so.

Or maybe we should truncate the ellipsis (...) somehow.

But that still does not answer my questions.

---

edit

My mentor told me this

$$\pi(x) + \pi(x/2) + \pi(x/3) + ... > x$$

if we take enough terms and if the following is true

$$\pi(x/a) > \frac{\pi(x)}{a}$$

(conjecture A)

It then follows that

$$f(x) + f(x/2) + f(x/3) + ... = x$$

implies that $f(x) < \pi(x)$

Conjecture A might have a name, it is a weaker version of the second hardy-littlewood conjecture :

$$\pi(x+y) - \pi(x) < \pi(y) + 1$$

Conjecture A makes sense for large $x$ and small $a$ for sure if we use a sharp version of the prime number theorem somewhat like the good asymptotic

$$\frac{x}{\ln(x) - 1 + \frac{1}{\ln(x)}}$$

But I am unsure if conjecture A has been proven.

Again this might be overkill but it is another way of thinking about it.

see :

$f(x) + f(x/2) + f(x/3) + ... = x$ and conjecture A : $\pi(x/a) > \frac{\pi(x)}{a}$

---

edit

A similar looking idea seems to have occured here on page 8 :

http://www.math.columbia.edu/~goldfeld/ErdosSelbergDispute.pdf

although that is an integral instead of a sum.

I wanted to mention it. Maybe it has value to someone.

---

Answer 1

here is a some what formalization of my thoughts

$$f(x) + f(\frac{x}{p_n}) + f(\frac{x}{p_{n+1}}) + f(\frac{x}{b_i}) + ... = \frac x{a_n} \tag{1}$$

where $p_n$ is the $n-$th prime number, $p_1=2,p_2=3,...$

$b_i$ is either a prime number after $p_n$ or a multiple of some prime number $p_n,p_{n+1},p_{n+2},...$ such that no prime before $p_{n}$ divides those multiples i.e. the multiples left only have factors $p_k$ with $k \geq n$. The classic Eratosthenes' prime sieve.

for $n=1$ we have the classic relation, which leads to $a_1=1$.

Let's apply $x \rightarrow \frac x{p_n}$ on $(1)$:

$$f(\frac{x}{p_n}) + f(\frac{x}{p_n^2}) + f(\frac{x}{p_{n+1}p_n}) + f(\frac{x}{b_ip_n}) + ... = \frac x{p_na_n}$$

we covered all the multiples of $p_n$ that were in $(1)$ so:

$$f(x) + f(\frac{x}{p_n}) + f(\frac{x}{p_{n+1}}) + f(\frac{x}{b_i}) + ... -[f(\frac{x}{p_n}) + f(\frac{x}{p_n^2}) + f(\frac{x}{p_{n+1}p_n}) + ...]= \frac x{a_n} -\frac x{p_na_n}$$

so

$$f(x) + f(\frac x{p_{n+1}}) + f(\frac x{p_{n+2}}) + f(\frac{x}{c_j}) +... = \frac x{a_n} -\frac x{p_na_n} = \frac x{a_n}(1- \frac1{p_n}) = \frac x{a_{n+1}}$$

so we have:

$$\frac1{a_{n+1}} = \frac1{a_n}(1-\frac1{p_n}) \implies \frac1{a_{k+1}} = \prod_{n=1}^k (1-\frac1{p_n}) \iff \\ a_k = \frac1{\prod_{n=1}^{k-1}(1-\frac1{p_n})}$$

which I believe to be correct. Interestingly, we have the Euler Identity for Riemann's zeta function:

$$\zeta (s) = \prod_p \frac1{1-p^{-s}}$$

which goes to show that $\lim_{n \rightarrow \infty} a_n = \infty$. Which is interesting:

$$f(x) + \lim_{n \rightarrow \infty} (f(\frac x{p_n}) + f(\frac{x}{b_n,i}) + ...) = 0$$

Answer 2

A property:

$$f(x) + f(x/m) + f(x/m^2) + f(x/m^3) + \ldots = x \times a_m(x)$$

$$f(x/m) + f(x/m^2) + f(x/m^3) + f(x/m^4) + \ldots = \frac{x}{m} a_m(x/m)$$ Hence, $$f(x) = x\left(a_m(x) - \frac{a_m(x/m)}{m}\right)$$

We also have, $$f(x/m) = x\left(\frac{a_m(x/m)}{m} - \frac{a_m(x/m^2)}{m^2}\right)$$

Hence $$x a_m(x) = f(x) + f(x/m) + f(x/m^2) + \ldots $$ $$= x \left(a_m(x) - \frac{a_m(x/m)}{m}\right) + x\left(\frac{a_m(x/m)}{m} - \frac{a_m(x/m^2)}{m^2}\right) + \ldots $$ $$= x\left(a_m(x) - \lim_j \frac{a_m(x/m^j)}{m^j}\right)$$

Hence $$\lim_j \frac{a_m(x/m^j)}{m^j} = 0$$

$$f(x) = x \lim_m \left(a_m(x) - \frac{a_m(x/m)}{m}\right)$$

Hence we have, $$f(x) = x \lim_m \left(a_{m}(x) - \frac{a_{m}(x/m)}{m}\right) = x g(x)$$ $$g(x) = \lim_m \left(a_{m}(x) - \frac{a_{m}(x/m)}{m}\right) $$ With, $$\lim_j \frac{a_m(x/m^j)}{m^j} = 0$$

$$\lim_j f(x/m^{j}) = \lim_j \frac{x}{m^j} \left(a_m(x/m^j) - \frac{a_m(x/m^{j+1})}{m}\right) = 0$$

So atleast $f(x)$ is continuous at $0$ w.r.t to sequence certain subsequences. An idea based on this is as follows:

One Idea is if we show the behaviour in a neighbourhood of $x = 0$ then: Let us assume $\sum_i f(x/m^i)$ converges absolutely, then, assume $f(x)$ is analytic in a neighbourhood of $x = 0$ then, $$f(x) = \sum_{i \geq 1} a_i x^i$$ Let $b_m(x) = x \times a_m(x)$ then, $$\sum_j f(x/m^j) = b_m(x)$$ $$\sum_j \sum_{i \geq 1} \frac{a_i}{m^{ji}} x^i = b_m(x)$$ Now by the assumption that $\sum_i f(x/m^i)$ converges absolutely we interchange the order of summation, $$\sum_{i \geq 1} \sum_j \frac{a_i}{m^{ji}} x^i = b_m(x)$$ $$\sum_{i \geq 1} \frac{a_i \times m^i}{m^i-1} x^i = b_m(x)$$ Hence we have an analytic expression for $b_m(x)$, $$b_m(x) = \sum_{i \geq 1} \frac{a_i \times m^i}{m^i-1} x^i $$

The series converges for all points the original series $f(x) = \sum_{i \geq 1} a_i x^i$ conveges. Hence, both $f(x)$ and $b_m(x)$ has same region of convergence.

So knowing $b_m(x)$ with assumption its analytic for any $m$, we can determine $f(x)$ in the neighbourhood of $x = 0$.

If you put $m = 1$, $b_1(x)$ diverges i.e., its region of convergence is $x = 0$, hence the assumption that $\sum_i f(x/m^i)$ converges absolutely is not true for $m = 1$ and might be true for $m > 1$.

Hopefully you will find some of the above ideas useful or able apply or use it. Cheers !

Answer 3

Let $x$ be a positive integer.

I considered thinking about estimating

$$\pi(x) + \pi(x/2) + \pi(x/3) + ... $$

The idea is simple.

we take primes $p_i < x$.

and we take $2 p_i < x$

and in general

$$n p_i < x $$

Then we naively expect all numbers between $1$ and $ x $ to be of these forms.

Hence the unions of $ n p_i$ should give about $x-1$.

However consider say $6 = 2*3$ and $35 = 5*7$.

$6$ is both of the form $ 2 p_i $ and $ 3 p_i $. Similar for $35$ ; $ 5 p_i$ and $ 7 p_i$.

So it appears

$$\pi(x) + \pi(x/2) + \pi(x/3) + ... = \sum_1^x fact(i) = T(x)$$

where $fact(x)$ is the number of factorizations of $x$

Since

$$f(x) + f(x/2) + f(x/3) + ... = x$$

This implies (even without conjecture $A$ but obviously with conjecture $A$)

$$ \pi(R(x) - 1) < f(x) < \pi(x+1) $$

Where $R(x)$ is the functional inverse of $T(x)$

Based on the average number of factorizations for a random $x$, $R(x)$ has as asymptotic $\frac{x}{\ln(\ln(x))}$,

therefore

$$f(x) = \frac{\pi(x)}{\ln(\ln(x))}$$

is a good asymptotic.

This implies the limit from the OP is not true. What was probably the reason the idea got abandonned.

As for getting better estimates I see two routes at the moment:

1. Improve the estimate of $R$ or $T$.

2. Use the estimate for $f(x)$ together with the ratio's $\pi(x/m)/f(x/m)$ and the functional equation, and repeat that idea for a sequence of increasing improved asymptotics.

( The previous answer was wrong and I will delete it )