If we define

$$\lVert a_n\rVert = \lim_{N\to\infty}\frac{1}{N}\sum_{k=0}^{N-1}a_k$$

To be the mean of a sequence, and we let $a_n$ be a bounded sequence of integers where not only does $\lVert a_n\rVert$ converge but $\lVert a_{pn}\rVert$ converges for all prime numbers $p$, where

$$\lVert a_{pn}\rVert=\lim_{N\to\infty}\frac{1}{N}\sum_{k=0}^{N-1}a_{pn}$$

Is the mean of the elements that are multiples of $p$.

If $\lVert a_{pn}\rVert>0$ $\forall$ primes $p$, then it feels only natural that $\lVert a_n \rVert>0$ as well, but I simply cannot seem to prove it. I have tested this property on several sequences and it seems to hold, but if this theorem does not hold and someone out there has a counterexample that would be greatly appreciated as well.

I think that this theorem has some deep consequences about the distribution of primes since from this theorem one can reasonably easily deduce the PNT.

  • 1
    $\begingroup$ In your definition of "$\lVert a_n\rVert = \lim_{N\to\infty}\frac{1}{N}\sum_{k=0}^{N-1}a_k$", with the RHS, if the limit exists, it would be a single value. However, on the LHS, you have $a_n$, which doesn't match unless you want all of your $a_i$ to be the same value. $\endgroup$ Commented Apr 24, 2020 at 18:58
  • $\begingroup$ The LHS of my equation, $\lVert a_n \rVert$, is not meant to be dependent on $n$ and only includes $n$ since it is a sequence. Should I replace it with $\lVert a \rVert$ to make this more clear? $\endgroup$
    – Milo Moses
    Commented Apr 24, 2020 at 19:01

1 Answer 1


Say that a function $\chi_q: \mathbf{Z} \rightarrow \{-1,0,1\}$ is suitable if:

  1. $\chi_q(n)$ only depends on $n \pmod q$.
  2. $\chi_q(0) = 1$.
  3. $\displaystyle{\sum_{n=0}^{p-1} \chi_q(n) = 0}$.

If $\chi_q(n)$ is suitable, then so is $\chi_q(pn)$ for any prime $(p,q) = 1$. The average of any suitable $\chi_q$ along an arithmetic progression with difference prime to $q$ is zero. Moreover, the partial sums of $\chi_q(pn)$ along any such progression are bounded in absolute value by $q$, since they are just sums over sequences which repeat every $q$ terms. Finally, $\chi_q(qn) = 1$ for any $n$.

There certainly exist many suitable functions, let us fix an explicit such function by the relations:

  1. $\chi_q(n)$ only depends on $n \pmod q$.
  2. If $0 \le n \le p-2$, then $\chi_q(n) = (-1)^n$.
  3. If $n = p-1$, then $\chi_q(n) = 0$. That is, the values of $\chi_q(n)$ for $n = 0, \ldots p-1$ are $$\{1,-1,1,-1,1,-1,\ldots,1,-1,0\}$$

We now let $b_{n,q}$ denote the following sequence:

$$b_{n,q} = \begin{cases} \chi_q(n) & 2^{q} \| n, \\ 0 & \text{otherwise} \end{cases}$$

We have $\|b_{n,q}\| = 0$, since we are averaging $\chi_q(n)$ over the arithmetic progression $2^q \pmod {2^{q+1}}$. Similarly, $\|b_{np,q}\| = 0$, since we are averaging $\chi_q(pn)$ over the same progression. Finally, $b_{nq,q} = 1$ whenever it is non-zero, so $$\|b_{nq,q}\| = \frac{1}{2^{q+1}}.$$ We now let $$c_n = \sum_{q > 2} b_{n,q}$$ Note that each $n$ is divisible by a unique power of $2$, and so $c_n \in \{-1,0,1\}$. We claim that $\|c_n\| = \|c_{2n}\| = 0$, and $\|c_{qn}\| = 2^{-q-1}$ for any odd prime $q$. This would follow immediately from the identities $$\|c_{n}\| = \sum \|b_{n,q}\|, \quad \|c_{np}\| = \sum \|b_{np,q}\|,$$ but we need to be slightly careful as the sums are not absolutely convergent. Still, it is relatively easy. Consider the case of $\|c_n\|$. If we sum up to $X$, the only contributions we see are coming from $b_{n,q}$ with primes $q$ such that $2^q < X$, or $q < \log(X)$ (in base $2$ but I don't want to bother writing subscripts). The partial sums for each $b_{n,q}$ are always bounded by $q$, and thus the partial sums of the $c_n$ up to $X$ are bounded by $$\sum_{q < \log(X)} q < \sum_{q < \log(X)} \log(X) < (\log(X))^2 = o(X).$$ In particular, the average of the $a_n$ is certainly tending to zero. The other cases are similar. Finally, to get the contribution at $2$ to work out, we can let $$a_n = c_n + \begin{cases} 1 & n \equiv 2 \mod 4 \\ -1 & n \equiv 3 \mod 4 \end{cases} $$ which is still valued in $\{-1,0,1\}$ and has the same averages as before except now $\|a_{2n}\| = 1/2$.

  • $\begingroup$ It looks like you are missing some $i$-s in your $\exp$ terms. $\endgroup$
    – aschepler
    Commented Apr 24, 2020 at 23:56
  • $\begingroup$ @user760870 once again this is a really cool example and I'm glad you put it, but it was for cases like this that I put the condition that $a_n$ is a sequence of integers. $\endgroup$
    – Milo Moses
    Commented Apr 25, 2020 at 1:02
  • $\begingroup$ @MiloMoses Happy now? $\endgroup$
    – user760870
    Commented Apr 25, 2020 at 3:09
  • $\begingroup$ @user760870 yes! thank you, this is exactly the type of example I was looking for $\endgroup$
    – Milo Moses
    Commented Apr 25, 2020 at 5:30

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