Evolution of partial sum of a sequence of induced Dirichlet characters

Question

Let's consider the Dirichlet Character $\chi_3(n)$ modulo 3 given by $\chi_3(1)=1$, $\chi_3(2)=-1$ and $\chi_3(3)=\chi_3(0)=0$.

Lets consider the sequence of induced characters $\chi^{P_N} $ obtained from $\chi_3$ having modulo $3.P_1...P_N$ where $P_N$ is the $N$-th prime number.

So at each step to define the new Character we replace by zero the value of multiple of $P_N$, we have :

$\chi_3 = (0,1,-1,0,1,-1,0,1,-1,...)$

$\chi^{P_1} =(0,1,0,0,0,-1,0,1,0,0,...)$ (all multiple of 2 have been put ot zero)

$\chi^{P_3} =(0,1,0,0,0,0,0,1,0,0,...)$ (all multiple of 5 have been put to zero)

So we obtain a sequence of induced characters with increasing modulo. My question is on the behavior of the maximum of the partial sum of the $\chi^{P_N}$ as $N$ increases :

$$S(\chi^{P_N},x) = |\sum_{n=1}^{x} \chi^{P_N}(n)|$$

$$Max_x(S(\chi^{P_N},x)$$

How this maximum will evoluate ?

My feeling is that this maximum will go up and down when $N$ goes to infinity and so that for a fixed constant $M$ (big enough) we can find an infinity of induced characters with this maximum lower than $M$. But is it true ? How to prove this is right or wrong ? Any reference on similar problem or idea ? Sieve theory can solve this kind of problem ?

(I already made a post on the subject in a more general way here : On partial sum of non-primitive Dirichlet characters)

Answer 1

The problem you are looking at can be stated in terms of the sum $$ S(x,y):= \sum_{\substack{n\le x \\ P^-(n)>y}} \chi_3(n) , $$ where $P^-(n)$ denotes the smallest prime divisor of $n$, with the convention that $P^-(1)=+\infty$. If $y\ge3$, then this sum is $Q$-periodic, where $$ Q=\prod_{p\le y} p = e^{(1+o(1))y}. $$ Because the modulus is very smooth, the Polya-Vinogradov bound can be improved quite a bit using the Eratosthenes-Legendre sieve: $$ S(x,y)= \sum_{n\le x} \chi_3(n) \sum_{d|(n,Q)} \mu(d) = \sum_{d|Q} \mu(d)\chi_3(d) \sum_{m\le x/d}\chi_3(m) . $$ The inner sum is $O(1)$ by the 3-periodicity of $\chi_3$, so that $$ S(x,y) \ll \sum_{d|Q} 1 = 2^{\pi(y)} = Q^{(\log 2+o(1))/\log\log Q} , $$ which is $Q^{o(1)}$ as $y\to\infty$.

It is even possible to go further: note that \begin{align} \sum_{n\le N} \chi_3(n) =\sum_{\substack{k\ge0 \\ 1+3k\le N}} 1 - \sum_{\substack{k\ge0 \\ 2+3k\le N}} 1 &= \left\lfloor \frac{N-1}{3} \right\rfloor - \left\lfloor \frac{N-2}{3} \right\rfloor \\ &= \begin{cases} 1 &\mbox{if $N$ is in $[1,2)$ mod 3},\\ 0 &\text{otherwise} \end{cases}, \end{align} so that $$ S(x,y) = \sum_{d|Q,\, d\in D} \mu(d)\chi_3(d), $$ where $$ D := \bigcup_{n\in\mathbb{Z}_{\ge0}} \left( \frac{x}{3n+2}, \frac{x}{3n+1}\right]. $$ This can be now estimated using smooth number technology. In particular, the interesting case is when $x$ is of size $Q$, in which case lattice point count heuristics start becoming accurate. What you could try and use now is either the saddle point method (see Tenenbaum's book "Introduction to Analytic and Probabilistic Number Theory"), or to try and adapt recent work of R. de la Breteche and G. Tenenbaum (see http://iecl.univ-lorraine.fr/~Gerald.Tenenbaum/PUBLIC/Prepublications_et_publications/Psi-resM.pdf.)

Answer 2

By the Pólya-Vinogradov inequality, $|S(\chi_N,x)|\ll (p_1\cdots p_N)^{1/2}$, whose order you can estimate using the prime number theorem.