In the framework of classical motives, there is no such thing as a motive $\mathbb Q(-\tfrac 12)$, i.e. a tensor root of $\mathbb Q(-1)$. There is one, however, in a more general setting of "exponential motives", over any number field containing $\sqrt{-1}$. There is even a motive $M$ over $\mathbb Q$ whose tensor square is $\mathbb Q(-1) \otimes H$, where $H$ is the Artin motive of the quadratic character associated with $\mathbb Q(i)|\mathbb Q$. These exponential motive have attached $L$-functions to them, and my question is about the $L$-function of $M$. Here it is, first as an Euler product: $$L(s) = \prod_p\frac{1}{1-g_pp^{-s}}$$ with $$g_p = \begin{cases} 0 & \mbox{if $p = 2$}\\ \sqrt p & \mbox{if $p\equiv 1 \bmod 4$}\\ i\sqrt p & \mbox{if $p\equiv 3 \bmod 4$.} \end{cases}$$ These $g_p$ arise as Gauss sums. The Euler product converges absolutely for $\Re(s)>\tfrac 32$. Here is the same $L$-function written as a Dirichlet series: $$L(s) = \sum_{n\geq 1 \:\mathrm{odd}}\frac{i^{r(n)}}{n^{s-1/2}}$$ where $r(n)$ stands for $$r(n) = \sum_{p \equiv 3 \:(4)}v_p(n),$$ the number of prime divisors $p\equiv 3\:(4)$ of $n$. The first question to ask about $L(s)$ is whether it has a singularity at $s=\tfrac32$ (it probably does), and if so, what kind. So I tried to figure out how much "cancellation" there is in the not-absolutely-converging series $$\sum_{n\geq 1 \:\mathrm{odd}}\frac{i^{r(n)}}{n}$$ We need to understand the class of $r(n)$ modulo $4$. So, for $r = 0,1,2,3$ and $x>0$, define the following sets of odd integers: $$A(r,x) = \{n <\:\mathrm{odd}\:|\:\: r(n)\equiv r \bmod 4\}$$ If for large $x$ these four sets have about the same cardinality, the sum might converge. I have no doubt that for $r=0,1,2,3$, the limit $$A(r) := \lim_{x\to \infty}\frac {2 \#A(r,x)}x$$ exists, and because of the parity of $r(n)$ depending only on $n \bmod 4$ we have $A(0)+A(2)=\tfrac 12$ and $A(1)+A(3)=\tfrac 12$. Now (lack any better ideas) I have done a count by computer, and to my amazement it seems that $A(0)$ is not $\tfrac 14$. Specifically, I counted the number of odd integers $n$ between $N$ and $N+10^8$ for which $r(n)$ is congruent to $0, 1, 2, 3$. The numerical calculation suggests the following approximate values $$\begin{array}{l|llll} & 4A(0) & 4A(1) & 4A(2) & 4A(3) \\ \hline N=10^9 & 0.8201 & 1.3632 & 1.1799 & 0.6368\\ N=10^{10} & 0.8114 & 1.3375 & 1.1886 & 0.6625\\ N=10^{11} & 0.8048 & 1.3143 & 1.1952 & 0.6857\\ N=10^{12} & 0.8000 & 1.2937 & 1.2000 & 0.7063 \end{array}$$ Equidistribution would mean that all entries in the table are close to $1$. My question can now be formulated most eloquently as:

What the heck is going on here?

Why are there fewer odd integres with $r(n)\equiv 0\bmod 4$ than there are with $r(n)\equiv 2\bmod 4$? And why does the situation get worse for large integers, when intuitively the contrary should happen?

  • $\begingroup$ Maybe this paper helps? " BIG BIASES AMONGST PRODUCTS OF TWO PRIMES" (Dummit, Granville, Kisilevsky, 2016 Mathematika): "We show that substantially more than a quarter of the odd integers of the form $pq$ up to $x$ with $p,q$ prime have $p\equiv q\equiv 3$ (4)." They also give data for finite levels of $x$ (as per your table). cambridge.org/core/journals/mathematika/article/… $\endgroup$ – just-a-guest Feb 1 '18 at 8:17
  • $\begingroup$ Should that be $p\equiv3\bmod4$ in the title? $\endgroup$ – Gerry Myerson Feb 1 '18 at 21:45

For your first question, note that one can write your $L$-series as $$ L(s) = \zeta\big(s-\tfrac12\big)^{(1+i)/2} L\big(s-\tfrac12,\chi_{-4}\big)^{(1-i)/2} M(s), $$ where $\chi_{-4}$ is the nontrivial character modulo $4$, and $M(s)$ is an Euler product that converges absolutely for $\Re s>1$. Therefore $L(s)$ does have a singularity at $s=\frac32$, one that looks like $c/(s-1)^{(1+i)/2}$ (that is, a logarithm-style singularity).

As for the eloquent question: remember that the average value, for $n\le x$, of $r(n)$ is about $\frac12\log\log x$, and the variance of $r(n)$ is also about $\frac12\log\log x$ (by the analogue of Hardy-Ramanujan's theorem about $\omega(n)$). For numbers $x$ between $10^9$ and $10^{12}$, the quantity $\frac12\log\log x$ is between $1.51$ and $1.66$. Therefore it it simply much more likely for numbers in this range to have $1$ or $2$ prime factors that are $3\pmod4$ than $0$ or $\ge3$ such prime factors. Of course the convergence to this average value (and indeed to a normal distribution, by a standard Erdös–Kac argument) is very slow; but I suspect you would see the opposite behavior in your histograms for $x$ near $e^{e^{7}} \approx 10^{476}$.

Ultimately, one can prove an analogue of the Selberg-Sathe theorem, which concerns $\#\{n\le x\colon \omega(n) = k\}$ uniformly for $k<1.5\log\log x$ say, for $\#\{n\le x\colon r(n) = k\}$, using the Selberg-Delange method (as in Tenenbaum's book for example). From this it follows, simply by sorting (according to their residue class modulo $4$) the integers $k$ near the average value of $\omega(n)$ or $r(n)$, that each of your $A(r)$ does converge to $\frac14$. It's just that the convergence is extremely slow, and does oscillate macroscopically along the way.


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