nt.number theory – Why is it at all reasonable that $0leqsum_{Ssubseteq[Q]^*}frac{(-1)^{|S|}}{mathrm{lcm}(S)}{|S|choose (|S|+sum_{iin S}mu(i))/2}leq1$?

Let $(Q)^*$ denote the set of all squarefree natural numbers $leq Q$, and let $mu(n)$ be the Mobius function. I recently discovered the identity

$$0leqsum_{Ssubseteq(Q)^*}frac{(-1)^{|S|}}{mathrm{lcm}(S)}{|S|choose left(|S|+sum_{iin S}mu(i)right)/2}leq1tag{1}$$

And it makes absolutely no sense to me. To facilitate discussion, we let $|hat{S}|:=frac{1}{2}|S|+frac{1}{2}sum_{iin S}mu(i)$. The reason I came upon this sum is that I was studying the function $A_Q(n):=sum_{substack{d|n \ d<Q}}mu(d)$, and I noticed the the probabilities

$$Pr_{ninmathbb{N}}(A_Q(n)=j)$$

seemed to converge as $Qtoinfty$, which is not at all obvious at first glace. Since $A_{Q+1}(n)$ and $A_Q(n)$ disagree if and only if $Q$ is squarefree and $Q|n$, and in this case $A_{Q+1}(n)=A_Q(n)+mu(Q)$, we notice that

$$Pr_{ninmathbb{N}}(A_{Q+1}(n)=j)=Pr_{ninmathbb{N}}(A_Q(n)=j)-frac{1}{Q}Pr_{ninmathbb{N}}(A_Q(Qn)=j)+frac{1}{Q}Pr_{ninmathbb{N}}(A_Q(Qn)=j-mu(Q))tag{2}$$

Using a slightly more general version of this recurrence formula along with the fact that $A_1(n)=0$ uniformly gets us that

$$Pr_{ninmathbb{N}}(A_Q(n)=j)=sum_{Ssubseteq(Q-1)^*}frac{(-1)^{|S|+j}}{mathrm{lcm}(S)}{|S|choose |hat{S}|-j}tag{3}$$

As a direct consequence, we can plug in $j=0$ to get (1). Which this proof is sound and satisfactory, it makes me wonder how on earth such cancelling can occur. Generally in these types of situations one can find groups of terms that pair up with each other and cancel, but I cannot find any method that yields a result better than the trivial bound.

If anybody has any insights about where the cancellation in the sums (2) is coming from, or even better a way of showing that they converge, then that would be greatly appreciated.

nt.number theory – Finite groups arising as Galois groups of maximal unramified extension of number fields

I was wondering if it is known for which number fields the maximal unramified (non-abelian) extension is of finite degree or do we know the finite groups that arise as the Galois groups of these finite degree maximal unramified extensions.

I have seen the trivial group and the group of two elements as these Galois groups but not beyond that.

Thanks in advance.

nt.number theory – Natural functions giving zero on integer partitions

I’m working with integer partitions $ eta$ of $w$, with parts of size at most $n$, so that $eta$ satisfies $sum_{i = 1}^{n} i, eta_i = w$.

I have a set of functions $c_{n,w}(eta)$ on integer partitions in computer algebra in Mathematica for a given $n$ and $w$, and I want to try to reconstruct the form of the functions for general $n$ and $w$. The functions are valued on rational numbers, ie. $c_{n,w}(eta) in mathbb{Q}$.

For a given $n$ and $w$ some of the $c_{n,w}(eta)$ are zero, and I’m trying to work out the zeros first.
For example, for some $n$ and $w$ I have a $mathbb{Z}_2$ symmetry which forces the $c_{n,w}$ to be zero, depending on whether $sum_{i = 1}^{n} eta_i$ is even or odd. So I know that in those cases, I have that

$$
c_{n,w}(eta) sim 1+(-1)^{sum_{i = 1}^{n} eta_i}.$$

or $$
c_{n,w}(eta) sim 1-(-1)^{sum_{i = 1}^{n} eta_i}.$$

I also have some other cases with $c_{n,w}(eta) = 0$, but in those cases it’s not so clear to me to see what the pattern is. I’m looking for suggestions about natural functions $c_{n,w}(eta)$ which have zeros, additional to the ones I gave above.

I’ve also thought of $$c_{n,w}(eta) = sum_{i = 1}^{n} (-1)^i eta_i,$$ and $$c_{n,w}(eta) = 1-(-1)^{f(eta)}$$ for some different functions $f$, maybe such as $f(eta) = sum_{i = 1}^{n} i^a (eta_i)^b$ for $a,b in mathbb{N}.$ Neither of those seem to work for my current problem. I also considered taking the sums up to some $m < n$ in the functional forms I gave above so that I consider only some of the parts of the partition, but that didn’t seem to help either, and I’m not really sure how natural it would be to do that either.

I’m looking for functions constructed of only addition, subtraction, multiplication and integer powers of the componenets of $eta$. (So combinatorial functions like factorial or binomial coefficients would also be fine). In the end I sum the $c_{n,w}(eta)$ over $eta$ and multiply by $prod_{i=1}^{n}a_i^{eta_i}$ to construct a polynomial with a well-defined scaling weight $w$, if that helps at all in suggesting relevant functions.

As an example, I was looking at a case with $w = 9$, $n=6$. In that case, I have all of the $c_{n,w}(eta) = 0$ when $sum_{i = 1}^{n} eta_i$ is even, so I know I need the relevant function above as a factor (I also have the $mathbb{Z}_2$ symmetry in this case). I was surprised to find that there are also two additional zeros, when $eta = (3,0,2,0,0,0)$ and when $eta = (1,1,0,0,0,1)$, so I think I need to multiply in an additional factor which is zero on these two $eta$, non-zero on $eta$ where $sum_{i = 1}^{n} eta_i$ is odd, and which I don’t have any information about when $sum_{i = 1}^{n} eta_i$ is even. Does this give enough information for someone to suggest a relevant function in this particular situation?

I appreciate that this is unlikely to fully specify the function that I’m interested in. My philosophy is to try some of the more simple possible such functions, and see if they match against the form I need in Mathematica. Hopefully this way I’ll be able to reconstruct the full functional form that I’m looking for, and this should make it easier to go back and prove that form later on. (Also I enjoy doing it this way!)

Any references which may be relevant for me to read about this would also be very welcome. I have some basic understanding of integer partitions, but I’ve not worked with them before.

nt.number theory – Given a point $P$ on a genus-$1$ curve over $mathbb{Q}_p$, is there an $R$ such that $2 mid [P – R]$ and $x(overline{P}) neq x(overline{R})$?

Let $p in mathbb{Z}$ be prime, and let $f in mathbb{Z}_p(x)$ be a quartic polynomial with nonzero discriminant. Let $C/mathbb{Q}_p$ be the genus-$1$ curve with affine equation $y^2 = f(x)$. Let $overline{C}/mathbb{F}_p$ denote the mod-$p$ reduction of $C$, and for any $P in C(mathbb{Q}_p)$, let $overline{P} in overline{C}(mathbb{F}_p)$ denote the mod-$p$ reduction of $P$.

Suppose that $C(mathbb{Q}_p) neq varnothing$, and fix $O in C(mathbb{Q}_p)$. Let $E$ be the Jacobian of $C$, and let $iota colon C(mathbb{Q}_p) to E(mathbb{Q}_p)$ be the map sending $P mapsto (P – O)$.

Question: Let $P in C(mathbb{Q}_p)$ be integral, meaning that $x(overline{P}) neq infty$. Does there exist $Q in C(mathbb{Q}_p)$ such that $iota(P) + 2 cdot iota(Q) = iota(R)$, where $R in C(mathbb{Q}_p)$ is integral and $x(overline{P}) neq x(overline{R})$?

Partial Answer: Suppose that $C$ has good reduction modulo $p$ (i.e., suppose that $p$ does not divide the discriminant of $C$), and let $overline{E}/mathbb{F}_p$ denote the mod-$p$ reduction of $E$. Then $#overline{C}(mathbb{F}_p) gg p$ by the Hasse bound, so the set of points of the form $iota(overline{P}) + 2 cdot iota(overline{Q}) in overline{E}(mathbb{F}_p)$ has size $gg p$. But the set of points $overline{R} in overline{C}(mathbb{F}_p)$ such that $x(overline{R}) in {infty, x(overline{P})}$ has size $ll 1$, so the answer to the question in this case is yes if $p$ is sufficiently large. I’m not sure how to make the above argument work in the case where $E$ does not have good reduction modulo $p$, because I can’t talk about the group $overline{E}(mathbb{F}_p)$.

nt.number theory – Automorphy factor and the determinant of the Jacobian matrix

There are two cocycles defining the automorphy factors.

Let $D$ be the bounded symmetric domain (due to Harish-Chandra). Then there is a canonical automorphy factor $J:Gtimes Dto K_{mathbb{C}}$ where $G$ is a real Lie group with a maximal compact subgroup $K$, $D=G/K$ and $K_{mathbb{C}}$ is the complexification of $K$.

Meanwhile, due to Baily and Borel, there is also an automorphy factor $J_0:Gtimes Dto mathbb{C}^{*}$ coming from the determinant of the Jacobian matrix, i.e. the Jacobian of the action of $g$ at $zin D$.(Baily, Walter L., and Armand Borel. “Compactification of arithmetic quotients of bounded symmetric domains.” Annals of mathematics (1966): 442-528.)

Is there any relation between these two? The first one seems more general to me since we can add the representation of $K_{mathbb{C}}$ while the second one is always a nonzero complex number. And with the first one, we can transfer the automorphic forms (vector-valued) on Lie groups to the ones defined on the domain.

nt.number theory – Products of short elements in a field

Consider a field $F$ of characteristic zero. Let $L=F[alpha]$ be an extension of degree $d.$ We call an element
$$
x=a_0 + a_1 alpha +ldots+ a_{d-1}alpha^{d-1}in L
$$

short if $a_{d-1}=0.$
Under which conditions on $alpha$ every element in $L^times$ is a product of short elements?

It is easy to see that every element of $L$ is a quotient of two short elements for $dgeq 3$.

nt.number theory – What did Terence Tao mean here?

At 35:42 of this lecture, Terence Tao claimed that the following is a consequence of the generalized Elliott-Halberstam conjecture:

If $N$ is a sufficiently large multiple of $6$, there exists $n$ such that at least two of $n,n+2,N-n$ are prime.

Of course, the above statement is obviously (uncondintionally) true, by taking $n=3$. You can see in the lecture that Terence Tao by no means was saying the statement as a joke and claimed that either the twin prime conjecture or something very close to the Goldbach conjecture would follow from it. I cannot for the life of me figure out what the statement should be.

nt.number theory – Tweaking the Catalan recurrence and $2$-adic valuations

Among many descriptions of the Catalan numbers $C_n$, let’s use the recursive format $C_0=1$ and
$$C_{n+1}=sum_{i=0}^nC_iC_{n-i}.$$
Then, the $2$-adic valuation of $C_n$ is computed by $nu_2(C_n)=s(n+1)-1$ where $s(x)$ denotes the number of $1$’s in the $2$-ary (binary) expansion of $x$. In particular, $C_n$ is odd or $C_nequiv 1mod 2$ iff $n=2^k-1$ for some integer $k$.

Now, let’s tweak this a little so as to generate the sequence $u_0=1$ and
$$u_{n+1}=sum_{i=0}^nu_i^2u_{n-i}^2.$$

QUESTION. Is the following true? $nu_2(u_n)=(C_nmod2)+2s(n+1)-3$.
Equivalently,
$$nu_2(u_n)=begin{cases} 2s(n+1)-2 qquadtext{if $n=2^k-1$} \ 2s(n+1)-3 qquadtext{otherwise}. end{cases}$$

nt.number theory – Counting odd entries of Catalan numbers in a square array

The number of odd binomial coefficients in the $n^{th}$-row of the Pascal triangle equals to $2^{s(n)}$, where $s(n)$ denotes the number of $1$’s in the $2$-ary (binary) expansion of $n$.

Let $C_k=frac1{k+1}binom{2k}k$ be the familiar Catalan numbers. I’m interested in the enumeration of odd terms inside a square arrangement. In detail,

QUESTION. what is the number $O_n$ of odd Catalan numbers $C_{i+j}$ if $0leq i, jleq n-1$? It amounts to asking how many entries of the $ntimes n$ matrix $M_n=left(C_{i+j}right)_{i,j=0}^{n-1}$ are odd? I believe it equals $2n-1$. Can you provide a proof?

NOTE. It’s known that $det(M_n)=1$.

nt.number theory – Density of the set of numbers whose sum of digits is prime

Let $A$ be the set of numbers whose sum of digits is prime (http://oeis.org/A028834).

I would like to know if $A$ has zero natural density, that is, if $$lim_{n to +infty} frac{A(n)}{n} = 0,$$ where $A(n)$ is the number of elements of $A$ which are less than or equal to $n$.

Numerical experiments seems to indicate that $A(n) / n$ goes to zero, but very slowly.

$$begin{matrix}n & A(n)/n \
10 & 0.400 \
10^2 & 0.370 \
10^3 & 0.340 \
10^4 & 0.301 \
10^5 & 0.267 \
10^6 & 0.249
end{matrix}$$

Graph of $A(n) / n$

Thanks