New answers tagged

1

Here is a comment following Max Alekseyev's resolution. It has to do with his generating function for $t_n$ and working out directly on the Taylor's expansion (Binomial Theorem). Namely, $$\frac14\left(\frac{1+x}{\sqrt{1-6x+x^2}}-1\right) =\sum_{n=1}^{\infty}\frac{h(n-1)+h(n)}4\,x^n$$ where $$h(n)=\sum_{k=\lfloor\frac{n}2\rfloor}^n(-1)^{n+k}\binom{2k}k\binom{...


8

The answer is Yes. The generating function for $t_n$ is $$\sum_{n\geq 0} t_n x^n = \frac14\big(\frac{1+x}{\sqrt{1-6x+x^2}}-1\big).$$ Correspondingly, $$\sum_{n\geq 0} t_n x^n \equiv \frac{1+x}{\sqrt{1+x^2}}-1 \pmod{3}.$$ It follows that for $n>0$, $$t_n \equiv \binom{-1/2}{\lfloor n/2\rfloor}\equiv (-1)^{\lfloor n/2\rfloor}\binom{2\lfloor n/2\rfloor}{\...


1

Strong staircase conjecture follows from a the following version of the $k$-tuple conjecture: for any admissible tuple $T$ and a finite set $S$ disjoint from it, there are infinitely many integers $n$ such that $n+T$ contains only primes, and $n+S$ contains only composite numbers. This version follows from the Dickson's conjecture, by an argument similar to ...


2

Ford's methods provide lower bounds for this "asymmetric" multiplication table problem that match the lower bounds for the standard multiplication table problem. Define $H(x,y,z) := \#\{n \le x : \exists d \in (y,z] \text{ with } d \mid n\}$. Ford proved that $$H(x,y,cy) \asymp \frac{x}{(\log Y)^\delta (\log\log Y)^{2/3}}$$ whenever $c>1$ and $\...


3

Here are two counterexamples to the specific question at the end. Let $\alpha$ be the unique real solution to $xe^x=1$. This number is also called the omega constant. It is transcendental by the Lindemann-Weierstrass Theorem which (in particular) says that if $x$ is nonzero and algebraic then $e^x$ is not algebraic. So if $\alpha$ were algebraic then LW ...


3

Last time I checked, the entire function $f \colon \mathbf C \to \mathbf C \colon z \mapsto a\exp(iz)$ was transcendental for every non-zero $a \in \mathbf C$. So, taking $\alpha = -a = 2\pi$ yields a negative answer for the "$\alpha + f(\alpha)$" case; and taking $\alpha = a^{-1} = 2\pi$ yields a negative answer for the "$\alpha f(\alpha)$&...


4

Assuming the GRH for L-functions attached to modular forms, if $D$ is a cuspidal eigenform of weight $k$ with rational coefficients such as $\Delta$ and $\Lambda(s)=(2\pi)^{-s}\Gamma(s)L(s)$ the completed $L$-function (so that $\Lambda(k-s)=(-1)^{k/2}\Lambda(s)$), then the Hadamard product immediately implies that $$\sum_{\rho}\dfrac{1}{|\rho|^2}=\dfrac{2}{k}...


3

For what it's worth, it should not be possible to make $P$ and $nP$ integral on a quasi-minimal equation if $n$ is very large. More precisely, the following theorem holds. It quantifies what Chris Wuthrich said in the comments. Theorem: If Lang's height conjecture is true (or if the $ABC$-conjecture is true), then there is an absolute constant $C$ such if $E/...


4

Rational points which are hard to find are those of large height, and in particular large denominator. This method will only find rational points with denominator u when you scale the equations by u, which requires knowing u in advance or looping over all possible u. Try this curve: https://www.lmfdb.org/EllipticCurve/Q/294504803/d/1


1

It depends on what your notion of "closed" is. A quick calculation in Mathematica shows that your expression equals $x{n-2\choose k-1}\cdot {}_2 F_1(1,1+k-n;2-n;x)$, where ${}_2 F_1$ denotes the Gaussian hypergeometric function. These are well-understood in many contexts.


9

As observed in comments, we have $f(n) = \lfloor g(n) \rfloor$ where $g(n) = \frac{\alpha^n + \alpha^{-n}}{4}$ and $\alpha = 2 + \sqrt{3}$. From the recurrence $g(n+1) = 4 g(n) - g(n-1)$ we see that $g(n)$ is a half-integer when $n$ is even and an integer when $n$ is odd. In fact we see from induction that for even $n$ we have $g(n) = \frac{1}{2} \hbox{ ...


4

Too long for a comment. If you are restricting to $n$ being an integer, unless I made a mistake, the problem can be rephrased as Let $$a_{n+1}=4a_n-a_{n-1} \\ a_1=1 \\ a_0=\frac{1}{2}$$ Then, show that $\lfloor a_n \rfloor$ is prime implies that $n$ is prime. Here are some notes, I didn't check the details so there may be many mistakes. Note 1: I think that $...


1

Let us define $j=J+744=\frac{1728E_4^3}{E_4^3-E_6^2}$. Then, as noted by OP, the question is equivalent to positivity of coefficients of $$ f=q\frac{d}{dq}J+E_2(J+24)=q\frac{d}{dq}j+E_2j-720E_2. $$ Next, by this answer, we have $$ q\frac{d}{dq}j=-\frac{E_6}{E_4}j, $$ therefore $$ f=j\left(E_2-\frac{E_6}{E_4}\right)-720E_2. $$ On the other hand, by Ramanujan ...


7

It is widely believed that there indeed exists a prime between $n(n+1)$ and $(n+1)(n+2)$ for every integer $n\geq 1$. This appears to be beyond the scope of existing methods, however. Even the easier problem of find a prime between $n^3$ and $(n+1)^3$ for all integers $n\geq 1$ appears to be beyond the scope of existing methods. Dudek recently proved that ...


2

So, the sum can be rewritten as $$\sum_{0\leq k\leq k_1\leq\dots\leq k_j} k^m=\sum_{k=0}^{k_j} k^m \sum_{k\leq k_1\leq\dots\leq k_j} 1 = \sum_{k=0}^{k_j} k^m \binom{k_j-k+j-1}{j-1}$$ As explained in my other answer, this sum can be further reduced to $$\sum_{t=0}^m \left\langle m\atop t\right\rangle \binom{k_j+j+m-1-t}{j+m},$$ where the binomial ...


4

The question is not new. See my old question and its answer available from Is it true that $\{x^4+y^2+z^2:\ x,y,z\in\mathbb Z[i]\}=\{a+2bi:\ a,b\in\mathbb Z\}$?


1

Just for the record, I noticed that if we take a common denominator for $a_n$ as coming out of the terms $n+k$ then one would get $\prod_{j=1}^n(n+j)=\frac{(2n)!}{n!}$. Hence, we may rewrite the given sequence as follows: \begin{align*} \sum_{k=1}^n\binom{n}k\frac{k}{n+k} &=\frac{n!}{(2n)!}\sum_{k=1}^nk\binom{n}k\frac{\prod_{j=1}^n(n+j)}{n+k} \\ &=\...


13

Authors of the paper you linked actually define $f(z)$ differently. They have $$ f(z)=\left(\frac{1}{1-z}-1\right)\zeta\left(\frac{1}{1-z}\right), $$ so your $f(z)$ is their $f(-z^2)$ and every $\alpha$ in their formula corresponds to $\pm i\alpha$ from yours, so the sum on the left should actually be $\frac12$ of what is in the question, so $$ \frac{1}{2}\...


18

First we notice that \begin{split} a_n & = n \int_0^1 x^n (1+x)^{n-1}{\rm d}x \\ & = n \int_0^1 (1-x)^n (2-x)^{n-1}{\rm d}x \\ & = n\sum_{k=0}^{n-1} \binom{n-1}{k}2^k (-1)^{n-1-k} \int_0^1 (1-x)^n x^{n-1-k}{\rm d}x \\ &= \sum_{k=0}^{n-1} 2^k (-1)^{n-1-k} \binom{2n}{k} / \binom{2n}n. \\ \end{split} Now, the numerator in the last expression is ...


4

Since the ring of integers of $\mathbb{Q}[\sqrt{-163}]$ is a PID, it follows that a rational prime $p \neq 163$ may be expressed in the form $x^{2} + xy + 41y^{2}$ for rational integers $x$ and $y$ if and only if $p$ is a quadratic residue (mod $163$).(This is well-known). But, as you point out yourself, your question is comparable to asking how many primes ...


3

For $G$ a group, let $X(G)$ be the set of $n\ge 1$ such that $n$ divides the index of some finite-index normal subgroup of $G$. Then "normal" can be skipped in the definition (since every finite-index subgroup contains a normal one). In particular, if $H$ has finite index in $G$ then $X(H)\subseteq X(G)$. The question is If an infinite finitely ...


3

$$k(k^2+2)=3(k-1)e^2+3d^2\tag{1}$$ We derive the quadratic equation for $k$ to get infinitely many integer solutions. Let $d = k-n$, then we get $3e^2 = (k^3-3k^2+6kn+2k-3n^2)/(k-1).$ $k^3-3k^2+6kn+2k-3n^2$ is divisible by $k-1$ if $-3n^2+6n=0.$ Hence if $n=0$ and $d=k$ or $n=2$ and $d=k-2$ then we get $3e^2= k^2-2k$ or $3e^2=k^2-2k+12.$ These equations have ...


6

First of all, as mentioned by Random above, this is a very strong conjecture, because it is stronger than Legendre's conjecture. As far as I know, it is not known even if we assume the truth of Riemann Hypothesis and also some reasonable conjectures on distribution of imaginary parts of zeros, such as the Montgomery's pair correlation conjecture. However, it ...


7

This identity is false. To see that, notice first that $$ \sum_{p\leq \sqrt{2n-3}}a_p(n)=2\sum_{p\leq \sqrt{2n-3}}1-\sum_{\substack{p\leq \sqrt{2n-3}\\ p\mid n}}1=2\mathrm{ord}_C(n)-O(\ln n), $$ so $$ N_2^{eq}(n)=n\prod_{p\leq \sqrt{2n-3}}\left(\frac{\pi(n)}{n}\right)^{a_p(n)/\mathrm{ord}_C(n)}=n\left(\frac{1+o(1)}{\ln n}\right)^{2+O(\ln^2 n/\sqrt n)}=\frac{...


3

A mindless gp search over all odd $d,e,k < 2048$ takes about 2 minutes to find the following triples $[d,e,k]$: [1, 1, 1] [3, 1, 3] [3, 3, 5] [7, 5, 9] [1, 9, 15] [11, 11, 19] [27, 15, 27] [63, 37, 65] [41, 41, 71] [115, 67, 117] [1, 69, 119] [153, 153, 265] [363, 209, 363] [309, 251, 435] [131, 271, 469] [45, 293, 507] [443, 385, 667] [237, 413, 715] [...


28

One doesn't need to look at the genus here. We have $$x^2+y^2+z^2= (x+iy) (x-iy) + z^2.$$ Plugging in $z=0$, $x=1+iy$, we obtain $$1 (1+2iy) + 0^2= 1+ 2iy$$ which gets every number with real part odd and imaginary part even. Doing the same with $z=1$ gets every number with real and imaginary parts even. So we can represent every number with imaginary part ...


10

Yes, every Gaussian integer with even imaginary part is the sum of three squares of Gaussian integers. This was proved by Ivan Niven in his paper Integers of quadratic fields as sums of squares (Transactions of the AMS 48 (1940), 405-417). For comparison, those with odd imaginary part are not sums of any number of squares, because $(a+bi)^2 = a^2-b^2+2abi$ ...


2

Not an answer but Philipp Habegger has done some interesting work that could be characterized as evidence in support of the (folklore) polynomial conjecture which you can find at: https://arxiv.org/abs/1611.07287 The work is exposited in his talk at: https://www.youtube.com/watch?v=dqIVzojueww


18

One heuristic is to replace the $n^{th}$ roots of unity by $n$ iid elements $\zeta_1,\dots,\zeta_n$ of the unit circle, drawn uniformly at random. For any sum $\zeta_{i_1} + \dots + \zeta_{i_k}$ of $k$ of these with $i_1 < \dots < i_k$, a standard Fourier analytic calculation (Esseen concentration inequality) shows that if $k \geq 5$ (to make the $k$-...


3

You don't say that $c\in\mathbb Z$, but you imply that it is. Since we can absorb cube factors of $c$ into $z$, we may assume that $c$ is cube-free. Then your assertion that $c$ cannot be a power of $2$ is simply the assertion that $c$ cannot be 1, 2, or 4. In general, let $E_c$ be the elliptic curve $$ E_c : x^3 + y^3 = cz^3. $$ You are asking for a general ...


1

This is just a rough draft. One example which has infinitely many coprime integer soutions using obvious soution. $$A^3+B^3=cC^3\tag{1}$$ Let $x=A/C$, $y=B/C$ then we get equation $(2).$ $$x^3+y^3 = c\tag{2}$$ Cubic curve $(2)$ can be transformed to ellipric curve $(3).$ $$Y^2 = X^3-432c^2\tag{3}$$ $X = \frac{\large{12c}}{\large{x+y}}$ $Y = \frac{\large{36c(...


6

We will use the notation $e_p(t)=\exp\left(\frac{2\pi it}{p}\right)$. First, let us show that for any $1\leq m\leq \frac{p-1}{2}$ there is an eigenvector of your matrix $A_p$ with eigenvalue $$ \lambda_m=\sqrt{p}\cos\frac{2\pi m^2}{p}. $$ To do so, for any $m\in \mathbb Z$ denote by $v_m$ the vector from $\mathbb C^{p-1}$ with $k$-th coordinate $e_p(2mk)$ ...


3

I think the action is $q\star z = -q^{-1} z$, so that you get the "correct" basic theta function. See p. 128 in Fresnel-van der Put or Roquette's book.


8

Yes, those are the only solutions. To see this note that $$1+3+9+27 \cdots 3^m = \frac{3^{m+1}-1}{2}.$$ So we are looking for solutions of $\frac{3^{m+1}-1}{2}=2^n$, or equivalently looking for solutions of $3^{m+1} -1 = 2^{n+1}$. But this equation has only the obvious solutions, a result which one can prove with a little modular arithmetic. (Historical note:...


1

Note that if there are only finitely many non-Weiferich primes, then that would imply that there is such a $k$. If there were only finitely many non-Weiferich primes, one could make a sequence based on essentially your procedure and one would be forced to eventually have all the primes which divide the number be accounted for, either from the sequence, or ...


14

Many number-theoretic functions have error bounds contingent on the supremum of the real parts of zeroes of the Riemann zeta function. Call this $\Theta$. If we write $f(x)=\Omega_{\pm}(g(x))$ for the notion that \begin{align*} \limsup_{x\to\infty}\frac{f(x)}{g(x)}&>0,\text{ and}\\ \liminf_{x\to\infty}\frac{f(x)}{g(x)}&<0, \end{align*} then (e....


16

If this were proven, it would be a huge breakthrough in number theory. The most direct improvement would of course be a power savings in the error term of $\lvert \pi (x) - \mathrm{Li} (x) \rvert$, but there are many more applications for such things. For example, this would imply that $\zeta \left( \sigma + i t \right) = \mathcal{O} \left( \lvert t \rvert^{\...


2

There are a number of things one can say about this. First, some of the papers you mention do also cover the case of $p$-adic fields. For example, see Bergé's paper Sur l’arithmétique d’une extension diédrale Annales de l’institut Fourier, tome 22, no 2 (1972), p. 31-59. On page 32 you'll see that A can be a PID satisfying conditions 1), 2) & 3) listed ...


17

Here is the more elementary way that one would have solved this before LLL. I will still use the computer for some calculations, but technically one could do all this by hand. Worse, I won't even do all of the steps, but I hope the part I show explains the method, for those that may be interested. Maybe someone wants to complete it or finds a shorter way. ...


1

Collecting comments into a community wiki answer: Regarding why $L(ζ)=K(ζ)$: $L(ζ)⊆K(ζ)$ is the compositum of the subextensions $L⊆K(ζ)$ and $F(ζ)⊆K(ζ)$. Now (changing notation) when $F⊆K$ and $K'⊆L$ are subextensions of a Galois $L/F$, the compositum $K⋅K'⊆L$ has $\operatorname{Gal}(L/K⋅K')=\operatorname{Gal}(L/K)∩\operatorname{Gal}(L/K')$ (an element of $\...


3

An answer that applies when $f$ is a non-CM newform with trivial nebentypus and integral Fourier coefficients can be found in Theorem 1.5 of a paper by Thorner and Zaman (arXiv version). As you suggested, this result uses techniques that descend from work on bound the least prime in an arithmetic progression. If $\ell\nmid N$ is a sufficiently large prime (...


9

According to SAGE, there are 6 integral points on the curve. sage: EllipticCurve([0,0,1,0,1]).integral_points(both_signs=True) [(-1 : -1 : 1),(-1 : 0 : 1),(1 : -2 : 1),(1 : 1 : 1),(11 : -37 : 1),(11 : 36 : 1)]


2

I'll answer (2) and (3) together. To elaborate from the many suggestions in the comments, Dedekind sums arise in the transformation law of the Dedekind eta function, which are in turn related to periods of Eisenstein series. The latter are connected to special values of $L$-functions, so it is natural that Dedekind sums are used to parametrized these special ...


3

The post that you quoted contains a proof that $$\pi(x)-S(x)=\Omega_\pm(x^c)\quad\text{for any}\quad c<3/4.$$ Equivalently, for any $c<3/4$ and any $C>0$, both $\pi(x)-S(x)\leq Cx^c$ and $\pi(x)-S(x)\geq -Cx^c$ are false.


17

$\DeclareMathOperator\SL{SL}$The same argument (due to Siegel, in a classical form, of course), adelized, gives the analogous computation for any number field, and, yes, the corresponding Dedekind zeta appears. (To know that the adelic analogue computes the same thing, rather than the adelic quotient for $\SL_n$ being larger than the classical one, probably ...


2

1)Pick a sequence of elements $p^{1/p^n}\in \mathcal{O}_{\overline{K}}$ such that $(p^{1/p^{n+1}})^p=p^{1/p^n}$. The ideal $\ker\phi$ is in fact principal and is generated by the element $p^{\flat}:=(\dots, p^{1/p^2},p^{1/p},0)$ (where we view $p^{1/p^n}$ as elements in the reduction $\mathcal{O}_{\overline{K}}/p$). Indeed, any element of $\ker(\phi)$ has ...


3

I answer because Federico Poloni suggested it in a comment. This sequence is in the oeis: oeis.org/A088217 I don't think the question is very well defined, what does "explicit function of $n$" mean? And what is a "meaningless function"? The number is computable of course, for example this sage code could be considered a "function&...


18

Actually, this happens for all natural $b$. Notice that $$ \frac{1}{b}=\frac{1}{b+1}+\frac{1}{b^2+b} $$ and iterate this identity. You will get $$ \frac{1}{b}=\frac{1}{b+1}+\frac{1}{b^2+b+1}+\frac{1}{(b^2+b)(b^2+b+1)+1}+\ldots, $$ i.e. $$ \frac{1}{b}=\sum_n \frac{1}{a_n}, $$ where $a_1=b+1$ and $a_{n+1}=a_n^2-a_n+1$. This is similar to Sylvester's sequence ...


8

Let $n\geq 1$. The $L$-function of an automorphic representation of $\mathrm{GL}(n)$ is either (1) entire, or (2) holomorphic away from a pole of order $\leq n$ at $s=1+i\tau$ for some fixed $\tau\in\mathbb{R}$. Kim (Theorem B) proved that if $\pi$ is a cuspidal automorphic representation of $\mathrm{GL}(2)$, then $L(s,\pi,\mathrm{Sym}^4)$ is the $L$-...


3

$\newcommand{\bQ}{\mathbb{Q}}\newcommand{\cO}{\mathcal{O}}\newcommand{\bC}{\mathbb{C}}\newcommand{\bZ}{\mathbb{Z}}\newcommand{\bF}{\mathbb{F}}$The open subgroup $Gal(\overline{\bQ}_p/\bQ_p(\mu_p))$ acts trivially on the mod $p$ reduction of the cyclotomic character so the invariants in question can be computed as $$H^0(G_{\bQ_p},\cO_{\bC_p}/p(1))=((\cO_{\...


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