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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}{\...

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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 ...

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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 ...

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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 ...

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