# Tag Info

7

$a_n$ is composite for $4 \le n \le 2016$. $a_{2017}$ appears to be prime (it passes a strong pseudoprime test). I have not tried to certify that it is prime (this would take a while as the number has 5789 digits).

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$\newcommand\an{\lfloor a n \rfloor}$ Let $a:=\alpha\in(0,1)$. By induction on $m=0,1,\dots$, $$\sum_{k=0}^m \binom nk(-1)^k\Big(1-\frac k{a n}\Big) \\ =(-1)^{m+1} (a+m-a n)\frac{m+1}{an (n-1)}\,\binom n{m+1}.$$ So, letting $S_n$ denote the sum in question, we have $$S_n\sim(-1)^{\lfloor a n \rfloor+1}(a-\{a n\}) \,M_n,$$ where $\{a n\}$ is the fractional ...

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Of course: for fixed $\varepsilon>0$ choose $M$ such that $\sum_{n>M} a_n<\varepsilon$, then for $n>M$ we have $$\sum_{k=1}^n \sqrt{a_k}=\sum_{k=1}^M \sqrt{a_k}+\sum_{k=M+1}^n \sqrt{a_k} \leqslant \sum_{k=1}^M \sqrt{a_k}+\sqrt{(n-M)\varepsilon}$$ by Cauchy–Schwarz. Dividing by $\sqrt{n}$ and taking limsup we get \limsup \frac1{\sqrt{n}} \... 2 To evaluate \begin{align} c_{n,m}&=\int_0^{1/2}\frac{x(x-1/2)}{\sin^2(2\pi x)}\, \sin(2\pi(2m+1)nx)\,dx\\ &=-\frac{1}{8}\int_0^{1}\frac{x(1-x)}{\sin^2(\pi x)}\, \sin(\pi(2m+1)nx)\,dx \end{align} we first notice by changing x\to 1-x, that the integral vanishes if n is even. In the following c_{2n+1,m} is calculated. We apply twice a result ... 2 \newcommand{\si}{\sigma} By the Irwin--Hall formula, your first displayed ratio is \begin{equation} f_n(x)=\frac{P(S_{n-1}\le an-x)}{P(S_n\le an-x)}=\frac{P(S_{n-1}\le a(n-1)-(x-a))}{P(S_n\le an-x)}, \end{equation} where a:=\alpha\in[0,1], x\ge0, S_n:=X_1+\dots+X_n, and X_1,\dots,X_n are iid random variables each uniformly distributed on [0,1]... 2 Call the two series S_1, S_2. Start out by letting g_1=1. Whatever we do afterwards, this makes sure that S_1\ge 1. Next, fix an M such that u_M e^{-1\cdot u_1}\ge 2, and then give g_2, \ldots, g_M a common small value that will give us e^{-\sum_{j=2}^M g_j u_j}\ge \frac{1}{2} . $$This guarantees that S_2\ge 1. Now just continue in this way.... 2 We write \mu=2^k \lambda with k \ge -3. The standing assumption \mu \ge 1 becomes 2^k \lambda \ge 1. The RHS of the sequence of equalities you reproduce is missing a factor \lambda^s; is this typo in the original paper? The sequence of equalities you included should thus be written (adding L on the left) as$$ L:= \sum_\lambda \lambda^s\sum_{\...

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(Extended comments in reply to Matt's comment to me.) The answer to your MO question was provided in the MSE question couched in terms of polynomials expressed as truncated power series, or ordinary generating functions (o.g.f.s) and their reciprocals. Here you use truncated Taylor series, or exponential generating functions (e.g.f.s). The series for the ...

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