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).


5

$\newcommand\an{\lfloor a n \rfloor}$ 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 ...


4

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_{\...


1

(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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