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Not an answer, but this may help with asymptotics: According to Maple the o.g.f. for $H^s_n$ is $$ \sum_{j=1}^\infty j^{-s} (-1)^{j-1} \sum_{n=j}^\infty {n \choose j} x^n = {\frac {1}{-1+x}{\it polylog …

There are really two separate cases: $$\eqalign{a' + 1 &= +\sqrt{1 - k^2 a^2}\cr a' + 1 &= -\sqrt{1 - k^2 a^2}\cr} $$ You could switch continuously from one to the other only on the curve …

$\log f(z) = z \log(1+1/z) = \sum_{k=0}^\infty \frac{(-1)^k}{k+1} z^{-k}$ as $z \to +\infty$, so $f(z)$ is the exponential of this sum. See http://oeis.org/A055505 for the numerators and http://oeis. …

Let $N(k) = \max \{n \in \mathbb N: \; {n \choose k} 2^{1-{k\choose 2}} < 1 \}$. Note that $$\frac{{n+1 \choose k}}{{n \choose k}} = \frac{n+1}{n+1-k}$$ so it suffices to prove that $N(k)/k \to \infty …

Maybe for some family of initial conditions (depending on the $O(1/r^4)$ term). But note that $f(r) = -1/r + c/r^2$ is a solution to $f'(r) = f(r)^2 - c^2/r^4$ with initial condition $f(1) = c-1$.

Using the gfun package in Maple, it appears that $$ M(x) = -\frac{1}{4}-{ \frac {3\,x}{28}}-6 (x-7)\,\int_{0}^{x}\!{\frac {{{\rm e}^{t \left( 4\,\sqrt {3}+7 \right) }}}{ \left( t-7 \right) ^{2}}{\it …

If $\beta$ is an integer $\ge 3$, it can be written as a hypergeometric $$ \mbox{$_0$F$_{\beta-1}$}(\ ;\, 1,\ldots,1;,\exp(c\beta))$$ where there are $\beta-1$ ones.

Let $X = \sum_{i=1}^n a_i Z_i^2$. If $m = \min(a_1,\ldots,a_n)$ and $M = \max(a_1,\ldots,a_n)$, we have $m A \le X \le M A$ where $A$ has $\chi^2$ distribution with $n$ degrees of freedom. Thus $$\ …

Of course it is possible, but much caution is required because we don't know about uniformity in $k$ for the estimates corresponding to the $\sim$ statements. Counterexamples are very easy to const …

According to Maple, the finite product $f(n) = \frac{\Gamma(n - (1 + \sqrt{1-4n})/2) \Gamma(n-(1-\sqrt{1-4n})/2) \Gamma(1-\sqrt{-n}) \Gamma(1+\sqrt{-n})} {\Gamma(n-\sqrt{-n}) \Gamma(n+\sqrt{-n}) \Gamm …

If $P_k(t) = \sum_{m=0}^k a_{m,k-n} t^m$ is the generating function of an ascending antidiagonal, we have $$P_k(t) = \frac{t^k-t}{t-1} + \frac{1+t}{2} P_{k-1}(t), \ P_0(t) = 0 $$ and this can be solv …

If you want the smallest, try $$x = -LambertW(-\epsilon)/\epsilon = 1+\epsilon+{\frac{3}{2}}{\epsilon}^{2}+{\frac{8}{3}}{\epsilon}^{3}+{\frac{125}{24}}{\epsilon}^{4} +O \left( {\epsilon}^{5} \right) …