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Your data is compatible with the more refined estimates proved by Vaughan in Ramanujan J. 15 (2008), 109–121. His Theorems 1 and 2 (together with his (1.9)) reveal that $$\log(A000607(n)) = 2 \pi \sqrt{\frac{n}{3 \log n}}\left(1+\frac{\log\log n}{\log n}+O\left(\frac{1}{\log n}\right)\right).$$ For $n=50000$, we have $$\log(A000607(n)) \approx 252.663$$ $$... 22 I'll show that$$ \log a(n) \sim \frac{(\log n)^2}{\log 4} \approx 0.7213\ldots (\log n)^2. $$So the range for the constant given in the conjecture is false, but an asymptotic of that general shape holds. One can obtain more precise asymptotics by working harder with the argument below. Roughly speaking what the argument says is that a(n) behaves (... 21 For the asymptotic case: Let t_1=n \log n - Cn and t_2 = n \log n + Cn, where C is slowly tending to infinity. It is a classic result that as C tends to infinity the probability all coupons are collected at time t_1 tends to 0, and the probability all coupons are collected at time t_2 tends to 1. So the number being asked for in the ... 19 I sketch the arguments for C(x), the arguments for L(x) are essentially the same. The specific form of the sum suggests probabilistic arguments. Let X_x be a \mathrm{Poiss}(x^2)-distributed random variable and note that$$C(x)\,e^{-x^2}=\mathbb{E}\frac{x}{\sqrt{X_x}}\,1_{\{X_x\geq 1\}}$$It is known that \frac{X_x -x^2}{x}\longrightarrow N(0,1) (... 19 The paper entitled Euler constant as a renormalized value of Riemann zeta function at its pole by Andrei Vieru contains a derivation of the first formula in the OP (Benoȋt Cloitre's formula), and a method to obtain variations thereof, such as as well as similar iterates in terms of the Dirichlet function \lambda(s)=\sum_{n=0}^\infty(2n+1)^{-s}, 18 In the main theorem of this recent paper of Pintz, Zhang's method is used to show that (for k large enough), there are \gg_{\mathcal H} \frac{x}{\log^k x} values of n \le x such that two of the n+h_i are prime and the remaining n+h_j are almost prime (have O_k(1) prime factors, all of which are \gg x^{c_k} for some c_k>0 independent of x)... 15 Since you mentioned Cayley's theorem for spanning trees, I believe one important example is its higher dimensional analogue due to Kalai. Indeed the number of simplicial spanning trees of the k-skeleton of an n-simplex is$$f(n)=n^{\binom{n-2}{k}} \implies \lim_{n\to \infty}\frac{\log\log f(n)}{\log n}=k.$$Some further work has been done in this direction, ... 15 For Q1 the answer is known to be \sim C_1C_2^n n^{-5/2}  for C_1\approx 0.5349496061... and C_2\approx 2.9955765856.... This can be found in Flajolet and Sedgewick's "Analytic Combinatorics" (see p.481) with the main ingredients being singularity analysis and the relation$$I(z)=H(z)-\frac{1}{2}\left(H(z)^2-H(z^2)\right)$$where I is the ... 14 No, even in the most favorable case (X_i)_{i\geqslant 0} iid with \mathbb P(X_i=1)=\mathbb P(X_i=-1)=1/2. Denoting F_n the cumulative distribution function of n^{-1/2}S_n, we have by symmetry$$F_{2n}(0)=\frac 12(1+\mathbb P(S_{2n}=0)).$$Since \mathbb P(S_{2n}=0)=\binom{2n}n2^{-2n}, denoting \Phi the cdf of the standard normal distribution,$$\...

14

See Erick Wong's response here. In particular, Kevin Ford proved (in more precise form) that $$f(n) = \frac{n}{\log n} \exp\left(O(\log \log \log n)^2\right),$$ whence $f(n)/n$ tends to zero. The same consequence also follows from an earlier result of Pillai (1929), available online here.

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The first formula is trivial. $$f(s)= \frac1{s-1}+\gamma +O(s-1)$$ $$g(z)=1+2^{-z}+3^{-z}+4^{-z}+O(5^{-z})=1+2^{-z}(1+(3/2)^{-z}+(4/2)^{-z}+O(5/2)^{-z})$$ $$f(g(z)) = \frac1{2^{-z}(1+(3/2)^{-z}+(4/2)^{-z}+O(5/2)^{-z})}+\gamma+O(2^{-z})$$ $$=2^z \left(1-(3/2)^{-z}-(4/2)^{-z}+O(5/2)^{-z}+(O(3/2)^{-z})^2\right)+\gamma + O(2^{-z})$$ $$=2^z- (3/4)^{-z}-1+O(9/8)... 13 Let us just consider the case \alpha=2 where there is an elegant answer: There exists a constant \lambda such that for all n we have u_n = \lceil \lambda^{2^n}\rceil. First note that by induction it is easy to see that u_n \ge (n+2) for all n\ge 0. Define$$ \lambda= 2 \prod_{n=1}^{\infty} \Big(1-\frac{n}{u_{n-1}^2}\Big)^{\frac{1}{2^n}}. $$... 13 This is a binary additive divisor problem (also known as the shifted convolution problem) and related questions are to understand asymptotics for \sum_{n\le x} d(n) d(n+k), or \sum_{n\le x} \lambda_f(n) \lambda_f(n+k) where \lambda_f denotes the Fourier coefficients of a cusp form. For your particular problem, Ingham first showed the asymptotic ... 13 We can attack this problem by using Fourier transforms (i.e. characteristic functions). I'll consider the example in the problem where X is a random variable taking the value 0 with probability 1/3, and 4^{k} and -4^{k} with probability 1/4^{k+1} (for k=0, 1, \ldots). I'll show that the probability that |X_n|/n >M is about a ... 12 I'll address Joel's edited question of getting good asymptotics on RH. The argument below is essentially due to Selberg, but this is not quite what he does and I haven't seen it presented this way in the literature. The natural problem is to consider the coefficients of (\log \zeta (s))^k/k! rather than numbers with exactly k prime factors. Note that ... 12 Gradshteyn & Ryzhik equation 3.773.1 gives (for q>0)$$\frac{1}{q}B(q)=\int_0^\infty \frac{\sin x}{x(q^2/4+x^2)^{1/2}}\,dx=\frac{1}{q}G^{21}_{13}\left(\frac{q^2}{16}\biggl|^1_{1/2,1/2,0}\right)\qquad\qquad=1-\gamma+\log 4-\log q+{\cal O}(q^2)=1.80908-\log q+{\cal O}(q^2).$$The asymptotic result desired by the OP is B(q)/q=\pi\gamma-\log q=1.... 12 The asymptotic formula is true for even dimensions k\geq 2. We can prove this by induction on k, inspired by Rodrigo's observation on Eisenstein series. The case k=2 is classical and addressed in the OP's previous post that he linked. Now it suffices to show that if the formula is true for even dimensions k,\ell\geq 2, then it is also true for ... 12 The asymptotic you want does not hold just because the "last-term fluctuation"$$ g(n)-g(n-1) = 2\sum_{a=1}^n \gcd(a,n)-n $$is too large. Indeed, denoting the sum in the right-hand side by \sigma(n), we have$$ \sigma(n) =\sum_{d\mid n} d\varphi(n/d) = n \sum_{d\mid n} \prod_{p\mid(n/d)} \Big(1-\frac1p\Big) = n \sum_{d\mid n} \prod_{p\mid d} \Big(1-\...

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Write $V(a)$ for the determinant $\prod_{0\leq i<j\leq n-1} |a_i-a_j|$. Selberg's formula tells you that $$\int_0^1 \cdots \int_0^1 V(a)^{2\beta} \prod_{i=0}^{n-1} da_i= n! \prod_{j=0}^{n-1} \frac{(\Gamma(1+j\beta))^2 \cdot \Gamma((j+1)\beta)} {\Gamma(2+(n+j-1)\beta)\cdot \Gamma(\beta)}=:A(n,\beta)$$ Thus the asymptotics you seek are given by $\lim_{\... 11 It is not hard prove the bounds you want by purely real variable techniques. First note that the$a_n$are non-negative for all$n$. For a general non-negative sequence$a_n$, and real numbers$N>0$, put $$F(N) = \sum_{n=0}^{\infty} a_n e^{-n/N},$$ and assume that there are constants$\alpha >1$, and positive constants$c_1$and$c_2$such that ... 11 We may just write down$\varphi(k)=k\cdot \prod_{p|k} (1-1/p)$,$p$runs over the primes which divide$k$, then $$\frac{\left(\prod_{k\leqslant n}\varphi(k)\right)^{1/n}}{n}=\frac{\sqrt[n]{n!}}n\prod_{p} (1-1/p)^{\frac{[n/p]}{n}}.$$ First multiple tends to$1/e$by weak version of Stirling's formula, and in the product we may replace each exponent$\frac{[...

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I think your final goal follows by taking the logarithmic derivative of the functional equation: $$\frac{\zeta'}{\zeta}(s)+\frac{\zeta'}{\zeta}(1-s)=\log\pi-\frac{1}{2}\frac{\Gamma'}{\Gamma}\left(\frac{s}{2}\right)-\frac{1}{2}\frac{\Gamma'}{\Gamma}\left(\frac{1-s}{2}\right).$$ Applying this with $s=it$ and using the familiar asymptotic expansion of $\Gamma'/\... 11 The answer is no, even in the smooth case. Take for example: $$f(x) = \frac{2}{x} + \frac{\cos(\log(x))}{x}$$ Alter it on a small neighborhood of$0$in such a way that there is no singularity there, preserving smoothness (this will be irrelevant for the asymptotics). This function is decreasing and, for$t$sufficiently large, we have $$\int_0^t f(x) ... 11 \newcommand{\ep}{\varepsilon} Let X be any nonnegative random variable (r.v.) with finite mean \mu>0 and variance \sigma^2<\infty. For any real u>0, we have \ln\frac xu\le\frac xu-1 for all real x>0, whence x\ln x\le\frac{x^2}u+x\ln\frac ue and \begin{equation*} EX\ln X\le\frac{EX^2}u+EX\ln\frac ue=\frac{\sigma^2+\mu^2}u+\mu\... 10 Uniformity on compact sets interior to the domain The boundary condition has no real impact on the small-time asymptotics of the heat kernel for interior points. More precisely, consider a Riemannian manifold \mathbb{M} and \Omega a relatively compact smooth open domain in \mathbb{M}. Denote by p_\Omega the heat kernel associated with a self-... 10 De Bruijn's "Asymptotic methods in analysis" is an excellent book for beginners. You'll need to work through it diligently to learn everything but no advanced a priori knowledge is required. Also, you can easily download it from many online places that do not worry too much about copyright and, even if you decide to stay law-abiding, it goes for under ten ... 10 I don't think a nice asymptotic formula like the one you've mentioned from Tutte's work is available for higher g to the best of my understanding; it is entirely possible that someone who regularly deals with such things will come along and show us otherwise. Meanwhile, the state-of-the-art on counting triangulations of genus g surfaces is the lovely ... 10 Assuming that |x+y|<1 and 4|xy| \le 1, here's a proof of the decay. First suppose that |x|> |y|. The desired sum is$$ \le \binom{2n}{n} |xy|^n \sum_{j=0}^{n} |y/x|^j \le \binom{2n}{n} |xy|^n \frac{1}{1-|y/x|}.$$Since$\binom{2n}{n}$is of size about$4^n/\sqrt{n}$, the desired decay follows in this case. Next suppose$|x| <|y|$. ... 10 Let$A(x)$denote the formal generating function of$\{ a_n\}$. The recurrence relation can be written as$A(x)=e^x A(x^2)$. Applying this repeatedly, we find$A(x) = e^x e^{x^2} \cdots = e^{f(x)}$, where$f(x) = \sum_{i \ge 0} x^{2^i}$. Asymptotics of$[x^n] e^{P(x)}$where$P$is a polynomial with positive coefficients are "easy", for example they follow ... 10 OK, let's see if I can put my money where my commenting mouth is. Let me say at the outset that I have no idea where such an integral comes from, but I claim that it doesn't matter to answer the question. First of all,$\mathbf e$is nothing mysterious: it is the 2-form dual to a hyperplane section of$M$. The equation$\int_M \mathbf e^3=5\$ is just a fancy ...

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