# Tag Info

Accepted

### A challenging (for me) limit calculation

This limit converges to $\frac{\sqrt3}2$. The idea is that $\sin(x) = x - \frac{x^3}6 + O(x^5)$, so we start with $\frac1{\sqrt n}$ and repeatedly subtract $\frac{x^3}6$. We can approximate this ...
• 1,555

### A new way of approaching the pole of the Riemann zeta function - and a new conjectured formula

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 ...
• 172k
Accepted

### Asymptotic expansion of $\sum\limits_{n=1}^{\infty} \frac{x^{2n+1}}{n!{\sqrt{n}} }$

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 ...
• 3,120
Accepted

### How does the number of trees on $n$ vertices *up to isomorphism* grow as $n \to \infty$?

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&...
• 84.5k
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• 3,315
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### Density of the set of numbers whose sum of digits is prime

Yes, $A(n)$ has zero natural density. It suffices to prove this for $n$ which is a power of $10$. and it is possible to make this more precise. To see this, first let $n=10^k$ and note that for $X$ ...
• 13.8k
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### Mertens-like theorem

This lies beyond Mertens, in the sense that this variant actually implies the Prime Number Theorem, as will be explained below, while Mertens' theorem is weaker than the PNT. I sketch below a complex ...
• 13.4k
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### Distinct exponents in the factorization of the factorial, a problem of Erdős

The primes $p \leq \sqrt{n}$ can be ignored as the number of them is $$\approx \sqrt{n} / \log (\sqrt{n} )= 2 \sqrt{n}/\log n = o (1) \cdot \sqrt { n/\log n}$$ For primes $p> \sqrt{n}$, the ...
• 131k
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• 172k
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• 98.7k
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• 95k
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### The number of representations of an integer as the inner product of integral lattice points

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 ...
• 95k
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• 98.7k
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### Asymptotics of $\int_0^\infty \frac{x^{2z}}{\Gamma(1+z)}\,dz$ for large $x$

$\newcommand\Ga\Gamma$Let $g(x)$ denote your integral. Then $$g(x)\sim e^{x^2} \tag{1}\label{1}$$ as $x\to\infty$. Proof: $$g(x)=\int_0^\infty dz\,\frac{x^{2z}}{\Ga(1+z)}. \tag{1.5}\label{1.5}$$ So, ...
• 110k

### Asymptotic behavior of a certain oscillatory integral

$$I(x):=\int_{0}^{\infty}\frac{e^{i r}}{r^{\frac{1}{2}}}\int_{0}^{\infty}\frac{e^{-s}}{s^{\frac{1}{2}}}\frac{r}{sx+\sqrt{sxr}+r}ds dr.$$ To aid the asymptotic analysis, I regularize $I(x)$ by ...
• 172k
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### Asymptotics for pairs of positive integers whose harmonic (resp. geometric) mean is an integer

[CORRECTED 8/28] I think the asymptotic ratio is $(4/3)*\ln(2)\approx 0.924196$. $$G(n) = (6/\pi^2)*n*\ln(n) + O(n)$$ $$H(n) = (8\ln(2)/\pi^2)*n*\ln(n) + O(n)$$ The proof for the geometric mean goes ...
Accepted

### For a round-robin tournament, what is the favorite's least favorite size?

I can show that $N(\epsilon)$ is equal to $\epsilon^{-2}$ up to a log factor on each side. The strategy I'll use is to give an upper bound for $\pi(1/2+\epsilon,n)$. Optimizing it, we obtain an upper ...
• 131k
Accepted

### Differentiating an integral that grows like log asymptotically

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 ...
• 3,143