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I think the number in question is $10^{10^{10}}+10^{11}\ln^4(10)$ plus a tiny positive number. That is, it starts with a digit $1$, followed by $10^{10}-13$ zeros, then by the string $2811012357389$, …

Golomb (1970) proved that for all $x\geq 1$ we have $$\frac{\zeta(3/2)}{\zeta(3)}x^{1/2}-3x^{1/3}\ \leq\ \left|\mathrm{POW}(x)\right|\ \leq\ \frac{\zeta(3/2)}{\zeta(3)}x^{1/2}.$$ Note that $$\frac{\ze …

It is like cooking. A recipe can be easy to follow, but figuring it out from the resulting meal can be hard (cf. secret recipe).

The answer is yes, and this follows from known results concerning the Waring-Goldbach problem.

This is a good question around an important issue. The first thing that came to my mind is Popa's article "Central values of Rankin $L$-series over real quadratic fields", which is about a related pro …

Yes, the graph $\mathcal{G}_n$ is connected, and its diameter is at most $n2^n$. To see this, write $s$ for $n2^{n-1}$, and fix any two vertices $a,b\in\mathbb{N}^*$. By Wright's solution of Waring's …

Using $$\frac{10^c-1}{9}=\sum_{m=0}^{c-1} 10^m,$$ the product in question equals $$\sum_{n=0}^{2a+2b+c-1}r(n)10^n,$$ where $r(n)$ is the number of times $n$ occurs among the numbers (counted with mult …

Let $G$ be the ideal class group, and let $H$ be the subgroup generated by the prime ideals of norm at most $3\log^2(d^2)=12\log^2(d)$. Assume that $H$ is a proper subgroup of $G$. Then there is a non …