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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 $2811012357389$, …

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

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 …

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 …

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).

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 …

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 …