All Questions

Recently I had a curious discovery. Namely, I have made the following conjectures. Conjecture 1. We have the identity $$\sum_{k=0}^\infty\frac{(10k-1)\binom{3k}k\binom{6k}{3k}}{(2k+1)512^k}=0.\label{1}...

Recently, I found a (conjectural) new series for $\sqrt3\pi$: $$\sum_{k=1}^\infty\frac{(8k-3)\binom{4k}{2k}}{k(4k-1)9^k\binom{2k}k^2}=\frac{\sqrt3\pi}{18}.\label{1}\tag{1}$$ The series converges fast ...

Is there a triple of nonzero even integers $(a,b,c)$ that satisfies the following infinite system of congruences? $$ a+b+c\equiv 0 \pmod{4} \\ a+3b+3c\equiv 0 \pmod{8} \\ 3a+5b+9c\equiv 0 \pmod{16} \\ ...

Let $p>3$ be a prime, and let $(\frac{\cdot}p)$ be the Legendre symbol. Then $$14\left(\frac 3p\right)-\left(\frac p3\right)-12=\begin{cases}1&\text{if}\ p\equiv1\pmod{12}, \\-25&\text{if}\ ...

On Jan. 18, 2012 I conjectured that for any prime $p>3$ we have $$R_p^2\equiv\frac1{10}\left(512\left(\frac{10}p\right)-27\left(\frac{-15}p\right)-475\right)\pmod p,$$ where $(\frac{\cdot}p)$ ...

For each $n=0,1,2,\ldots$, the harmonic number $H_n$ is given by $$H_n:=\sum_{0<k\le n}\frac1k.$$ In 2016 I conjectured that $$\sum_{k=1}^\infty\frac{3H_{k-1}^2+4H_{k-1}/k}{k^2\binom{2k}k}=\frac{...