All Questions

Let $wt(n)$ be A000120, number of $1$'s in binary expansion of $n$ (or the binary weight of $n$) and $$n=2^{t_1}(1+2^{t_2+1}(1+\dots(1+2^{t_{wt(n)}+1}))\dots)$$ Then we have an integer sequence given ...

Stirling numbers of the second kind can be expressed by means of a simple hypergeometric (considering $n$ fixed) sum $$S_2(n,k) = \frac{1}{k!}\sum_{j=0}^{k}(-1)^{k-j}{k \choose j} j^n. \qquad (1)$$ ...

After noticing that the determinant of an $n \times n$ matrix $A_n$ with elements $a_{i,j}=i^j$, $1 \le i \le n$, $1 \le j \le n$, is the superfactorial (product of the first $n$ factorials), I wanted ...

Past days I've been trying to prove that certain polynomials have positive coefficients. After a lot of thinking, I came up with a formula for each coefficient individually, and they are not that ugly....

My previous question was solved in a very elegant way, hopefully this (seemingly more complicated) case is also easy for experts. I need the inequality $\Big(\prod^r_{i=1}p_i\Big)\sum^n_{j=0}(-1)^j\...

Let $u(k,j) = 1$ if $j=0$, $0$ if $j > k$, or else it is $j*u(k-1,j-1) +(j+1)*u(k-1,j) $. Prove that $ \sum_{i=0}^{2k} {n \choose i+1} u(2k,i) +\sum_{i=0}^{2k} {-n-1 \choose i+1} u(2k,i)=0. $ ...