# Tag Info

• 42.7k

### A conjectural trigonometric identity

Denote $\omega=e^{2\pi i/n}$, then $\Omega=\{\omega^k:k=0,\ldots,n-1\}$ is the set of roots of the polynomial $f(t):=t^n-1$, $-\Omega=\{-\omega^k:k=0,\ldots,n-1\}$ the set of roots of $g(t):=t^n+1$. ...
• 92k

• 7,262
Accepted

### A "quantum" identity: in search of a proof -Part II

Both sides are equal to $\binom{x+y+1}{n}_q$. This enumerates lattice paths in an $n\times (x+y-n+1)$ rectangle, according to the area statistic. We will assume that these paths start at $(0,0)$ and ...
Accepted

### Identity with binomial coefficients and k^k

Let's denote $T(x)=\sum_{n\geq 1}\frac{n^{n-1}x^n}{n!}$ the exponential generating function of labeled rooted trees, and $U(x)=\sum_{n\geq 1}\frac{n^{n-2}x^n}{n!}$ the corresponding function for ...
Accepted

### An interesting identity: in search of a proof -Part I

This is known as Jensen's identity and dates back to 1902. See here an overview of this identity and related ones, and a proof: https://arxiv.org/abs/1005.2745, a paper by Victor Guo.
Accepted

• 27.6k

### Products and sum of cubes in Fibonacci

This is just the following identity: $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca).$$ Since $$F_n+(-F_{n-1})+(-F_{n-2})=0,$$ your formula follows.
Accepted

### Products and sum of cubes in Fibonacci

$F_n$ is the number of compositions (ordered partitions) of $n-1$ into parts equal to 1 or 2. The number of triples $(a,b,c)$ of such compositions is $F_n^3$. The number such that $a,b,c$ all begin ...
• 46.1k
Accepted

### Reference for exponential Vandermonde determinant identity

This looks like a special case of a formula by Samson Shatashvili related to the HCIZ integral as mentioned in Ryan's answer. Compare, in particular the two ways of computing $\langle 1\rangle$ given ...

• 51.1k