Euler characteristic of the structure sheaf I started to study vector bundles on the spaces, then I have my first contact with the instanton bundles, (bundles that are cohomology of the $0 \to  \mathcal{O}^{k}_{\mathbb{P}^n}(-1) \to \mathcal{O}^{2(k+1)}_{\mathbb{P}^n} \to \mathcal{O}^{k}_{\mathbb{P}^n}(1)\to 0)$. So to find the Hilbert polynomial of the instanton bundle we just need to know the Hilbert polynomial of the structural sheaf. But I'm having problem with it, because I known that the Hilbert polynomial is $P_{\mathcal{O}_{P^n}}(t) = \chi(\mathcal{O}_{P^n}) = \sum_i(-1)^ih^i(P^n,\mathcal{O}_{P^n}(t))$.
My question is if there exists any closed formula to this, because despite we know the cohomology of $P^n$, we don`t have control on the variable t.
Thank you so much in any advice!
 A: The Hilbert polynomial of $\mathcal{O}_{\mathbb{P}^n}$ is defined to be the polynomial $p(t)\in\mathbb{Q}[t]$ such that for all $m\in\mathbb{Z}$, $p(m)=\sum_{i}(-1)^ih^i(\mathbb{P}^n,\mathcal{O}_{\mathbb{P}^n}(m))$. This may be where your confusion comes from, since you only have to think of $\mathcal{O}_{\mathbb{P}^n}(m)$ with $m$ as an integer and not as a variable (which could possibly take on any value). Using Serre vanishing, this is the same as considering $h^0(\mathbb{P}^n,\mathcal{O}_{\mathbb{P}^n}(m))$ for sufficiently large $m$. Therefore, since the global sections of $\mathcal{O}_{\mathbb{P}^n}(m)$  consist precisely of homogeneous polynomials of degree $m$ in $n+1$ variables, we obtain that for $m$ sufficiently big, 
$$p(m)=\sum_{i}(-1)^ih^i(\mathbb{P}^n,\mathcal{O}_{\mathbb{P}^n} (m))=h^0(\mathbb{P}^n,\mathcal{O}_{\mathbb{P}^n}(m))=\binom{m+n}{n}.$$
Note that since the right hand side is a polynomial (if we replace $m$ by a variable $t$), then the Hilbert polynomial is $p(t)=\binom{t+n}{n}=\frac{1}{n!}(t+n)(t+n-1)\cdots(t+1)$.
