Consider $T=k[x\_1,\ldots,x\_n]$ ( $k$ alg. closed and of char $k=0$), and consider the ideal $$I=(x\_1,x^{a\_2}\_2,\ldots,x^{a\_n}\_n)$$ 
with $2\leq a\_2 \leq\ldots\leq a\_n$. I want to prove that $$\sum\_{i=0}^t HF(T/I,i)=\prod\_{i=2}^n a\_i.$$
I know that $I$ is a complete intersection ideal, so $T/I$ is Artinian and Gorenstein, so there exists $\tau$ s.t. $\dim\_k [T/I]\_{\tau}=1$ (and of course I can cut this sum after the $\tau$ index), but I can't see why I can pass from this summatory to a product of esponents. T tried some combinatorial proofs without much success. 


This question is taken by these two paper (resp. page 8 and 2, both at the bottom), which I link you:


<https://www.sciencedirect.com/science/article/pii/S0021869312003730>


<https://arxiv.org/abs/1110.0745>


Can anyone help me? Thanks in advance.



Relation between Hilbert function and complete intersection ideals
The sum of values of the Hilbert function is equal to the vector space dimension of the algebra:
$$
 \sum\_{i=0}^{\infty} HF(T/I,i) = \dim(T/I).
$$
In this case, the vector space $T/I$ has a basis consisting of monomials $x\_1^{b\_1} \dotsm x\_n^{b\_n}$ with $0 \leq b\_i \leq a\_i-1$ for each $i$ (where I am taking $a\_1=1$). So the dimension is $\prod\_{i=1}^n a\_i$. Or, this follows from general facts about complete intersections.


