can be recursively calculated as follows:

1.  $\Omega(0,f,t)=1$ and $\Omega(d,f,0)=1$, for any $d\geq 1$, $f$ and $t$;

2.  Write $n_t\triangleq \Omega(d,f,t)$ for $t=0,1,\cdots,f(1)$, then they can be successively calculated by $n_0=0$ and for $t\geq 1$, $$n_t=n_{t-1}+\Omega(d-1,f_{+n_{t-1}},f(n_{t-1}+1)-f(1)+1).$$ Here $f_{+m}$ is a function defined as $f_{+m}(n)=f(m+n)$.

*Proof:* It is equivalent to show that the recursive function $\Omega$ defined by the calculation procedure above is the same as the one defined by equation (\[equ:Omega\]); namely, if we compute a number $k=\Omega(d-1,f,t)$ by the calculation, then $\hat{H}_k(f(k))$ should be as in equation (\[equ:Omega\]). On the other hand, for any $k\geq 1$, decompose $\hat{H}_k(f(k))$ as $\hat{H}_k(f(k))=(a_1,a_2,\cdots,a_r)_{f(k)}$. It can be verified by equation (\[equ:H\_exp\]) that $\hat{H}_{k+1}(f(k+1))=(a_1,a_2,\cdots,a_r)_{f(k+1)}-1$. So $\hat{H}_k(f(k))$ can also be recursively calculated from $\hat{H}_1(f(1))$. Then it is easy to prove the result by induction on $d$ and $t$.$\Box$

For instance, let $f(n)=2\times 3^{n-1}$, then $\Omega(2,f,2)=4382$ by Theorem \[thm:Omega\], which is exactly the number of monomials in Example \[exm:4382\].

Reduction from the general case
-------------------------------

Now we remove the restriction (\[equ:special\]) and consider the length of a general $f$-generating sequence of monomials $m_1,\cdots,m_n$ in $\mathcal{M}[{\pmb{x}}]$. Our method is to reduce this general case to the homogenous case. Specifically, we introduce a new variable $x_{d+1}$ (for which 