My own research led to the following result...
Knowing that there are $A$ days in the year (typically $A=365$), the probability $P(A, M, n)$ that at least $n$ children have their birthday the same day within a class of $M$ children is:
$\boxed{ P(A, M,n) = 1 - \dfrac{ K_n(A, M) }{ A^M } }$
where $K_n(A, M)$ represents the number of configurations in which one cannot find $n$ children (or more) having their birthday the same day, and can be computed by recurrence as follows:
$\forall n\ge 2, \quad \boxed{ K_{n+1}(A, M) = \sum_{0\le k\le \left\lfloor{\frac{M}{n}}\right\rfloor} \dfrac{ \binom{A}{k} \; (M)_{nk} \; K_n(A-k, M-nk)}{ (n!)^k} }$
with the following initialization: $\boxed{ K_2(A,M)=(A)_M }$
and where $(n)_k$ stands for the decreasing factorial : $(n)_k = n(n-1)...(n-k+1)$
Numerically:
Within a class of $M=30$ children, knowing that the year counts $A=365$ days...
The probability that at least $n=2$ children have their birthday the same day is:
$P(365,30,2)\simeq 70,6\%$
The probability that at least $n=3$ children have their birthday the same day is:
$P(365,30,3)\simeq 2,85\%$
The probability that at least $n=4$ children have their birthday the same day is:
$P(365,30,4)\simeq 0,0532\%$
Nicolas