如何计算第二类斯特林数三角形第n行中不被素数x整除的元素个数？

Great question! Let's dive into this—since you’re familiar with counting non-divisible elements in Pascal’s triangle (using tools like Lucas’s theorem), it’s helpful to start by contrasting that with the trickier behavior of Stirling numbers of the second kind (denoted ( S(n,k) )).

In Pascal’s triangle, Lucas’s theorem gives a clean rule: ( \binom{n}{k} \equiv 0 \mod p ) (for prime ( p )) if any digit in ( k )’s ( p )-base expansion is larger than the corresponding digit in ( n ). This makes counting non-divisible elements trivial by multiplying valid digit counts across each ( p )-base position.

Stirling numbers of the second kind don’t have this luxury, because their recurrence includes a multiplicative factor that breaks the direct analogy:
S(n,k) = S(n-1,k-1) + k·S(n-1,k)

That said, here are practical approaches and known results to tackle your problem:

1. Direct Computation for Small n #

If ( n ) is small, you can compute the entire ( n )-th row using the recurrence relation, then iterate through each element to count how many aren’t divisible by your prime ( x ). This is straightforward but inefficient for large ( n ).

Example for ( x=2 ), ( n=3 ):

• Row 3: [0, 1, 3, 1]
• Non-divisible elements: 1, 3, 1 → count = 3

2. Recursive p-base Expansion Method #

For primes ( x=p ), we can use recursive properties tied to the ( p )-base expansions of ( n ) and ( k ). First, write ( n = n_0 + n_1p + ... + n_mp^m ) and ( k = k_0 + k_1p + ... + k_mp^m ) (pad with leading zeros to match lengths).

A key result from combinatorial number theory (built on work by Carlitz) lets us compute ( S(n,k) \mod p ) recursively:
Split ( n = a·p + b ) and ( k = c·p + d ) where ( 0 ≤ b,d < p ). Then:
S(a·p + b, c·p + d) ≡ C(a, c) · S(b, d) · d!^{a - c} mod p
(This holds when ( d ≤ b ); for other cases, additional adjustments are needed. Binomial coefficients ( C(a,c) ) can be computed via Lucas’s theorem, and factorials modulo ( p ) are precomputable.)

Using this recursion, you can build up the count of non-divisible elements by validating combinations of ( p )-base digit pairs from ( n ) and ( k ).

3. Special Cases and Observed Patterns #

For small primes like ( x=2 ), there are observed patterns in non-divisible counts, though no simple universal formula exists:

• If ( n )’s binary representation has ( m ) 1s, the count relates to combinations of these 1s, but it’s not linear. For example:
  • ( n=3 ) (binary 11, ( m=2 )) → count = 3
  • ( n=4 ) (binary 100, ( m=1 )) → count = 3
  • ( n=5 ) (binary 101, ( m=2 )) → count = 4

Similar pattern analysis works for primes like ( x=3 ), but results are case-specific.

4. Generating Function Approach #

The generating function for Stirling numbers of the second kind is:
sum_{k=0}^n S(n,k) (t)_k = t^n
where ( (t)_k ) is the falling factorial: ( t(t-1)...(t-k+1) ).

Modulo ( p ), use Lucas’s theorem to expand ( t^n ) into its ( p )-base components, then equate coefficients with the sum of ( S(n,k)(t)_k \mod p ). This lets you deduce when ( S(n,k) \not\equiv 0 \mod p ) by checking if the coefficient of ( (t)_k ) in the expanded ( t^n ) is non-zero.

内容的提问来源于stack exchange，提问作者user1608