Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I'm wondering whether the following polynomial of a single indeterminate has been studied: take the (partial) Bell polynomial $B_{n,k}$, which is a polynomial in indeterminates $x_1$, $x_2$, …, $x_{n-k+1}$, and replace each indeterminate $x_i$ by the falling factorial $(x)_i=x(x-1) \dots (x-i+1)$. Call this polynomial $N(n,k)$.

I conjecture the following matrix identity. Assemble the polynomials into an infinite lower-triangular matrix $N$. Let $s$ be the lower-triangular matrix of Stirling numbers of the first kind. Let D be the diagonal matrix with entries $x$, $x^2$, $x^3$, ….

Conjecture: $Ns=sD$

share|improve this question
2  
Perhaps the following remark will be useful. One can show by a generating function argument that $$ N(n,k) = \frac{1}{k!}\sum_{i=1}^k(-1)^{k-i} \binom ki\binom{ix}{n}. $$ –  Richard Stanley Feb 10 '12 at 0:31
add comment

1 Answer

As pointed out by Richard Stanley in the comments, from the generating function of the Bell polynomials one finds $$N(n,k) = \frac{1}{k!}\sum_{i=1}^k(-1)^{k-i}\binom ki (ix) _n.$$ Now use the fact that $$(ix) _n=\sum _{j=0}^n s(n,j)i^j x^j,$$ and that $$\frac{1}{k!}\sum _{i=0}^k (-1)^{k-i}\binom{k}{i} i^j=S(j,k),$$ to obtain $$N(n,k)=\sum _{j=k}^n s(n,j)x^jS(j,k).$$ In other words, $N=sDS$, where $s,S$ are the lower triangular matrices of Stirling numbers of first and second kind, respectively. To finish off, notice that $s^{-1}=S$ by Stirling number duality.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.