# About $t$-analogue of the Euler polynomials.

A certain way to define the $t$-analogue of the Euler polynomials $C_n(x)$ is by $$C_n(x,t)=\sum_{\pi\in S_n}x^{\text{des}(\pi)+1}t^{\text{maj}(\pi)}$$ where $des(\pi)$ is the descents in $\pi$, and $maj(\pi)$ is the major index of $\pi$.

Based on this, how can one derive the general formula $$\sum_k (k)_t^nx^k=\frac{C_n(x,t)}{(1-x)(1-tx)\cdots(1-t^nx)}?$$ Thank you.

I thought it would be more approachable to rewrite as $$\sum_k (k)_t^nx^k(1-x)(1-tx)\cdots(1-t^nx)=C_n(x,t)$$ and compare coefficients. With two variables and an unwieldy product of terms, I couldn't find a nice form for the coefficients of the left hand side above.

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What approaches have you tried or thought about? – anon Feb 13 '12 at 8:47
@anon I tried rewriting the formula as $$\sum_k (k)_t^nx^k(1-x)(1-tx)\cdots(1-t^nx)=C_n(x,t).$$ My idea was to somehow find the coefficent of a general monomial $x^it^j$ for some $i$ and $j$, in hopes that the coefficients would match up. With two variables especailly, I had a tough time finding an explicitly form for them. – Clara Feb 13 '12 at 18:19