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An interesting polynomial congruence is given by $$A_n(t^m)\equiv \left(\frac{1+t+\cdots+t^{m-1}}m\right)^{n+1}A_n(t) \qquad \mod (t-1)^{n+1}, \tag1$$ where $A_n(t)$ are the Eulerian polynomials with generating function $$\sum_{n=0}^{\infty}\frac{A_n(t)}{(1-t)^{n+1}}\,\frac{x^n}{n!}=\frac1{1-t\,e^x}.$$ This result have been proved and reproved (for example, see the more recent preprint by Ira Gessel).

QUESTION. Let $f_n(t)=(t+1)^n$. Is there a similar (and non-trivial) congruence between $f_n(t^m)$ and $f_n(t)$, as in (1)?

Remark. My motivation is that both set of polynomials are symmetric and bimodal, so the question might be feasible.

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  • $\begingroup$ You want a congruence between $f_n ( t^m)$ and a multiple of $f_n(t)$? $\endgroup$
    – Will Sawin
    Commented Jan 26, 2021 at 22:22
  • $\begingroup$ @WillSawin: thanks, I fixed accordingly. $\endgroup$
    – T. Amdeberhan
    Commented Jan 26, 2021 at 23:04
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    $\begingroup$ If $m$ is odd, then there is a trivial identity. $f_n(t^m)= (t^{m-1} - t^{m-1} + \dots + t^2 - t + 1)^n f_n(t)$. I don't see how you could have a meaningful non-trivial congruence while that identity exists. But maybe if $m$ is even you could do something cool. $\endgroup$
    – Will Sawin
    Commented Jan 26, 2021 at 23:23

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