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Define the congruence "modulo m" on exponential Taylor series following the previous post (A conjecture about the degrees of special polynomials)

It has been conjectured, that if we define the sequence of polynomials $Q_{n,k}(x)$: $$ Q_{n,k}(x):= \left(x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+...+\frac{x^{p_n-1}}{p_n-1}\right)^k \mod p_n! $$ then $\forall n\in \mathbb{N}, 2<k<p_{n-1}:$ $$ \boxed{\deg Q_{n,k}(x)=k\cdot p_{n-1}} $$ (where $p_n$ is the n-th prime number and so $p_{n-1}$ is the previous one)

It is also conjectured to be true for $k=2$ and all $n\neq 1,2,5~(p_n\neq 2,3,11)$.

I still have no idea, why the degrees of polynomials of this kind have such "number theoretical" properties and would appreciate, if you could give a proof of this fact, maybe share philosophical speculations about the nature of this structure or post some references. Thank you.

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  • $\begingroup$ I think you meant to say: $\deg Q_{n,k}(x)=k\cdot(p_n-1)$ instead of $\deg Q_{n,k}(x)=k\cdot p_{n-1}$. Right? $\endgroup$ May 23, 2017 at 20:06
  • $\begingroup$ @T.Amdeberhan No, the calculations show that the degree is $k\cdot p_{n-1}$. It is $p_{n}-1$ only for $k=1$. $\endgroup$ May 24, 2017 at 9:49

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