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Suppose I have the following generating functions:

$$\frac{x^ke^{\left(z-\frac{1}{N}\right)x}}{N^{k-1}k!\sum_{j=0}^{N-1}w_N^{-jk}e^{\frac{w_N^jx}{N}}}=\sum_{j=0}^\infty H_{N,k,j}(z)\frac{x^j}{j!}$$ where $0\le k\lt N$ and $w_N=e^{2i\pi/N}$

Is there a way to find a particular combination of arguments such that

$$H_{N,0,n-i}(z)=f(n)\sum_{j=1}^{N-1}H_{N,j,n}(a_{i,j}z+b_{i,j})$$ or for a particular fixed $k$ independent of the left hand side, $$H_{N,0,n-i}(z)=f(n)\sum_{j=1}^{N-1}H_{N,k,n}(a_{i,j}z+b_{i,j})$$

where $n,i\in \mathbb{N}$ and $a_{i,j}, b_{i,j} \in\mathbb{C}$.

This happens in terms of Euler polynomials as a sum of Bernoulli polynomials of different arguments. These generating functions above can be shown to be Euler polynomials and Bernoulli polynomials (Bernoulli when $N=2, k=1$ and Euler when $N=2, k=0,)$. This case being that

$$E_{n-1}(z)=\frac{2^n}{n}\left[B_n\left(\frac{z+1}{2}\right)-B_n\left(\frac{z}{2}\right)\right]$$

This is analogous to the second case where $k$ is fixed inside the summation. One of the difficulties that I've found in this is finding ways of transforming the denominators of the generating function to the appropriate combination.

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  • $\begingroup$ $f(n)$ would be a factor in $n$ dependent on the sum. Note the Euler / Bernoulli Polynomial equation in my post, the $f(n)=2^n/n$. I'm about to fix the sums as well too. $\endgroup$ Commented Nov 21, 2019 at 18:23
  • $\begingroup$ I think I fixed it. I haven't looked at this problem in a while, so I had to figure out what it was. The coefficients $a,b$ in the argument of $H$ are dependent on both $i$ and $j$. The main difference in the two expressions is that the top is the sum of the individual $H$ over $j$ where the bottom is not, only over some (any) fixed $k$ $\endgroup$ Commented Nov 21, 2019 at 18:46

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