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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

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

where $n,i\in \mathbb{N}$ and $a_i, b_i\in\mathbb{C}$.

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.

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.

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

$$H_{N,0,n-i}(z)=f(n)\sum_{j=1}^{N-1}H_{N,j,n}(a_jz+b_j)$$ or $$H_{N,0,n-i}(z)=f(n)\sum_{j=1}^{N-1}H_{N,k,n}(a_jz+b_j)$$

where $n,i\in \mathbb{N}$ and $a_i, b_i\in\mathbb{C}$.

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.

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

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

where $n,i\in \mathbb{N}$ and $a_i, b_i\in\mathbb{C}$.

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.

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

$$H_{N,0,n-i}(z)=f(n)\sum_{j=1}^{N-1}H_{N,j,n}(a_iz+b_i)$$ or $$H_{N,0,n-i}(z)=f(n)\sum_{j=1}^{N-1}H_{N,k,n}(a_iz+b_i)$$

where $n,i\in \mathbb{N}$ and $a_i, b_i\in\mathbb{C}$.

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.

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

$$H_{N,0,n-i}(z)=f(n)\sum_{j=1}^{N-1}H_{N,j,n}(a_jz+b_j)$$ or $$H_{N,0,n-i}(z)=f(n)\sum_{j=1}^{N-1}H_{N,k,n}(a_jz+b_j)$$

where $n,i\in \mathbb{N}$ and $a_i, b_i\in\mathbb{C}$.

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.