Euler numbers, $E_{n}$, $n∈ℕ,$ are integers defined by the Maclaurin series $\frac{1}{\cosh x}=∑_{n=0}^{∞}\frac{E_{n}}{n!}x^n.$
Euler polynomials, $E_{n}(x), n∈ℕ,$ are defined by the generating function identity $\frac{2e^{xt}}{e^{t}+1}=∑_{n=0}^{∞}\frac{E_{n}(x)}{n!}x^n$ so that we have $E_{n}=2^nE_{n}(\frac{1}{2})$.
There are some very complicated explicit formulas for $E_{n}$ in https://en.wikipedia.org/wiki/Euler_numbers.
There is also a a kind of duality between Bernoulli numbers and Euler numbers.
Bernoulli polynomials and numbers can be defined respectively by the umbral calculus identities $B_{n}(x)=(B+x)^n↓$ for all $n∈ℕ$ and $B_0=1$, $(B+1)^n=B^n↓, n≥2$.
Long time ago, after complications and mistakes, I derived the identity $$E_{n}=∑_{k=0}^n\binom{n}{k}(2^{k+1}-2^{2k+2})\frac{B_{k+1}}{k+1}$$ for all $n∈ℕ$. I think this identity I found is correct. I am also wondering the formula for $B_{n}$ in terms of Euler numbers. Are there some faster ways to derive these identities?
My second question is the following: Are the umbral calculus identities $E_{n}(x)=(E+(2x-1))^n↓$ for all n∈ℕ and $(E-1)^n↓=(2^{n+1}-2^{2n+2})\frac{B_{n+1}}{n+1}$ for $n≥1$, correct?
My third question is about the connections between Euler numbers and L-functions, specifically Rieaman Zeta Function. Can Euler numbers be connected with some special values of such functions? I have not much experience in analytic number theory.
I would have defined Euler numbers using the coefficients of the power series of $1/\cos x$, not $1/\cosh x$: $$ \frac{1}{\cos x} = \sum_{n \geq 0} \frac{\mathcal E_n}{n!}x^n, $$ where I write $\mathcal E_n$ to avoid clashing with your notation $E_n$. We have $\mathcal E_n= 0$ for odd $n \geq 0$ since $\cos x$ is an even function, just as $E_n = 0$ for odd $n \geq 0$.
Since $\cos x = (e^{ix} + e^{-ix})/2$ and $\cosh x = (e^x + e^{-x})/2$, $\cosh(ix) = \cos x$, so $i^nE_n = \mathcal E_n$ for all $n$. This is boring for odd $n$, since both sides are $0$. For even $n \geq 0$, $(-1)^{n/2}E_n = \mathcal E_n$.
You ask about a connection between Euler numbers and $L$-functions. The even-indexed Bernoulli numbers are related to the Riemann zeta-function at positive even numbers $2k$: $$ \zeta(2k) = \frac{(-1)^{k-1}B_{2k}}{2(2k)!}(2\pi)^{2k}. $$ Since $\zeta(2k) > 0$, $(-1)^{k-1}B_{2k} > 0$, so $$ \zeta(2k) = \frac{|B_{2k}|}{2(2k)!}(2\pi)^{2k}. $$ In a similar way even-indexed Euler numbers are related to the $L$-function of the nontrivial character mod $4$ at positive odd numbers: for $k \geq 0$, $$ L(2k+1,\chi_4) = \frac{\mathcal E_{2k}}{2(2k)!}\left(\frac{\pi}{2}\right)^{2k+1} = \frac{(-1)^kE_{2k}}{2(2k)!}\left(\frac{\pi}{2}\right)^{2k+1}. $$ For example, taking $k = 0$ we get $L(1,\chi_4) = (E_0/2)(\pi/2) = \pi/4$, which is the famous Leibniz series evaluation. Since $L(2k+1,\chi_4) > 0$, $(-1)^kE_{2k} > 0$, so $$ L(2k+1,\chi_4) = \frac{|E_{2k}|}{2(2k)!}\left(\frac{\pi}{2}\right)^{2k+1}. $$ That's the analogue of the formula for $\zeta(2k)$ in terms of $|B_{2k}|$.
There are also formulas for the analytic continuations of $\zeta(s)$ and $L(s,\chi_4)$ at negative integers: for $n \geq 2$, $$ \zeta(1-n) = -\frac{B_n}{n} $$ (this is not correct at $n = 1$: $\zeta(0) = -1/2$ while $-B_1/1 = -(-1/2) = 1/2$) and for $n \geq 1$, $$ L(1-n,\chi_4) = (-1)^{(n-1)/2}\frac{\mathcal E_{n-1}}{2} = \frac{E_{n-1}}{2}. $$ (In this last formula, all terms are $0$ when $n$ is even, just like $\zeta(1-n)$ and $B_n$ are $0$ when $n$ is odd.)