Decomposition of even symmetric polynomials and Euler numbers

Question

Let's denote the even part of a polynomial $p$ by $E[p]$, which means only taking into account the monomials in $p$ which are even in all the arguments. Now let's consider the even part of the symmetric polynomial $(x_1+x_2+\cdots+x_n)^{2k}$. For example, $E[(x_1+x_2+x_3)^{4}] = x_1^4+x_2^4+x_3^4+6x_1^2x_2^2+6x_2^2x_3^2+6x_3^2x_1^2$. It's clear that $E[(x_1+x_2+\cdots+x_n)^{2k}]$ is a symmetric polynomial in terms of $x_1^2, x_2^2, \cdots, x_n^2$, which can be decomposed into elementary symmetric polynomials $e_0(x_1^2, x_2^2, \cdots, x_n^2)$, $e_1(x_1^2, x_2^2, \cdots, x_n^2)$, $\cdots$, $e_n(x_1^2, x_2^2, \cdots, x_n^2)$, by the fundamental theorem of symmetric polynomials. For example, \begin{align} E[(x_1+x_2+x_3)^{4}] =& x_1^4+x_2^4+x_3^4+6x_1^2x_2^2+6x_2^2x_3^2+6x_3^2x_1^2\\ =& (x_1^2+ x_2^2+ x_3^2)^2 + 4 (x_1^2x_2^2+x_2^2x_3^2+x_3^2x_1^2)=e_1^2+4e_2. \end{align} I further worked out the first few cases: \begin{align} E[(x_1+x_2+\cdots+x_n)^{2}] =& e_1\\ E[(x_1+x_2+\cdots+x_n)^{4}] =& e_1^2+4e_2\\ E[(x_1+x_2+\cdots+x_n)^{6}] =& e_1^3 + 12 e_1e_2 + 48 e_3\\ E[(x_1+x_2+\cdots+x_n)^{8}] =& e_1^4 + 24 e_1^2 e_2 + 16 e_2^2 + 256 e_1 e_3 + 1088 e_4\\ E[(x_1+x_2+\cdots+x_n)^{10}] =& e_1^5 + 40 e_1^3 e_2 + 80 e_1 e_2^2 + 800 e_1^2 e_3 + 640 e_2 e_3 + 9280 e_1 e_4 + 39680 e_5 \end{align} Now my question is what is the general formula for this decomposition, like the Newton's Identites, which can be written in terms of generating functions or determinants?

Some observations I have made include: The coefficients of terms $e_1^{k-2i}e_2^{i}$ are $2^{2i}\binom{k}{2i}$. The coefficients of $e_k$ are A024255, which are related to the alternating permutations, Euler numbers and Genocchi numbers.

Answer 1

It is easy to express $E[(x_1+\dots+x_n)^{2k}]$ in terms of monomial symmetric polynomials of $x_1^2,\dots,x_n^2$: $$E[(x_1+\dots+x_n)^{2k}] = \sum_{\lambda \vdash k} \frac{(2k)!}{(2\lambda_1)!(2\lambda_2)!\cdots} m_\lambda(x_1^2,\dots,x_n^2).$$ Then it amounts to expressing $m_\lambda$ in terms of elementary symmetric polynomials to get an answer to your question. Unfortunately there seems to be no simple formula for the transition coefficients between the two types of polynomials.

Still, this problem can solved computationally quite easily for small values of $k$. Here is a sample Sage code that computes the required expression for $k=1,\dots,10$.