Variant Generating Function related to Euler Numbers

Question

The generating function $$\frac{2e^x}{e^{2x}+1}=\sum_{n\ge 0}E_k\frac{x^k}{k!}$$ counts the number of alternating permutations of a set with an even number of elements. My question is this, if we alter the denominator slightly: $$\frac{2e^x}{e^{2x}+1+2x}=\sum_{n\ge 0}E_{2,k}\frac{x^k}{k!}$$ what would this particular variant generating function count exactly? What effect does the addition of $2x$ do to the number of alternating permuations of a set of elements? How can I intuitively "see" what this functions counts (if anything)?

Answer 1

Best do a series expansion around $x=0$. In your case, your favorite computer algebra system yields: $$ \frac{2 e^x}{e^{2x}+1+2x} = 1-x+3/2 \cdot x^2 -5/2 \cdot x^3+ 31/8 \cdot x^4+ O(x^5) $$ This is before multiplying each coefficient by $n!$. Since the series is alternating, I don't think it actually "counts" anything. Maybe the sum of this series and another has a meaning.
Btw, had all coefficients been positive, the next step would have been to search for an entry in OEIS.