4
$\begingroup$

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)?

$\endgroup$
  • $\begingroup$ Where did you get your original series from? I just found that it also alternates. $\endgroup$ – john_leo May 21 '15 at 8:07
  • $\begingroup$ What was the source of the variant series, and is there any reason behind specifically the '2' in the '2x' factor? It seems unlikely that the coefficients of that series would have any particularly clean form... $\endgroup$ – Steven Stadnicki May 21 '15 at 8:19
  • $\begingroup$ The reason for the $2x$ is, if we expand $e^{2x}$, we get $1+2x+4x^2/2!+8x^3/3!+...$. It is just something I am interested in. $\endgroup$ – Iceman May 21 '15 at 12:19
  • $\begingroup$ @john_leo, the original series is that of Euler Numbers. $$\frac{1}{\cosh{x}}=\frac{2}{e^x+e^{-x}}=\frac{2e^x}{(e^x+e^{-x})e^x}=\frac{2e^x}{e^{2x}+1}$$ $\endgroup$ – Iceman May 21 '15 at 12:22
  • 1
    $\begingroup$ You must be careful. Those numbers are strongly related to the number of alternating permutations, but strictly speaking, the alternating permutations are not counted by $1/\cosh(x)$ (as gf may not have neg. coefficients). Best re-read wikipedia's article on alternating permutations. $\endgroup$ – john_leo May 21 '15 at 12:34
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I'm using the exponential generating functions, so my sequence is $1,-1,3,-15,93,-725,...$ $\endgroup$ – Iceman May 21 '15 at 12:27
  • $\begingroup$ I know, yet I don't know of a series with negative coefficients that "counts" something. $\endgroup$ – john_leo May 21 '15 at 12:29
  • $\begingroup$ That was my confusion with the Euler Numbers (check the link above). It definitely says on the Wikipedia page that they count what I mentioned above. Perhaps if it were unsigned.... $\endgroup$ – Iceman May 21 '15 at 12:31
  • $\begingroup$ And those numbers definitely don't appear in the OEIS anywhere. $\endgroup$ – Iceman May 21 '15 at 12:31
  • $\begingroup$ Related, not the same. $\endgroup$ – john_leo May 21 '15 at 12:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.