I have derived an explicit formula for the Euler zigzag numbers, the number of alternating permutations for n elements:

$$A_n = i^{n+1}\sum _{k=1}^{n+1} \sum _{j=0}^k {k\choose{j}} \frac{(-1)^j(k-2j)^{n+1}}{2^ki^kk}$$

For details, please refer to my article in Voofie:

An Explicit Formula for the Euler zigzag numbers (Up/down numbers) from power series

I would like to ask, if my formula is new, or is it a well known result? Since I can't find it in Wikipedia or MathWorld. If it is an old formula, can anyone give me some reference to it? Also, if possible, can anyone provide some reference to other explicit formula for $A_n$?

  • $\begingroup$ Could you please put your formula in display mode so that it is legible? Also, \binom{n}{k} is a much nicer binomial coefficient. $\endgroup$ – JBL Jul 26 '10 at 12:38
  • $\begingroup$ Also, what is $i$? $\endgroup$ – JBL Jul 26 '10 at 12:39
  • 2
    $\begingroup$ Your formula is too tiny to be really new! :-) Jokes aside, the wolframworld page gives the link to OEIS, research.att.com/~njas/sequences/A000111, from which I see many compact explicit formulas for your (and Euler's, of course) numbers. $\endgroup$ – Wadim Zudilin Jul 26 '10 at 12:45
  • $\begingroup$ @JBL: $i=\sqrt{-1}$ :-) $\endgroup$ – Wadim Zudilin Jul 26 '10 at 12:45
  • 1
    $\begingroup$ I by no means want to diminish your result, but I'd say it could be new, or known, or already discovered and forgotten several times. Is 8837*93934=830094758 a new result or not? $\endgroup$ – Pietro Majer Jul 26 '10 at 12:50

Dear Ross,

It looks that you don't really wish to see known formulae for your zigzag numbers. Otherwise I don't understand why you found my search insufficient.

The OEIS A000111 gives the formula $$ A_m=2^m\biggl|E_m\biggl(\frac12\biggr)+E_m(1)\biggr| $$ where $E_m(x)$ are the Euler polynomials which can be generated by the following explicit expansion $$ E_m(x)=\sum_{n=0}^m\frac1{2^n}\sum_{k=0}^n(-1)^k\binom nk(x+k)^m, $$ a double sum as in your case. Even if this formula is not exactly the same as yours (although it looks pretty similar), this is a known double sum expression for $A_m$. There is a lot of room for playing with this double sum and producing many other (useful and useless) formulae for the zigzag numbers.

And don't forget: I've never seen this specific sequence before.

Best wishes, Wadim

  • $\begingroup$ Wadim, you are right. It is really an explicit formula. I didn't notice it since I don't know Euler polynomials has explicit expansion. However, the formula you provided has an absolute sign. I think it will make all the manipulation and computation inconvenient? $\endgroup$ – Ross Tang Jul 27 '10 at 14:00
  • 6
    $\begingroup$ Ross, who care about having another formula for the same sequence if this is just a new formula without applications or with applications which can be achieved by a known one?! $\endgroup$ – Wadim Zudilin Jul 27 '10 at 23:46
  • $\begingroup$ $$ A_m=2^m\biggl|E_m\biggl(\frac12\biggr)+E_m(1)\biggr| $$ In this case, $A_0 =2^0\biggl|1+1\biggr|=2≠1$, Is there wrong with my arithmetic? $\endgroup$ – user94635 Jul 4 '16 at 7:06
  • $\begingroup$ Wadim, can you check out the last comment? Also, I'm not able to find your claimed formula under oeis.org/A000111 Thanks. $\endgroup$ – Todd Trimble Jul 4 '16 at 7:36

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.