Explicit formula for Euler zigzag numbers(Up/down numbers) - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T09:44:52Z http://mathoverflow.net/feeds/question/33387 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/33387/explicit-formula-for-euler-zigzag-numbersup-down-numbers Explicit formula for Euler zigzag numbers(Up/down numbers) Ross Tang 2010-07-26T12:22:51Z 2010-07-27T10:50:10Z <p>I have derived an explicit formula for the <a href="http://mathworld.wolfram.com/EulerZigzagNumber.html" rel="nofollow">Euler zigzag numbers</a>, the number of alternating permutations for n elements:</p> <p><code>$$A_n = i^{n+1}\sum _{k=1}^{n+1} \sum _{j=0}^k {k\choose{j}} \frac{(-1)^j(k-2j)^{n+1}}{2^ki^kk}$$</code></p> <p>For details, please refer to my article in <a href="http://www.voofie.com/" rel="nofollow">Voofie</a>:</p> <p><a href="http://www.voofie.com/content/117/an-explicit-formula-for-the-euler-zigzag-numbers-updown-numbers-from-power-series/" rel="nofollow">An Explicit Formula for the Euler zigzag numbers (Up/down numbers) from power series</a></p> <p>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 $A_n$?</p> http://mathoverflow.net/questions/33387/explicit-formula-for-euler-zigzag-numbersup-down-numbers/33453#33453 Answer by Wadim Zudilin for Explicit formula for Euler zigzag numbers(Up/down numbers) Wadim Zudilin 2010-07-26T23:35:35Z 2010-07-27T08:54:05Z <p>Dear Ross,</p> <p>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.</p> <p>The OEIS <a href="http://www.research.att.com/~njas/sequences/A000111" rel="nofollow">A000111</a> gives the formula $$A_m=2^m\biggl|E_m\biggl(\frac12\biggr)+E_m(1)\biggr|$$ where $E_m(x)$ are the <a href="http://en.wikipedia.org/wiki/Bernoulli_polynomials" rel="nofollow">Euler polynomials</a> 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 <em>known</em> 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.</p> <p>And don't forget: I've never seen this specific sequence before.</p> <p>Best wishes, Wadim</p>