Explicit formula for Euler zigzag numbers(Up/down numbers)

[ross_tang]

I have derived an explicit formula for the http://mathworld.wolfram.com/EulerZigzagNumber.html" , the number of alternating permutations for n elements:
A_j=i^{j+1}\sum _{n=1}^{j+1} \sum _{k=0}^n \frac{C_k^n(n-2k)^{j+1}(-1)^k}{2^ni^nn}

For details, please refer to my article in http://www.voofie.com" :

http://www.voofie.com/content/117/an-explicit-formula-for-the-euler-zigzag-numbers-updown-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?

 

[arildno]

Well, mathworld mentions something called the "Entringer numbers", in its article:
http://mathworld.wolfram.com/EulerZigzagNumber.html
It doesn't seem to be quite the same thing??

 

[CRGreathouse]

There are many formulas at Sloane's http://www.research.att.com/~njas/sequences/A000111 though I don't see yours, at least directly. But it may be there in disguised form.

Computationally, your formula is not competitive with some of the others listed there. For example, given

a(n)=local(v=[1], t); if(n<0, 0, for(k=2, n+2, t=0; v=vector(k, i, if(i>1, t+=v[k+1-i])));v[2])
A(j)=I^(j+1)*sum(n=1,j+1,sum(k=0,n,binomial(n,k)*(n-2*k)^(j+1)*(-1)^k/2^n/I^n/n))

your code A(500) takes four times longer than Somos' code a(500). I don't know if yet more efficient code exists.

 

[ross_tang]

@arildno
Thank you for the link. I haven't read the "Entringer numbers" before. But I don't found explicit formula, and seems the Euler zigzag number is it's special case.

@CRGreathouse
I have read the link you send me before. Thank you. First, I found out the formula, not because it is computationally efficient. I am interested in the explicit form of it. Since the zigzag number sequence is solution to a few number of recurrence relation. And the Somo's code you mentioned used recurrence relation too. But I found none explicit form in the link.