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.

 

[CellCoree]

Thank you for sharing your work on the explicit formula for Euler zigzag numbers. It is always exciting to see new developments in mathematics.

As a scientist, it is important to acknowledge that the discovery of new formulas or results is an ongoing process and it is possible that your formula may have already been discovered by someone else. However, if you have not found it in Wikipedia or MathWorld, it is possible that it is a new result.

I would recommend reaching out to other mathematicians or publishing your work in a peer-reviewed journal to get feedback and to see if your formula has been previously discovered. It is also important to provide a clear explanation of your derivation and any assumptions made in your formula to ensure its validity.

Regardless of whether your formula is new or not, it is an impressive accomplishment and I encourage you to continue exploring and contributing to the field of mathematics. Keep up the good work!