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Jul 3, 2016 at 12:10 comment added Todd Trimble It says "Your website is not available" when I click on the second and third links.
Jul 27, 2010 at 11:48 comment added Ross Tang Dear everyone, I found the above formula when I am trying to solve a partial difference equation as an exercise. mathoverflow.net/questions/33498/… The Euler zigzag numbers are secant and tangent number respectively for the even term and odd term, and it is given by $A_0^n$ from the sequence $A_k^n$ from the above link. I would be very please if anyone can give me some insight in solving the partial difference equation. Thank you.
Jul 26, 2010 at 23:35 answer added Wadim Zudilin timeline score: 2
Jul 26, 2010 at 15:51 history edited Robin Chapman CC BY-SA 2.5
Tex made display
Jul 26, 2010 at 15:37 history edited Ross Tang CC BY-SA 2.5
added 90 characters in body
Jul 26, 2010 at 15:23 history edited Ross Tang CC BY-SA 2.5
added 12 characters in body
Jul 26, 2010 at 15:18 comment added Ross Tang @Wadim Zudilin: At which session do u refer to? I found none explicit formula in the Formula session. Thank you for your reference.
Jul 26, 2010 at 12:50 comment added Pietro Majer 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?
Jul 26, 2010 at 12:45 comment added Wadim Zudilin @JBL: $i=\sqrt{-1}$ :-)
Jul 26, 2010 at 12:45 comment added Wadim Zudilin 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.
Jul 26, 2010 at 12:39 comment added JBL Also, what is $i$?
Jul 26, 2010 at 12:38 comment added JBL Could you please put your formula in display mode so that it is legible? Also, \binom{n}{k} is a much nicer binomial coefficient.
Jul 26, 2010 at 12:37 history edited JBL
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Jul 26, 2010 at 12:22 history asked Ross Tang CC BY-SA 2.5