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I'm interested in finding a nice exponential generating function, $g$, for $$\frac{1}{2-\tan{x}-\sec{x}}.$$ The sequence listed over here on OEIS is actually... very nasty and hard to compute. Surely, this cannot be the best possible EGF out there? Perhaps a different way may be to perform Taylor expansion, but I'm not sure how to best approach that method. Can someone give me some advice on how I can find a better generating function?

Arkyter
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  • ouch..... i see. I'm going to try my best with taylor expansion, but if that doesn't work I suppose I'll have to accept it as is. – Arkyter Jul 20 '22 at 19:19
  • Do you mean that you want a formula for the coefficients of that function? – JBL Jul 25 '22 at 12:02
  • yeah, but something neater than OEIS if possible. – Arkyter Jul 25 '22 at 19:57
  • Well I mean this is necessarily more complicated than the Euler numbers https://oeis.org/A000111 so how good an answer are you expecting? Certainly we have $T_n = \sum_{k = 1}^n \binom{n}{k} E_k \cdot T_{n - k}$ where $(T_n)$ is your sequence and $(E_k)$ is the Euler numbers. – JBL Jul 25 '22 at 22:06
  • I apologize for not being clear enough - I was originally wondering if there was a closed exponential generating function that's neater than the Euler numbers used in the first formula listed in OEIS. If possible, I want to avoid any kind of recursion! Though, I don't know if that's doable. – Arkyter Jul 26 '22 at 02:17
  • A "closed exponential generating function" for what? The function $\frac{1}{2 - \tan(x) - \sec(x)}$ that appears in the question is the exponential generating function of the sequence $1, 1, 3, 14, 87, \ldots$ whose OEIS page you linked. Apparently this is not what you want; so what do you want? – JBL Jul 26 '22 at 16:39

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