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Good morning. I am working with higher powers of the Generalized Euler Number generating function

$$\left[\frac{n}{\sum_{j=0}^{n-1}\exp{\left(w_n^jx\right)}}\right]^\alpha$$

where $w_n=\exp{\left(2i\pi/n\right)}$. This can be written in terms of what are called Olivier functions, where

$$\Phi_{n,k}(x)=\frac{1}{n}\sum_{j=0}^{n-1}w_n^{-jk}\exp{\left(w_n^jx\right)}$$

and thus I can rewrite the exponential generating function as

$$\left[\frac{1}{\Phi_{n,0}(x)}\right]^\alpha=\sum_{j=0}^\infty{A_{n,0}^{(\alpha)}\frac{x^j}{j!}}$$

My goal is to generate the first fifty or so numbers for varying $\alpha$. So I use the following definitions to help in Mathematica:

 w[n_] := E^((2*i*\Pi)/n)
 Phi[n_, k_, x_] := (1/n)*Sum[(w[n]^(-j*k)*E^(x*w[n]^j)), {j, 0, n - 1}]
 NEuler[a_, n_, x_] := (1/Phi[n, 0, x])^a
 NENumber[a_, n_, r_, M_] :=
   FullSimplify[Coefficient[r! Normal[Series[NEuler[a, n, x], {x, 0, M}]], x, r]]

My problem here is when I go to evaluate say

 Table[NENumber[2, 3, j, M], {j, 0, M}]

If $M>20$ it seems that the calculation never actually evaluates, or if it will, it takes way too long. Is there another method in order to generate these numbers much more quickly and efficiently on my computer? My math skills are better than my Mathematica skills, as I have definitely never had formal instruction using Mathematica, and everything that I've learned I've learned by myself. Thank you for your help.

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  • $\begingroup$ I and i are two very different things. Thus: OlivierPhi[n_, k_, x_] := Sum[Exp[x Exp[2 π I j/n] - 2 π I j k/n], {j, 0, n - 1}]/n; With[{n = 2, a = 3}, Table[j! SeriesCoefficient[(1/OlivierPhi[n, 0, x])^a, {x, 0, j}] // RootReduce, {j, 0, 20}]] $\endgroup$
    – J. M.'s slightly less busy ♦
    Jul 24, 2016 at 14:05
  • $\begingroup$ I usually use the symbolic i on the palettes. So I assume then that >I is the imaginary number? $\endgroup$
    – Eleven-Eleven
    Jul 24, 2016 at 14:13
  • $\begingroup$ Yes, I is the imaginary unit in InputForm[]; the "fancy i" should have automatically pasted as I here if you copied things right. $\endgroup$
    – J. M.'s slightly less busy ♦
    Jul 24, 2016 at 14:14
  • $\begingroup$ I actually didn't cut and paste, I just recalled from memory and typed it in. And ha ha to "fancy i"... Thanks I'm going to try this now. $\endgroup$
    – Eleven-Eleven
    Jul 24, 2016 at 14:15
  • $\begingroup$ I am running into problems with $n=3,a=2$ and table from 0 to 40. Since $n=3$ yields complex roots of unity, is this the reason? It has been evaluating for about 5 minutes now... $\endgroup$
    – Eleven-Eleven
    Jul 24, 2016 at 14:29

1 Answer 1

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It turns out the Olivier function is already built-in, but in a disguised form:

With[{n = 3, a = 2},
     Table[j! SeriesCoefficient[1/MittagLefflerE[n, x^n]^a, {x, 0, j}],
           {j, 0, 40}]]
   {1, 0, 0, -2, 0, 0, 58, 0, 0, -6218, 0, 0, 1630330, 0, 0, -847053482, 0, 0,
    766492673914, 0, 0, -1106653345942538, 0, 0, 2392407356983116538, 0, 0,
    -7379716460043385709162, 0, 0, 31273642050397659221050618, 0, 0,
    -176597979883794828225332275658, 0, 0, 1295782854051144759324367927117690, 0, 0,
    -12096399366143533384617425907661145642, 0}

If you are using an older version that does not have the Mittag-Leffler function, here is some alternative code:

With[{n = 3, a = 2}, 
     Table[j! SeriesCoefficient[1/HypergeometricPFQ[{}, Range[n - 1]/n, (x/n)^n]^a,
                                {x, 0, j}], {j, 0, 40}]]
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