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I have to find the sequence generated when I use this function as exponential generating function:

f(x)=$3x^{2x}$

I don't know how to expand this function into a series.

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  • $\begingroup$ First you have to shift $3 (x+1)^{2 (x+1)}$ then MacLaurin who doesn't give a nice answer $$\left(3,6,9,9,8,\frac{11}{2},\frac{37}{10},\frac{19}{10},\frac{118}{105},\frac{167}{420},\frac{47}{168}\ldots\right)$$ I can't find a closed form... $\endgroup$ – Raffaele Sep 26 '17 at 11:02
  • $\begingroup$ @Raffaele, Shifting $x \mapsto x+1$ gives you a generating function for a different sequence. $\endgroup$ – Marcus M Sep 26 '17 at 13:36
  • $\begingroup$ @MarcusM but $3x^{2x}$ has no MacLaurin expansion! $\endgroup$ – Raffaele Sep 26 '17 at 13:42
  • $\begingroup$ @Raffaele, Yeah, which suggests perhaps there's an issue with the problem. Kishan, where did this problem come from? $\endgroup$ – Marcus M Sep 26 '17 at 14:00
  • $\begingroup$ @MarcusM I found this problem while solving discrete mathematics by Kenneth Rosen. The answer given for the sequence is $3*2^n$. I suppose this is the solution for $3e^{2x}$ which means there is some printing mistake. $\endgroup$ – Kishan Kumar Sep 26 '17 at 15:36

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