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May 7, 2017 at 8:10 comment added Johann Cigler @Amdeberhan: Thank you for the details. But I still have difficulties if $n=6m+3$.
May 6, 2017 at 19:06 history edited T. Amdeberhan CC BY-SA 3.0
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May 6, 2017 at 16:17 history edited T. Amdeberhan CC BY-SA 3.0
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May 6, 2017 at 12:45 comment added Johann Cigler @Amdeberhan: Could you please mention the details of your proof?
May 6, 2017 at 9:31 comment added Jeffrey Shallit And for $p = 2$ it is $\nu_2(L_n (2)) = (n \bmod 2)-\nu_2 (n!)$.
May 6, 2017 at 8:36 comment added Jeffrey Shallit Right! And then the generalization to odd $p$ (mentioned below) is $\nu_p (f(2n)) = \nu_p ({{2n} \choose n})$ and $\nu_p (f(2n+1)) = \nu_p(p(2n+1){{2n} \choose n})$. Very nice. Thanks so much.
May 6, 2017 at 8:36 vote accept Jeffrey Shallit
May 6, 2017 at 6:30 history answered T. Amdeberhan CC BY-SA 3.0