Is it possible to further simplify the expression

$$\sum\limits_{n = 0}^{\left\lfloor {\frac{M}{2}} \right\rfloor } {\frac{{M!}}{{{2^n} \times \left( {n!} \right) \times \left( {M - 2n} \right)!}}},$$

where $M$ is a positive integer?

  • 1
    $\begingroup$ it depends ... if you think that the holonomic sequence, or the hypergeometric function are a simplification $\endgroup$ Sep 29 '20 at 14:23
  • 2
    $\begingroup$ Notice that this is the number of involutions on permutations of $M$ elements. $\endgroup$
    – Phicar
    Sep 29 '20 at 14:33
  • $\begingroup$ General strategy ... calculate the first few values & then look up the sequence in OEIS. $\endgroup$ Sep 29 '20 at 20:31

Making the problem more general $$S_M=\sum\limits_{n = 0}^{\left\lfloor {\frac{M}{2}} \right\rfloor } {\frac{{M!}}{{ {n!} \, \left( {M - 2n} \right)!}}}x^n$$ $$S_{2m}=(-1)^m (4x)^m \,U\left(-m,\frac{1}{2},-\frac{1}{4 x}\right)$$ $$S_{2m+1}=(-1)^m (4x)^m \,U\left(-m,\frac{3}{2},-\frac{1}{4 x}\right)$$ where appears the confluent hypergeometric function.

If $x=\frac 12$, these generate the sequences $$\{2,10,76,764,9496,140152,2390480,46206736,997313824,23758664096\}$$ which correspond to $$S_{2m}=2^n n!\, L_n^{-\frac{1}{2}}\left(-\frac{1}{2}\right)$$ and $$\{4,26,232,2620,35696,568504,10349536,211799312,4809701440,119952692896\}$$ which correspond to $$S_{2m+1}=2^n n!\, L_n^{\frac{1}{2}}\left(-\frac{1}{2}\right)$$ where appear the generalized Laguerre polynomials.

Have a look at $OEIS$ for very interesting informations.

  • $\begingroup$ Could you kindly confirm that $U\left( {} \right)$ represent the Tricomi's (confluent hypergeometric) function U(a, b, z) (introduced by Francesco Tricomi (1947)), which is a solution to Kummer's equation. Or also known as the confluent hypergeometric function of the second kind ? and $L_n^\alpha \left( x \right) = \frac{{{{\left( { - 1} \right)}^n}}}{{n!}}U\left( { - n,\alpha + 1.x} \right)$ $\endgroup$ Sep 30 '20 at 11:07
  • 1
    $\begingroup$ @TuongNguyenMinh. Yes, it is $\endgroup$ Sep 30 '20 at 11:25

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