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Let $H_k(x)$ be the Hermite polynomial, which can be written in terms of Laguerre polynomial such as \begin{align*} H_{k} \left( x \right) = H_{2k} \left( x \right) + H_{2k+1} \left( x \right)\qquad \forall k\ge1 \end{align*} \begin{align*} H_{2k} \left( x \right) = \left( { - 1} \right)^k 2^{2k} \left( k \right)!\sum\limits_{j = 0}^{k} {\left( { - 1} \right)^j \frac{{x^{2j} }}{{j!}}\left( {\begin{array}{*{20}c} {k - \frac{1}{2}} \\ {k - j} \\ \end{array}} \right)} \end{align*} and \begin{align*} H_{2k + 1} \left( x \right) = \left( { - 1} \right)^k 2^{2k + 1} \left( k \right)!\sum\limits_{j = 0}^{k} {\left( { - 1} \right)^j \frac{{x^{2j + 1} }}{{j!}}\left( {\begin{array}{*{20}c} { k + \frac{1}{2}} \\ {k - j} \\ \end{array}} \right)}. \end{align*} I tried to find a general formula for $H_k(1)$ but I couldn't do that!. The problem deals on how one can simplify a combinations with factorials. Any help is appreciated. Thanks

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The values of $H_k(1)$ corresponds to sequence $A062267$ in $OEIS$ and corresponds to the recurrence

$$a_n = 2\left(a_{n-1} - (n - 1)\,a_{n-2}\right)\qquad (a_0=1,a_1=2)$$

According to this

$$H_{2k}(1)=(-1)^k2^k (2 k-1)\text{!!}\, \, _1F_1\left(-k;\frac{1}{2};1\right)$$

$$H_{2k+1}(1)=(-1)^k 2^{k+1} (2 k+1)\text{!!}\, \, _1F_1\left(-k;\frac{3}{2};1\right)$$

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  • $\begingroup$ yes, thank you very much. I was seeking a closed from. it's ok thanks $\endgroup$ Jan 9 '17 at 11:43

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