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


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

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

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.