Generating function for the sequence

Question

$$ 2, 2\sqrt 2, \sqrt 2 ( \sqrt 2 -1 ); \frac 1 3 \sqrt 2( \sqrt 2 - 1 )( \sqrt 2 - 2), \frac 1 {3*4} \sqrt 2( \sqrt 2 - 1)(\sqrt 2 - 2)(\sqrt 2 - 3),... $$ We can derive summation: $$ \sum_{i=0}^{\infty}x^i \frac 2 {i!} \prod_{j=0}^i (\sqrt 2 - j) $$ $e^x$ expansion is pretty close but I am stuck beyond this.

Answer 1

$$S_{n+1} = \frac{\sqrt{2} - n}{n+1}S_n$$ $$S_0 = 2$$

But also note:

$$\binom{x}{n+1} = \frac{x-n}{n+1}\binom{x}{n}$$

Thus we can conclude $S_n = 2\binom{\sqrt{2}}{n}$. Using the generating function for the binomial coefficient we find:

$$\sum_{n=0}^\infty S_nx^n = 2(x+1)^\sqrt{2}$$

Answer 2

Hint: The series expansion implies that, if our function is $f(x)$, then

$$f^{(n)}(0) = 2\prod_{j=0}^i (\sqrt{2}-j).$$

Do you know any functions that have iterated derivatives of this form, a product of terms that decrease by $1$ each time?