I'm wondering if there is a general solution for $$S_a =\sum_{n=0}^{\infty} \frac{n^a}{n!}$$ with $a \in \mathbb{Z}$ and $a > 0$. From Mathematica: $$S_1 = 1e$$ $$S_2 = 2e$$ $$S_3 = 5e$$ $$S_4 = 15e$$ $$S_5 = 52e$$ $$S_6 = 203e$$
Asked
Active
Viewed 93 times
0
-
2You may be interested in this: http://oeis.org/search?q=1%2C2%2C5%2C15%2C52%2C203&language=english&go=Search – Antonio Vargas Oct 14 '14 at 20:37
-
It might work better if you compute $S_a(x) = \sum_{n=0}^\infty \frac{n^a}{n!}x^n$. (I haven't tried, so maybe it doesn't help.) – Daniel Fischer Oct 14 '14 at 20:38
-
$S_0$ is in line with the sequence Antonio points to as well. – 2'5 9'2 Oct 14 '14 at 20:41
-
@DanielFischer Your suggestion would appear to help, consult this MSE link. – Marko Riedel Oct 14 '14 at 21:27
