Is there a closed-form for $\sum_{k=0}^{\infty} \frac{1}{(k!)^2}$?

Question

As the title says, I'm wondering whether there is any known closed-from for the following series:

$$\sum_{k=0}^{\infty} \frac{1}{(k!)^2}$$

Trying on WolframAlpha, I get the value $2.279585302336067267437204440811533353...$ and the continued fraction begins like this $[2; 3, 1, 1, 2, 1, 3, 7, 4, 3, 1, 2, 2, 1, 2, 1, 1, 2, 7, 8, 1, 1, 21, 1, 16, 2, 1, 8, 1, 1, 8, ...]$

If you have any reference which focuses on this number, I would also be interested. I found nothing so far!

Thank you very much!

Answer 1

Well you might be interested in looking at the Modified Bessel function. Hence you sum is just

$$\sum_{k=0}^{\infty} \frac{1}{(k!)^2} = I_0(2)$$

I don't think it simplifies further.