[QUOTE=ericw;465329][url]https://oeis.org/A100289[/url]
Numbers n such that (1!)^2 + (2!)^2 + (3!)^2 +...+ (n!)^2 is prime.
a(19) = 32841 from Serge Batalov, Jul 29 2017[/QUOTE]

I'll probably leave that one alone. I haven't working on it.

Sieving completed a couple of days ago. I've tested to n=30000 and have verified known results. I should reach Serge's find by the end of the week.

I've tested to n=40000 and have verified known results.

I've tested to n=50000 and have verified known results, including Serge's find. Continuing.

I've tested to n=60000. No new PRPs.

I've tested to n=70000. No new PRPs.

I don't know how I missed this one. The 25th term is for n=59961 which is 260447 digits (if I calculated that correctly).

The search is around n=74000. I'm not certain how much further I'm going to search, but I have sieved to n=100000.

After a hiatus, I've tested to n=80000. No new PRPs.

Completed to 90000. No new PRPs. Continuing.

[QUOTE=rogue;468033]I don't know how I missed this one. The 25th term is for n=59961 which is 260447 digits (if I calculated that correctly).[/QUOTE]

Great! (I verified that it's 3-prp.) I see that you've added it to [url=https://oeis.org/A001272]A001272[/url].

[QUOTE=rogue;468033]I don't know how I missed this one. The 25th term is for n=59961 which is 260447 digits (if I calculated that correctly).[/QUOTE]

(Actually 260448 decimal digits.)

Congratulations again :)