There are two 16-digit repeated sequences in the first 180 million digits:
2994 9096 2576 2240 (at positions 23,974,220 and 152,251,556)
6346 5019 1472 1352 (at positions 132,314,276 and 179,728,264).
There’s one 15-digit repeat within the first 30 million (2768 5455 1127 715), which is probably the one Bailey found.
I wouldn’t want to look for a 24-digit repeat without a lot more memory than I currently have.
Which is nicely consistent with Bailey’s formulae: the expectation is that you should find 1.6 repeats that long in those many digits.
What an interesting problem! Using the 200 million digits from the University of Tokyo and a little program I wrote, I’ve found the minimum number m of digits required for a repeat of length n. [ol]
[li] ‘1’ in 4 digits [] ‘26’ in 23 digits [] ‘592’ in 64 digits [] ‘0582’ in 136 digits [] ‘60943’ in 556 digits [] ‘949129’ in 1302 digits [] ‘8530614’ in 4608 digits [] ‘52637962’ in 15442 digits [] ‘201890888’ in 33853 digits [] ‘4392366484’ in 240489 digits [] ‘89879780761’ in 694410 digits [/ol]That’s as far as I’ve got in about 20 hours of cpu-time. (Time to think of a faster algorithm. 
)[/li]
The scaling is m ~ 2 n[sup]3[/sup]. This is different than the one described by bonzer, presumably because I’m finding the minimum number of digits for a repeat to exist.
Oops, I meant the scaling is m ~ 2 3[sup]n[/sup]. Very different!