March 2019 - Solution
N1 is approximately 6.24E18 and N2 is 1.76E10. The solution we aimed for is n1=53 and n2=477, where BAT(sqrt(n1),N1)=BAT(sqrt(n1),N1)=5929223246159194801 and BAT(sqrt(n1),N2)=BAT(sqrt(n1),N2)=17217982601, both of which are prime numbers.
The answer to the bonus question is BAT(sqrt(n1),10000)=BAT(sqrt(n2),10000)=9100, which is the number of patents IBM had in 2018.
Uoti Urpala sent us a really nice solution, which does not require more than a pen and paper:
n1 = 10000000000000000000000000000000000000000000000000100000000000000000000000000000000000000000000000001
n2 = 10000000000000000000000000000000000000000000000000100000000000000000000000000000000000000000000000002
(first integers after (1e50 + 0.5)**2)
BAT(sqrt(n), L) will be 2 for all L up to about 1e49, as the square roots are very close to 1e50 + 1/2.
Oscal Volpatti sent us a solution for all the last 26 years of IBM patents numbers:
year 1993: 40610066,162440264
year 1994: 82555214,742996926
year 1995: 63363131,253452524
year 1996: 6894703,27578812
year 1997: 976405495,8787649455
year 1998: 2842190725,71054768125
year 1999: 47036688,423330192
year 2000: 251952448,6298811200
year 2001: 44692204,178768816
year 2002: 608115615,15202890375
year 2003: 16982390,67929560
year 2004: 95759246,861833214
year 2005: 17758607,71034428
year 2006: 15182737,60730948
year 2007: 16727176,66908704
year 2008: 108575937,977183433
year 2009: 303117418,7577935450
year 2010: 230388515,2073496635
year 2011: 177366413,8690954237
year 2012: 442656442,3983907978
year 2013: 10627194,42508776
year 2014: 83928763,755358867
year 2015: 3248882,12995528
year 2016: 568907081,14222677025
year 2017: 17107172,68428688
year 2018: 53,477