Would have been nice to include what happens as n gets large. That series is going to be about log(sqrt(n)) which means the ratio of the number of loops to number of initial strands goes to 0 pretty fast. I think that is correct anyway, certainly the conclusion
Would have been nice to include what happens as n gets large. That series is going to be about log(sqrt(n)) which means the ratio of the number of loops to number of initial strands goes to 0 pretty fast. I think that is correct anyway, certainly the conclusion
yes, the actual answer is ln2 + c/2 + ln(sqrt(n)) which is almost 1 + ln(sqrt(n)), where c = euler's constant
This is also a STEP 3 question! It'd be good to also look at that one because there's also a bit more to it.