Let there be a sequence with a₁=3/2 such that a_n+1=1+n/a_n. Find n such that 2020≤a_n<2021.

a_n exceeds 1 when n=1, 2 when n=3, 3 when n=7, 4 when n=13, and so on.  

So I conjecture that the digit k is exceeded on the (k^2 - k + 1)th term.

For the problem where k=2020, n= 2020^2 - 2020 + 1 = 4078381

I have no idea what is going on.

**Kenny M got here first.

**The answer to Jer's question would seem to be 2021^2 -2021.

Posted by xdog on 2020-10-28 11:51:59