For Fun

Find the fifth term in the innocent looking sequence: 1,4,9,16, besides 25. List your sequence generally.

"I'm not that innocent."

The only one I can come up with is 21.  The idea being that each alternate pair adds to 10, 20, 30etc. ( 9 + 21 = 30). Oh, this just occurred to me:  If each number is the previous number with the next odd prime added to it  (only need to specify "odd" is because 4 is 1+3, not 1+2) then the next number would be 26, or 25+11 (not 25 since 9 is not prime.)

Let f(n):= a(n-1)(n-2)(n-3)(n-4)+x^2, where a is a parameter.  This gives the correct values for the first four terms, whereas f(5)=24a+25.  So choosing a=(your mother-25)/24, the fifth term in the sequence is clearly your mother.

I think 21 looks right.

Yes, I would say that 21 works. Another more general formula would be:

n^4-10n^3+39n^2-50n+24 all divided by 4 yields 31 as the 5th term.

Good work guys, and nice your mama joke too.