As you can see there are three ways to reach 96 (yes, 2 and 3 are basically the same but with a slightly different thought process). And I withdraw my previous 116 solution because of an addition mistake! Thanks OP for posting!
Yeah, I was hesitant to accept your 116.
Actually all your solutions for 96 can be boiled down to the same right hand sequence s(n) = n(n+4). You can verify that your first solution generates the same sequence 5, 12, 21, 32, 45, 60, 77, 96.
1) Starting from your first solution which is f(x,y) = xy +x which is equivalent to my f(x,y) = x(y+1). We can generate the sequence by constraining y to be x+3. So s(n) = f(n, n+3) = n((n +3) + 1) = n(n+4).
2) Your solutions 2 and 3. Since each line simply adds new terms to the preceding line's results we observe that the n'th line can be expressed as s(n) = (1 + 2 + ... + n) + [(1 + 2 + ... + n) + 3n]. Note that the square brackets contain the sum of all the second terms. Okay so far? But we know that (1 + 2 + ... + n) is n(n+1)/2. Since there are two of them we can drop the division by two and then we just need to add the 3n. So s(n) = n(n+1) + 3n = n(n+4).
Nifty eh?
I also liked the 201 base 3 solution. I'd like to see a Rube Goldberg solution using trig functions but that makes my head hurt. Basically any non linear function that hits the first three data points works. The result for the fourth equation is immaterial.
