Fun fact: Mathemetica has a method FindSequenceFunction which does exactly what you describe. In my experience, however, it generally requires at least 4 terms.

I'm not totally sure how the math behind it works (maybe it's similar to Eureqa?) but the results speak for themselves and are rather incredible.

For example, if I run FindSequenceFunction on this input:

    {0, 1, 3, 8, 19, 43, 94, 201, 423, 880}

Which is the number of 0,1 sequences of length n that contain two adjacent 1s

Mathematica produces the result:

    1/10 (5 2^(1 + x) - 5 (1/2 - Sqrt[5]/2)^x + 
    3 Sqrt[5] (1/2 - Sqrt[5]/2)^x - 5 (1/2 + Sqrt[5]/2)^x - 
    3 Sqrt[5] (1/2 + Sqrt[5]/2)^x)

Which, astonishingly, is correct for all the values I've tried. So apparently Mathematica understands more about this sequence than I do, and I know its definition.

Another party trick is to use the input

    {-(1/6), 2/15, -(13/140), 23/315, -(83/1386), 305/6006, -(2269/
     51480), 4259/109395, -(16103/461890), 30616/969969}

Which is the integral x^n (1 - 2 x)^n for x from 0 to 1, for n = 0..<10. Here it seems 10 numbers are required. This yields the solution

    (2^(-2 - 3 x)
    x! (Sqrt[\[Pi]] (1 + x)! + 
    3 (-1)^x 2^(
    2 + 3 x) (1/2 (1 + 2 x))! Hypergeometric2F1[1, 3/2 + x, 
    2 + x, -8]))/((1/2 + x)! (1 + x)!)

Which as far as I can tell, is a closed-form solution (!) to the integral. A solution it worked out to an integral it has never seen, but only the first 10 elements in the sequence.

So it's safe to say Mathematica knows a lot more about math than I do.