For a bonus, what is the equation?

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You can fit infinitely many equations to the points, so there are infinitely many sequences that can start with 1, 5, 19, 49. Here are some more obscure ones I found....

3^n - (n^3 - 12n^2 + 5n)/3. For n = 0, 1, 2, and 3 it gives you 1, 5, 19, and 49. For n = 4, 5, 6 the next three terms are 117, 293, 791.

2^n + (5n^3 + 12n^2 + n)/6. For n = 0, 1, 2, and 3 it gives you 1, 5, 19, and 49. For n = 4, 5, 6 the next three terms are 102, 187, 317.

Since 49 = 2*19 + 2*5 + 1*1, you could also fit the linear recursive sequence A(n) = 2*A(n-1) + 2*A(n-2) + A(n-3) to the points. Then the next three terms would be

2*49 + 2*19 + 5 = 141

2*141 + 2*49 + 19 = 399

2*399 + 2*141 + 49 = 1129

You can actually work out an explicit equation for this, but it invoves finding the roots of the polynomial x^3 - 2x^2 - 2x - 1 = 0.

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Very awesome!

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The next three numbers in the series are: 101, 181, & 295.

The equation is [n(n+1)^2]+1

Thanks for the brain teaser.

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This works. A simpler version of the equation is n^3 - n^2 + 1, and starting at n=1 instead of n=0, as you did. Interestingly, the two equations, though not entirely mathematically equivalent, produce the same series.