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### Best Answer TR Smith says

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.

Very awesome!

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### CWanamaker says

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

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

Thanks for the brain teaser.

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.