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Any set of numbers that you choose can be the next number. It is easy to find a rule based on a polynomial of order 6 such that the first six numbers are as listed in the question followed by your chosen next number. A polynomial of order 9 will fit the above six and any three following numbers. There are also non-polynomial solutions. Short of reading the mind of the person who posed the question, there is no way of determining which of the infinitely many solutions is the "correct" one.

The simplest solution here, besed on a polynomial of order 5 is:

U(n) = (221*n^5 - 3845*n^4 + 24725*n^3 - 72055*n^2 + 94994*n - 41760)/120 for n = 1, 2, 3, ...

and, accordingly,

U(7) = 464,

U(8) = 2154 and

U(9) = 6867.

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9y ago

Any three number that you choose can be the next three numbers. It is easy to find a rule based on a polynomial of order 9 such that the first six numbers are as listed in the question followed by the chosen three numbers. There are also non-polynomial solutions. Short of reading the mind of the person who posed the question, there is no way of determining which of the infinitely many solutions is the "correct" one.

The simplest polynomial solution is


U(n) = (221*n^5 - 3845*n^4 + 24725*n^3 - 72055*n^2 + 94994*n - 41760)/120 for n = 1, 2, 3, ...and, accordingly, the next three numbers are: 464, 2154 and 6867. Check it out!

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