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20 1 18 4 13 SEQUENCE? - Answers

There are infinitely many polynomials of order 5 that will give these as the first five numbers and any one of these could be "the" rule. 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.A rule, based on a polynomial of order 4 is:U(n) = (121n^4 - 1478n^3 + 6275n^2 - 2830n + 6312)/24 for n = 1, 2, 3, ...



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20 1 18 4 13 SEQUENCE? - Answers

https://math.answers.com/math-and-arithmetic/20_1_18_4_13_SEQUENCE

There are infinitely many polynomials of order 5 that will give these as the first five numbers and any one of these could be "the" rule. 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.A rule, based on a polynomial of order 4 is:U(n) = (121n^4 - 1478n^3 + 6275n^2 - 2830n + 6312)/24 for n = 1, 2, 3, ...



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https://math.answers.com/math-and-arithmetic/20_1_18_4_13_SEQUENCE

20 1 18 4 13 SEQUENCE? - Answers

There are infinitely many polynomials of order 5 that will give these as the first five numbers and any one of these could be "the" rule. 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.A rule, based on a polynomial of order 4 is:U(n) = (121n^4 - 1478n^3 + 6275n^2 - 2830n + 6312)/24 for n = 1, 2, 3, ...

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      20 1 18 4 13 SEQUENCE? - Answers
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      There are infinitely many polynomials of order 5 that will give these as the first five numbers and any one of these could be "the" rule. 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.A rule, based on a polynomial of order 4 is:U(n) = (121n^4 - 1478n^3 + 6275n^2 - 2830n + 6312)/24 for n = 1, 2, 3, ...

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