Not necessarily. The domain could well be restricted and, in that case, so will the range.
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There are infinitely many sequence that will fit the above four numbers so there are infinitely many possible answers. For example, the sequence could be based on the rule Un = (-7n4 + 70n3 -233n2 + 338n - 168)/12 for n = 1, 2, 3, ... It is probably based on Un = n*(n+1) for n = 1, 2, 3, ...
A: Un+1 = Un + d is recursive with common difference d.B: Un+1 = Un * r is recursive with common ratio r.C: The definition seems incomplete.A: Un+1 = Un + d is recursive with common difference d.B: Un+1 = Un * r is recursive with common ratio r.C: The definition seems incomplete.A: Un+1 = Un + d is recursive with common difference d.B: Un+1 = Un * r is recursive with common ratio r.C: The definition seems incomplete.A: Un+1 = Un + d is recursive with common difference d.B: Un+1 = Un * r is recursive with common ratio r.C: The definition seems incomplete.
Given any 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 and the next is the given number. There are also non-polynomial solutions.The simplest rule, based on a polynomial of order 5, isUn = (-4n5 + 75n4 - 510n3 + 1665n2 - 2306n + 1440)/120 for n = 1, 2, 3, ...Another possibility is a Fibonacci-like sequence withU1 = 3, U2 = 4 and Un+2 = Un + Un+1 for n = 1, 2, 3, ...
Infinitely many. For example: Un+1 = Un + 3 or Un+1 = 2*Un - 1 or Un+1 = 3*Un - 5 or, more generally, Un+1 = k*Un + 7 - 4*k where k is any number. Each one of them will be different from the third term onwards. These are linear patterns. There are quadratic and other recursive relationships.