There are infinitely many rules. It is easy to find a polynomial of order 5 (or higher) such that the first five numbers are as listed in the question followed by any number that you wish. 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 of order 4 is:
T(n) = (n^4 - 6n^3 + 15n^2 - 10n + 4)/4 for n = 1, 2, 3, ...
The formula is an=3+(n-1)(4).
158 One possible sequence is t(n) = 0.5*(19n2 - 7n + 40) for n = 1, 2, 3, ...
57 can be divided by 3 and 19 with no remainder. 51 can be divided by 3 and 17 with no remainder.
No. Neither one are. 51 = 17 x 3 57 = 19 x 3
It can be: 3*17 = 51 or 1*51 = 51
1, 3, 17, 19, 51, 57, 323, 969.
The factors of 51 are: 1, 3, 17, 51The factors of 57 are: 1, 3, 19, 57
Factors of 38 are 1, 2, 19, and 38.Factors of 51 are 1, 3, 17, and 51.
1, 3, 19, 57 1, 3, 37, 111
The formula is an=3+(n-1)(4).
1, 3, 19, 57 1, 2, 3, 6, 17, 34, 51, 102
158 One possible sequence is t(n) = 0.5*(19n2 - 7n + 40) for n = 1, 2, 3, ...
1, 3, 17, 51
1, 3, 17 and 51 are the factors of 51.
1, 3, 17, 51
Factors of 3: 1, 3 Factors of 51: 1, 3, 17, 51
n = 1, 2, 3, 4, 5 a1 = 3, a2 = 7, a3 = 11, a4 = 15, a5 = 19, ..., an = 4n -1