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, ...
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
51 - 19, D16
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
1, 3, 19, 57 1, 2, 3, 6, 17, 34, 51, 102
Oh, isn't that a happy little question! Let's take a look at the factors of 51 together. The factors of 51 are 1, 3, 17, and 51. Each of these numbers can be multiplied together to give you 51, like little pieces coming together to create a beautiful painting.
1, 3, 17, 51
1, 3, 17 and 51 are the factors of 51.
Factors of 3: 1, 3 Factors of 51: 1, 3, 17, 51
158 One possible sequence is t(n) = 0.5*(19n2 - 7n + 40) for n = 1, 2, 3, ...
n = 1, 2, 3, 4, 5 a1 = 3, a2 = 7, a3 = 11, a4 = 15, a5 = 19, ..., an = 4n -1
The nth term is 4n-1 and so the next term will be 19