Given any set of k number it is easy to find a rule based on a polynomial of order 4+k such that the first five numbers are as listed in the question and the next k are the given numbers. That is, it is easy to find a rule so that any number can be the "next" in the sequence.
So, suppose you want the next two numbers to be (5, 15) again then you define the rule as
Un = (24n5 - 400n4 + 2460n3 - 6890n2 + 8736n - 3915)/3
The simplest polynomial rule is
Un = (20n3 - 90n2 + 160n - 75)/3
and according to that, the next three numbers are 325, 655, 1165.
Alternatively, you could go for a power sequence
Un = 5*3n-1 which implies the next three terms are 405, 1215, 3645.
An order 5 polynomial, order 3 polynomial, geometric progression (plus others). There is no objective way to choose between them.
Each number is three times the previous number.
141
1 x 135, 3 x 45, 5 x 27, 9 x 15
To answer this question you first need to break 135 down into its prime factors: 135 = 3x3x3x5 In order to find two numbers that multiply to make 135, one of them must be a product of a certain combination of these numbers (and 1) and the other must have the rest. Therefore the possible pairs are: 1 and 135 3 and 45 5 and 27 9 and 15
Must be the 2 odd numbers, 9 & 15 (P = 135, S = 24)
There are no such numbers but if you meant -135 then the numbers are -9 and 15
Each number is three times the previous number.
135
Numbers such as 135, 270, 405, and other multiples of 135.
No, but these numbers can: 1, 3, 5, 9, 15, 27, 45, 135.
No, but these numbers can: 1, 3, 5, 9, 15, 27, 45, 135.
15 and 270 do.
No, just these numbers can: 1, 3, 5, 9, 15, 27, 45, 135.
Yes, by these numbers: 1, 3, 5, 9, 15, 27, 45, 135,
141
425
These numbers go evenly into 135: 1, 3, 5, 9, 15, 27, 45, 135.