Double the previous number
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t(n) = 3(n-1) + 1, for n = 1, 2, 3, etc
+2, +3, +2
Add 3n-1. For example 1 + 30 = 2 2 + 31 = 5 5 + 32 = 14 14 + 33 = 41
The gaps are 5, 10, 15, 20, ..... The rule becomes add 5(n-1) to the previous number So that the second number is found by adding 5 times (2-1) to 1 = 5+1 = 6 and the third number is found by adding 5 times (3-1) to 6 = 10+6 = 16 and the fourth number is found by adding 5 times (4-1) to 16 = 15+15 = 31 and so on....
There are many possible answers. One possible rule is: Un = (3n4 - 18n3 + 69n2 - 54n + 120)/8 for n 1, 2, 3, ...