A1 = 1, a (n + 1) (subscript) = 3an + 2, find an

A1 = 1, a (n + 1) (subscript) = 3an + 2, find an

a(n+1)=3a(n)+2
If 1 is added on both sides, then a (n + 1) + 1 = 3A (n) + 3
a(n+1)+1=3[a(n)+1]
Let a (n) + 1 = B (n), then B (n + 1) = 3B (n), B (1) = 2
So B (n) is an equal ratio sequence with 2 as the first term and 3 as the common ratio, B (n) = 2 * 3 ^ (n-1)
So a (n) = B (n) - 1 = 2 * 3 ^ (n-1) - 1

A1 = 2, (n + 1) a (n + 1) (subscript) = Nan, find an

2a2=a1
3a3=2a2
4a4=3a3
.
nan=(n-1)an-1
The two sides of the equal sign are added and offset to get Nan = A1, so an = 2 / n

Level 0 to 1: 101
Levels 1 to 2: 116
Grade 2 to 3: 166
Level 3 to 4: 278
Grade 4 to 5: 475
Grade 5 to 6: 783
Grade 6 to 7: 1225

If your requirements are not very accurate, you can fit a cubic curve
y=4.1111n^3-6.5357n^2+5.2817n+98.2857
N is 1:7. After calculation, the sum of squares of the deviation of the fitting curve is 0.42. If you want to know the real formula, you have to ask the person who proposed the sequence, ha ha