Let's put the values in order:
x |
y |
1 |
28 |
4 |
13 |
5 |
41 |
7 |
25 |
13 |
58 |
The y values do not increase as x increases, and the x values do not increase in steps of 1.
If we write y=f(x) then f(5)=f(1)+f(4) because 41=28+13. This could mean f(a+b)=f(a)+f(b). If we apply this to find f(3), we get f(4)=f(1)+f(3), so f(3)=f(4)-f(1)=13-28=-15. f(2)=f(1)+f(1)=56 and f(5)=f(3)+f(2)=-15+56=41. f(7)=f(4)+f(3)=13-15=-2, which isn't true. Also f(7)=f(5)+f(2)=41+56=97, not true. So the rule fails. f(a+b)=f(a)+f(b) where a not equal to b. This is a modification which might work. Apply it: f(6)=f(1)+f(5)=28+41=69. But it's clear this can't work because f(7) doesn't equal f(1)+f(6). Try another rule.
The next try is to expand the table so that we have a column for x^2 and x^3 (to start with) so that we can consider polynomials.
x |
x^2 |
x^3 |
y |
1 |
1 |
1 |
28 |
4 |
16 |
64 |
13 |
5 |
25 |
125 |
41 |
7 |
49 |
343 |
25 |
13 |
169 |
2197 |
58 |
More...