searching for $f(x)$ when knowing $f(2^{2x})$ I have a function that works for powers of 2
This is only for INTEGERS
is there a way to calculate any integer x?
$$f(2^{2x})=\frac{4^x+2}3.$$
$f(x)$=?
here are the first 100 values of f(x)
1, 2, 2, 2, 2, 4, 4, 6, 6, 8, 8, 6, 6, 8, 8, 6, 6, 8, 8, 6, 6, 8, 8, \ 14, 14, 16, 16, 14, 14, 16, 16, 22, 22, 24, 24, 22, 22, 24, 24, 30, \ 30, 32, 32, 30, 30, 32, 32, 22, 22, 24, 24, 22, 22, 24, 24, 30, 30, \ 32, 32, 30, 30, 32, 32, 22, 22, 24, 24, 22, 22, 24, 24, 30, 30, 32, \ 32, 30, 30, 32, 32, 22, 22, 24, 24, 22, 22, 24, 24, 30, 30, 32, 32, \ 30, 30, 32, 32, 54, 54, 56, 56, 54 is there a pattern to calculate any f(x)?
thanx

$$f(2^{2x})=\frac{4^x+2}{3}=\frac{(2^2)^x+2}{3}=\frac{2^{2x}+2}{3}$$ so $$f(x)=\frac{x+2}{3}$$
