There are functions with f (n) = f (2n) which can't be calculated
Cook up some encoding of Turing machines so that if $n$ codes a machine, $2 n$ codes "the same" (perhaps use binary numbers ending in 1 as starting points, and 0s at the end are disregarded, thus making that odd $n$ and $n \cdot 2^k$ represent the same machine). Then use that e.g. $\operatorname{HALT}(M)$ (does $M$ halt if started on an empty tape?) isn't computable.