Whoops, I missed something. My claim was true about odd positive integers, but is actually false for all even ones. The argument goes like this:
((2^n)+1) is divisible by 3 iff n is odd. To show this, one can use a pretty straightforward induction argument built around the fact that 1+2+...+2^(n-1)=2^n-1.
Nc((2^n+1)/3)=(2^(n-1))(2^n-1) for all odd integers n. (2^n+1)/3 is an integer iff n is odd, so (2^(n-1))(2^n-1) is definitely a centered nonagonal number for odd n. However, Nc, when the domain is extended to [1, infinity) in the natural way, is injective, so the only pre-image of (2^(n-1))(2^n-1) is (2^n+1)/3, which isn't an integer for even n.