Supplementary question to interesting problem post

[jdp]

Supplementary question to "interesting problem" post

If f(2x)=f(f(x))
and f(2x+1)=f(2x)+1

then for what value n such that n is in the set of natural numbers could f(0) equal 2^n.

also for what value n does f(0) equal 2^n +2?

 

[Bacle2]

Why don't you get some data by working with simple values for x, like 1,0, etc?

You get : f(0)=f(f(0))
f(1)=f(0)+1

Use x=-1/2 , then f(-1)=f(f(-1/2)
f(0)=f(-1)+1

for x=-1 , f(-1)=f(-2)+1 ...