The function is given as
$$f(x)\geq \sum_{i=1}^{[x/2]}f(i)+1$$
The boundary condition is $f(0)=0$.
What I can get is this function grows faster than any polynomial function, and grows slower than any exponential function.
Apparently this function is monotonic, estimate in the integral form gives
$$f(x)\geq \int_{i=0}^{x/2-1}f(t)dt+1$$ must be satisfied.
I also tried put $\geq$ as equality and solve the ODE obtained, which looks like $f(x)=2f'(2x)$. But I could not go any further.
So is there some way to estimate the growth of this function? More specifically, I would like know how to find some asymptotic lower bound which is better than polynomials. Thanks
With $\ge$, this does not define the function uniquely. It might grow exponentially or faster, e.g. $2^x$ satisfies the inequality.
For the version with $=$, see http://oeis.org/A018819
EDIT: This is the version of http://oeis.org/A000123 with terms repeated. Precise asymptotics are given in N. G. de Bruijn, "On Mahler's partition problem", Indagationes Mathematicae, vol. X (1948), 210-220.