# How to asymptotically estimate a lower bound of this function?

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

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Doesn't $f(x) = e^{x/2}$ satisfy the equation? –  Ron Gordon Jan 18 '13 at 3:41
It satisfies the inequality, not the version that is an equation. –  Robert Israel Jan 18 '13 at 3:43

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