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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
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1 Answer 1

up vote 1 down vote accepted

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.

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Sorry I was not clear enough before. Actually I was looking for some good lower bound for this function. –  Jiajun Wu Jan 18 '13 at 4:55
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