Looking for a function $f$ such that $f(i)=2(f(i-1)+f(\lceil i/2\rceil))$ I'm looking for a solution $f$ to the difference equation $$f(i)=2(f(i-1)+f(\lceil i/2\rceil))$$ with $f(2)=4$. Very grateful for any ideas. 
PS. I've tried plotting the the initial values into "Sloan", but it doesn't seem to recognize the sequence.

Since $f$ is always positive, $f(i) \gt 2f(i-1)$ and so by induction $f(n) \gt 2^n$.  $f(\lceil i/2\rceil)$ is then exponentially smaller than $f(i)$, so $2^n$ is the dominant term.  Divide out $f$ by the exponential and define $g(n) = 2^{-n}f(n)$; then $g(i) = g(i-1) + 2^{1-\lfloor i/2\rfloor}g(\lceil i/2\rceil)$, with $g(2)=1$.  It's easy to see by induction that $g(n) \lt n$ and in fact that $g(n) = O(n^\epsilon)$ for any $\epsilon$ (the differences for a series of $\Theta(n^\epsilon)$ are $\Theta(n^{\epsilon-1}) = {n^\epsilon\over n}$, which is eventually larger than $n^{\epsilon}\over2^\epsilon2^{n/2}$ for any $\epsilon$).  In fact, the form of the series loosely suggests that $g(n)$ may be some exponentially damped constant, roughly $C+\Theta(2^{-kn})$ for some $k$ and $C$; you might try from that perspective...
