I met a recurrence equation for my algorithm $$ f(n) = 2\cdot \left( f(n-1) + f(\frac{n}{2}) \right)$$with $f(1)=1$, $f(2)=4$, $f(3)=10$. I guess it is $\Theta((2+\epsilon)^n)$, where $\epsilon$ ban be arbitrarily close to 0. I want to have an asymptotic formula of $f(n)$. I do not know how to prove or disprove my guess. 
Bounding the recurrence $f(n)=2f(n-1)+2f(n/2)$
asymptotics;recurrence relation
For concreteness, let's assume that your recurrence is really$$ f(n) = 2f(n-1) + f(\lfloor n/2 \rfloor) + f(\lceil n/2 \rceil). $$Define $g(n) = f(n)/2^n$. Then$$ g(n) = g(n-1) + 2^{\lfloor n/2 \rfloor + 1 - n} g(\lfloor n/2 \rfloor) + 2^{\lceil n/2 \rceil + 1 - n} g(\lceil n/2 \rceil). $$In particular,$$ g(n) \leq (1 + 2^{\lfloor n/2 \rfloor + 1 - n} + 2^{\lceil n/2 \rceil + 1 - n}) g(n-1). $$Taking logarithms,$$ \log g(n) \leq \log g(n-1) + O(2^{\lfloor n/2 \rfloor + 1 - n} + 2^{\lceil n/2 \rceil + 1 - n}). $$We conclude that$$ \log g(n) \leq O(1) + O(1) \sum_{m=c}^n [2^{\lfloor m/2 \rfloor + 1 - m} + 2^{\lceil m/2 \rceil + 1 - m}] = O(1), $$where $c$ is some arbitrary starting point.Therefore $\log g(n)$, and so $g(n)$, is bounded. Since clearly $g(n)$ is monotone increasing, we deduce that $g(n)$ tends to some constant $C$. We conclude that$$ f(n) \sim C 2^n. $$