Is this algebra simple? I have this algebra $A$, defined on the elements of the basis $ \left\{ L_{k}\right\} _{k\in\mathbb{Z}} $ with a bilinear product such that
$$L_{a}\circ L_{b}=\frac{1}{2}\left(L_{\phi(a,b)+2a-b}+L_{-\phi(a,b)+2b-a}\right),$$
where $\phi(a,b)=-\phi(b,a)$ is an integer number such that $\phi(a,b)\sim -2(a-b)/3$. Unfortunately $\phi(a,b)$ does not have a definite law, the only thing that is known is that the integer is bounded by $$−2(a-b)/3−3≤\phi(a,b)≤−2(a-b)/3+3 $$ and that letting $a,b$ grow to infinity we have $\phi(a,b)\sim -2(a-b)/3$.
Essentially you can think of $\phi(a,b)$ as $\phi(a,b)=2(b−a)/3$ plus or minus something that makes it an integer
Now, I have reasons to say that this algebra is simple since suppose you have an ideal $I$ then $L_{a}\circ I$ has to be contained in $I$ for all  $L_{a}$ and it seems to me that this will always end up implying $I=A$. Nevertheless I cannot make this formulation precise. Does anybody have an idea on how to proof that?
Remark: note that if
$$L_{a}\circ L_{b}=\frac{1}{2}\left(L_{n}+L_{m}\right).$$
Then $a+b=n+m$. This identity was useful for other things, so even is almost trivial I wanted to pointed out.
 A: The conditions on $\phi$ stated are NOT enough to show that the algebra is simple. We can pick, see below, our own $\phi$ (within the constraints) for which $A$ has an ideal. Now I get the impression that you are dealing with a concrete (though hard to compute) $\phi$ coming from an application, so below results do not rule out that $A$ is simple in your case. However it means that you would need to extract more of its secrets before you can start hoping for a simple $A$ again.

Following the notation in your useful formula we define
$$n(a, b) = \phi(a, b) + 2a - b; \quad m(a, b) = -\phi(a, b) + 2b - a$$
so that
$$L_a \circ L_b = 1/2(L_{n(a, b)} + L_{m(a,b)}).$$
As remarked in the comments we have that $m(a, b) = n(b, a)$ so that
$$L_a \circ L_b = L_b \circ L_a$$
and $A$ is commutative.
This in turn means that the right ideal $L_0A$ is in fact a two sided ideal.

If $A$ is simple the two-sided ideal $L_0A$ must be all of $A$, hence contain the element $L_{17}$. In other words, there must be a solution $X$ of the equation
$$L_0 \circ X = L_{17} \qquad \qquad {(1)}$$
Any $X$ is of the form $\sum_{i = 1}^k \lambda_i L_{b_i}$ for a finite set of $b_i \in \mathbb{Z}$ and $\lambda_i \in \mathbb{C}$. The left hand side of $(1)$ is then a finite linear combination of the basis elements $L_{m(0, b_i)}$ and $L_{n(0, b_i)}$ and so in order to have a solution of (*) we  at the very least must have that one $m(0, b_1), \ldots, m(0, b_k)$, $n(0, b_1), \ldots, n(0, b_k)$ equals 17.

Hence, in order to create a counter-example, all we need to do is device a $\phi$ so that no $m(0, b)$ and no $n(0, b)$ equals 17. Since there are at most 7 (positive) values of $b$ for which there is any hope anyway that $m(0, b) = 17$ and at most 7 (negative) values of $b$ for which we might dream that $n(0, b)$ is 17, this is rather easily achieved.
We choose the values of these 14 $\phi(0, b)$ carefully so that no $n(0, b)$ and no $m(0, b)$ equals 17 and fill in the other values of $\phi(a, b)$ as we please. Then the $\phi$ we constructed yields an algebra $A$ in which the ideal $L_0A$ is proper.

The above suggests an easy necessary condition for $A$ to be simple: for each $z$ (playing the role of 17 in the above) there must be a $b \in \mathbb{Z}$ so that $m(0, b) = z$ or $n(0, b) = z$. However, this condition only guarantees that the ideal $L_0A$ contains an element of the form
$$L_{17} + \textrm{ something.}$$
If we want it to contain $L_{17}$ itself we get a more conditions. Staring a bit at the situation we conclude that in order for $X \mapsto L_0 \circ X$ to be surjective, we need that each $z \in \mathbb{Z}$ appears as $m(0, b)$ or $n(0, b)$ in at least two ways.

Moreover, the above only talks about the ideal $L_0A$. But we also need $L_1A, L_2A$ etc to be all of $A$ in order for $A$ to be simple. Packing everything together we conclude that in order for $A$ to be simple we must at least satisfy the following
NECESSARY CONDITION:
For each $a, z \in \mathbb{Z}$ there are $b, c \in \mathbb{Z}$ (possibly equal) such that at least one of the following three scenarios is fulfilled:

*

*$m(a, b) = m(a, c) = z$

*$m(a, b) = n(a, c) = z$

*$n(a, b) = n(a, c) = z$
I do not claim this necessary condition is also sufficient.
