If $f$ is a cusp form of weight $k$ and level $\Gamma_1(N)$. For simplicity, suppose that the Fourier coefficients of $f$ at $\infty$ are in $\mathbb{Z}$. Let $\ell$ be a prime which does not divide the level $N$. Then it is a classical result of Serre that almost all (outside a set of density 0) Fourier coefficients are divisible by $\ell$.

Question :Is there any known results on the smallest index $n$ such that $\ell | a_n(f)$?

Of course I am expecting an answer in terms of the weight and level of $f$.

**How I approached the problem :** Serre's proof is basically an application of the Chebatrov density theorem and some efficient counting.

If $f$ were an eigen form, then the smallest Fourier coefficient that I am looking for is the smallest prime $p$ such that $\ell | a_p(f)$ is the smallest prime in the Chebatrov sequence (analogous to smallest primes in arithmetic progressions), for which there are really strong conditional upper bounds. But the problem here is that these upper bounds depend on the number field $K(f)$ generated by the Fourier coefficients of $f$.

So, I tried to write the cusp form $f$ as a linear combination of eigenforms and use a variant of the argument given in the previous paragraph. But as mentioned above, this again leads to considering the field $K(f_i)$ for every eigen form $f_i$ of that particular weight and level, on which I have very limited control.

Any help is greatly appreciated.

Thank you