Let $f$ be a primitive form of an even weight $k\geq 2$ for the full modular group $SL_2(Z)$ and $\lambda_f(n)$ be the $n$-th normalized Fourier coefficient. Define a multiplicative function $g$ by $g(p)=2$ if $\lambda_f(p)\neq 0$ and $g(p^\nu)=0$ whenever $\nu \geq 2$ or $\lambda_f(p)=0.$ Can someone give me an upper bound in terms of $\log{x}$ for the following product: $$\prod_{p\leq x} \left(1+\frac{g(p)}{p}+\frac{g(p^2)}{p^2}+...+\frac{g(p^\nu)}{p^{\nu}}\right)=\prod_{\substack{p\leq x\\ \lambda_f(p)\neq 0}} \left(1+\frac{2}{p}\right)?$$ Many thanks.
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4$\begingroup$ What kind of bound are you looking for? Certainly one can just bound the product above by $\prod_{p \leq x} (1 + 2/p)$. It seems that this is the best one can hope for in general; if $f$ does not have CM then the set of primes $p$ for which $\lambda_f(p) = 0$ has density $0$. $\endgroup$– Daniel LoughranCommented Aug 9, 2016 at 14:00
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$\begingroup$ @Daniel Loughran By bound I mean that this product is less than an expression of $\log{x}.$ $\endgroup$– Khadija MbarkiCommented Aug 9, 2016 at 14:37
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1$\begingroup$ An argument similar to the proof of Mertens' theorem should give you what you want $\endgroup$– Stanley Yao XiaoCommented Aug 9, 2016 at 16:31
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3$\begingroup$ What Daniel Loughran and Stanley Yao Xiao are saying is the following. $\prod_{p\leq x}(1+2/p)$ is trivally an upper bound for your product, and it is asymptotically a constant times $(\log x)^2$ by standard estimates. Moreoever, this is the best one can hope in general, i.e. the condition $\lambda(p)\neq 0$ usually does not make the product much smaller. $\endgroup$– GH from MOCommented Aug 9, 2016 at 21:49
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1$\begingroup$ @KhadijaMbarki: No, this fails already for $x=2$ and $a=1$ (left hand side is $3/2$, right hand side is $1$). $\endgroup$– GH from MOCommented Sep 16, 2016 at 23:06
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