I did some computations but I am stuck in finding the exression of the sum $$\lambda_f(n^2)+\lambda_f(n)^2 $$ in terms of $\lambda_f(n),$ where $f$ is a modular form for the full modular group. Any help is appreciated.
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- 1$\begingroup$ I would think that for $n$ composite it is not generally possible to do this. $\endgroup$paul garrett– paul garrett2015-08-15 22:01:28 +00:00Commented Aug 15, 2015 at 22:01
- $\begingroup$ @paulgarrett is it possible when $n=p^r$ is a power of prime number? $\endgroup$Khadija Mbarki– Khadija Mbarki2015-08-15 22:12:38 +00:00Commented Aug 15, 2015 at 22:12
- $\begingroup$ For prime powers $p^\ell$, the corresponding eigenvalue is of the form (a trace, thinking of Shintani-Cassleman-Shalika... formulas) $(a^{\ell+1}-b^{\ell+1})/(a-b)$, where the product $ab$ is normalized in some way, without much loss of generality to $ab=1$ (just depending on the "weight"). So whatever identities follow from this... Does this respond to the question? $\endgroup$paul garrett– paul garrett2015-08-15 22:18:26 +00:00Commented Aug 15, 2015 at 22:18
- $\begingroup$ @paulgarrett Thank you for your remark but it would be better to think about multiplicativity. I shall explain to you the main problem is to transform the following sum: $$\sum_{n=1}^{+\infty}\lambda_f(n)\times[\lambda_f(n^2)+\lambda_f(n)^2]$$ to a product .So if the function $\lambda_f(n)\times[\lambda_f(n^2)+\lambda_f(n)^2]$ is multiplicative then I can express this sum as an Euler product and this respond to my question. $\endgroup$Khadija Mbarki– Khadija Mbarki2015-08-15 22:28:12 +00:00Commented Aug 15, 2015 at 22:28
- 1$\begingroup$ It seems to me that the sum in each term is a problem. $\endgroup$paul garrett– paul garrett2015-08-15 23:00:04 +00:00Commented Aug 15, 2015 at 23:00
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