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Oct 5 at 16:02 vote accept Khadija Mbarki
Oct 5 at 1:39 comment added Gerry Myerson "My question is ... how to find a good bound for the sums on the right side ...." There is no sum on the right side. Presumably, "right" should be "left".
Oct 4 at 15:16 answer added Ofir Gorodetsky timeline score: 5
Oct 3 at 15:41 comment added Khadija Mbarki See mathoverflow.net/questions/479974/sum-over-three-squares
Oct 3 at 15:15 comment added Steven Clark I haven't been able to find an explicit or asymptotic formula, but the growth of $$\sum\limits_{\substack{n\leq x \\ n=a^2+b^2}} 1$$ and $$\sum\limits_{\substack{n\leq x \\ n=a^2+b^2+c^2}} 1$$ both seem to be nearly linear.
Oct 2 at 21:25 comment added Khadija Mbarki @steven Clark is there an explicit or asymptotic formula for the sum $\sum_{n\leq x, n=a^2+b^2+c^2} 1$ ?
Oct 2 at 15:48 comment added Steven Clark OEIS Entry A071377 gives number of positive integers $\le n$ which are the sum of $3$ squares, which is the partial sum of OEIS Entry A071374 (with its zeroth term omitted). OEIS Entry A071374 is defned as $0$ iff $n$ is of the form $4^a\, (8 k+7)$, otherwise 1.
Oct 2 at 11:52 comment added Khadija Mbarki @StevenClark have you an idea how to express the sums over integers written as sum of three squares?
Oct 1 at 22:11 history edited LSpice CC BY-SA 4.0
Link to paper; deleted "thanks"
Oct 1 at 21:43 comment added Steven Clark I believe $\sum\limits_{n \leq x} \beta(n)$ corresponds to OEIS entry A102548 which gives links to a few more related papers.
Oct 1 at 21:34 comment added Steven Clark @OfirGorodetsky Thanks, I'm looking forward to reading your answer!
Oct 1 at 21:02 comment added Ofir Gorodetsky @StevenClark Yes and yes. One can write $F(s)$ as $\sqrt{\zeta_{\mathbb{Q}(i)}(s)}$ times $G(s)$, where $\zeta_{\mathbb{Q}(i)}(s)$ is the Dedekind zeta function of $\mathbb{Q}(i)$, itself factoring as $\zeta(s)L(s,\chi_{-4})$, and $G$ is given by a Dirichlet series converging absolutely and bounded in $\Re(s)\ge \tfrac{1}{2}+\varepsilon$. In fact, one can write $G(s)$ as an infinite product involving $\zeta(2^i s)$ and $L(2^i s, \chi_{-4})$ with $i\ge 1$. I intend to post an answer later on.
Oct 1 at 19:16 comment added Steven Clark @OfirGorodetsky Did you mean $F(s)=\sum\limits_{n\ge 1} \frac{\beta(n)}{n^s}$ (i.e. $\beta(n)$ versus $\beta(s)$)? Do you know if $F(s)$ can be expressed in terms of the Riemann zeta function $\zeta(s)$?
Oct 1 at 14:53 comment added Ofir Gorodetsky The function $\lambda(s)$ is given by $\sqrt{s F(1-s)}/(\pi (1-s))$ where $F(s)=\sum_{n\ge 1} \beta(s)/n^s$ (this definition relies on $F(s)\sqrt{1-s}$ having an analytic continuation to, say, $|s|<1$). The integral on the RHS enjoys an asymptotic expansion that starts with $Kx/\sqrt{\log x} (1+O(1/\log x))$ for a positive constant $K$ known as the Landau-Ramanujan constant. This constant is essentially $\lambda(0)$ (maybe divided by a simple factor such as $\pi$).
Oct 1 at 13:48 history edited Daniele Tampieri CC BY-SA 4.0
Minor Math Jaxing and grammar improvement.
Oct 1 at 13:12 history asked Khadija Mbarki CC BY-SA 4.0