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How may we estimate $$\sum_{m=1}^n \Big(d\big(m^2\big)\Big)^2$$ where for every positive integer $m$ , $d(m)$ denotes the number of positive divisors of $m$ ?

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Are you looking for an asymptotic formula precise upper/lower bounds, or just an order of magnitude? –  Erick Wong Apr 24 '13 at 5:52
    
@ErickWong: Either an asymptotic formula , or to write the sum as $f(n) + O(g(n))$ , where $f$ and $g$ are some known functions. –  Souvik Dey Apr 24 '13 at 5:57
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Did you try Perron's formula? –  user27126 Apr 24 '13 at 6:05
    
@Sanchez: To effectively apply Perron's formula I would need a simple looking form of the generating function of $(d(n^2))^2$ , but I cann't recall anything like that. –  Souvik Dey Apr 24 '13 at 6:19
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@SouvikDey, It's definitely multiplicative. Write the generating L function as a product of some zeta function and something with analtyic continuation pass the pole at 1. This allows you to get the main term by moving the contour pass 1. –  user27126 Apr 24 '13 at 6:20

2 Answers 2

In what follows, we calculate the answer for the similar function $f(n)=d\left(n\right)^{2}$. These methods can be extended to $d(n^2)^2$ without much change. I originally misread the question, and as a result, did the calculation for $d(n)^2$. A less explicit approach for the function $d(n^2)^2$ is outlined in this answer: http://math.stackexchange.com/a/383602/6075

On prime powers, $f(p^{k})=(k+1)^{2},$ and most importantly, $f(p)=4$, and $f(p^k)$ is not too large. Let $$g(n)=\left(f*\mu*\mu*\mu*\mu\right)(n).$$ Then $g(p)=O\left(\frac{1}{p}\right),$ and so $$\sum_{n=1}^{\infty}\frac{|g(n)|}{n}=\prod_{p}\left(1+\frac{|g(p)|}{p}+O\left(\frac{1}{p^{2}}\right)\right)<\infty.$$ Now, since $g*1*1*1*1(n)=f(n),$ it follows that $$\sum_{n\leq x}f(n)=\sum_{abcde\leq x}g(a)=\sum_{a\leq x}g(a)\sum_{bcde\leq\frac{x}{a}}1.$$ Since $$\sum_{bcde\leq\frac{x}{a}}1=\sum_{n\leq\frac{x}{a}}d_{4}(n)=\frac{x}{a}\log^{4}\left(\frac{x}{a}\right)+O\left(\frac{x}{a}\log^{3}\left(\frac{x}{a}\right)\right),$$ we have that $$\sum_{n\leq x}f(n)=x\sum_{a\leq x}\frac{g(a)}{a}\log^{4}\left(\frac{x}{a}\right)+O\left(x\log^{3}(x)\sum_{a\leq x}\frac{|g(a)|}{a}\right),$$ and so $$\sum_{n\leq x}f(n)\sim cx\log^{4}(x)$$ for some constant $c>0$. consider the Dirichlet series $$F(s)=\prod_p\left(1+\frac{f(p)}{p^s}+\frac{f(p^2)}{p^{2s}}+\cdots\right).$$ Then $$\frac{F(s)}{\zeta(s)^4}=\prod_p\left(1+\frac{4}{p^s}+\frac{9}{p^{2s}}+\frac{16}{p^{3s}}+\cdots\right)\left(1-\frac{1}{p^s}\right)^4.$$ We consider $\zeta(s)^4$ since the coefficient of $p^s$ is $4$. Letting $x=\frac{1}{p^s}$, we are trying to find a nice form for $$g(x)=(1-x)^4\sum_{k=0}^\infty (k+1)^2 x^k.$$ Using generating series, we have that $$g(x)=\frac{d}{dx}\left(x\frac{d}{dx}\sum_{k=0}^{\infty}x^{k+1}\right)=\frac{d}{dx}\left(x\frac{d}{dx}\frac{x}{1-x}\right)=\frac{x+1}{\left(1-x\right)^{3}}.$$ Hence $g(x)=1-x^{2},$ and so $$F(s)=\zeta(s)^{4}\prod_{p}\left(1-\frac{1}{p^{2s}}\right)=\frac{\zeta(s)^{4}}{\zeta(2s)}.$$

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$f(p^k)=(2k+1)^2$ not $(k+1)^2$ , right ? –  Souvik Dey May 6 '13 at 5:43
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@SouvikDey yes it should be $(2k+1)^2$ –  Ethan May 6 '13 at 6:07
    
@Ethan: Yup , so you noted it too? –  Souvik Dey May 6 '13 at 6:10

Let $g(n)=d(n^2)^2$. Then we can work out the Dirichlet series $$G(s)=\sum_{n=1}^\infty g(n)n^{-s}$$ by using the same methods as was done in my previous answer regarding $f(n)=d(n)^2$. This allows us to obtain

$$G(s)=\zeta(s)^{9}H(s)$$ where $$H(s)=\prod_{p}\left(1-\frac{20}{p^{2s}}+\frac{64}{p^{3s}}-\frac{90}{p^{4s}}+\frac{64}{p^{5s}}-\frac{20}{p^{6s}}+\frac{1}{p^{8s}}\right).$$ The function $H(s)$ is extremely well behaved, and converges absolutely for all $\text{Re}(s)>\frac{1}{2}$, so it does not affect the contour integration. This allows us to prove that $$\sum_{n\leq x } d(n^2)^2\sim c x\log^9 x$$ where $c$ is a complicated constant that depends on the derivatives of the function $H(s)$, as well as the Stieltjes constants.

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