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I have an expression for the sum-of-divisors function defined as $$\sigma(n)=\sum_{d\mid n}d.$$ However I do not know how nontrivial or practical it actually is.

Let us define $$A=\frac{b}{2}-\frac{b}{\pi}\sum_{n=1}^\infty{\frac{\sin{\frac{2\pi nx}{b}}}{n}}=x\pmod b$$ where, when $\frac{x}{b}$ is a whole number, $A=\frac{b}{2}$

Now define $$B=\lim \limits_{n\to\infty}\frac{b}{2}-\frac{b}{\pi}\tan^{-1}\left(\cot\left(\frac{\pi x}{b}+\frac{1}{n}\right)\right)=x\pmod b$$ where, when $\frac{x}{b}$ is a whole number, $B=0$. The limit helps avoid division by zero in certain cases, but can be excluded in theory.

Now $$\sum_{b=1}^x A-\sum_{b=1}^x B=\frac{\sigma(x)}{2}.$$ So then $$\sigma(x)=2\left(\sum_{b=1}^x A-\sum_{b=1}^x B\right).$$

Can't this be considered a generating function for the $\sigma(x)$? If so, why hasn't it been seen before? Is this function somehow trivial?

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  • $\begingroup$ I'm not sure that that's what "generating function" means. A(n ordinary) generating function of $\sigma(n)$ would be the function $\sum_{n=0}^\infty\sigma(n)x^n$. (There are other types of generating function, such as exponential or Dirichlet (di-ri-CLAY, sometimes di-ri-SHLAY).) $\endgroup$ Aug 31, 2015 at 18:45
  • $\begingroup$ This classical Dirichlet series is a generating function for $\sigma_a$ (with $a=1$ in your case) :$$\zeta(s)\,\zeta(s-a)=\sum_{n=0}^\infty\frac {\sigma_a(n)}{n^s}$$ $\endgroup$ Aug 31, 2015 at 19:26

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A Generating function can be defined as a generating function is a formal power series in one indeterminate, whose coefficients contain information about a sequence of numbers $a_n$ which is indexed by the natural numbers. What you gave is not a formal power series and hence not a generating function. $\sigma(n)$ is an arithmetic function , its formal power series is the Dirichlet series which for an arithmetic function $f(n)$ is of the form $$ \ F(s)=\sum_{n=1}^{\infty}\frac{f(n)}{n^s}$$ And hence as Raymond Manzoni mentioned in his comment the generating function for $\sigma(n)$ is $$\zeta(s)\,\zeta(s-1)=\sum_{n=0}^\infty\frac {\sigma_1(n)}{n^s}$$

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