2
$\begingroup$

Given that the sum of divisors has the form:

$$\large \sigma(n) = \sum _{k=1}^n \lim_{s\to 0} \, \left(\frac{(s+1) (-1)^{\frac{2 n}{k}}+s-1}{k \cdot s \cdot 2}\right)^{-1}$$

$$1, 3, 4, 7, 6, 12, 8, 15, 13, 18, 12, 28,...$$

and a harmonic number has the form:

$$\large H_n = \sum _{k=1}^n \lim_{s\to 0} \, \frac{(s+1)(-1)^{2 n}+s-1}{k \cdot s \cdot 2}$$

$$1,\frac{3}{2},\frac{11}{6},\frac{25}{12},\frac{137}{60},\frac{49}{20},\frac{363}{140},\frac{761}{280},\frac{7129}{2520},\frac{7381}{2520},\frac{83711}{27720},\frac{86021}{27720}...$$

can we say anything about Lagarias version of the Riemann hypothesis?

$$\sigma(n) \leq H_n + \exp (H_n) \log (H_n)$$ for $n \geq 1$

Or is the exponent $\frac{2n}{k}$ in the $\sigma(n)$ expression a dead end?

And does the formula for $H_n$ have too much form and too little content?

My hope is that one could reduce this expression down to a relationship between the exponents $\frac{2n}{k}$ and $2n$.

Edit 27.9.2014:

$$\sigma(n)=\sum _{k=1}^n k \left(\lim_{s\to 0} \, \frac{2 n}{\frac{k \left((s-1)^{\frac{2 n}{k}-1}+s+1\right)}{s}}\right)$$

$$H_n=\sum _{k=1}^n \frac{\lim_{s\to 1} \, \frac{2 n}{\frac{k \left((s-1)^{\frac{2 n}{k}-1}+s+1\right)}{s}}}{n}$$

Monitor[a = 
  Table[Sum[
    Limit[2*n/k/(((s - 1)^((2*n/k - 1)) + s + 1)/s), s -> 1]/n, {k, 1,
      n}], {n, 1, 12}], n]
Monitor[a = 
  Table[Sum[
    Limit[2*n/k/(((s - 1)^((2*n/k - 1)) + s + 1)/s), s -> 0]*k, {k, 1,
      n}], {n, 1, 12}], n]
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.