Generalised Divisor Function

Number Theory Level 5

$\large A = \sum _{k=1}^{\infty }\dfrac{M_{10}(k)}{k^3}$

Given that $M_{n}(k)$ is the sum of all divisors of $k$ that is greater than $n$ , find $\left\lfloor 10000A \right\rfloor$ .

Notation : $\lfloor \cdot \rfloor$ denotes the floor function .

The answer is 1143.

1 solution

Samrat Mukhopadhyay
Sep 12, 2016

Note that $\sum_{k\ge 1}\frac{M_n(k)}{k^3}=\sum_{k\ge 1}\frac{1}{k^3}\sum_{d|k:d>n}d\\=\sum_{d\ge n+1}\sum_{m\ge 1}\frac{d}{(dm)^3}\\=\sum_{d\ge n+1}\frac{1}{d^2}\sum_{m\ge 1}\frac{1}{m^3}\\=\zeta(3)\left(\frac{\pi^2}{6}-\sum_{k=1}^n \frac{1}{k^2}\right)$ Noting that $\zeta(3)\approx 1.2021$ and putting $n=10$ , we get the answer $\boxed{1143}$ .

Generalised Divisor Function

The answer is 1143.

1 solution

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