Divisor function: Difference between revisions

Revision as of 04:08, 8 October 2023

Revision as of 08:22, 8 January 2024

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:<math>\begin{align}

\sigmasigma_1(n) &= \sigmasigma_1(n-1)+\sigmasigma_1(n-2)-\sigmasigma_1(n-5)-\sigmasigma_1(n-7)+\sigmasigma_1(n-12)+\sigmasigma_1(n-15)+ \cdots \\[12mu]

&= \sum_{i\in\N} (-1)^{i+1}\left( \sigmasigma_1 \left( n-\frac{1}{2} \left( 3i^2-i \right) \right) + \sigmasigma_1 \left( n-\frac{1}{2} \left( 3i^2+i \right) \right) \right),

\end{align}</math>

where <math>\sigmasigma_1(0)=n</math> if it occurs and <math>\sigmasigma_1(x)=0</math> for <math>x < 0</math>, and <math>\tfrac{1}{2} \left( 3i^2 \mp i \right)</math> are consecutive pairs of generalized [[pentagonal numbers]] ({{OEIS2C|A001318}}, starting at offset 1). Indeed, Euler proved this by logarithmic differentiation of the identity in his [[pentagonal number theorem]].

For a non-square integer, ''n'', every divisor, ''d'', of ''n'' is paired with divisor ''n''/''d'' of ''n'' and <math>\sigma_{0}(n)</math> is even; for a square integer, one divisor (namely <math>\sqrt n</math>) is not paired with a distinct divisor and <math>\sigma_{0}(n)</math> is odd. Similarly, the number <math>\sigma_{1}(n)</math> is odd if and only if ''n'' is a square or twice a square.{{sfnp|Gioia|Vaidya|1967}}