Timeline for given prime factors of a number, what is the fastest way to calculate it's proper divisors? [closed]
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Oct 1, 2012 at 18:29 | vote | accept | S L | ||
| Oct 1, 2012 at 10:56 | history | closed |
Peter Taylor Gareth Matt boothby Ventero |
off topic | |
| Sep 30, 2012 at 8:23 | answer | added | o_o | timeline score: 1 | |
| Sep 27, 2012 at 21:03 | review | Close votes | |||
| S Oct 1, 2012 at 10:56 | |||||
| Sep 27, 2012 at 21:03 | answer | added | DavidC | timeline score: 1 | |
| Sep 27, 2012 at 18:24 | comment | added | mellamokb |
I don't know Matlab, but given factors a_0^k_0 * a_1^k_1 * ... * a_n^k_n, the sum of the proper divisors is given as (a_0^0 + a_0^1 + ... + a_0^k_0) * (a_1^0 + a_1^1 + ... + a_1^k_1) * ... * (a_n^0 + a_n^1 + ... + a_n^k_n), which should be a pretty fast formula to calculate manually.
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| Sep 27, 2012 at 17:33 | review | First posts | |||
| S Oct 1, 2012 at 10:56 | |||||
| Sep 27, 2012 at 17:29 | history | asked | S L | CC BY-SA 3.0 |