Message boards : Random stuff : Confused as to how the list from Jan. 23

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EBM Send message Joined: 24 Jan 17 Posts: 3 Credit: 23,010 RAC: 0 |
Hello. I recently read the list that was posted on the 23rd of January (https://sech.me/ap/log/2017/2017-01-23/VOLUNTEERS%40BOINC.txt), And for the numbers that are listed, their factors (as listed) do not add up to equal the other number. Now, clearly I read something from the linked Wikipedia article wrong but I don't see what. Can someone explain how the numbers on that text documents are Amicable? Thanks. |

Sergei Chernykh Project administrator Project developer Send message Joined: 5 Jan 17 Posts: 461 Credit: 72,451,573 RAC: 0 |
How exactly do you "add up" factors? Can you show me it with numbers 220=2^2*5*11 and 284=2^2*71? All pairs on that list are pairs of amicable number, that I'm sure of. |

Bryan Send message Joined: 23 Jan 17 Posts: 17 Credit: 278,854,007 RAC: 0 |
The factors of 220 are: 1, 2, 4, 5, 10, 11, 20, 22, 44, 55, 110The sum of the factors is 284The factors of 284 are: 1, 2, 4, 71, 142The sum of the factors is 220 |

Bryan Send message Joined: 23 Jan 17 Posts: 17 Credit: 278,854,007 RAC: 0 |
Double post |

EBM Send message Joined: 24 Jan 17 Posts: 3 Credit: 23,010 RAC: 0 |
Thanks for replying. I saw the 220 and 284 example. 220=1,2,4,5,10,11,20,22,44,55,110 (added together, they equal 284) 284=1,2,4,71,142 (added together, they equal 220) For example, on the text document (exactly as copied from the page; I do see that they are first written with a number raised to a power in some places but not here) 66 "BOINC: shiva, Bryan" 2017 1908798622056988970=2*5*7*11*23*11057*19661*495791 (added together, these equal 526557) 2561728891117125334=2*11*31*43*193*227*36307*54917 (added together, these equal 91731) Shouldn't 190879...'s factors add to equal 256172, and visa versa? Thanks. |

EBM Send message Joined: 24 Jan 17 Posts: 3 Credit: 23,010 RAC: 0 |
Is it that these are simply what you MULTIPLY to get to those numbers, and not their entire factor list? |

Bryan Send message Joined: 23 Jan 17 Posts: 17 Credit: 278,854,007 RAC: 0 |
It isn't showing ALL of the factors. If you look at the Wiki page you referenced earlier then drop down from the basic formula and look at the form of the Euler's rule. The displayed result is a amicable pair but just shown in a different format. I'm sure there are more modern and complete theorems being used now because both models mentioned by Wiki aren't all inclusive. Wiki states; "While these rules do generate some pairs of amicable numbers, many other pairs are known, so these rules are by no means comprehensive." |

Sergei Chernykh Project administrator Project developer Send message Joined: 5 Jan 17 Posts: 461 Credit: 72,451,573 RAC: 0 |
66 "BOINC: shiva, Bryan" 2017 No, sum of proper divisors (a.k.a. aliquot sum) is calculated differently: https://en.wikipedia.org/wiki/Divisor_function#Definition If you're given factorization 1908798622056988970=2*5*7*11*23*11057*19661*495791 then s(1908798622056988970) = 3*6*8*12*24*11058*19662*495792 - 1908798622056988970 = 2561728891117125334 s(2561728891117125334) = 3*12*32*44*194*228*36308*54918 - 2561728891117125334 = 1908798622056988970 |

Message boards : Random stuff : Confused as to how the list from Jan. 23

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