Message boards : Random stuff : Amicable pairs and what must be a stupid question

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Henrik Nilsson Send message Joined: 27 Apr 17 Posts: 9 Credit: 16,243,969 RAC: 0 |
Hi, Pardon me for asking what must be a stupid question. I suppose I'm getting the notation wrong. Regarding the amicable pair: 22559194686624505630=2*5*7*23*61*99643*2305266221 26660358080018237666=2*7*13*159773*523681*1750751 The factors of 22559194686624505630 do not sum up to 2666035808001823766, and the factors of 26660358080018237666 do not sum up to 22559194686624505630. What, then, makes this pair amicable? |

Sergei Chernykh Project administrator Project developer Send message Joined: 5 Jan 17 Posts: 455 Credit: 72,451,573 RAC: 0 |
You need to sum up all proper divisors of a number, not just factors: https://en.wikipedia.org/wiki/Aliquot_sum To quickly check that it's indeed an amicable pair: http://www.wolframalpha.com/input/?i=sigma(22559194686624505630)-22559194686624505630 http://www.wolframalpha.com/input/?i=sigma(26660358080018237666)-26660358080018237666 |

Henrik Nilsson Send message Joined: 27 Apr 17 Posts: 9 Credit: 16,243,969 RAC: 0 |
I see. Thanks! Is there, then, an easy way to list all the proper divisors of these numbers? (But yes, I can see why you choose to list only the factors.) |

Sergei Chernykh Project administrator Project developer Send message Joined: 5 Jan 17 Posts: 455 Credit: 72,451,573 RAC: 0 |
I see. Thanks! There is a formula to calculate sum of all proper divisors based on factorization: https://en.wikipedia.org/wiki/Divisor_function#Properties For your case, s(22559194686624505630) = (2+1)*(5+1)*(7+1)*(23+1)*(61+1)*(99643+1)*(2305266221+1) - 22559194686624505630 = 26660358080018237666 |

Message boards : Random stuff : Amicable pairs and what must be a stupid question

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