Amicable pairs and what must be a stupid question

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Henrik Nilsson

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Message 572 - Posted: 11 Jul 2017, 22:22:58 UTC

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?
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Sergei Chernykh
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Message 573 - Posted: 12 Jul 2017, 6:12:47 UTC
Last modified: 12 Jul 2017, 6:13:43 UTC

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Henrik Nilsson

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Message 574 - Posted: 12 Jul 2017, 9:40:43 UTC - in response to Message 573.  

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.)
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Sergei Chernykh
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Message 575 - Posted: 12 Jul 2017, 10:08:16 UTC - in response to Message 574.  
Last modified: 12 Jul 2017, 10:10:56 UTC

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.)

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
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