X11 amicable pairs

fish224 · 2023-05-10, 05:37

Today I found two amicable pairs who have the "X11" type. Is it special?

I tried to find X11 amicable pairs in the largest database (https://sech.me) but it shows that there are no 13, 14, 15 and 16 digit pairs!

The pairs I found are

6675857113019558789105248923=3^10*7*13^2*31*53*61*107*953*3851*2428243
6712540199486419700739413349=3^10*7*13*31*53*61*107*953*3851*31740617

and

16382975165472341870170867413=3^6*7*13^2*31*53*61*137*547*953*1093*2428243
16472997777461355562410834219=3^6*7*13*31*53*61*137*547*953*1093*31740617

Proof that no one has found them before: https://sech.me/ap/log/2023/2023-05-...IUKUKRAINE.txt

So, did I do something special?

P. S. Sorry if it's spamming, I am new at this forum

Andrew Usher · 2023-05-10, 23:51

No, I wouldn't call this spam at all. There's probably no better place to post it.

And X11 pairs, I would say, are notable! I'd point out, though, that that is really only one discovery - the two are related by a very familiar substitution. And now I wonder if you could try for an even one - that's even harder, but evidently possible as the database shows (are there any more differing in their 2-power? That is not obvious.).

By the way, do you know if there's anywhere to see the progress of the exhaustive search? The numbers show the search to 10^21 must be mostly complete.

garambois · 2023-05-11, 17:31

Quote:

Originally Posted by Andrew Usher

By the way, do you know if there's anywhere to see the progress of the exhaustive search? The numbers show the search to 10^21 must be mostly complete.

To my knowledge, the calculations are done with BOINC.
You can see here the progress for 10^21 : we will soon be at 69% for part 2 and the end of the calculations for this part 2 must take place in July 2024.

But that doesn't mean anything, because I think there will be a part 3 even longer than the two previous ones, see here from post 1369.

Andrew Usher · 2023-05-13, 12:02

Thanks for the link. Evidently part 3, though longest, will find the fewest pairs (which is reasonable given the definition). To see how close it is by number of pairs, calculate the ratio of those known with each successive number of digits:

10-11 2.27
11-12 2.25
12-13 2.30
13-14 2.21
14-15 2.184
15-16 2.172
16-17 2.168
17-18 2.161
18-19 2.154
19-20 2.150
20-21 1.679 so far

The cube root of 10 is 2.154, so a conjecture that proportionality with the cube root is being approached would seem to be a bit high, but I would think there must be some limit >1.

2023-05-10, 05:37	#1
fish224 "IvanBorysiukUkraine" May 2023 Switzerland 15₈ Posts	X11 amicable pairs Today I found two amicable pairs who have the "X11" type. Is it special? I tried to find X11 amicable pairs in the largest database (https://sech.me) but it shows that there are no 13, 14, 15 and 16 digit pairs! The pairs I found are 6675857113019558789105248923=3^10713^231536110795338512428243 6712540199486419700739413349=3^10713315361107953385131740617 and 16382975165472341870170867413=3^6713^231536113754795310932428243 16472997777461355562410834219=3^6713315361137547953109331740617 Proof that no one has found them before: https://sech.me/ap/log/2023/2023-05-...IUKUKRAINE.txt So, did I do something special? P. S. Sorry if it's spamming, I am new at this forum

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2023-05-10, 23:51	#2
Andrew Usher Dec 2022 3·13² Posts	No, I wouldn't call this spam at all. There's probably no better place to post it. And X11 pairs, I would say, are notable! I'd point out, though, that that is really only one discovery - the two are related by a very familiar substitution. And now I wonder if you could try for an even one - that's even harder, but evidently possible as the database shows (are there any more differing in their 2-power? That is not obvious.). By the way, do you know if there's anywhere to see the progress of the exhaustive search? The numbers show the search to 10^21 must be mostly complete.

2023-05-13, 12:02	#4
Andrew Usher Dec 2022 3·13² Posts	Thanks for the link. Evidently part 3, though longest, will find the fewest pairs (which is reasonable given the definition). To see how close it is by number of pairs, calculate the ratio of those known with each successive number of digits: 10-11 2.27 11-12 2.25 12-13 2.30 13-14 2.21 14-15 2.184 15-16 2.172 16-17 2.168 17-18 2.161 18-19 2.154 19-20 2.150 20-21 1.679 so far The cube root of 10 is 2.154, so a conjecture that proportionality with the cube root is being approached would seem to be a bit high, but I would think there must be some limit >1.