I've been working on a new method to find 23 pairs and have developed a fairly effective approach to do so. Within two hours, it found over 1000 unique 23 pairs. 15 of which were undiscovered. Currently, I'm planning on running it until I find a few 100 new 23 pairs. Then attempt to find 24 pairs or something. If anyone wants to copy the equations I used to solve them, they are listed below. It hard to explain how they setup but a Pair is (A1*P1*P2,A1*P3*P4*P5), and the equations apply limits to those primes that as longs as you assume P1<P3<P4<P5<P2 is the order of the primes and that P2 is so larger that it forces P1 to hit it limit it max size and that the limit of P4 as P3 approaches infinity is between P3 minimum and upper bound. Which, if true, pretty much guarantees that for the A1 value you choose, you find every possible 23 pair that shares it as a common number.

A2 = divisor_sigma(A1)
P5 = ((A1*P1**2 + A2*P1 - ((A1 - A2)*P1 - A2)*P3 - ((A1 - A2)*P1 + ((A1 - A2)*P1 - A2)*P3 - A2)*P4 + A2)/((A1 - A2)*P1 + ((A1 - A2)*P1 - A2)*P3 + ((A1 - A2)*P1 + ((2*A1 - A2)*P1 + A1 - A2)*P3 - A2)*P4 - A2))
P2 = (((A1*P3**2 + A1*P3)*P4**2 + A2*P1 + (A2*P1 + A2)*P3 + (A1*P3**2 + A2*P1 - ((A1 - A2)*P1 - A2)*P3 + A2)*P4 + A2)/((A1 - A2)*P1 + ((A1 - A2)*P1 - A2)*P3 + ((A1 - A2)*P1 + ((2*A1 - A2)*P1 + A1 - A2)*P3 - A2)*P4 - A2))
P1_Max = A2/(2*A1-A2)
P4_Max = -((A1 - A2)*P1 + ((A1 - A2)*P1 - A2)*P3 - A2 - sqrt((A1**2 - A1*A2)*P1**3 - A1*A2*P1 + (A1**2 - 2*A1*A2)*P1**2 - ((A1**2 - A1*A2)*P1**2 - A1*A2 + (A1**2 - 2*A1*A2)*P1)*P3**2 + ((2*A1**2 - A1*A2)*P1**3 + A1*A2*P1 + (2*A1**2 - A1*A2)*P1**2 + A1*A2)*P3))/((A1 - A2)*P1 + ((2*A1 - A2)*P1 + A1 - A2)*P3 - A2)

P4_Limit = -((A1 - A2)*P1 - A2 - sqrt(-A1**2*P1**2 + A1*A2*P1**2 - A1**2*P1 + 2*A1*A2*P1 + A1*A2))/((2*A1 - A2)*P1 + A1 - A2)

P3_Max = -((A1 - A2)*P1 + ((A1 - A2)*P1 - A2)*P4_Limit - A2 - sqrt((A1**2 - A1*A2)*P1**3 - A1*A2*P1 + (A1**2 - 2*A1*A2)*P1**2 - ((A1**2 - A1*A2)*P1**2 - A1*A2 + (A1**2 - 2*A1*A2)*P1)*P4_Limit**2 + ((2*A1**2 - A1*A2)*P1**3 + A1*A2*P1 + (2*A1**2 - A1*A2)*P1**2 + A1*A2)*P4_Limit))/((A1 - A2)*P1 + ((2*A1 - A2)*P1 + A1 - A2)*P4_Limit - A2)

Good work! This is a similar method to what I used for finding all 21-digit pairs with the largest prime factor of the smaller number being > 10¹⁴. I just took all pairs of numbers from certain ranges and tried to match them using the similar equations to what you used for P2 and P5 (but in general form). You probably already noticed that 22-, 23-, 24- and bigger-digit known amicable numbers (up to around 36-digit) have a lot of amicable numbers with large prime factors in them. This is a by-product of that search.

P.S. Keep searching using your method - this is also a good independent check of the 10²¹ search, because your method is also exhaustive.

So I just want to say the max value I used for P1 is wrong though you can still get an overestimate for what it is using a similar method to how I defined P3. Which Is I took the equation for the limit of P4 showed that as P3 approaches infinity P4 approaches -((A1 - A2)*P1 - A2 - sqrt(-A1**2*P1**2 + A1*A2*P1**2 - A1**2*P1 + 2*A1*A2*P1 + A1*A2))/((2*A1 - A2)*P1 + A1 - A2) and since all the primes being used are interchangeable we can also say as P4 approaches infinity the max value of P3 approaches that, and to get the actual max value of P3 we plug in the lowest maximum which as long as it bigger then the largest minimum of P3 when plug into P4 will give us a number that an over estimate of the maximum value of P3. So P1 what we do is the same but we compare P3_Limit and P1 and then do the same step as before but again. we get.
P3_Limit = (A2-A1+sqrt(A1*(A2-A1)))/(2*A1-A2)
P1_Max = -((A1 - A2)*P3_Limit - A2 - sqrt(-A1**2*P3_Limit**2 + A1*A2*P3_Limit**2 - A1**2*P3_Limit + 2*A1*A2*P3_Limit + A1*A2))/((2*A1 - A2)*P3_Limit + A1 - A2)
Which as I said is an overestimate for the value of P1 but it better to overestimate it then to underestimate it.

Still think I'll stick to using P1_Max = A2/(2*A1-A2) as for every k value up to at least 500 it does not increase the amount of 2,3 pair it returns so the false starting equation I used A2*(P1+1)*(P2+1) < 2*A1*P1*P2 must be close enough to the true limit that it some just works, or maybe it is true for all 2,3 pair that A2*(P1+1)*(P2+1) < 2*A1*P1*P2 holds true. You can prove though that A2< 2*A1 it just that how do you prove that the size difference between those two is large enough that (P1+1)*(P2+1) being larger then P1*P2 doesn't break the equation.

Okay so the last two post I made were wrong I just forgot where I got A2*(P1+1)*(P2+1) > 2*A1*P1*P2 from. You still have to assume that floor((A2*P2 + A2)/(2*A1*P2-A2*P2 - A2)) = (A2/(2*A1-A2)), but in the case what you could do is solve (A2*P2 + A2)/(2*A1*P2-A2*P2 - A2)=P2 to get P2 = (A2+sqrt(2*A2*A1))/(2*A1-A2) and since you can interchange any value you can say P1 < (A2+sqrt(2*A2*A1))/(2*A1-A2). Which after checking (A2/(2*A1-A2)) isn't true but out of all know pairs (26696254263573912381,26696257801944167619) is the only one false which is only false by like a small margin, 178.996752 is limit and P1 is 179, but the finder I wrote doesn't work with floats so it just take the ceiling of the value it given to be safe so it would be able to find all know 23 pairs.

New BOINC project or a continuation of Amicable Numbers in the works here?