Currently @ 7960 with 2^3*3 driver (ouch!). Well, at least, it isn't the 2^3*3*5 driver.

I'm still whacking away at it, now that factordb is back up (and being careful not to put too much load on the site myself).

Any predictions/estimates for how long the 2^3*3 driver will persist?

[QUOTE=ryanp;442314]I'm still whacking away at it, now that factordb is back up (and being careful not to put too much load on the site myself).

Any predictions/estimates for how long the 2^3*3 driver will persist?[/QUOTE]

well until we get a factorization that doesn't lead to a multiple of 24 would be the logical place to start. I don't know the conditions for such to happen but I bet someone does.

[QUOTE=ryanp;442314]
Any predictions/estimates for how long the 2^3*3 driver will persist?[/QUOTE]
The 3 can not disappear. That is because all the terms of the sigma which are not multiple of 3 are 1, 2, 4, 8, and F* times 1, 2, 4, 8, where F* is all possible combinations of factors in F, which are not 2 or 3. (i.e. the factorization 2^3*3^x*F, where x>0 and F is a product of primes not 2, 3).
You can lose a 2 or gain a 2 (or more) in special cases, which depend of the size of the numbers. For 10 digits in F, you have about 18% to lose the exponent 3 of 2^3. For 20 digits numbers, your chances go close to 10% (below it).
For 30 digits numbers in F, the chances to get a sigma 2^x*3^y*F' with x!=3 and y>=1 are about 7%.
If you go higher, your chances to lose the 2^3 get slimmer.

If you limit y to 1, your chances to lose 2^3 are about 3.5%, at 20 digits.
If y=2, your chances to lose 2^3 are about 22% at 20 digits.
If y=3 or larger, you get again, 10% at 20 digits.
They also get slimmer for larger numbers.

[QUOTE=science_man_88;442316]well until we get a factorization that doesn't lead to a multiple of 24 would be the logical place to start. I don't know the conditions for such to happen but I bet someone does.[/QUOTE]Not exactly true, this driver usually breaks by getting a multiple of 48, which is also a multiple of 24, see my comment above. You can not lose the 3 here.

edit: I know the numbers because I am in kinda' the same shit with my [URL="http://factordb.com/sequences.php?se=1&eff=2&aq=225900&action=last20&fr=0&to=100"]225900[/URL], to which I am currently working actively, see its evolution in the last weeks.

[QUOTE=ryanp;442314]

Any predictions/estimates for how long the 2^3*3 driver will persist?[/QUOTE]

I don't have a prediction or estimate.

But I feel that a YouTube video from The Sound of Music (1965) of Peggy Wood as Mother Abbess, dubbed by Margery McKay, singing "Climb Ev'ry Mountain" is appropriate.

[url]https://www.youtube.com/watch?v=EoCPuhhE6dw[/url]

That could be taken the wrong way for a huge number of reasons, but it's meant in the spirit of encouragement.

[QUOTE=LaurV;442328]Not exactly true, this driver usually breaks by getting a multiple of 48, which is also a multiple of 24, see my comment above. You can not lose the 3 here.

edit: I know the numbers because I am in kinda' the same shit with my [URL="http://factordb.com/sequences.php?se=1&eff=2&aq=225900&action=last20&fr=0&to=100"]225900[/URL], to which I am currently working actively, see its evolution in the last weeks.[/QUOTE]

ah but you use usually that's the thing that means it doesn't always break it because it hasn't escaped being a multiple of 24 yet.

[QUOTE=LaurV;442325]The 3 can not disappear. That is because all the terms of the sigma which are not multiple of 3 are 1, 2, 4, 8, and F* times 1, 2, 4, 8, where F* is all possible combinations of factors in F, which are not 2 or 3. (i.e. the factorization 2^3*3^x*F, where x>0 and F is a product of primes not 2, 3).[/QUOTE]

I don't know anything about the math behind Aliquot Sequences, but it seems like it has lost a 3 several times before near the peaks in the graph?

Are these situations special because of the 2^6 and 2^2 infront of the 3? Is it only the 3 after 2^3 that cannot be lost?

[CODE]Step 4064: 2^6 · 3^3 · 11^2 · 2280345883<10> · 17383086933815011<17> · 155927685066589177241898435529<30>
Step 4065: 2^6 · 67 · 17207650673<11> · 539744611637742767<18> · 72407948320983644687137288253<29>

Step 5306: 2^2 · 3 · 23 · 43 · 2113 · 1124214194939693956866751<25> · 1021217636...83<169>
Step 5307: 2^2 · 3^3 · 139 · 877 · 1614463 · 966231493 · 1532527068...71<71> · 1365298954...49<108>
Step 5308: 2^2 · 17 · 53 · 5519 · 20879 · 23912690842817779000064290538036192241983<41> · 3832885252...49<63> · 1822375450...77<86>

Step 6691: 2^2 · 3 · 1297 · 43920156157693<14> · 1500555007933194276969235039038229<34> · 64626509745706158780051904878300730111033<41>
Step 6692: 2^2 · 2212646575...89<92>

Step 7245: 2^2 · 3^2 · 7 · 43 · 1181 · 68053 · 2850476466709771<16>
Step 7246: 2^2 · 3 · 7 · 37 · 1489 · 226564231 · 4637314011607<13>
Step 7247: 2^2 · 7 · 11 · 1619 · 32810003 · 517277073001771<15>[/CODE]

Yes. This is due to sigma(2^3)=15=3*5
It is impossible for us to gain or lose a 5. Our only hope is to change the power of 2.

Ok, but if the power of 2 changes then we can loose the 3 later.

I misunderstood it as it was never going away again, which did not match the previous steps.

[QUOTE=mshelikoff;442335]I don't have a prediction or estimate.

But I feel that a YouTube video from The Sound of Music (1965) of Peggy Wood as Mother Abbess, dubbed by Margery McKay, singing "Climb Ev'ry Mountain" is appropriate.

[url]https://www.youtube.com/watch?v=EoCPuhhE6dw[/url]

That could be taken the wrong way for a huge number of reasons, but it's meant in the spirit of encouragement.[/QUOTE]
One can also "climb ev'ry mountain" in this problem - [url]https://projecteuler.net/problem=569[/url]
...again, in the spirit of encouragement. :rolleyes: