Reserved for MF - Sequence 4788

[axn]

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

[ryanp]

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?

[science_man_88]

Originally Posted by ryanp

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?

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.

[LaurV]

Originally Posted by ryanp

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

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.

[LaurV]

Originally Posted by science_man_88

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.

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 225900, to which I am currently working actively, see its evolution in the last weeks.

[mshelikoff]

Originally Posted by ryanp

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

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.

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

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

[science_man_88]

Originally Posted by LaurV

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 225900, to which I am currently working actively, see its evolution in the last weeks.

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.

[ATH]

Originally Posted by LaurV

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).

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?

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>

[henryzz]

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.

[ATH]

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.

[Batalov]

Originally Posted by mshelikoff

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.

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

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

One can also "climb ev'ry mountain" in this problem - https://projecteuler.net/problem=569
...again, in the spirit of encouragement.