Aliquot sequences that start on the integer powers n^i

[Happy5214]

Originally Posted by garambois

Yes, we wrote our posts at the same time, I added point 6) in my post #1307 !
Can you just tell me if the word "modulo" is correct please and if the proofs of conjectures 82, 103 and 116 are written in an acceptable way ?

I would just write "mod" rather than modulo, though "modulo" would be the correct pronunciation of "mod", and I would parenthesize it (i.e. "= 1 (mod #)"). It really should use an equivalence operator (three lines) instead of an equals, but that might be too much work. And don't forget the <var> tags around "k" (that's where the italics come from). You may want to consider using centered dots instead of periods. And don't forget about the similar conjecture 45!

As an aside, I'd eventually like to reformat the conjecture statements to use proper superscripts and <var> tags.

[Happy5214]

210^56 terminates at a prime. I'll let you know if I can get to any others before Monday.

[garambois]

Originally Posted by Happy5214

I would just write "mod" rather than modulo, though "modulo" would be the correct pronunciation of "mod", and I would parenthesize it (i.e. "= 1 (mod #)"). It really should use an equivalence operator (three lines) instead of an equals, but that might be too much work. And don't forget the <var> tags around "k" (that's where the italics come from). You may want to consider using centered dots instead of periods. And don't forget about the similar conjecture 45!

OK, thank you very much for your help !
I think I've corrected all that except for what you say :
"You may want to consider using centered dots instead of periods."
I don't understand what that means ?

Originally Posted by Happy5214

As an aside, I'd eventually like to reformat the conjecture statements to use proper superscripts and <var> tags.

This is a job that requires much more time !
I can't get into it right now...

[Happy5214]

Originally Posted by garambois

OK, thank you very much for your help !
I think I've corrected all that except for what you say :
"You may want to consider using centered dots instead of periods."
I don't understand what that means ?

Eventually, I want to convert everything to TeX using MathJax, so I can do it then.

Originally Posted by garambois

This is a job that requires much more time !
I can't get into it right now...

That can actually be done using a regex-capable text editor (or sed) relatively quickly. I think I did most of the <var> tags already, so it would just leave the superscripts.

[Happy5214]

I completed the trivial sequences for the two remaining Lehmer five bases up to 100 digits, so they'd be in by the Monday deadline.

[henryzz]

Originally Posted by garambois

OK, excellent !
Don't forget to let me know if you've finished initializing this base by Monday, August 9 at 6:00 PM (GMT) !

I have taken 720 up to >100 digits and 90 digit composites for n up to 40.

[garambois]

Thank you all for your recent (and older !) calculations.

I've recapped everything for tonight's big FactorDB scan that will start after 6 PM GMT.
Here is the input table for my program that will be considered for this scan :

basepuis=[[2,559],[3,335],[5,250],[6,210],[7,190],[10,160],[11,160],[12,150],[13,145],[14,140],[15,140],[17,140],[18,140],[19,140],[20,100],[21,100],[22,100],[23,100],[24,100],[26,100],[28,100],[29,100],[30,100],[31,100],[33,100],[34,100],[35,100],[37,100],[38,100],[39,100],[40,100],[41,100],[42,100],[43,100],[44,100],[45,100],[46,100],[47,100],[48,100],[50,100],[51,100],[52,100],[53,100],[54,100],[55,100],[58,100],[62,100],[72,100],[74,100],[75,100],[79,100],[98,100],[105,90],[162,80],[200,80],[210,80],[211,80],[220,80],[231,80],[242,70],[276,70],[284,70],[288,60],[338,60],[385,60],[392,60],[439,60],[450,60],[496,60],[552,60],[564,60],[578,60],[660,60],[720,60],[722,60],[770,60],[882,60],[966,60],[1155,50],[2310,50],[8128,40],[12496,40],[14288,40],[14316,40],[14264,40],[14536,40],[15472,40],[15015,40],[19116,40],[30030,40],[510510,30],[9699690,25],[33550336,25],[82589933,20],[223092870,20],[6469693230,20],[8589869056,20],[10^10+19,15],[200560490130,14],[7420738134810,14]]

There are exactly 100 input bases, the 90 listed on the project page, plus 10 new bases that you are calculating or that you have initialized very recently : 51, 52, 54, 55, 276, 552, 564, 660, 720 and 966.
Of course, these 10 new bases will also be added on the project page, hopefully by the end of the week, if I have time !

Please check if the last bases you have calculated will be taken into account for the work this summer, and if no, let me know before the date and time specified above !

I will scan these 100 bases once.
My program will also check for all sequences to see if any are broken.
I hope I won't have such a bad surprise !!!

[garambois]

I just want to ask two questions about the demonstrations of the conjectures stated around this project.

1) I am totally unable to do it myself, but wouldn't it be possible to prove conjecture (10) by a method similar to the one used for conjecture (2) ? (see henryzz's post #476 and warachwe's post #480).
Indeed, 2^12-1 = 3^2*5*7*13 and we do have the 13 which preserves the 7 for the second iteration, except perhaps for some values of k that I did not examine, but the whole problem is precisely there, isn't it ?

2) At the end of the quote below, Alexander says: "(the composite ones will be harder)".
But does this mean that no one knows how to do such demonstrations yet, or does it just mean that we know how to do them, but that it is very long and tedious ?

Originally Posted by Happy5214

I updated the conjectures page to add some of the more trivial proofs. Specifically, I added proofs for index 1 conjectures for bases 2, 3, and 5, which all use algebraic factors. You could probably use that template and FactorDB to fill in the rest of the index 1 conjecture proofs for the remaining prime bases (the composite ones will be harder).

[bur]

I finished (2*18^2)^n = (648^2)^n up to n = 50.

For 50 < n < 60 the inial term as has 144-169 digits and since every odd n has factor 3 they decrease very slowly. So I think I'll stop on that base now that it's at a nice n=50.

[Happy5214]

Originally Posted by garambois

Thank you all for your recent (and older !) calculations.

I've recapped everything for tonight's big FactorDB scan that will start after 6 PM GMT.
Here is the input table for my program that will be considered for this scan :

[...]

Please check if the last bases you have calculated will be taken into account for the work this summer, and if no, let me know before the date and time specified above !

I did the trivial sequences for the remaining three-digit start values that merge into the Lehmer five sequences, up to a maximum exponent of 30. I don't know if you want to include those too:

306, 396, 696, 780, 828, 888, 996

[warachwe]

Originally Posted by garambois

1) I am totally unable to do it myself, but wouldn't it be possible to prove conjecture (10) by a method similar to the one used for conjecture (2) ? (see henryzz's post #476 and warachwe's post #480).
Indeed, 2^12-1 = 3^2*5*7*13 and we do have the 13 which preserves the 7 for the second iteration, except perhaps for some values of k that I did not examine, but the whole problem is precisely there, isn't it ?

The problem is that when the factor 13 are with even power, it does not preserve the 7 for the second iteration.
This only happen when k is multiple of 13, for example 2^(12*13)-1 =3^2*5*7*13^2*53*79*...

This is why conjecture 34 ( 3^(18*37) ), 35 ( 3^(36*37) ), and 106 ( 11^(6*37) ) are false.

But this doesn't mean all similar conjecture are false, as there maybe others prime(s) that preserve p.

When we try to 'get rid of' those primes, there maybe yet another that will preserve p instead. Since the size of first iteration grow very quickly, it is hard to find other contradiction this way.

If some of those are true, I imagine the proof might be similar to the proof of conjecture (2).