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2021-03-04, 09:08   #936
garambois

"Garambois Jean-Luc"
Oct 2011
France

2×32×31 Posts

Quote:
 Originally Posted by Happy5214 You know, you don't have to copy the forum posts exactly. You might want to reorganize it. Maybe put the conjectures in one section and the remarks and other stuff in a separate section.

I will probably put the page on my website as shown in post # 925, unless you still make some changes in the layout or in the tags, area where I don't understand much.
I will also have less time in the next few days, because unfortunately my vacation is going to end. I can always edit the page later.

2021-03-04, 09:24   #937
garambois

"Garambois Jean-Luc"
Oct 2011
France

10001011102 Posts

Quote:
 Originally Posted by Happy5214 All of those powers seem have the factors 3*5*7*13 in the first term. While not abundant on its own (3*5*7*13 = 1365, while its aliquot sum is 1323), it is very nearly so. I wasted nearly 3 hours trying to find a formula to derive the sequence of first terms from the previous powers to write an inductive modular proof that all base 38 powers that are multiples of 12 have those factors, with no success. The rule does hold to 38^300, though, and thus they are likely to be abundant.

Thank you for this demonstration attempt !
Me, I am focusing on the end of remark 2 of post #921.
OK, now the calculations have been done for base 38.
But why do we have nothing for bases 6, 12, 24, 30, 72 ...?
These are bases which have the factor 3 in their decomposition.
But there are others who have the factor 3 and who have something: 18, 42, 882 ...
I look at the data over and over again !

2021-03-04, 11:54   #938
Happy5214

"Alexander"
Nov 2008
The Alamo City

24016 Posts

Quote:
 Originally Posted by garambois I will probably put the page on my website as shown in post # 925, unless you still make some changes in the layout or in the tags, area where I don't understand much. I will also have less time in the next few days, because unfortunately my vacation is going to end. I can always edit the page later.
Oh, I tried to mock up an example (not full) layout, but forgot to post it. It's attached here.

PS Regarding above, I did end up finding the formulas, but still couldn't complete the inductive proof. I'll post the formulas here if anyone wants to try their hand at it:

Code:
x = ((19^(12*(n+1)+1))-((19^(12*n+1))))/18
y = 2^(12*(n+1)+1)-2^(12*n+1)
z = (((19^(12*(n+1)+1))-((19^(12*n+1))))/18) * (2^(12*(n+1))-2)
w = (2^(12*(n+1)+1)-2^(12*n+1)) * (((19^((12*n)+1))-19)/18)
v = 2^(12*(n+1)) * (((19^(12*(n+1)))-((19^(12*n+1))))/18)
prev = 2^(12*n) * 19^(12*n)
old = <previous sum>
new = old+prev+x+y+z+w+v
where n is the multiple of 12. Finding that it's always 0 mod 1365 is a good start (it proves my claim above about the factors), but not sufficient to prove that it's always abundant.
Attached Files
 conjectures.tar.gz (1.6 KB, 11 views)

Last fiddled with by Happy5214 on 2021-03-04 at 12:00 Reason: Adding formulas for 38^12n

 2021-03-04, 13:58 #939 RichD     Sep 2008 Kansas 63648 Posts In keeping with the spirit of investigating base 2*p, next up is base 46. Preliminary results show a similar phenomena. Additionally, 46^6n shows an abundance of (3 * 5 * 7) along with 46^12n (3 * 5 * 7 * 13) to advance the sequence.
2021-03-04, 15:45   #940
warachwe

Aug 2020

2·32 Posts

Quote:
 Originally Posted by Happy5214 where n is the multiple of 12. Finding that it's always 0 mod 1365 is a good start (it proves my claim above about the factors), but not sufficient to prove that it's always abundant.
38^(12*19) has next term as 3*5*7*13*229*457*C353. As the composite has no small factor, this is not abundant.

2021-03-04, 18:51   #941
garambois

"Garambois Jean-Luc"
Oct 2011
France

10568 Posts

Quote:
 Originally Posted by Happy5214 Oh, I tried to mock up an example (not full) layout, but forgot to post it. It's attached here.

Thanks a lot for your help !
I will try to modify the html code like in your example.
But I'm running out of time.
I think it takes hours of work to rearrange everything like in your example. I hope I am wrong.
I don't know when I will be able to publish the page or both pages if I separate the two posts.
I will keep you posted...

2021-03-04, 19:03   #942
garambois

"Garambois Jean-Luc"
Oct 2011
France

10001011102 Posts

Quote:
 Originally Posted by RichD In keeping with the spirit of investigating base 2*p, next up is base 46. Preliminary results show a similar phenomena. Additionally, 46^6n shows an abundance of (3 * 5 * 7) along with 46^12n (3 * 5 * 7 * 13) to advance the sequence.

OK, seen.
Thanks for the base 46.
You talk about an abundance of (3 * 5 * 7), didn't you want to write (3^3 * 5 * 7) instead ?

2021-03-04, 19:06   #943
garambois

"Garambois Jean-Luc"
Oct 2011
France

2·32·31 Posts

Quote:
 Originally Posted by warachwe 38^(12*19) has next term as 3*5*7*13*229*457*C353. As the composite has no small factor, this is not abundant.

Well seen !

Thank you very much !

2021-03-04, 21:57   #944
RichD

Sep 2008
Kansas

22·829 Posts

Quote:
 Originally Posted by garambois You talk about an abundance of (3 * 5 * 7), didn't you want to write (3^3 * 5 * 7) instead ?
Ah, yes, poor choice of words on my part. I should have used presence of 3 * 5 * 7 or use abundance as you stated above.

2021-03-04, 22:09   #945
henryzz
Just call me Henry

"David"
Sep 2007
Cambridge (GMT/BST)

10110111000102 Posts

Quote:
 Originally Posted by warachwe 38^(12*19) has next term as 3*5*7*13*229*457*C353. As the composite has no small factor, this is not abundant.
What is the upper bound on the abundance contribution of the C353? 3*5*7*13*229*457 already has an abundance ratio of 1.982=s(3*5*7*13*229*457)/3*5*7*13*229*457

2021-03-04, 22:22   #946
henryzz
Just call me Henry

"David"
Sep 2007
Cambridge (GMT/BST)

2·29·101 Posts

Quote:
 Originally Posted by henryzz What is the upper bound on the abundance contribution of the C353? 3*5*7*13*229*457 already has an abundance ratio of 1.982=s(3*5*7*13*229*457)/3*5*7*13*229*457
Assuming no factors below 1e20 then the maximum abundance contribution of the C353 can be upper bounded by ((1e20+1)/1e20)^(floor(353/20))= (1+1e-20)^17= 1.00000000000000000017 or there about.

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