[QUOTE=Batalov;442359]One can also "climb ev'ry mountain" in this problem - [URL]https://projecteuler.net/problem=569[/URL]
...again, in the spirit of encouragement. :rolleyes:[/QUOTE]
What's the scale and how tall is Tenzing? I add just under 2 meters to those peaks if I stand on top...:smile:

Lost the 3 in step 8051 and now it starts going down in steps 8051-8054:

[QUOTE=ATH;442346]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]
Sure, otherwise we won't need work any aliquot sequence, if we would have a proof that 3 never disappears, we would have a sequence which goes up to infinity. Fortunately (or unfortunately) this is not the case. If, for example, the power of 2 will switch from 2^3 to 2^2, and the sequence do not get a 7, then we will have larger chances to lose the 3:

Assume we have t=2^2*3^y*F, with s=sigma(t), then the terms in s which do not contain 3 (i.e. the divisors of t/(3^y) are 1, 2, 2^2, F*, 2F*, 4F*, where F* is all combinations of primes in F. Therefore, s\3 (s without the terms with 3) is (1+2+4)(1+F*). As all terms in F* do not contain a 3, (but their sum can contain one) we must have F*=1 (mod 3) or F*=0 (mod 3) to lose the 3.

Example,

(1) F=p (a single prime):
t=2^2*3^y*p, t\3=2^2*p, s=1+2+4+p+2p+4p+(terms with 3), s\3=7(1+p). As p can not be a multiple of 3, it is either 1 or 2 (mod 3), in this case the sequence will lose a 3 if p is 1 (mod 3), which happens in 50% of the cases, regardless of the size of t.

(2) F=p*q (product of 2 primes):
t=2^2*3^y*p*q, t\3=2^2*p*q, s\3=1+2+4+p+2p+4p+q+2q+4q+pq+2pq+4pq, so s\3=7(1+p+q+pq)=7(1+F*). As p, q can not be a multiple of 3 (they are primes) F* needs to be 0 or 1 (mod 3), for the sequence to lose a 3, which happens in 25% of the cases, regardless of the size of t.
That because we can have either p=q=1 (mod 3) or p=q=2 (mod 3) or p=1, q=2 (mod 3), (or viceversa, which we may ignore due to symmetry). Therefore we have, respectively (mod 3):
[CODE]
[FONT=Courier New]p q | pq F*=p+q+pq 1+F*
1 1 | 1 3=0 1
1 2 | 2 5=2 0
2 1 | 2 5=2 0
2 2 | 1 5=2 0[/FONT]
[/CODE]which means that the parenthesis is divisible by 3 always, except when all primes in F are 1 (mod 3). You can try as an exercise with 3 or 4 primes. Then put all together to see for the general case.

This case (i.e t=2^[COLOR=Red][B]2[/B][/COLOR]*3^y*F) is not dependent of the size of t, and not much of the y (power of 3). But when 2^3, and general odd power of 2, the size of t counts (due to distribution of primes, they get rarer upper, so F=pq, F=pqr, etc, is more often than F=p).

[QUOTE=ATH;442388]Lost the 3 in step 8051 and now it starts going down in steps 8051-8054:[/QUOTE]
2^4 without 31 is "steady", it does not go down. To go down needs the next prime after 2 to be higher than 31 always, and there are a lot of random odd numbers which are multiples of either 3, 5, 7, 11, 13, 17, 19, 23, or 29. With 2^4, the threes can freely come and go. So the sequence oscillates up and down in small limits, but average is "steady" until it changes the power of 2, or gets a 31.

[QUOTE=ATH;442342]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?
[/QUOTE]

You would make an excellent guinea pig for my all-about-the-theory-of-Aliquot-sequences page! (Ernst, you watching this thread? :razz:)

[url]https://www.rechenkraft.net/aliquot/intro-analysis.html[/url]

Admittedly there's a substantial rewrite available that hasn't been pushed to the link above, but it should still be something at least.

(ChristianB, could you pull the most recent changes from my GitHub? Edit: You can view the most recent version [URL="https://htmlpreview.github.io/?https://github.com/dubslow/MersenneForumAliquot/blob/master/website/html/intro-analysis.html"]here[/URL])

It is very long winded though, and works slowly in order to aid development of intuition properly motivate each new idea.

Edit: Btw, for the record, what was the lowest number reached (so far) since 200 digits? EditEdit: Looks like the following is the smallest, coming on the second primary downdriver run:

[code]Checked 7188 16 (show) 3974341414523362<16> = 2 · 1987170707261681<16>
Checked 7189 16 (show) 1987170707261684<16> = 2^2 · 7 · 2897 · 24497888299<11>[/code]

[QUOTE=Dubslow;442397]You would make an excellent guinea pig for my all-about-the-theory-of-Aliquot-sequences page! (Ernst, you watching this thread? :razz:)

[URL]https://www.rechenkraft.net/aliquot/intro-analysis.html[/URL]

Admittedly there's a substantial rewrite available that hasn't been pushed to the link above, but it should still be something at least.

(ChristianB, could you pull the most recent changes from my GitHub? Edit: You can view the most recent version [URL="https://htmlpreview.github.io/?https://github.com/dubslow/MersenneForumAliquot/blob/master/website/html/intro-analysis.html"]here[/URL])

It is very long winded though, and works slowly in order to aid development of intuition properly motivate each new idea.

Edit: Btw, for the record, what was the lowest number reached (so far) since 200 digits? EditEdit: Looks like the following is the smallest, coming on the second primary downdriver run:

[code]Checked 7188 16 (show) 3974341414523362<16> = 2 · 1987170707261681<16>
Checked 7189 16 (show) 1987170707261684<16> = 2^2 · 7 · 2897 · 24497888299<11>[/code][/QUOTE]

When I go to either referenced page (with IE) I find things like:
[CODE] we also have to include $2 \cdot (2^n - 1)$, $2^2 \cdot (2^n - 1)$, ..., $2^{n-1} \cdot (2^n - 1)$. (If $2^n - 1$ were not prime[/CODE]which is somewhat difficult to understand. What is needed to read the page as intended?

You may have a "mathjax dollar" problem, the forum itself here had one in the past :)

Your IE may not recognize the $ sign as matjax delimiter, or it is not enabled on the destination site. The owner of the site (Dubslow) has to enable matjax to render the single dollars as math delimiters.

Can you see equations which are not "inside text", i.e. they are single on a text line? Then right-click on one of them and see your mathjax settings.

Edit: same here, with (newest) Firefox. The problem is not from you, but from Mathjax. The forum solved the issue by substituting the inline dollars with "\ (" and respectively "\ )" for inline math (with no quotes and no spaces between, grrr...)

Looks like a problem with the HTML preview site. When I look at rechenkraft (or a local git clone), the inline math displays correctly.

Antonio, you sure it happens at rechenkraft too, not just the HTML preview thingy?

I don't see ANY math at rechenkraft (only see dollars everywhere!)
I see standalone equations on the other site, but not the inline, which are displayed unrendered (with dollars).

If I copy the file from [url]https://github.com/dubslow/MersenneForumAliquot/blob/master/website/html/intro-analysis.html[/url], select all, ctrl+c, ctrl+v, save locally to aaa.html, open it with the browser (any browser) then [B]everything looks fine[/B].

The problem is that prevview/github has not enabled the single dollar in matjax rendering. There is a lot of arguing about those mathjax dollars on the web.

[QUOTE=Dubslow;442403]Looks like a problem with the HTML preview site. When I look at rechenkraft (or a local git clone), the inline math displays correctly.

Antonio, you sure it happens at rechenkraft too, not just the HTML preview thingy?[/QUOTE]

I looked again , and it magically sorted it's self on both sites when I told IE to display insecure content on the rechenkraft site (didn't complain about this on my previous visit, as far as I recall).

I think we might all be suffering from caching issues. Browsers trying to be smart and just making life more difficult.

At any rate, I had changed the MJ link to http months ago because RKN didn't used to be https. Git has it now back to https, so when ChristianB pulls the changes they should be in agreement again (and hopefully less issues as a result).