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#2674 | |
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Apr 2014
27 Posts |
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I understand it is not the place of this forum to offer help to those who are experiencing symptoms of mental illness, but I think that we have an opportunity that we don't actively acknowledge. These posts on the forum are frequently disruptive, but how disrupted is the life of the individual to be engaging 'like a child' the way they do with our community? I've used this forum for similar reasons when I was personally going through an acute psychotic episode. It's terrifying beyond anything I can describe and the ordered structure of numbers was incredibly comforting -- annoying to everyone here else because it was nonsense, but comforting to me. Some of the worst times I had during this period were when I would post something here I thought was useful and insightful -- it was not, it was pure numerology, pure gibberish -- only to be demeaned and dismissed by the community (their perspective was 100% correct and mine 100% distorted). I used this forum as a way to engage socially when I was unable to do so in daily life, to find a sense of belonging and talk about things I thought were a part of some deeper underlying beauty that resonated with my underlying condition. I apologize for placing this in this thread, but I have been waiting for an example of this type of interaction. Mental illness and math based apophenia seem to go hand-in-hand, but ostracizing people, for something they feel is incredibly important, in this state can do serious harm. I do like when the forum has given disruptive members their own thread so they can meander, posting as much gibberish as they please. There is no easy solution, I'm not even saying there is a solution or that anything should, or needs, to change. But I have not seen any conversation on here that has addressed this and the possible good, or harm, our community contributes in these situations. Apologies again for posting in this inappropriate setting. |
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#2675 | |
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P90 years forever!
Aug 2002
Yeehaw, FL
2×53×71 Posts |
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#2676 |
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Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
23×3×5×72 Posts |
I had failed to consider that. Apologies Raman.
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#2677 |
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Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 -89<O<-88
160658 Posts |
Gah!
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#2678 |
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"Forget I exist"
Jul 2009
Dumbassville
26×131 Posts |
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#2679 |
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Einyen
Dec 2003
Denmark
35·13 Posts |
Episode V: The Driver strikes back?
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#2680 |
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Jun 2003
5,051 Posts |
These are not the drivers you're looking for!
[2^2*3 is not a driver. 3 is independent of 2^2, and can come and go as it pleases without any change in the power of 2] Last fiddled with by axn on 2016-09-24 at 03:01 |
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#2681 |
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Einyen
Dec 2003
Denmark
35·13 Posts |
Yeah I saw on the intro analysis link. It should be a 7 with 2^2 for it to be a driver.
But maybe 3 does not know it's not supposed to be a driver, at least it has been "driving" it up higher for a while now (step 10565-10574 and 10586-10605). Last fiddled with by ATH on 2016-09-24 at 11:04 |
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#2682 | |
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Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
23×3×5×72 Posts |
Quote:
The only way out of this is to change the power of 3. The 3 can be eliminated if there are no factors of \(x \equiv 2 mod 3\). Each factor has a 50% chance of meeting this condition so the chance of losing the 3 is 1 in 2^a where a is the number of factors >3. As n gets larger on average the number of factors increases so it is less likely that the 3 will be lost. The other alternative is to raise the power of the 3. This requires \(\sigma(2^2*3*x)\equiv 2^2*3*x mod 9\). Only one of the factors of x can be 2 mod 3 otherwise \(9|\sigma(2^2*3*x)\). I believe that there is a 50% chance of increasing the power of 3 if only one of the factors of x is 2 mod 3. As you can see it isn't easy to loose the 3. We have been lucky so far. \(\sigma(2^2*3)=28 >24=2*(2^2*3)\) which means it will always rise while we have it. |
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#2683 |
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"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
947710 Posts |
Argh. Close but no cigar.
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#2684 | |
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"Forget I exist"
Jul 2009
Dumbassville
100000110000002 Posts |
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