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Duh, what am I missing here?
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compositeVolunteer tester Send message
Joined: 16 Feb 10 Posts: 1098 ID: 55391 Credit: 957,538,116 RAC: 993,618

Here's a headscratcher.
Coming back from a wedding party last weekend and seeing the lull in postings on the PG message boards I thought to liven things up by trying to prove the lowest conditional bound of Waring's Problem and post it here. See also the bound for equation 1.1 in this PDF at Penn State U between "provided that" and "if this fails".
It looks easy enough to prove, so why is it still a conjecture?
The first term of the left hand side is the "fractional part" of 3^k / 2^k multiplied by 2^k and second term is the integer part of the same fraction. That's right, the first term is exactly 3^k mod 2^k. But that doesn't matter.
By some obvious conventions let's call the first term R_k (for "remainder") and the second term Q_k (for "quotient"), and S_k for the actual numerical sum.
After a week in the rabbit hole chasing 3^k  R = 2^k Q and getting nowhere fast with a nice recursive expression for Q, I came out of the hole and the sun shone and the birds sang and there it was, sitting in plain sight...
We restate the conjecture as
(1): S_k = R_k + Q_k <= 2^k for k >= 0
According to the references in the links above, relation (1) holds for k <= 417,600,000. So I think we have some material to start an inductive proof. (* sound of rusty gears creaking *)
Just to be sure let's repeat some of that work in my "handy" browser spreadsheet.
k 2^k 3^k Q_k R_k S_k S_k<=2^k
0 1 1 1 0 1 TRUE
1 2 3 1 1 2 TRUE
2 4 9 2 1 3 TRUE
3 8 27 3 3 6 TRUE
4 16 81 5 1 6 TRUE
5 32 243 7 19 23 TRUE
6 64 729 11 25 36 TRUE
7 128 2187 17 11 28 TRUE
Proof:
Assume relation (1) holds for k+1 and we know it holds for k. Let's assess the difference of these sums as k goes to k+1:
(2): S_(k+1)  S_k <= 2^(k+1)  S_k
In general we can express S_k = 2^k  D_k where 0 <= D_k <= 2^k
(3): S_(k+1)  S_k <= 2^(k+1)  2^k + D_k
But 2^(k+1)  2^k = 2^k so
S_(k+1) <= 2^k + D_k
and D_k <= 2^k so
S_(k+1) <= 2^k + D_K <= 2^k + 2^k = 2^(k+1)
Then S_(k+1) <= 2^(k+1)
QED.
Is it really that simple? Please refrain from comments about bricks and marbles.  


This is not likely to be provable by induction.
Proof:
Assume relation (1) holds for k+1 and we know it holds for k. Let's assess the difference of these sums as k goes to k+1:
(2): S_(k+1)  S_k <= 2^(k+1)  S_k
I do not understand. You cannot "assume" the relation holds for k+1 (when we know it holds for k).
You must prove the relation holds for k+1, given that it holds for k. For that you can either (a) assume nothing more and arrive at the desired relation for k+1, or (b) assume the negation of the relation for k+1 and arrive at some kind of contradiction.
/JeppeSN  

compositeVolunteer tester Send message
Joined: 16 Feb 10 Posts: 1098 ID: 55391 Credit: 957,538,116 RAC: 993,618

Thanks for the refresher and the positive feeding.
I'll see what I can do.  

compositeVolunteer tester Send message
Joined: 16 Feb 10 Posts: 1098 ID: 55391 Credit: 957,538,116 RAC: 993,618

Waring's conjecture is proven, about 2 hours ago.
Full details to follow after the proof is peerreviewed and published.
I was hoping to use this for a MSc thesis in Computer Science at the local university, but the Department became rather shallow in combinatorics since my former professors retired, and the one I spoke with last week as potential supervisor already has enough grad students for this year. What I showed him wasn't complete, but some of the ideas were correct even though I was barking up the wrong tree at the time. He suggested that I think about choosing another topic, given the high likelihood of not succeeding. I wrote him an email 2 days later saying that I wouldn't be intruding on his time until I solve the problem or I give up and choose something else. Computer Assisted Algebra might be interesting  I started to use one of these packages last weekend to verify my handwritten work and I probably recouped the time to learn it.
I had also spoken with an Associate Dean for a recommendation on a supervisor. He thought this proof would be more appropriate for a thesis in Mathematics than in Computer Science. I'm not keen on doing a math degree, computers are really my area of interest and math courses would burn my brain (but a few whacks on the head might reduce the density sufficiently for higher maths to make sense). He also said, if I have a proof to a named problem, why bother with the degree, just submit it to a peerreviewed journal. I guess some degrees are overrated.  

compositeVolunteer tester Send message
Joined: 16 Feb 10 Posts: 1098 ID: 55391 Credit: 957,538,116 RAC: 993,618

Withdrawing claim of proof.
Independent review turned up a problem.
Not sure it can be rescued.  

compositeVolunteer tester Send message
Joined: 16 Feb 10 Posts: 1098 ID: 55391 Credit: 957,538,116 RAC: 993,618

The proof has been fixed. Sending it in for another review today.
 

compositeVolunteer tester Send message
Joined: 16 Feb 10 Posts: 1098 ID: 55391 Credit: 957,538,116 RAC: 993,618

Another day, another problem. Working on fixing that.
Technically, it's not Waring's conjecture, it's someone else's.
Waring's conjecture suggested that there was any answer at all for every k, which Hilbert proved in 1909.
The problem has been distilled into a set of equations and conditions, and the conjecture is that just one of these equations is needed.
Work on Waring's problem has a very good math pedigree (arxiv.org, pages 1618).
Pillai's work on this problem has been described posthumously as "almost certainly his best piece of work and one of the very best achievements in Indian Mathematics since Ramanujan". (archive.org)
The Associate Dean of Science I spoke with (who's specialty is number theory) is right to be skeptical of any claim of a proof, and he's been correct so far.
The conjecture posits that g(k) = 2^k + floor(3^k/2^k)  2 is the exact answer, rather than merely the only answer we have been able to obtain so far (subject to a condition).
Everyone believes it is true but no one has a proof that the condition is unnecessary.
Furthermore a supercomputer search with a CM1 in 1989 up to k = 471,600,000 by Kubina and Wunderlich failed to find any exceptions (thereby proving that the equation is valid for k up to this much).
From my work on it, I also believe the condition is unnecessary.
This is what I have been working on since August 18, eliminating the condition.
I didn't know what I was walking into when I tackled on this problem on a lark. But I think I'm getting there.
The experience has been like entering A MAZE OF TWISTY LITTLE PASSAGES, ALL DIFFERENT.
But the nature of this maze is that the goal is always visible and always just out of reach.
 

compositeVolunteer tester Send message
Joined: 16 Feb 10 Posts: 1098 ID: 55391 Credit: 957,538,116 RAC: 993,618

Hey, folks, in case you think I'm writing a short story, maybe I am.
It's been pretty boring up to now, and I can't say much for the title.
Here's the latest installment...
ROFL. I really hit the jackpot this time.
This proof will be perfectly raunchy to a number theorist
because the show is over when the second veil is dropped.
The reviewer said to me, pointing out the mistake in the last iteration,
something like "if you can prove this then you are done immediately".
Looking it over I wondered what brain fart led me to make a statement
that was patently false. And of course my subconscious had been yelling
at me the whole time I as was writing up that drivel,
"HERE!! HEY!!! OVER HERE!!!! IT'S OVER HERE!!! THE PROOF IS OVER HERE, DUMMY!".
The subconscious is never subtle. It communicates through the limbic brain, so I was
really happy when I submitted the proof, which I FELT was correct because, hey,
I eliminated the shenanigans from the first round, and the remaining pieces fit together
so nicely in Lemma 2 and Lemma 3 that I was frankly amazed.
Then a week went by before it was reviewed. Obviously the reviewer wasn't
placing a priority on it after my first failed attempt, saying he might get around
to it on the weekend. And he signed his review more formally, maybe to remind
me that he knows a thing or two more than me and that I was wasting his time.
At least he said that Lemma 2 and Lemma 3 were correct, but they depended
on the result from Lemma 1. Doh! Who knows, he may already have the correct
proof in hand and he is getting pissed off that academic integrity prevents him
from publishing the result himself. What should I do, offer joint authorship, and
at least get my Erdos number reduced to 3 as a consolation prize? He doesn't
have a formal obligation to me. I'm not even registered as a student yet.
But then after a second day and a fitful sleep I got out of bed with the machinery
to make the proof work. Cognizant that I might be wearing down his patience,
I wrote the reviewer a note, asked if this proposed approach could make the
proof work. Encouragingly, he quickly replied yes, as long as the proof is rigourous.
So now it's do or die. He will probably read up to the point where the proof is completely
screaming obvious and tell me that it looks good up to there but that the rest of the proof
is poop because I already provided the answer. I guess I'm just too attached to the small
successes to let the good math go quietly.
 

dukebgVolunteer tester
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Hey, folks, in case you think I'm writing a short story, maybe I am.
It's been pretty boring up to now, and I can't say much for the title.
Here's the latest installment...
Heh, I would say it's actually an interesting short story, not a boring one. Keep the spirits up!  

compositeVolunteer tester Send message
Joined: 16 Feb 10 Posts: 1098 ID: 55391 Credit: 957,538,116 RAC: 993,618

Hey, folks, in case you think I'm writing a short story, maybe I am.
It's been pretty boring up to now, and I can't say much for the title.
Here's the latest installment...
Heh, I would say it's actually an interesting short story, not a boring one. Keep the spirits up!
You know, the story could easily turn into a conspiracy thriller. Feel free to contribute any facts you can turn up.
For some odd reason, the NSA supported Kubina and Wunderlich's effort with 240 hours of supercomputer time. (semanticscholar.org)
In that paper, these guys, possibly not the first ones to do so, admitted that they "bend the interpretation of history"
in calling the restricted problem at the heart of this investigation, Waring's conjecture.
Was the real purpose of the investigation to probe Waring's hint that this works for polynomials as well?
You know, the Russians have an interest in this too.
Linnik gave an elementary proof of the HilbertWaring Theorem in 1940. (University of Warsaw)
What was the point of that? It was already proven in 1909.
It was wartime, and the Soviets had a secret pact with Nazi Germany to divide up Poland. (wikipedia.org)
The Soviets couldn't afford to spend resources on pure research in wartime. Were they onto something big? (arxiv.org)
And if you Google for Linnick Waring you get NOTHING.
Ya, ok I was sloppy with this one, his name is Linnik and it was a Wikipedia search. (wikipedia.org)
It's not a conspiracy theory unless you make it one.
On the other hand was Linnik any relation to a highranking member of the US State Department,
the one who released the 83page report which distanced them from Hillary Clinton's email practices? (wikipedia.org)
It could be a juicy thriller.
And what if my proof is censored at publication?
Do I get suddenly get a "job" in a certain national security agency?
Alright, so what if I told you that Hardy and Littlewood were out to lunch on this one?
This proof needs a deadman's switch which dumps it to the dark web.  

compositeVolunteer tester Send message
Joined: 16 Feb 10 Posts: 1098 ID: 55391 Credit: 957,538,116 RAC: 993,618

Was the real purpose of the investigation to probe Waring's hint that this works for polynomials as well?
I hear a chorus of silence asking, "So what's all the fuss about polynomials? That's gradeschool stuff."
Yes, well there is one particularly simple polynomial in 2 variables that has a name.
It's called an elliptic curve. (wikipedia)  

compositeVolunteer tester Send message
Joined: 16 Feb 10 Posts: 1098 ID: 55391 Credit: 957,538,116 RAC: 993,618

If you want a good thriller, it has to be plausible and interesting. On the other hand was Linnik any relation to a highranking member of the US State Department,
the one who released the 83page report which distanced them from Hillary Clinton's email practices? The similarity in surnames is just a coincidence, and an intergenerational spy ring wouldn't be this obvious.
The Zimmerman Telegram incident (Wikipedia) in the First World War provides a much better story template.
The parallel here would have the Clinton email server hacking as the cover story which hides
the interception of internet traffic and successful decryption of a secure transport mechanism.
The Russians presumably would know if it wasn't them but couldn't be sure they were not a
scapegoat for another hacker like the DPRK. This plotline would follow the theme of this thread.  

compositeVolunteer tester Send message
Joined: 16 Feb 10 Posts: 1098 ID: 55391 Credit: 957,538,116 RAC: 993,618

Now here's an interesting concept.
Write a fictional story, but embed in it a bona fide firsttime proof of Waring's conjecture.
Is there a precedent for such a thing?  

compositeVolunteer tester Send message
Joined: 16 Feb 10 Posts: 1098 ID: 55391 Credit: 957,538,116 RAC: 993,618

I've been diddling and dawdling, sitting on the completed proof, and trying to find a shorter one. I may just have to leave this endeavour to someone else, Proving the bound on the conjecture is such a minor result, it's probably not worth trying to find a shorter proof. The real value in any proof is when it introduces new ideas that open up new solution space territory.  


I've been diddling and dawdling, sitting on the completed proof, and trying to find a shorter one. I may just have to leave this endeavour to someone else, Proving the bound on the conjecture is such a minor result, it's probably not worth trying to find a shorter proof. The real value in any proof is when it introduces new ideas that open up new solution space territory.
Rather than not submitting anything, submit what you have and let someone else figure out how to make it shorter.
The real value of any proof is that it changes "we think" to "we know". It adds a stable block on which future knowledge can be built.  

compositeVolunteer tester Send message
Joined: 16 Feb 10 Posts: 1098 ID: 55391 Credit: 957,538,116 RAC: 993,618

I've been diddling and dawdling, sitting on the completed proof, and trying to find a shorter one. I may just have to leave this endeavour to someone else, Proving the bound on the conjecture is such a minor result, it's probably not worth trying to find a shorter proof. The real value in any proof is when it introduces new ideas that open up new solution space territory.
Rather than not submitting anything, submit what you have and let someone else figure out how to make it shorter.
The real value of any proof is that it changes "we think" to "we know". It adds a stable block on which future knowledge can be built.
It's getting close to the 250 year anniversary of Waring's original problem statement. I'll write up the proof for publication in the new year.
I spent well over 100 hours on this, and have around 760 singlespaced typed lines of notes including a lot of deadend ideas.
I suspect professional mathematicians didn't get this result because they quickly decided they would be more productive working on something else.
The proof exploits 2 ideas I haven't seen used before. You can see the timeline in the previous messages of this thread.
At the beginning I avoided reading previous papers on this problem, to tackle it with a fresh mind and be unaffected by groupthink.
After I had assured my independent approach, I read some of the papers to see what they did.
The first good idea in early September set up a useful framework for the problem.
I was so excited at developing this framework that I was awake and alert for over 40 hours straight.
I managed to relax with a 90 minute massage, but I wasn't sleepy or tired when I forced myself to go to bed a few hours later.
The second idea from early October created the machinery essential to proving a key assertion.
The concept for this machinery emerged while I in the sauna at the public pool I frequented.
This wasn't nearly as exciting as the initial framework, as by now my enthusiasm was tempered by the previous error in the proof I had submitted.
Now here's some added context you haven't seen.
One month before starting this project, I whacked the back of my head pretty hard on the water doing a backflip from a 1 meter springboard.
So I had intermittent headaches for around 3 months, the last two covering the period during which the proof was developed.
During these few months, I speculate there was extra bloodflow augmenting my brain's normal operating level.
Twice I took an antiinflamatory pill to quell the headaches, but I decided to tolerate the headaches after that.
And there may have been some neurological rearrangement from mild shocks to the head, from diving off the 5 metre platform around 80 times since starting in June.
I'm saying there was a process, and maybe it's come to an end (or maybe not).
It's clear from the first post that I wasn't anywhere near a solution when I started.
I haven't felt like doing much work on the proof since the headaches stopped.
I've also recovered from being hit by a car (quite hard) as I was crossing the street in November.
As precaution, I gave my reviewer permission to publish the proof jointly in the event that I somehow become incapacitated (dead) before I can publish it.
It's also become clear to me that although I've long wanted to achieve some result like this, it's not really such an important thing to have done, in the big picture.
I agree with you that the result adds a stable block for future knowledge.
But I double down and assert the framework and machinery in this proof are likely more significant than the result.
Whether there is further use for them in other proofs remains to be seen.  

compositeVolunteer tester Send message
Joined: 16 Feb 10 Posts: 1098 ID: 55391 Credit: 957,538,116 RAC: 993,618

Ah, "framework" is a bit oversold. It's a construction.  

compositeVolunteer tester Send message
Joined: 16 Feb 10 Posts: 1098 ID: 55391 Credit: 957,538,116 RAC: 993,618

The proof is getting shorter.
I should retract any claim about machinery. It was a means to an end.
A more fundamental consideration of the problem makes the machinery redundant.
That's the stupid thing about mathematical proofs.
Everyone gets to see the oohahh version without the cruft,
and no clue is revealed about the process that leads to the piÃ¨ce de rÃ©sistance.
I should give the writeup a go now.  

compositeVolunteer tester Send message
Joined: 16 Feb 10 Posts: 1098 ID: 55391 Credit: 957,538,116 RAC: 993,618

I wrote it up and went to bed, thinking... there it was.
The next day I read it over and decided to tweak a few statements for clarity.
Then I tried the unthinkable. What happens if I invert the logic in the key part?
Oh *&#*! The saga continues. The short proof isn't any good.
Opposite statements work equally well  which means the proof doesn't work at all.
Nice catch. Back to the long version. And maybe that's wrong too.  

compositeVolunteer tester Send message
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I found the booboo in the short proof, so I think I saved it.
Thankfully, it's near the end, so there's not a lot to change.  

compositeVolunteer tester Send message
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I contacted the reviewer and told him "I have something now, I just have to write it up", and "see you soon".
He replied that he would be excited to see it.
A week went by and as I tried to clean up the proof, I was stuck at that same old confounding problem,
because the workaround didn't work. This was after I fixed the booboo.
So for the last couple of days I considered eating crow and sending him another message that I didn't have it.
But then this morning, arguably more in desperation than as obsession, I was turning over the problem
in my head for an hour in bed, without the benefit of a notepad, reiterating the facts as they slipped in
and out of focus, still being nagged by the subconcious about a stupid claim that I know doesn't work.
And TADA!! I saw it. The answer was clear. I did the math in my head. Yup, it worked.
I rushed to the computer and added a couple of columns to my spreadsheet to verify this numerically.
Of course it works. And now I know the reason and I can explain it. The thing that was patently false,
is false for a reason  a reason that I had missed, the reason that makes the proof work. I felt pretty good.
At lunch time I batted the birdie with my buddie in the gym, as we do on Fridays, and I challenged him to try this
wicked general formula with any numbers he chooses, providing they met the criteria I laid out. He's a funny guy.
He used to have a local TV show with a cult following  "Math with Marty". Ya, that's the guy. I've known him since
university days. He's quick with numbers, and he likes to talk about odd theories, but he labours with symbolic
math in his head as much as anyone. There's too much going on. He needs a blackboard to stay organized.
There's one at the gym, but no luck, there's no chalk. So I couldn't explain it to him. He didn't want to listen to me.
He wanted to see it. I'll buy some chalk and bring it next week.
Although he tried it successfully with numerical examples, he couldn't believe that this thing works.
He doesn't want to believe it, but the formula doesn't lie. We sounded like Doc Brown and Marty McFly,
incredulous that no one saw this thing in 3000 years of math history.
No  actually more. Babylonian maths was active from 5th to 3rd millenia BC. Maybe this thing was lost in antiquity.
Realistically though, Babylonian mathematicians probably didn't have it, as they were missing certain tools that
children are now taught in elementary school.
So now the proof is ludicrously short. It has been STARING AT ME FOR NEARLY 6 MONTHS and I didn't see it
for what it was. Kudos to the reviewer, who said to me, that if I could prove this assertion, then I'm done right there.
That was the challenge. Of course my assertion was wrong, but I clung to it because I needed it for a particular
approach to the proof. That approach might still be viable, but it's pointless  the approach has been shortcircuited.
If this proof wasn't such a minor thing, you guys would poop your pants for how simple it is.
(*** Distant echos of Pierre de Fermat's scoffing laughter. ***). Ahhh, just wait for it.
Next up: Goldbach's Conjecture? Yes, I took a stab at this one a few years ago. I had a discussion with a math
professor who looked at me like I'm a kook (I'm starting to see pattern LOL), about an idea he didn't understand.
It's not his field, obviously. I would try to make this idea work, if I ever get a round tuit. Shutting down the GC
distributed computing project would be a nice little contribution to reducing greenhouse gasses.
However, breaking bitcoin would be a much better project for saving the environment.  

compositeVolunteer tester Send message
Joined: 16 Feb 10 Posts: 1098 ID: 55391 Credit: 957,538,116 RAC: 993,618

Oh, man! Seriously?
I had flipped a sign, so the short proof is messed up.
There has to be a trick in here somewhere, or Pierre wins another round.  

compositeVolunteer tester Send message
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I found the missing link for the short proof. I just have to plug it in.
Threatening checkmate, Pierre!
EDIT: And she's a beauty. This proof will be extremely satisfying. I guarantee it.
EDIT EDIT: Hmmm.
Doubts are creeping up because I haven't been able to plug it in yet.
I might have tricked myself into repeating an old fact in a new way that still doesn't resolve the problem.
Right  just because I see something a new way doesn't mean I can prove it must be that way.  

compositeVolunteer tester Send message
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Direct proof is not going to happen.
The easy case is  easy. And the hard case is  impossible.
I told the reviewer I have abandoned this approach.
So now believe it or not, I'm going to resort to proof by induction, trying to show that the problem is completely covered by just the easy case.
I have some induction formulae and I need to transform the initial problem into these formulae to complete the proof.
I know  this sounds backwards, but I've been working on this thing long enough to already know where all the handles are located.  

compositeVolunteer tester Send message
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**bbzz* ff ** radio static* something weird is going on here, my program is very slowly converging to the sequence of PolyBernoulli numbers B_n^(k) with k=2
Given half a million years, it could converge on the values for n up to 12. Perl is horrible, time to switch to C.  

compositeVolunteer tester Send message
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... and the C program computed in around 2 minutes what the Perl program did in 32 days. That's the difference between using Perl arrays and hashes, versus bit string operations in C, and being able to reuse some intermediate computations in the C version. This includes recompiling the C version with O3, which accounted for a 4.9 times speedup over the unoptimized version, and a modest speedup of 1.3 times by switching from linear search to bisection search (which was penalized by using insertion sort for keys). I think I could do better timewise and storagewise by building a tree.
The benefit of having a Perl implementation for reference is that it was easy to determine the existence of program bugs in the C version, especially in memory addressing. It took a few hours to write the Perl version and over 2 weeks (with sporadic attention) to debug the C version. And I even noted a minor issue in output from the Perl version (index value off by 1, correctible after the fact).
Alas, all it did was confirm a pattern I had previously observed for k up to 250 thousand, extending that range to 30 million in 842 minutes. The arrays gobbled up 10.2 GB of RAM. I could make it more efficient memorywise but there's no point to repeating or extending the computation. It's an interesting result. The hard part is figuring out why I'm seeing this, and if it has any consequence to what I am trying to prove.  

compositeVolunteer tester Send message
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It seems odd to be dispassionate now about saying that I might have a proof of the conjecture, "this time" a proof by contradiction. It was a long and hard journey, and along the way I learned to distrust my gut feeling, after several exciting interludes that amounted to very little progress.
The last time, on the first anniversary of starting to work on the problem, I pulled an allnighter which I felt went somewhere, that I was at the end of the tunnel. But I held back from saying anything about it, especially here, until I confirmed that it worked. As usual, reflecting on it the next day uncovered an impasse. The only reliable conclusion that I could reach was that logical sensibility was shutting down with prolonged periods of avoided sleep.
But it was refreshing to have another tool on the board. Actually, I was using the new visualization, the product of that night's toil, when I came up with this proof. When I placed a new fact on the diagram where the formula told me to put it, the label on the adjacent fact was begging for a contradiction. That could lead to only one conclusion. So I went for it and did the math.
Ya! It's beautiful. Now I'll give it a couple of days to see if it stands up to more scrutiny.  

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Oh oh. I found that I flipped a sign along the way, which means the proof is garbage.
It's time to quit when it stops being fun, and it's juuuust about there.
As sometimes happens with these kinds of explorations, something interesting pops out along the way. I now have a C program that slowly enumerates polybernoulli numbers with k=2, by partitioning these numbers in some way. The set of numbers adding up to each PolyBernoulli number is symmetric, and OEIS doesn't deliver a clue of what they are, so it will be an interesting challenge to figure out what describes them.
Briefly, I produce a set of keys of various lengths by concatenating the results of a sequence of calculations to produce each key (calculations related to Waring's Problem). Then I sum the parts of the keys together and drop the keys into bins labelled with the sums, while tracking the number of unique keys that appears in each bin. That's where the surprises start.
Surprise #1. For N parts each ranging from 1 to X, there are only N+X1 distinct sums. For unrelated parts, I expected N*XN+1 distinct sums. So the parts have some kind of internal parity that binds them tightly to the sequence, even though the parts seem to appear random individually.
Surprise #2. Sorting the bins according to the bin's label, the contents of the bins are symmetric, in the way that binomial coefficients are symmetic. Actually not so surprising, because the middle sums have more degrees of freedom, more ways of achieving the sum for that bin.
Surprise #3. Adding together the contents of the N+X1 bins produces the Nth PolyBernoulli number with k=log(X). On reflection, this probably has something to do with the requirement for the uniqueness of the keys; the prior work points to that.
So the only really surprising fact is #1.  

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Waking up from a long sleep, I decided to check my understanding of these surprises with a sequence of keys that I understand better: a full keyspace of X^N keys.
With X=100 and N=3, produce the set of all 1'000'000 unique 6digit (base 10) keys.
Treat these keys as 3 concatenated 2digit values corresponding to the "results of a sequence of calculations" referred to previously. Unlike the sequence generated from Waring's Problem, these "results" are uniformly distributed (from 0 to 99).
The sum of the 3 values in a key produces a bin number ranging from 0 to 297.
The number of bins is N*XN+1 as expected for the full keyspace.
Assigning keys to these 298 bins according to bin number, let the count of keys in a bin be called the contents of the bin.
The total contents of all the bins is clearly 1'000'000 because all we've done is partitioned the set of keys into bins according to a rule. Stepping through the bins in order of ascending bin number, we see the contents are 1, 3, 6, 10, 15, 21, ..., 4851, 4950, 5050, 5148, 5244, 5338, ..., 7494, 7498, 7500, 7500, 7498, 7494, ..., 5244, 5148, 5050, 4950, ..., 21, 15, 10, 6, 3, 1. The sequence is symmetric about the maximum value.
The first difference of that sequence shows us what's going on. This is [1], 2, 3, 4, 5, ... , 98, 99, 100, 98, 96, 94, 92, ..., 6, 4, 2, 0, 2, 4, 6, ..., 96, 98, 100, 99, 98, 97, ..., 6, 5, 4, 3, 2, 1. With an initial [1] as an extension for symmetry, the difference sequence is one cycle of a triangle wave with negative slope twice the positive slope.
Now, do this again, but modify the keyspace by throwing away all keys containing a "00" for one or more of the "results". This removes 10'000 + 10'000 + 10'000  100  100  100 + 1 keys from the keyspace, leaving 970'299 keys with "results" from 1 to 99. Now there are 295 bins, having lost bins numbered 0, 1, and 2 from the previous case. The sequence of bin contents is similar: 1, 3, 6, 10, ..., 4753, 4851, 4950, 5047, 5142, ..., 7347, 7350, 7351, 7350, 7357, ..., 5142, 5047, 4950, 4851, 4753, ..., 10, 6, 3, 1, except that the sequence starts at bin 3 and progresses such that the extrema of the difference sequence are 99 and 99 rather than 100 and 100. Again, the sequence is symmetric about the maximum value.
We can throw away a set of keys from the full keyspace according to more complex rules. If we use a keyspace that is a power of 2 rather than a power of 10, and we throw away keys according to rules that generate a keyspace whose size is the PolyBernoulli number with k=2, then we can get the observed sequence behaviour. Now it's not really so surprising that the sequence is symmetric and the number of distinct sums is so different from what I originally expected.
So what I've discovered is a connection between some recursion formulae I developed for Waring's Problem and Brewbaker's combinatorial interpretation of the PolyBernoulli numbers. Since I know the recursion formulae produce a restricted keyspace, I hope to map Brewbaker's combinatorial generator onto the allowed ranges for resides that makes Waring's conjecture true.
{edit} If that leads to the answer, then Waring simply didn't have a chance because he didn't have the tools 250 years ago.  

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I can hardly believe that I got it right this time. Maybe I didn't. I wasted so much effort on fruitless investigations that always led me to one version or another of a fact that I could not prove, even though I found some interesting stuff along the way for which I can say that I've gained a mathematical insight or two. I suppose the work has conditioned my thinking to try something else.
The proof is simple, continuing from the point where I ran into trouble on the first review. I basically found a contradiction for the difficult side where the proof bifurcates. Obviously my confidence in the proof is a lot higher this time, bolstered by the considerable brevity and readability. And by now it's not giving me the endorphin rush that kept me awake for 40 hours straight nearly 3 years ago.
I've sent it for review but the reviewer might ignore me by now. He probably thinks I'm a crank at worst, at best someone beset with a hopeless obsession. I'm not claiming to be a genius, I'm just being persistent and exploiting the luxury of not having to stake a reputation or put academic productivity on the line, although there's been an economic cost to it from the point of view that I could have been writing software for fun and profit. Luckily I've not sent him some of the letters I drafted which I later found to contain defects. My enthusiasm has been already tempered by second thoughts.
The thing I've tentatively proven is referenced in Chapter 3 of this article on S Chowla and S S Pillai at the Indian Statistical Institute, which reviews Pillai's contribution to Waring's Problem and summarizes Pillai's "ideal Waring conjecture". The Wikipedia article on S S Pillai quotes a glowing account of Pillai's careerdefining contribution to Waring's problem, so I would say that if successful I've done no more than peek beyond the horizon by standing on the shoulders of a giant.
The funny thing to me is that the problem is structurally more interesting than the result to be proven, since investigation of the carry bits in the recursive formulae for the quotient and the remainder put me into wierd PolyBernoulli numbers territory, even though it's unrelated to the proof. On a side note I amazed a mathematical friend with a formula to produce the remainder of a related division without doing the second division, which I demonstrated to him by leading him through the calculation. He assumed that remainders behave randomly within their allowed ranges, but that's true when you look at a statistical sample, not at specific instances.  

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Aw shucks, he checked it and a typo was fatal to the proof. Nice try though. Maybe I'll continue on that line, it looks promising.  

compositeVolunteer tester Send message
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Waring's Problem has seen much work by a lot of mathematicians, including some whose names are wellknown around here at PrimeGrid (Fermat, Wieferich). It's one of the two classical problems in Additive Number Theory, the other the presentday favorite Goldbach's Conjecture. I found an interesting Master's Thesis by Janne Suomalainen which covers some history and is sprinkled with short biographies of mathematicians who made substantial contributions. I highly recommend reading it for the entertainment value. Don't worry about the math. Like Fermat's Last Theorem, it's easy to get sucked in by the simple explanation of the problem, while the solution is anything but simple.
Before proofs really started to roll in on this problem, the first hundred+ years of work involved tabulating solutions, trying to reduce the upper bounds of various cases. This is analogous to the work of PrimeGrid  which is creating tables of ever larger primes without advancing the theory of it  and occasionally making progress on conjectures without ever knowing whether an answer even exists.
For reference, I've been working on a very limited aspect of a specific version of the problem (Ideal Waring's Problem), trying to prove that first condition in Theorem 1.13 in this paper applies to all n >= 2. The condition has stubbornly resisted proof for several decades. If proven it challenges the writer's statement of "much yet to be done before Ideal Waring's problem is completely solved".
I'm more convinced than ever that I have the solution but due to my basic education I find it difficult to state it in the simple terms that mathematicians would understand. LOL, I think that's something a crank would say.
I wasn't thinking about this earlier, but I wonder if a solution would qualify for the Elie Cartan prize, one of the minor prizes of France's Academie des Sciences. It is awarded for solving a difficult problem. At only 1500 Euros it would not compensate me for the time I put into this, but better than nothing and would look nice beside my amateur diving medals. I admire Perelman for his refusal to accept any awards (some of which he deserved). I suppose if you accept a gifthorse you could be expected to ride it.
Just now I was poking around the internets and there was some activity on this topic on math.stackexchange and mathoverflow several years ago. I saw that one of them actually started down one of the same deadend paths that I had looked at. It's also giving me an appreciation for the insight that I have for the solution.  

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I have achieved clarity. The proof is simple enough. Sent for review.  

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I have achieved clarity. The proof is simple enough. Sent for review. I've only managed to mathematically shut out an edge case that was intuitively false already.
Clearly the hard case remains unproven, and I have to look for a new attack on it.  

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A proof of 8 parts.
Hopefully, I've obliterated the mistakes.
I checked it in a spreadsheet and it works.
I'm crossing my fingers and ready to cheer!
As always, I don't have the last word.  

compositeVolunteer tester Send message
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I was pretty happy when the answer popped out, but after giving it some second thought several hours later, it's really creepy that it worked. A simple mistake led me to an invalid conclusion many times before. What cracked open the problem this time is applying an existing method in an unexpected context (I said I would try something new, right?). It didn't faze me at the time because I was just trying something different, but my skeptical self is surprised it worked. It's possible I've done something new because proof has been inaccessible to professional mathematicians for several decades, but my critical self accuses me of making a horrible mistake  again. So I'll say nothing more publicly for now about what I did, and let the professionals decide.  

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Just a thought. Humans like to celebrate success, and science is a human endeavour. Anytime we focus on a pinnacle of success, we ignore the fact that it sits upon a mountain of failures.  

Nick Send message
Joined: 11 Jul 11 Posts: 2149 ID: 105020 Credit: 8,006,239,534 RAC: 2,382,941

I think you are doing excellent
I was /am interested in Price's Theorem.
And so interested at each level you you go (even not very close to infinity)
It is always 0.5
It never gets that far away far rom 0.5 whatever the numbers!
What I am looking at is x
Then sqrt of x
Then x minus sqrt of x
Then the sqrt of that.
The difference is 0.5 (so close to it for numbers we are very use o!)
I may have got wrong what I am saying
I will correct tomorrow when I can see the TV
I don't know why.
Not a maths person.
Nick  

Nick Send message
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Also I have what I consider to be the exact probability of getting a small straight in Yahtzee  when at every role you are aiming for a straight in Yahtzee.  

compositeVolunteer tester Send message
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I am still waiting for the most recent review. At this point I am satisfied the proof is complete and it works (but that's just me). The version I submitted July 14 addressed a concern from the reviewer that I had not proven a certain point that I asserted on July 3, and I also replaced a handwaving argument which the reviewer had trouble following, with a rigorous mathematical analysis (which pleased me very much). The problem was I could see where the proof needed to go (because I was already there) while the proof did not lead the reader there. I also corrected an error, which meant a necessary final part was still missing on July 3.
That final part is the piece de resistance which incredibly closes the proof, employing a technique for which I am having trouble finding prior art. I'm not going to say what that is right now, call it a "trade secret" which might be useful for solving some advanced open problems. I did look briefly to see if I could apply it to the holy grail Re(s) = 1/2 which carries a hefty prize. You know the one.
Anyway I am now searching the literature for anything that resembles what I did. I would not be doing due diligence if I do not credit the related work of others even if I came up with the corresponding parts of the proof independently. So far I don't understand a lot of what has been published, and that's a good sign! If I had studiously held close to the literature I probably would not be thinking sufficiently out of the box to get to the end.
The proof may be longer than necessary if higher mathematics can substitute for some of what I had done laboriously, but that doesn't matter at this point  it would just guarantee a citation by someone looking to publish some lowhanging fruit.  

Nick Send message
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I hope it turns out well this time.
If Erdos was still with us, you would want him to say "straight from the book!'  

compositeVolunteer tester Send message
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It could be a while before I get the reviewer's response. Summertime is premium here.
So far 18 days with daily high temperature exceeding +30 C and we are still a month before the usual hottest period.
74.9 C temperature range this year (annual low 38.4 C Feb 13; annual high +36.5 C June 5).
Last winter had 30 days with daily low temperature below 20 C, for 11 of those days below 30 C.
There's also a drought happening in the prairies. Expect higher food prices.
Some immature crops have been salvaged as fodder with an early cut to prevent the grasshoppers from taking it all.  

Nick Send message
Joined: 11 Jul 11 Posts: 2149 ID: 105020 Credit: 8,006,239,534 RAC: 2,382,941

74.9 C temperature range this year (annual low 38.4 C Feb 13; annual high +36.5 C June 5).
Last winter had 30 days with daily low temperature below 20 C, for 11 of those days below 30 C.
I just reread that  75 degrees C difference between cold and hot?
I'll take my chances with ANY weather in Australia compared to that.
I am going to take a guess here  upper mid west?
Or in Anchorman 2  middle east?
Stay Classy!
Edit: You are in Canada  I thought I should see where you might be  very a lot upper mid west!  

compositeVolunteer tester Send message
Joined: 16 Feb 10 Posts: 1098 ID: 55391 Credit: 957,538,116 RAC: 993,618

74.9 C temperature range this year (annual low 38.4 C Feb 13; annual high +36.5 C June 5).
Last winter had 30 days with daily low temperature below 20 C, for 11 of those days below 30 C.
I just reread that  75 degrees C difference between cold and hot?
I'll take my chances with ANY weather in Australia compared to that.
I am going to take a guess here  upper mid west?
Or in Anchorman 2  middle east?
Stay Classy!
Edit: You are in Canada  I thought I should see where you might be  very a lot upper mid west!
On Google Maps, line up Dallas, TX with the right edge of the browser window. Then follow that window edge northward, waaayyy up North. Zoom in on that highway East of Winnipeg. You'll see the sign on Highway #1 that says "Longitudinal Centre of Canada" at 96 degrees 48 minutes 35 seconds West. Wikipedia has Dallas at 96 degrees 48 minutes 32 seconds West. That's a difference of about 40 metres at the latitude of the highway. In Dallas, that line would run through the Hyatt Regency's Reunion Ballroom, about 70 metres West of the Reunion Tower.  

Nick Send message
Joined: 11 Jul 11 Posts: 2149 ID: 105020 Credit: 8,006,239,534 RAC: 2,382,941

74.9 C temperature range this year (annual low 38.4 C Feb 13; annual high +36.5 C June 5).
Last winter had 30 days with daily low temperature below 20 C, for 11 of those days below 30 C.
I just reread that  75 degrees C difference between cold and hot?
I'll take my chances with ANY weather in Australia compared to that.
I am going to take a guess here  upper mid west?
Or in Anchorman 2  middle east?
Stay Classy!
Edit: You are in Canada  I thought I should see where you might be  very a lot upper mid west!
On Google Maps, line up Dallas, TX with the right edge of the browser window. Then follow that window edge northward, waaayyy up North. Zoom in on that highway East of Winnipeg. You'll see the sign on Highway #1 that says "Longitudinal Centre of Canada" at 96 degrees 48 minutes 35 seconds West. Wikipedia has Dallas at 96 degrees 48 minutes 32 seconds West. That's a difference of about 40 metres at the latitude of the highway. In Dallas, that line would run through the Hyatt Regency's Reunion Ballroom, about 70 metres West of the Reunion Tower.
I read 'Dallas'
Proper above 'Middle East'
I need to get to a map.
It is much simpler here in Australia.
And less metrics to explain where you are.
Crocodiles? Heat? Sand? Politicians?  

compositeVolunteer tester Send message
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On Google Maps, line up Dallas, TX with the right edge of the browser window. Then follow that window edge northward, waaayyy up North. Zoom in on that highway East of Winnipeg. You'll see the sign on Highway #1 that says "Longitudinal Centre of Canada" at 96 degrees 48 minutes 35 seconds West. Wikipedia has Dallas at 96 degrees 48 minutes 32 seconds West. That's a difference of about 40 metres at the latitude of the highway. In Dallas, that line would run through the Hyatt Regency's Reunion Ballroom, about 70 metres West of the Reunion Tower.
I read 'Dallas'
Proper above 'Middle East'
I need to get to a map.
It is much simpler here in Australia.
And less metrics to explain where you are.
Crocodiles? Heat? Sand? Politicians?
Ah, sorry 'bout the confusion, mate  has Google localized maps for Australians with South at the "top"? LOL  


It is much simpler here in Australia.
And less metrics to explain where you are.
Songlines are easy?
____________
Werinbert is not prime... or PRPnet keeps telling me so.
Badge score: 12x3 + 1x4 + 2x6 + 2x7 + 1x8 + 1x9 + 1x10 = 93  

Nick Send message
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Ah, sorry 'bout the confusion, mate  has Google localized maps for Australians with South at the "top"? LOL
You totally nailed it there  when I was in Europe  really funny stories  I just couldn't understand that the sun is in the south. Total fried my brain. I still am unable to get that. I would need the brain version of upside down glasses to travel north of Singapore  

compositeVolunteer tester Send message
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The reviewer is busy organizing a conference. That explains the delay.
A song? This is starting to sound like the lyrics. Here's the latest verse.
Setting aside the proof a week, I looked at it again.
Some minor fixes and clearer speak, good so far, my friend.
Were mathematics so watertight, the proof I thought was true?
I cast the doubt and checked it out, then shot a hole clean through!
I saved the reviewer the trouble this time, the claim of proof  withdrew.
 

Nick Send message
Joined: 11 Jul 11 Posts: 2149 ID: 105020 Credit: 8,006,239,534 RAC: 2,382,941

The reviewer is busy organizing a conference. That explains the delay.
A song? This is starting to sound like the lyrics. Here's the latest verse.
Setting aside the proof a week, I looked at it again.
Some minor fixes and clearer speak, good so far, my friend.
Were mathematics so watertight, the proof I thought was true?
I cast the doubt and checked it out, then shot a hole clean through!
I saved the reviewer the trouble this time, the claim of proof  withdrew.
I am so sorry.
It is awful.
You will get there!
And an excellent verse before the chorus:
It is done, I did it, yer yay ha!  

Nick Send message
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When I was 17 I had an idea
I know exactly where I was when I thought this.
What if gravity is the opposite?
It isn't matter that pulls
But space that pushes?
Not sure how this fits in with matter bending space/time  

compositeVolunteer tester Send message
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But space that pushes?
It's good to imagine possibilities, however theories have to be consistent with everything that we observe. That's science. The theory of special relativity accounted for results of experiments that didn't agree with the predictions of Newtonian (classical) mechanics, but the power of special relativity was confirmed by the accuracy of predictions it made for future measurements.
Two identical fermions (particles of halfinteger spin, like protons or electrons) have distinct quantum states (the Pauli exclusion principle) so they can't occupy the same position. That would be degenerate in Quantum Mechanics (QM)  multiple linearly independent eigenvectors having the same eigenvalue. One way of thinking of this is a "degeneracy pressure" that keeps them apart. In simple terms, particles which have different quantum states can't be merged together under normal circumstances.
There is one exception  black holes. They are an enigma from the point of view of QM because fermions appear to be compressed down to a single point in space, which is degenerate  the extreme gravity overcomes the quantum force that keeps them apart. But it's only degenerate from the point of view of 3 spacial dimensions. If other spacial dimensions are accessible in a black hole, QM may hold so that fermions retain their distinct quantum states in those other dimensions. Since we have no way of getting information from inside a black hole and this explanation doesn't arise from some theory that explains other things, this idea would not be considered seriously. Another imaginative idea, that black holes are a "gateway to another universe" isn't credible  the matter and energy that disappear into a black hole measurably exert their gravitational effect here in our 4dimensional spacetime. And actually, we don't have evidence that the contents of a black hole are concentrated at a single point; there could be a hard core of material well within the Schwarzschild radius. Gravitationally, they look the same from the outside.
In contrast to fermions, a BoseEintein Condensate (BEC) has multiple identical bosons (particles of integer spin, like helium nuclei) acquiring the same (lowest) quantum state by extreme cooling, so you can't tell where one gas particle starts and another ends  the collection of gas particles behaves statistically as if it's one particle. The Riemann Zeta function of (3/2) appears in the calculation of the critical temperature at which the collection of gas particles becomes a BEC.
Guess where else the Riemann Zeta function appears. The zeroes of the Zeta function have a connection with the distribution of prime numbers. If you can prove that the nontrivial zeroes all lie on the line in the complex plane with real part = 1/2, there's a million dollar prize waiting for you at the Clay Mathematics Institute.  

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The approach I took always ended in an insurmountable block one way or another whenever an error was corrected. So I've abandoned that approach and tried a different one, very different. Submitted for review. Let's see if this one sticks.  

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The reviewer didn't understand it. He thinks I used some incorrect logic with upper bounds. At first blush I took him at his word. Then I realized it's my fault for not explaining clearly enough that what I did concerns partitioning the conjecture to prove the parts separately. I clarified and told him not to bother rereading it with the new understanding, wait for the rewrite.
While waiting for that feedback on ideal Waring's conjecture, I took a first stab at proving the Collatz conjecture, it looks related (both conjectures involve multiplications by 3 and divisions by 2). It would be a gangbuster of a paper to prove both conjectures with one blow. I'd like to stop the waste of compute cycles and energy trying to find a counterexample to the Collatz conjecture (one click and you'll see I was guilty of that). A couple of interesting lemmas came out the first day of work, like: there are no length1 cycles except the trivial cycle, and there are no length2 cycles. Of course the state of the art says there are no nontrivial cycles of length less than 2^{68} (based on the state of the Collatz search). I'm not peeking at prior art so that I can continue thinking outside the box  actually, I don't even know where the box is because I haven't looked. But I'll pause for a while, there are other demands for my time.  

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I really don't see the big deal.
When I reduce the proof to the essential, this is what I have for the proof of the ideal Waring's conjecture.
The ideal Waring's conjecture posits that g(k) = 2^{k} + q_{k}  2 for k >= 2 is the exact formula for the original Waring's Problem,
by conjecturing that the condition q_{k} + r_{k} <= 2^{k} holds for k >= 2.
We proceed here to prove that this condition is satisfied.
By definition Euclidean division of positive integers, 3^{k} divided by 2^{k} has
a quotient q_{k} = FLOOR(3^{k}/2^{k}) and
a remainder r_{k} = 3^{k}  2^{k}q_{k} with 0 <= r_{k} <= 2^{k}1
3^{k} is coprime with 2^{k} for k>=1 so q_{k} < 3^{k}/2^{k} for k >= 1.
We know that 3^{k} < 3^{k} + 1 <= 4^{k} for k >= 1.
So 2^{k}q_{k} + r_{k} < 2^{k}q_{k} + (2^{k}1) + 1 <= 4^{k} for k >= 1
Using the right hand inequality
2^{k}q_{k} + 2^{k} <= 4^{k} for k >= 1
q_{k} + 1 <= 2^{k} for k >= 1
So q_{k} < 2^{k} for k >= 1 is a proven fact because we have used only known facts with the definitions of q_{k} and r_{k}.
The conjecture proposes that
q_{k} + r_{k} <= 2^{k} for k >= 2
Substituting the definition of r_{k}
q_{k} + 3^{k}  2^{k}q_{k} <= 2^{k} for k >= 2
Adding 2^{k}q_{k} to both sides
q_{k} + 3^{k} <= 2^{k} + 2^{k}q_{k} for k >= 2
We showed above that q_{k} < 3^{k}/2^{k} for k >= 1 then
q_{k} + 3^{k} <= 2^{k} + 2^{k}q_{k} < 2^{k} + 2^{k}(3^{k}/2^{k}) for k>=2
Simplifying the rightmost term
q_{k} + 3^{k} <= 2^{k} + 2^{k}q_{k} < 2^{k} + 3^{k} for k>=2
Taking the extreme left and right sides of the inequalities
q_{k} + 3^{k} < 2^{k} + 3^{k} for k>=2
q_{k} < 2^{k} for k>=2
Through substitution of known facts into the condition we have transformed the condition into a statement which matches a previously proven fact so the conjecture is true.
Is there something incorrect here?
 

Ravi FernandoProject administrator Volunteer tester Project scientist Send message
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Your logic goes in the wrong direction. You've assumed that
q_{k} + r_{k} <= 2^{k} for k >= 2
and proved under this assumption that
q_{k} < 2^{k} for k>=2.
You want to do the opposite: given the fact that q_{k} < 2^{k} (which you correctly proved), can you conclude that q_{k} + r_{k} <= 2^{k}? There is no obvious reason this should followit could be that r_{k} is very close to 2^{k} for some k, in which case q_{k} + r_{k} = floor(3^{k}/2^{k}) + r_{k}, which is approximately (3^{k}/2^{k}) + 2^{k}, and in particular greater than 2^{k}.
If k is large, this is very unlikelyI would expect q_{k} to behave like a "random" number in the interval [0, 2^{k}), and very few such numbers are close enough to 2^{k} to make the conjecture go wrong. But I for one have no idea how one could prove such a thing.  

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Your logic goes in the wrong direction. You've assumed that
q_{k} + r_{k} <= 2^{k} for k >= 2
and proved under this assumption that
q_{k} < 2^{k} for k>=2.
You want to do the opposite: given the fact that q_{k} < 2^{k} (which you correctly proved), can you conclude that q_{k} + r_{k} <= 2^{k}? There is no obvious reason this should followit could be that r_{k} is very close to 2^{k} for some k, in which case q_{k} + r_{k} = floor(3^{k}/2^{k}) + r_{k}, which is approximately (3^{k}/2^{k}) + 2^{k}, and in particular greater than 2^{k}.
If k is large, this is very unlikelyI would expect q_{k} to behave like a "random" number in the interval [0, 2^{k}), and very few such numbers are close enough to 2^{k} to make the conjecture go wrong. But I for one have no idea how one could prove such a thing.
OK, thanks.
I think you mean that r_{k} appears to be "random", not q_{k}, since the ratio q_{k+1} / q_{k} converges monotonically to 1.5, whereas r_{k} bounces around in the range [0, 2^{k}) somewhat chaotically.
How would I publish the surprising things I know about this in a serious journal without it costing me $3600? This is selffunded research.  


>How would I publish the surprising things I know about this in a serious journal
>without it costing me $3600? This is selffunded research.
http://revistas.ufro.cl/ojs/index.php/cubo/index
" CUBO, A Mathematical Journal is a scientific journal founded in 1985 by the Department of Mathematics and Statistics of the Universidad de La Frontera, Temuco, Chile.
...
Our Journal is entirely free of charge to both authors and readers. All publication fees are covered by Universidad de La Frontera. "  

compositeVolunteer tester Send message
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I mostly abandoned the spreadsheet and switched to wxMaxima, which carries multiprecision math far beyond 2^32.
It is comparatively a joy to use and permits direct entry of algebraic relations.
Autosave tripped me up once and accidentally wiped out a workbook, so I've turned that feature off in the configuration.
As for progress, it is so far away from status quo that barely any sequences I routinely use have hits in OEIS.
I saw one recently which proffered a "backoftheenvelope proof sketch" for Waring's Problem. OEIS A297446
Where it goes wrong is: "add the restriction that x is the common floor of (3^{n}  1) / (2^{n}  1) and 3^{n} / 2^{n}".
That's not an assumption we are allowed to make.
I initially went in nearly the same direction:
3^{k} = 2^{k}q_{k} + r_{k} with 0 <= r_{k} < 2^{k}
and
3^{k} = (2^{k}1)u_{k} + v_{k} with 0 <= v_{k} < 2^{k}1
Equating the two and solving for q_{k} + r_{k}
q_{k} + r_{k} = (2^{k}1)(u_{k}q_{k}) + v_{k}
In most cases u_{k} = q_{k} so q_{k} + r_{k} = v_{k} which we know is < 2^{k}1
The hard part is constraining v_{k} <= 1 when u_{k} = q_{k} + 1
which is like proving 3^{k} <= (2^{k}1)(q_{k}+1) + 1
Intuitively u_{k}  q_{k} decreases monotonically,
so it is only necessary to show that u_{k} = q_{k} for k > 2.
I could not prove this, whereas the proof sketch in OEIS simply denies this is possible by declaring x = u_{k} = q_{k}.
The above is more general than just 2's and 3's.
Pick any coprime N and D for N^{k}/D^{k}, and you will have
N^{k} = D^{k}q_{k} + r_{k} with 0 <= r_{k} < D^{k}
and
N^{k} = (D^{k}1)u_{k} + v_{k} with 0 <= v_{k} < D^{k}1
so
q_{k} + r_{k} = (D^{k}1)(u_{k}q_{k}) + v_{k} with 0 <= r_{k} < D^{k} and 0 <= v_{k} < D^{k}1
That's kids' play. What I have is wildly unexpected, which I've been trying to exploit for the proof.  

Nick Send message
Joined: 11 Jul 11 Posts: 2149 ID: 105020 Credit: 8,006,239,534 RAC: 2,382,941

This is so coolly unfathomable to me  


>How would I publish the surprising things I know about this in a serious journal
>without it costing me $3600? This is selffunded research.
http://revistas.ufro.cl/ojs/index.php/cubo/index
" CUBO, A Mathematical Journal is a scientific journal founded in 1985 by the Department of Mathematics and Statistics of the Universidad de La Frontera, Temuco, Chile.
...
Our Journal is entirely free of charge to both authors and readers. All publication fees are covered by Universidad de La Frontera. "
Wow puh32 there are some highly technical papers published in that journal. I'm surprised it is open access.
____________
 

compositeVolunteer tester Send message
Joined: 16 Feb 10 Posts: 1098 ID: 55391 Credit: 957,538,116 RAC: 993,618

We have a hypothesis to prove
a(x) <= b(x)
with no obvious solution in x
and
a(x) <= c(x) is true and has a solution in x
So then we solve
c(x) <= b(x)
for x to get
x >= constant (or conversely we get x<= constant, depending on the result)
Does this prove that a(x) <= b(x) for x >= constant ? (or conversely for x <= constant)
Or is this not allowed, and if not allowed then why not?
 


We have a hypothesis to prove
a(x) <= b(x)
with no obvious solution in x
and
a(x) <= c(x) is true and has a solution in x
So then we solve
c(x) <= b(x)
for x to get
x >= constant (or conversely we get x<= constant, depending on the result)
Does this prove that a(x) <= b(x) for x >= constant ? (or conversely for x <= constant)
Or is this not allowed, and if not allowed then why not?
Something that is valid is:
IF:
a(x) <= c(x) and c(x) <= b(x)
THEN:
a(x) <= b(x)
This is known as transitivity of the relation <=.
Your question is confusing as to whether you mean f(x) <= g(x) for all x (which is what I would think when you say "prove"), or for at least one x. Or if you just want to characterize the set of x for which it is true (which is what I would think when you say "solve").
Note that the set of solutions to an inequality of form f(x) <= g(x) can be more complicated than x >= constant, or x <= constant. For example, think of the solution to cos(x) <= sin(x).
/JeppeSN  

compositeVolunteer tester Send message
Joined: 16 Feb 10 Posts: 1098 ID: 55391 Credit: 957,538,116 RAC: 993,618

FINALLY! I got it, the proof we've all been waiting for,
after nearly 4 years of onandoff investigations and making a lot of dumb mistakes
(but I discovered some very interesting surprises in the process).
This puts to rest the last question about the "Classical" or "Ideal" Waring's Problem, which is
g(k) = 2^k + FLOOR(3^k/2^k)  2 for k > 0
being conditional upon 2^k{fractional part of 3^k/2^k} + FLOOR(3^k/2^k) <= 2^k
this condition having resisted proof since 1942. Unsurprisingly, the supercomputer search in 1988 found no exceptions.
2^k{fractional part of 3^k/2^k) is just the remainder of the division 3^k/2^k.
PROOF
[1] Define Q = FLOOR(3^k/2^k) the quotient
[2] Define R = 3^k  2^k Q the remainder, so 0 <= R < 2^k
Proof by contradiction of the condition
[3] Q + R <= 2^k for k > 0
For example, [3] is TRUE for k = 1, since Q = 1 and R = 1 and 2^k = 2.
Assume the opposite of [3]
Q + R > 2^k for some k >= 2
Equivalently we are assuming
[4] (Q + R)/2^k > 1
We know from [2] that 3^k = 2^k Q + R
so (2^k Q + R)/3^k = 1
Substitute this fact for the RHS of [4]
(Q + R)/2^k > (2^k Q + R)/3^k
Eliminating denominators by multiplying both sides by 6^k
3^k(Q + R) > 4^k Q + 2^k R
Rearranging to collect terms in Q and R
(3^k  2^k) R > (4^k  3^k) Q
Now from [2] substitute Q = (3^k  R)/2^k for Q in the RHS
(3^k  2^k) R > (4^k  3^k) (3^k  R)/2^k
Multiplying both sides by 2^k and distributing the multiplications
6^k R  4^k R > 12^k  4^k R  9^k + 3^k R
Simplifying
3^k R > 12^k  9^k
Dividing both sides by 3^k, the assumption becomes
R > 4^k  3^k
From [2] we know that R < 2^k so subtract 2^k from both sides
[5] R  2^k > 4^k  3^k  2^k
The RHS of [5] is dominated by the term 4^k, so it increases monotonically, and it's positive for k >= 2.
However the LHS < 0 for k > 0.
So then [5] is FALSE for k >= 2 contradicting our assumption.
Therefore [3] is TRUE for k >= 2. QED
If there are no errors, this proof will make the math news.
My name is Robert Lacroix, and you saw it here first, folks.
NB. The most interesting thing I discovered is a neat analytic formula for R.
It didn't play a role in this proof, but it did give me an idea that made this proof possible.  

compositeVolunteer tester Send message
Joined: 16 Feb 10 Posts: 1098 ID: 55391 Credit: 957,538,116 RAC: 993,618

I made another mistake, again, of course!
I found it myself while demonstrating it to someone.
6^k  3^k is not 3^k
Resuming from just before the error.
6^k R  4^k R > 12^k  4^k R  9^k + 3^k R
Simplifying
(6^k  3^k) R > 12^k  9^k
Dividing both sides by 3^k, the assumption becomes
(2^k  1) R > 4^k  3^k
Dividing both sides by (2^k  1)
R > (4^k  3^k)/(2^k  1)
And this is where a proof attempt runs into the usual trouble,
because the ratio on the RHS < 2^k so we can't contradict the assumption.
(2^k  1) R is not even monotonic.
I might be able to rescue it with the formula for R, but I doubt it.  

compositeVolunteer tester Send message
Joined: 16 Feb 10 Posts: 1098 ID: 55391 Credit: 957,538,116 RAC: 993,618

Here's a new idea, using Bezout's identity for the solution of the classic Waring's Problem conjecture.
The conjecture is q_{k} + r_{k} <= 2^{k} for k >= 2
where 3^{k} = 2^{k} q_{k} + r_{k} with 0 <= r_{k} < 2^{k} for k >= 0
First, recapping the old idea:
3^{k} = (2^{k}1) u_{k} + v_{k} with 0 <= v_{k} < 2^{k}1 for k > 0
Equating the two relations for 3^{k} and rearranging we get
q_{k} + r_{k} = (2^{k}1)(u_{k}q_{k}) + v_{k} for k > 0
and since v_{k} < 2^{k}1
q_{k} + r_{k} < 2^{k}1 when u_{k} = q_{k}
proving the case of the conjecture the when u_{k} = q_{k}
I can dig out an old proof that u_{k}  q_{k} < 3 and that u_{k}  q_{k} = 2 only at k = 1.
The problem is that I was never able to limit k in the case where u_{k}  q_{k} = 1
because the conjecture is also satisfied if v_{k} <= 1 but not if v_{k} > 1.
So u_{k}  q_{k} = 1 could happen anywhere for k > 2 and the conjecture would fail.
Now, the new idea:
Using a snippet from the constructive existence proof by induction of the Chinese Remainder Theorem,
(the case of two moduli), a solution of Bezout's Identity a_{k} 2^{k} + b_{k} (2^{k}1) = 1 for some integers a_{k} and b_{k} is
3^{k} = r_{k} + (v_{k}  r_{k}) a_{k} 2^{k}
and another solution is
3^{k} = v_{k} + (r_{k}  v_{k}) b_{k} (2^{k}1)
So then
(v_{k}  r_{k}) a_{k} = q_{k}
and
(r_{k}  v_{k}) b_{k} = u_{k}
so
u_{k}  q_{k} = (r_{k}  v_{k}) b_{k}  (v_{k}  r_{k}) a_{k}
For the case where u_{k}  q_{k} = 1
1 = (v_{k}  r_{k}) (a_{k} + b_{k})
Since 1/(a_{k} + b_{k}) = v_{k}  r_{k} is an integer
(a_{k} + b_{k}) = 1 or (a_{k} + b_{k}) = 1 are the only solutions in integers.
For the case (a_{k} + b_{k}) = 1
we have v_{k}  r_{k} = 1
Putting these into q_{k} = (2^{k}1)(u_{k}q_{k}) + v_{k}  r_{k} we get
q_{k} = 2^{k}  2
Since q_{k} and 2^{k} both increase monotonically this only happens at k = 2
For the case (a_{k} + b_{k}) = 1
we have v_{k}  r_{k} = 1
Putting these into q_{k} = (2^{k}1)(u_{k}q_{k}) + v_{k}  r_{k} we get
q_{k} = 2^{k} which is possible only for k = 0 but u_{k} is undefined for k = 0
So (a_{k} + b_{k}) = 1 doesn't happen at all.
So u_{k}  q_{k} = 1 happens only at k = 2
I think that this finally proves the conjecture.  

Nick Send message
Joined: 11 Jul 11 Posts: 2149 ID: 105020 Credit: 8,006,239,534 RAC: 2,382,941

I am not in the mathematical way to say good luck, it is in theatre  Chookas and break a leg! If breaking a leg is lucky, I wish you to break 2.
I wish you to get the designation "straight from the book" (I am not sure if people know what I mean  Paul ErdÅ‘s.)  

compositeVolunteer tester Send message
Joined: 16 Feb 10 Posts: 1098 ID: 55391 Credit: 957,538,116 RAC: 993,618

Interesting, I just saw today a quote of Ennio de Giorgi (accomplished Italian mathematician), where he said
If you can't prove your theorem, keep shifting parts of the conclusion to the assumptions, until you can.
This is basically where I got to, where I can't show a result by mathematical induction using the assumption x >= 0,
but it works if I assume in x >= y, where y is the result I got when I assumed x >= 0. I should take that advice.  

Nick Send message
Joined: 11 Jul 11 Posts: 2149 ID: 105020 Credit: 8,006,239,534 RAC: 2,382,941

I like the idea of mapping out how you get from A to Z.
You have a starting point and you have a goal  how do you construct the path?
I understand that mathematics is the only rigorous subject  if that is the correct word.
Everything else has limitations  or a level of 'dodgy' some monumentally more than others.  

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