Hands, fists
Sticks, stones
Spears, knives, swords
Bows, arrows
Guns, cannons
Planes, bombs
Missiles, rockets
Thermonuclear bombs, glory
Blood
I’m just a glorified ape, a glorified monkey
The piano is just a glorified harp
This glorified monkey plays a glorified harp
We’ve come a long way, baby.
The end? The beginning?
Consciousness has no name
Silence is forever.
withthesongstillinsidethem 
WELL, ALL OF A SUDDEN WORDPRESS CHANGED THE METHODS FOR WRITING A NEW POST, AND I CAN’T WRITE A NEW POST! THUS, I AM POSTING HERE AS A MERE “REPLY” TO MY POEM, A PROOF OF THE GOLDBACH CONJECTURE. Since it is impossible that I have succeeded in actually proving the Goldbach Conjecture (GC), this “proof” must be FATALLY FLAWED. However, it is now 827 am on Sunday, Jan 17, 2021 and I can not go through this “proof” in depth to find out what is wrong, so I’ll just post it here as a reply to the poem.
The “proof” occurred in an email to my friend Dr. Fred Strohm, along with comments on the pitiful and terrible state of my country, America. So, here it is:
GC SIEVICS Re: THE STROHM SIEVE
Alan DiCenzo
Sun 1/17/2021 8:04 AM
Hi Fred,
Well, since I awakened at 5 am, maybe I’ll try to correct a few mistakes from my previous sieving email. This won’t lead to the success we wish for, but it may lead at least to a more correct failure. So, please allow me, embarrassed and red-faced, to correct in red what I can find for now:
From: Alan DiCenzo
Sent: Saturday, January 16, 2021 11:31 PM
To: Fred Strohm
Subject: THE STROHM SIEVE
Hi Fred,
Yes, very, very sad.
As part of an effort to escape the pain of watching my country turn into yet another of the standard, boring Planet Earth despotic dictatorships, I started Goldbaching around again tonight, just now this minute in fact, at 953 pm, as I was beginning to attempt to fall asleep in my recliner chair. As you know all too well by now, this is how I commonly try to help myself to fall sleep: Thinking about some math problem, using only elementary techniques because these are often the only techniques I remember at this point in my life, if I’m lucky, after awhile I find myself drifting off into oblivion.
So, this time, no sooner had I started Goldbaching around, when I found myself next Strohming around. More specifically, all this sieve nonsense you have recently been pumping into my brain, which itself is probably a pretty leaky sieve already, caused me to hatch an idea for yet another crazy sieve, which naturally I am now calling the Strohm Sieve (Full name: The Fred Strohm Sieve Of Pairwise Prime Sums). It’s really a “double sieve,” which at one and the same time generates an increasing sequence of primes, and an increasing sequence of even integers.
As I continued along this crazy, zigzag path, I came to realize that mathematicians are not what we need here at all! No; what we need are instead CARPET SALESMEN AND CARPET INSTALLERS. Maybe that is why we’ve had such a problem with this GC for all these many years.
So, once again, I must request your patience, perhaps even more so this time, because in my new profession as a CARPET SALESMAN AND INSTALLER, I have but limited experience. May God forgive…..
The FSSOPWPS goes like this:
Our first sieve prime is 3.
So then we create our first pairwise sum, namely, 3+3=6.
Naturally, we did that, because so far we only have one prime, namely 3, so we could not generate any more pairwise sums beyond 3+3=6. We take the time to note then that we have used ONE prime, 3, so far, and with it, we have covered precisely one even integer of CARPET with a sum of two primes, namely, 6.
Entries so far in the Strohm Sieve:
3,(3+3),6
Next we bring in the next prime, 5. We then produce new sums in order as follows:
5+3=8 and 5+5=10
Notice that in producing our new sums, we “went back” and “re-used” all the primes we had used previously, in ascending order, to produce our new sums with our latest, newest prime. So far, we only had two primes to re-use, but we we the previous two primes appearing in ascending order as the second term of the new sums.
Thus, we now have these entries so far in the Strohm Sieve:
3,3+3,5+3,5+5
and we have covered in PLUSH CARPET the three even integers
6,8,10
with NO GAPS so far in our carpet of even integers.
Using more or less obvious bookkeeping, we have next in the Strohm Sieve, sums created with the next prime, 7:
3,3+3,5+3,5+5,7+3,7+5,7+7
which covers in plush carpet the even integers
6,8,10,12,14
Some even integers have already been covered twice in plush carpet, but we don’t care about that. What’s important is that again we have no GAPS in our carpet of primes.
I’ll give just one more SS entry, and then explain the general rule for the sieve:
3,3+3,5+3,5+5,7+3,7+5,7+7,11+3,11+5,11+7,11+11
which covers in plush carpet the even integers
6,8,10,12,14,16,18,22 (1)
Oh crap! A GAP in our carpet, at even integer 20.
But, what we can see is that our carpet has NO gaps at least up to TWICE the last preceding prime, in this case
2*7=14.
Now, the rule: I’ll bet you don’t even need the rule stated, but, here it is: Each time we bring in a new prime, namely the next higher prime, we produce all the new possible pairwise prime sums in order, STARTING AGAIN with the first prime, 3, but this time added to the LATEST NEW PRIME, and then we add in sequence all the other lower primes to this new, latest, highest prime.
I’ll be you can also see already that I am trying to hatch in my madcap, leaky-sieve brain an inductive formulation. And so I am:
Let me assume now inductively that at each stage in the SS, I have COVERED IN PLUSH CARPET WITH NO GAPS, all the even integers up to TWICE the value of the last previous prime. More precisely:
When I have brought in the kth prime, pk, and then produced all the possible pairwise sums with it, I note inductively that I have COVERED IN PLUSH CARPET WITH NO GAPS, all the even integers up to and including 2*(p(k-1)).
Thus for example, if you refer to (1) you can see that as I use the prime 11, I will have covered with NO gaps all the even integers up to and including 2*7=14, i.e., twice the last previous prime.
Wouldn’t it be nice if now, at this point, we could argue that when we introduce the next prime p(k+1) we will have covered with NO GAPS all the even integers up to and including the sum pk+pk=2*pk?
Uh….yes, it WOULD be nice, because that would PROVE GC, because we will have “chained inductively” the case of p(k-1)+p(k-1) to the case of pk+pk! Therefore, it seems reasonably certain that I will NOT be able to successfully argue my inductively chaining argument! But, being an arrogant fool who just wants to have fun, let me proceed anyway, until you can see with no longer any disguises what a fool I truly am:
So, what happens when we introduce a new prime, the new next prime, p(k+1)?
What happens is, we see that we have created an entirely new swatch of carpet, BUT, you can see that this new swatch of carpet is a perfectly accurate “xerox copy” of the earlier COMPLETELY COVERED SWATCH OF CARPET we had already previously created, which was completely covered from 3+3=6 onward, up to and including 2*(p(k-1)). But, this new, XEROX COPY of carpet is RIGHT-SHIFTED from the original by the amount of our newest prime, p(k+1).(!) (I used the parentheses so you would not be thinking of factorials.) In other words, with p(k+1) we have started a new swatch of carpet with a new “origin,” or “zero,” which looks just like the swatch just previously created by using pk, but moved to the right and thus occurring starting at the “new origin of” p(k+1).
Thus this new swatch covers completely, with no gaps, carpet starting at p(k+1) and going all the way out to 2*(p(k-1)) + p(k+1), due to the right shifting by p(k+1). That’s because we already knew (inductive assumption) that the no-gap covered carpet had to be at least 2*(p(k-1)) units wide, and it is right-shifted to a new “origin” at p(k+1). Is that good news?
I think the answer is yes, because of an ancient Strohm lemma which you introduced to me awhile back: Strohmian Lemma No. 1, still retained in DiCenzo sieve brain, which says: For any integer n, there is always at least one prime between n and 2n. (I probably got that wrong, but it may still be “close enough,” as we government workers and engineer-types are prone to say. For example, maybe the lemma requires n to be a prime, but that will work for us too.) To use this lemma, I take for the integer n, instead, the integer p(k-1). Here’s a “picture”:
______p(k-1)______2p(k-1)_________________
pk 2pk
p(k+1) 2p(k+1)
p(k+1)+2p(k-1)
(2)
So, if I got this right (Is there any chance of that?), we have from picture (2) and logic, the carpet from 0 to 2p(k-1) is completely covered. Then also, we know that the carpet BETWEEN p(k+1) and p(k+1)+2p(k-1) is completely covered. (We also know that the swatch of carpet created by using prime p(k+1) in the sieve, goes BEYOND p(k+1)+2p(k-1), but we’re not sure it’s completely covered with no gaps of even integers, in this “beyond” region.)
Note now that in picture (2) we have drawn pk between p(k-1) and 2p(k-1). That works because Strohmian Lemma 1 says that there must exist at least one such prime, so we know that that is either pk itself (the next prime after p(k-1)) or a later prime. Thus, naturally, we choose pk as the one we show in the picture (2). Similar reasoning applies to the placement of prime p(k+1) in picture (2).
Thus, as we have just explained, we know from Strohmian Lemma 1 that we may write the inequalities
2p(k-1)>pk
2pk>p(k+1) (3)
Then, from (3) and (2),
p(k+1)+2p(k-1)>p(k+1)+pk>2pk (4)
Now, at this juncture, the hope from (4) is that we have successfully covered carpet all the way out to 2pk, with NO gaps! (But, have we?) IF we have, then we have completed our inductive assumption.
So, let’s take a look.
Well, first, we are at least “no-gaps covered” (ngc) over a carpet strip from p(k+1) all the way out to p(k+1)+2p(k-1), again via our inductive assumption.
But now, the real question is, how can we guarantee we are ngc between 2p(k-1) and p(k+1)?
Ah, so close and yet so far!
It’s now 1125 pm and I’ve been thinking in real time and writing this, all starting at 953 pm, without going back or checking anything, so God only knows how many terrible, sloppy and fatal errors I’ve already made.
But, maybe you can correct my errors and fill in the “gaps.” (Pun not intended.)
First, you might verify whether I was right in saying I have two ngc regions namely (0,2p(k-1)) and the half-closed interval [p(k+1),p(k+1)+2p(k-1)).
Second, can you fill in the gap between these two ngc regions?
Re this pesky gap, if we could have just a slightly stronger version of Strohmian Lemma 1, namely that there are always at least TWO primes betwee n and 2n, then we’d know our pesky gap is covered, because then the second Strohmian Lemma prime in our case would HAVE TO BE p(k+1) and p(k+1) WOULD HAVE TO BE , there always exists at least one prime number with < there is always at least one prime such that < <. Another formulation, where is the -th prime, is for ≥ + 4 quintillion. Since these n are >Sandy’s Candy Crush score, I conclude that we are indeed talking only about large n. Therefore we may conclude that we always have at least TWO primes between n and 2n. Therefore, using for n our special case of n=p(k-1) we have that there are at least TWO primes between p(k-1) and 2p(k-1). But that has to mean that both p(k) and p(k+1) are between p(k-1) and 2p(k-1), which in particular proves that p(k+1)<2p(k-1) (5)
Eq (5) thus proves that we have ngc in the formerly worrisome region between 2p(k-1) and p(k+1), because in actuality this "region" is not a region at all, because p(k+1) is actually LESS than 2p(k-1).
Thus, (5) establishes that GC is true. QED.
Hmmm…..wait a minute. QE[WRONG]. There MUST be something wrong here. It is IMPOSSIBLE that I have just proven GC, the Goldbach Conjecture.
But it is 758 am and I need to go back to sleep. I am losing my country which used to shine at the top of a hill but which is now turning into just another standard despotic dictatorship, and this gives me a lot of pain. Further, I am likely to die in the next month or two of c19 because I can not get the damn vaccine owing to the stinky politics of how the vaccine is being prioritized. And finally, I have proven the Goldbach Conjecture? Um…..I don't think so. But I can't find the mistake this morning and I am going back to sleep.
IF (and only if) you are so motivated, ye author of Strohmian Lemma No. 1, please comb through my terrible mess above and find what is wrong, and then please show me. In the meantime, I hope we have had fun fooling around with this damn thing. And, I'll quit here in wishing that Sandy may soon improve her Candy Crush score to 4 quintillion.
Thankfulness to 40,000 quintillion (40 sextillion? I didn't want to introduce sex into the picture, but there it is….), aloha, Alan, your eternally nl0, 805 am.
But, third, even if we can NOT fill in this gap, I think we may be able to construct a similar inductive chain of thought by using as the ngc carpet region not the one all the way out to 2p(k-1), but maybe only out to 2p(k-2).
What thinkst?
I should really clean up this mess now instead of leaving this pile of crap in your hands, but I'm tired and going to go to sleep, or at least to bed, which means I am quitting. Please help if you can!
Thanks 40,000 quintillion, aloha, Alan
From: Fred Strohm
Sent: Saturday, January 16, 2021 9:11 PM
To: Alan DiCenzo
Subject: Re: CORRUPTION HUMAN RIGHTS FREEDOM
Hi Alan,
A very sad situation.
Power corrupting the corrupt.
All the best,
Fred
On Sat, Jan 16, 2021 at 8:42 PM Alan DiCenzo wrote:
Hello everyone who will read this,
THE LEVEL OF CORRUPTION IN AMERICA HAS REACHED A THRESHOLD WHERE THE PRESERVATION OF FREEDOM AND THE PROTECTION OF HUMAN RIGHTS ARE NO LONGER POSSIBLE. Meanwhile, as always when this happens, the great mass of the people are sleeping.
We are now no longer different from Communist China, the old Soviet Union, North Korea, Cuba and Venezuela.
How long will THIS VERY POST last on THIS social media? This is a TEST.
The courts no longer do their sacred duty.
We have a one-party system.
Big Tech and the Mass Media now run an INFORMATION MONOPOLY as tight and irresistible as anything in the world’s great despotic dictatorships.
It matters not that now the choke-hold on information is run by Big Tech and the Mass Media, supposedly private enterprise entities, as opposed to the government. Their stranglehold is every bit as murderous as anything our Founders could have ever envisioned, every bit as choking as anything in Nazi Germany or Communist Russia or Communist Korea.
America, America, if you don’t wake up right now and see this, and, despite party affiliation, IMMEDIATELY AND URGENTLY make laws and run challenges in the courts to overthrow these tech monopolies and bring some true COMPETITION INTO THE MARKET so that WE THE PEOPLE AND OUR DIVERSE LEADERS TOO can continue to communicate in openness and freedom, then nothing else will matter.
NOTHING ELSE WILL MATTER. THIS IS NO LONGER ABOUT TRUMP, BELIEVE ME.
Cyber Anguish
GOLDBACH CONJECTURE (GC) NEW COMMENT ON POSSIBLE RECURSION:
Here’s another, probably crazy and failed, attempt to prove GC.
GOLDBACH CONJECTURE: A STRANGE, UNEXPECTED LINE OF REASONING
Alan DiCenzo
Tue 1/19/2021 7:55 AM
Hi Fred,
What’s wrong with the following line of reasoning? Check:
Last night I kept trying to think of how I could revise my inductive scheme so that it would still take advantage of the Fred Strohm Sieve (the FS, to avoid any Nazi associations) and also be correct, while at the same time not violating the Strohm Principle Of Maximal Goldbach Prime Size.
After banging up against annoying constraints which seemed to railroad my mental wandering into a very narrow direction, the following strange line of reasoning finally occurred to me:
Suppose that we have a set Sk of the first k primes {p1,p2,….,pk} and the even integer Mk=pk+1. Suppose also that we constrain our choice of k so that the k primes cover all the even integers less than or equal to (leq) Mk, with no gaps, where “cover” means that all even integers up through Mk, i.e., including Mk, can be written as the sum of two primes in Sk. If the Goldbach Conjecture GC is to be true, then there would have to be at least one such interval and one such set of primes inside that interval, for, as you reminded me, any prime p(k+1) larger than Mk could not possibly contribute to a sum of two primes leq Mk.
Well, let’s highlight here, regardless of whether GC is true, we know empirically that we certainly have such intervals and prime sets galore, e.g., with sizes up to 4 quintillion. I started thinking of using one somehow in the inductive hypothesis (ih) in my new inductive scheme. But I wasn’t sure exactly how.
It was then I remembered how I’d previously communicated to you that that same set Sk would have to already include enough primes to cover Mk+2, without any need or use for larger primes than pk, because, for example, even if the odd integer Mk+1 were a prime, it could not help to generate a cover for the even integer Mk+2 because Mk+2 would be simply (Mk+1)+1 and 1 is not a prime in the sense we are considering. And, in fact, we know anyway that since pk=Mk-1 is a prime, then pk+3 is a sum of two primes which covers Mk+2.
As a matter of fact, we can clearly see that since Mk is just pk+1, Mk+2 must be simply pk+3, where 3 is just p1. But then I remembered that also, we would have to cover Mk+4, because that would just be pk+5. And we would of course cover Mk+6, which would just be pk+7. (We may clearly assume that Mk is so large that we have no trouble including tiny primes like 5 and 7.)
It was then I remembered in this context what I had not been thinking before in the context of Sk and the interval Ik=[3,Mk), because I had not yet created in my own mind the idea of FS, the Fred Strohm Sieve Of Pairwise Prime Sums: I knew from thinking in terms of FS how a prime pk uses all the primes <pk all over again in the sieve FS, when it is first introduced into the sieve FS, to create a "xerox copy" of the coverage of the interval up to (but not including) Mk, but now this "xerox copy" is RIGHT SHIFTED so that it starts right at pk, and not at zero.
In other words, it's as if we "shifted the origin 0 to a false new origin at pk," and then recreated the same covered interval we had before, but this time the covered interval (with no gaps again) runs from pk through pk+3,pk+5,pk+7,…,[pk+p(k-1)],pk+pk. That is, our "virtual copy" of coverage which comes from adding in pk to the FS, looks just like the coverage from 3 to Mk-1 we already had, since we chose our k, Mk and Sk empirically to satisfy the requirement of coverage.
(Note that the new collection of sums seems at this point to only cover as far as the even integer (pk+pk), which is only (2Mk-2).)
But now let's look very carefully at what we just said, and what we just demonstrated. We made no assumptions whatsoever other than the following: We took the first k primes, which we called the set Sk, where we chose k specifically such that our primes cover all the even integers up to and including Mk=pk+1. We know empirically that such a prime set Sk and even integer Mk exist, without any need to assume GC is true. And from ONLY these assumptions, we discovered that we also covered all even integers up to 2Mk-2.
In simple terms, we created an "if…then" where an empirical "if" created a "then" of a whole new extended interval of coverage all the way to about twice the original coverage. So, naturally, we want to do it again.
But can we do it again? What situation are we in now that we have found we are covered all the way to (2Mk-2)? Well, first, the bad news: (pk+pk) is not a prime, and we also do not know that (pk+pk-1) is a prime; in fact 2pk-1 is likely NOT a prime.
Reviewing, we had set up our inductive assumption to that a prime, namely pk, butts up right against Mk, and we needed that for everything else to follow (Did we really? I think we did.) And now, the "second time," we do not have that fortuitous event. So, are we lost, and unable to repeat what we did?
Maybe not. It's at this point where we may benefit if we recall Strohm-Bertrand (SB), which guarantees for us that there IS a prime, at least one of them, in the interval (Mk,2Mk-2); why not call it pj for now?
But, uh, wait a minute. Strictly speaking, SB only guarantees that we have such a prime in (Mk,2Mk) and we want it in the slightly shorter interval (Mk,2Mk-2). But, that's OK, because we can simply have recourse to the stronger version of SB, which says that there are many primes in this interval, and in particular certainly at least two. Thus, even if one of them is between 2Mk-2 and 2Mk, another has to be less than 2Mk-2. Good enough; that can be our pj. In fact, we might as well choose pj to be the prime closest to 2Mk-2 on its left.
This now leaves us with a new, potentially decent interval to use for an attempt at a second "go-around," namely the interval [3,pj+1). Let's call pj+1 simply Mj for now. We have now a new interval, [3,Mj). A little bookkeeping verifies that our new prime set Sj={p1=3,p2,…,pj} covers our new interval Ij=[3,Mj) completely with no gaps.
So, as best I can tell, we could do it again, and we did. But that means we can do it again, and again, and again, infinitely many times. But if we can keep doing this thing over and over, infinitely many times, then GC is true.
Uh oh. Now I need you again, Fred. Are you motivated enough to want to help by finding the mistake(s)?
At the moment, and I haven't checked my thought or anything else yet, the only risk I can think of is that somehow my succession of generated longer intervals "collapses" in the sense that I'm only adding an extension by one integer which somehow will not do the trick, but I don't see how that is possible. I think we're OK on that one. So, once again, the reasonable probabilities tell us that we need to find the mistake, the flaw, the fatal flaw(s). Either that, or I am in the carpet selling and installing business again, which may allow us to correct our mistake and come a little bit closer to GC. (???)
Thanks, aloha, Alan, nl0, still failing in the carpet business? Still crazy after all these years….
BUT, here is a better reply to the first reply, giving a better version of my creation of what MUST BE A FAULTY PROOF OF THE GOLDBACH CONJECTURE, but I haven’t yet been able to find the fault in my own proof:
GC CORRECT INDUCTIVE HYPOTHESIS? AND DUSART IS EVEN BETTER THAN PNT. Re: GC question
You replied on Sun 1/17/2021 11:06 PM
Alan DiCenzo
Sun 1/17/2021 10:51 PM
Hi Fred,
Ha! You’re not slow at all, and this one probably IS really complicated and hard to understand, so I’ll even give MYSELF a bit of a break for not having yet figured out how to communicate it clearly to you or anyone else. Also, that might even be the reason the real FATAL ERRORS are still hiding in there so that I can’t yet see them myself: Maybe I don’t fully understand even what I did myself. It FEELS natural at this point to me but it still doesn’t seem very simple. I wish it were simpler.
And it’s all YOUR fault anyway for managing to get my mind hooked with all your mentioning of sieves, and going backward in sieves, etc.. If it hadn’t been for you I never would have thought of all of this stuff at all.
Now, in trying to look at your questions below, I’m scratching my head trying to figure out how to understand them in turn. Let me consider:
“If there are gaps in the even integers less than
2p[k-1], why are these gaps filled in when we add p[k] to all the primes up to p[k-1]? Isn’t that supposed to happen?”
I’m confounded as I try to understand and address the question. I may be just spouting stupid things, so please guide me in the right direction if the following is meaningless in relation to your question: The “gaps” in the even integers less than 2p(k-1) are there ONLY when we restrict ourselves to considering pairwise sums which use NO primes as summands LARGER than p(k-1). But, when we then bring in a whole bunch of new sums by using p(k) as one of the summands, and thus considering all these new pairwise sums for the first time:
(3+pk),5+pk,…p(k-1)+pk,pk+pk
then we find that some of these NEW pairwise sums filled in for the first time the missing gaps in the even integers up to 2p(k-1)=p(k-1)+p(k-1).
But THAT is the inductive assumption. We do NOT know at all, generally speaking, that bringing in a bunch of new sums involving pk will somehow magically fill in the gaps occurring all the way up to p(k-1)+p(k-1).
As with all inductive proofs, we have a tendency at first to say, “But we don’t KNOW that this is true for k!” and that is always exactly true for an inductive proof. The whole point of an inductive proof, is to show that IF it WAS true for k, then you would not be able to avoid having it end up also being true for k+1. And inductive proof is never about the truth of the kth case, but rather always about the truth of the strength of the LINK between k and (k+1).
Once we verify the link is infallible, then we’re home free because the inductive assumption WAS known to be true for ONE SMALL EARLY, FIRST case, whatever that first case was.
Repeat of the inductive proof formulation:
1) First proof that the inductive hypothesis IS true for some appropriate FIRST case, say k=1, or k=3 or k=4 quintillion;
2) Then show that IF the inductive hypothesis is true for k, the ASSUMED truth for k forces the inductive hypothesis to be true for (k+1).
Part 2) of the inductive formulation is all about the strength of the chain LINK. Part 1) is all about proving that the FIRST chain link is really there to begin with.
So, OUR inductive assumption (which can always be anything you want, as long as you’ll be able to use it in your inductive formulation) is that “Bringing pk into the Strohm Sieve forces ALL even integers up to and including 2p(k-1) to be covered, whether or not they were already covered before.”
You’re right; we DON’T KNOW that the inductive assumption is true at k, for k. But our inductive proof involves proving the following for part 2):
IF this thing in quotes is true: “Bringing pk into the Strohm Sieve forces ALL even integers up to and including 2p(k-1) to be covered, whether or not they were already covered before,”
THEN it will be FORCED to be true that this SECOND thing in quotes is also true as well: “Bringing p(k+1) into the Strohm Sieve forces ALL even integers up to and including 2p(k) to be covered, whether or not they were already covered before.”
We can see that the red “IF THEN” has the exact structure required for an inductive proof part 2): IF bla bla k, THEN bla bla (k+1).
At that point, only ONE MORE THING is needed for the inductive proof to work, and that is to SHOW that this EXACT SAME INDUCTIVE HYPOTHESIS actually IS true, or actually WAS true (pick your favorite wording) for k=some initial value. But we DID previously show that our exact inductive hypothesis was true for say, k=4, where p1=3, p2=5, p3=7, and p4=11. Part 2 was so long that we may have forgotten part 1, but if you recall, part 1 was shown in (V1) and in words using the Strohm Sieve for k=4. We actually displayed the Sieve for k=4 and showed that the inductive hypothesis was true.
Then, you can see, part 2 says the chain link is strong, so k=4 forces k=5 to be true; k=5 forces k=6, etc..
Now, that all SEEMS correct to me, but it may STILL be the crux of where the ERROR lies. But, if so, I can’t exactly see why the exact inductive hypothesis I used is faulty, at least no today.
So much for that. I leave that mess for now, and look for a moment at the Prime Number Theorem. From wikipedia on the PNT, we see:
“Let π(x) be the prime-counting function that gives the number of primes less than or equal to x, for any real number x. For example, π(10) = 4 because there are four prime numbers (2, 3, 5 and 7) less than or equal to 10. The prime number theorem then states that x / log x is a good approximation to π(x) (where log here means the natural logarithm), in the sense that the limit of the quotient of the two functions π(x) and x / log x as x increases without bound is 1:
{\displaystyle \lim _{x\to \infty }{\frac {\;\pi (x)\;}{\;\left[{\frac {x}{\log(x)}}\right]\;}}=1,}
known as the asymptotic law of distribution of prime numbers. Using asymptotic notation this result can be restated as
{\displaystyle \pi (x)\sim {\frac {x}{\log x}}.}
This notation (and the theorem) does not say anything about the limit of the difference of the two functions as x increases without bound. Instead, the theorem states that x / log x approximates π(x) in the sense that the relative error of this approximation approaches 0 as x increases without bound.”
What the above quote shows is that the PNT is NOT a probabilistic formulation at all. It is an asymptotic one, with a definite limit equation given as x–>infinity. That means that there is NO chance that there are not at least two primes between n and 2n for n, say, = n=x=3 quintillion in the limit equation given. So we’re freed of worries about “probabilistic fogginess.”
But even THAT is not the end of the story. Further down in the same article we find:
“The prime number theorem is an asymptotic result. It gives an ineffective bound on π(x) as a direct consequence of the definition of the limit: for all ε > 0, there is an S such that for all x > S,
(1-\varepsilon ){\frac {x}{\log x}}<\pi (x)<(1+\varepsilon ){\frac {x}{\log x}}.
However, better bounds on π(x) are known, for instance Pierre Dusart's
{\frac {x}{\log x}}\left(1+{\frac {1}{\log x}}\right)<\pi (x)<{\frac {x}{\log x}}\left(1+{\frac {1}{\log x}}+{\frac {2.51}{(\log x)^{2}}}\right).
The first inequality holds for all x ≥ 599 and the second one for x ≥ 355991.[26]
A weaker but sometimes useful bound for x ≥ 55 is[27]
{\displaystyle {\frac {x}{\log x+2}}<\pi (x)<{\frac {x}{\log x-4}}.}
In Pierre Dusart's thesis there are stronger versions of this type of inequality that are valid for larger x. Later in 2010, Dusart proved:[28]
{\displaystyle {\begin{aligned}{\frac {x}{\log x-1}}&<\pi (x)&&{\text{for }}x\geq 5393,{\text{ and}}\\\pi (x)& 0, there is an S such that for all x > S,
{\frac {x}{\log x-(1-\varepsilon )}}<\pi (x)<{\frac {x}{\log x-(1+\varepsilon )}}."
May I give my heartfelt thanks to Dusart and the result in ref (27), for now we have removed all doubt that a lower bound on the number of primes between n and 2n is far, far above 2. We only need "2."
So, I guess we must still keep looking elsewhere, perhaps still for an error in the use of the statement for the inductive assumption which we actually used. But….so far I can't "break it."
Thanks again my infinitely patient friend, aloha, Alan, nl0
From: Fred Strohm
Sent: Sunday, January 17, 2021 10:02 PM
To: Alan DiCenzo
Subject: GC question
Hi Alan,
I’m sorry I’m slow at understanding your induction. Let me ask a question I have.
If there are gaps in the even integers less than
2p[k-1], why are these gaps filled in when we add p[k] to all the primes up to p[k-1]? Isn’t that supposed to happen?
When we do this with the first few small primes, it looks that way. But how do we know there won’t be any gaps left below 2p[k-1], after we add p[k] to all the primes up to p[k-1]?
Thanks for your help.
All the best,
Fred
***************** AND:
PROBABLE PROOF? Re: GC SIEVICS Re: THE STROHM SIEVE
Alan DiCenzo
Sun 1/17/2021 6:53 PM
Hi Fred,
“The PNT may only make this very highly unlikely. But does it actually make it impossible?”
Yes, I was aware of that. But, given the log(n) increase over time, the number of primes between n and 2n becomes so large that I don’t think there is truly at all a nonzero probability that the 10^10 primes between n and 2n by the time we get to 4 quintillion could ever become less than 2 again. But, yes, I don’t know the answer to that at the moment. The precise nature of the PNT is a bit foggy still to ME too! Still, I think we would have a “proof” stronger than a probability proof, if this “proof” weren’t also killed by more serious errors, which for the moment, I suspect it is.
In fact, I am about to look at your next email, where I will probably see the correctness of the fault discovery you have made in the “proof” and that fault discovery will be much more important than the question of whether our “proof” is ultimately a probability proof or not.
So, let me get to that valuable email now!
Thanks, aloha, Alan, nerd level 0
From: Fred Strohm
Sent: Sunday, January 17, 2021 1:39 PM
To: Alan DiCenzo
Subject: Re: GC SIEVICS Re: THE STROHM SIEVE
Hi Alan,
I have to spend some more time on your email to be sure I follow your reasoning that leads to the confirmation of GC if there are always at least 2 primes between any prime and 2 times that prime.
But the prime number theorem confuses me. It has been proven. But given its nature, which I don’t fully grasp, I am not sure that it implies there is never a gap between a prime and 2 times that prime with only one prime in the gap.
The PNT may only make this very highly unlikely. But does it actually make it impossible?
You have a better chance of judging this than I do.
All the best,
Fred
On Sun, Jan 17, 2021 at 8:04 AM Alan DiCenzo wrote:
Hi Fred,
Well, since I awakened at 5 am, maybe I’ll try to correct a few mistakes from my previous sieving email. This won’t lead to the success we wish for, but it may lead at least to a more correct failure. So, please allow me, embarrassed and red-faced, to correct in red what I can find for now:
From: Alan DiCenzo
Sent: Saturday, January 16, 2021 11:31 PM
To: Fred Strohm
Subject: THE STROHM SIEVE
Hi Fred,
Yes, very, very sad.
As part of an effort to escape the pain of watching my country turn into yet another of the standard, boring Planet Earth despotic dictatorships, I started Goldbaching around again tonight, just now this minute in fact, at 953 pm, as I was beginning to attempt to fall asleep in my recliner chair. As you know all too well by now, this is how I commonly try to help myself to fall sleep: Thinking about some math problem, using only elementary techniques because these are often the only techniques I remember at this point in my life, if I’m lucky, after awhile I find myself drifting off into oblivion.
So, this time, no sooner had I started Goldbaching around, when I found myself next Strohming around. More specifically, all this sieve nonsense you have recently been pumping into my brain, which itself is probably a pretty leaky sieve already, caused me to hatch an idea for yet another crazy sieve, which naturally I am now calling the Strohm Sieve (Full name: The Fred Strohm Sieve Of Pairwise Prime Sums). It’s really a “double sieve,” which at one and the same time generates an increasing sequence of primes, and an increasing sequence of even integers.
As I continued along this crazy, zigzag path, I came to realize that mathematicians are not what we need here at all! No; what we need are instead CARPET SALESMEN AND CARPET INSTALLERS. Maybe that is why we’ve had such a problem with this GC for all these many years.
So, once again, I must request your patience, perhaps even more so this time, because in my new profession as a CARPET SALESMAN AND INSTALLER, I have but limited experience. May God forgive…..
The FSSOPWPS goes like this:
Our first sieve prime is 3.
So then we create our first pairwise sum, namely, 3+3=6.
Naturally, we did that, because so far we only have one prime, namely 3, so we could not generate any more pairwise sums beyond 3+3=6. We take the time to note then that we have used ONE prime, 3, so far, and with it, we have covered precisely one even integer of CARPET with a sum of two primes, namely, 6.
Entries so far in the Strohm Sieve:
3,(3+3),6
Next we bring in the next prime, 5. We then produce new sums in order as follows:
5+3=8 and 5+5=10
Notice that in producing our new sums, we “went back” and “re-used” all the primes we had used previously, in ascending order, to produce our new sums with our latest, newest prime. So far, we only had two primes to re-use, but we we the previous two primes appearing in ascending order as the second term of the new sums.
Thus, we now have these entries so far in the Strohm Sieve:
3,3+3,5+3,5+5
and we have covered in PLUSH CARPET the three even integers
6,8,10
with NO GAPS so far in our carpet of even integers.
Using more or less obvious bookkeeping, we have next in the Strohm Sieve, sums created with the next prime, 7:
3,3+3,5+3,5+5,7+3,7+5,7+7
which covers in plush carpet the even integers
6,8,10,12,14
Some even integers have already been covered twice in plush carpet, but we don’t care about that. What’s important is that again we have no GAPS in our carpet of primes.
I’ll give just one more SS entry, and then explain the general rule for the sieve:
3,3+3,5+3,5+5,7+3,7+5,7+7,11+3,11+5,11+7,11+11
which covers in plush carpet the even integers
6,8,10,12,14,16,18,22 (1)
Oh crap! A GAP in our carpet, at even integer 20.
But, what we can see is that our carpet has NO gaps at least up to TWICE the last preceding prime, in this case
2*7=14.
Now, the rule: I’ll bet you don’t even need the rule stated, but, here it is: Each time we bring in a new prime, namely the next higher prime, we produce all the new possible pairwise prime sums in order, STARTING AGAIN with the first prime, 3, but this time added to the LATEST NEW PRIME, and then we add in sequence all the other lower primes to this new, latest, highest prime.
I’ll be you can also see already that I am trying to hatch in my madcap, leaky-sieve brain an inductive formulation. And so I am:
Let me assume now inductively that at each stage in the SS, I have COVERED IN PLUSH CARPET WITH NO GAPS, all the even integers up to TWICE the value of the last previous prime. More precisely:
When I have brought in the kth prime, pk, and then produced all the possible pairwise sums with it, I note inductively that I have COVERED IN PLUSH CARPET WITH NO GAPS, all the even integers up to and including 2*(p(k-1)).
Thus for example, if you refer to (1) you can see that as I use the prime 11, I will have covered with NO gaps all the even integers up to and including 2*7=14, i.e., twice the last previous prime.
Wouldn’t it be nice if now, at this point, we could argue that when we introduce the next prime p(k+1) we will have covered with NO GAPS all the even integers up to and including the sum pk+pk=2*pk?
Uh….yes, it WOULD be nice, because that would PROVE GC, because we will have “chained inductively” the case of p(k-1)+p(k-1) to the case of pk+pk! Therefore, it seems reasonably certain that I will NOT be able to successfully argue my inductively chaining argument! But, being an arrogant fool who just wants to have fun, let me proceed anyway, until you can see with no longer any disguises what a fool I truly am:
So, what happens when we introduce a new prime, the new next prime, p(k+1)?
What happens is, we see that we have created an entirely new swatch of carpet, BUT, you can see that this new swatch of carpet is a perfectly accurate “xerox copy” of the earlier COMPLETELY COVERED SWATCH OF CARPET we had already previously created, which was completely covered from 3+3=6 onward, up to and including 2*(p(k-1)). But, this new, XEROX COPY of carpet is RIGHT-SHIFTED from the original by the amount of our newest prime, p(k+1).(!) (I used the parentheses so you would not be thinking of factorials.) In other words, with p(k+1) we have started a new swatch of carpet with a new “origin,” or “zero,” which looks just like the swatch just previously created by using pk, but moved to the right and thus occurring starting at the “new origin of” p(k+1).
Thus this new swatch covers completely, with no gaps, carpet starting at p(k+1) and going all the way out to 2*(p(k-1)) + p(k+1), due to the right shifting by p(k+1). That’s because we already knew (inductive assumption) that the no-gap covered carpet had to be at least 2*(p(k-1)) units wide, and it is right-shifted to a new “origin” at p(k+1). Is that good news?
I think the answer is yes, because of an ancient Strohm lemma which you introduced to me awhile back: Strohmian Lemma No. 1, still retained in DiCenzo sieve brain, which says: For any integer n, there is always at least one prime between n and 2n. (I probably got that wrong, but it may still be “close enough,” as we government workers and engineer-types are prone to say. For example, maybe the lemma requires n to be a prime, but that will work for us too.) To use this lemma, I take for the integer n, instead, the integer p(k-1). Here’s a “picture”:
______p(k-1)______2p(k-1)_________________
pk 2pk
p(k+1) 2p(k+1)
p(k+1)+2p(k-1)
(2)
So, if I got this right (Is there any chance of that?), we have from picture (2) and logic, the carpet from 0 to 2p(k-1) is completely covered. Then also, we know that the carpet BETWEEN p(k+1) and p(k+1)+2p(k-1) is completely covered. (We also know that the swatch of carpet created by using prime p(k+1) in the sieve, goes BEYOND p(k+1)+2p(k-1), but we’re not sure it’s completely covered with no gaps of even integers, in this “beyond” region.)
Note now that in picture (2) we have drawn pk between p(k-1) and 2p(k-1). That works because Strohmian Lemma 1 says that there must exist at least one such prime, so we know that that is either pk itself (the next prime after p(k-1)) or a later prime. Thus, naturally, we choose pk as the one we show in the picture (2). Similar reasoning applies to the placement of prime p(k+1) in picture (2).
Thus, as we have just explained, we know from Strohmian Lemma 1 that we may write the inequalities
2p(k-1)>pk
2pk>p(k+1) (3)
Then, from (3) and (2),
p(k+1)+2p(k-1)>p(k+1)+pk>2pk (4)
Now, at this juncture, the hope from (4) is that we have successfully covered carpet all the way out to 2pk, with NO gaps! (But, have we?) IF we have, then we have completed our inductive assumption.
So, let’s take a look.
Well, first, we are at least “no-gaps covered” (ngc) over a carpet strip from p(k+1) all the way out to p(k+1)+2p(k-1), again via our inductive assumption.
But now, the real question is, how can we guarantee we are ngc between 2p(k-1) and p(k+1)?
Ah, so close and yet so far!
It’s now 1125 pm and I’ve been thinking in real time and writing this, all starting at 953 pm, without going back or checking anything, so God only knows how many terrible, sloppy and fatal errors I’ve already made.
But, maybe you can correct my errors and fill in the “gaps.” (Pun not intended.)
First, you might verify whether I was right in saying I have two ngc regions namely (0,2p(k-1)) and the half-closed interval [p(k+1),p(k+1)+2p(k-1)).
Second, can you fill in the gap between these two ngc regions?
Re this pesky gap, if we could have just a slightly stronger version of Strohmian Lemma 1, namely that there are always at least TWO primes betwee n and 2n, then we’d know our pesky gap is covered, because then the second Strohmian Lemma prime in our case would HAVE TO BE p(k+1) and p(k+1) WOULD HAVE TO BE , there always exists at least one prime number with < there is always at least one prime such that < <. Another formulation, where is the -th prime, is for ≥ + 4 quintillion. Since these n are >Sandy’s Candy Crush score, I conclude that we are indeed talking only about large n. Therefore we may conclude that we always have at least TWO primes between n and 2n. Therefore, using for n our special case of n=p(k-1) we have that there are at least TWO primes between p(k-1) and 2p(k-1). But that has to mean that both p(k) and p(k+1) are between p(k-1) and 2p(k-1), which in particular proves that p(k+1)<2p(k-1) (5)
Eq (5) thus proves that we have ngc in the formerly worrisome region between 2p(k-1) and p(k+1), because in actuality this "region" is not a region at all, because p(k+1) is actually LESS than 2p(k-1).
Thus, (5) establishes that GC is true. QED.
Hmmm…..wait a minute. QE[WRONG]. There MUST be something wrong here. It is IMPOSSIBLE that I have just proven GC, the Goldbach Conjecture.
But it is 758 am and I need to go back to sleep. I am losing my country which used to shine at the top of a hill but which is now turning into just another standard despotic dictatorship, and this gives me a lot of pain. Further, I am likely to die in the next month or two of c19 because I can not get the damn vaccine owing to the stinky politics of how the vaccine is being prioritized. And finally, I have proven the Goldbach Conjecture? Um…..I don't think so. But I can't find the mistake this morning and I am going back to sleep.
IF (and only if) you are so motivated, ye author of Strohmian Lemma No. 1, please comb through my terrible mess above and find what is wrong, and then please show me. In the meantime, I hope we have had fun fooling around with this damn thing. And, I'll quit here in wishing that Sandy may soon improve her Candy Crush score to 4 quintillion.
Thankfulness to 40,000 quintillion (40 sextillion? I didn't want to introduce sex into the picture, but there it is….), aloha, Alan, your eternally nl0, 805 am.
But, third, even if we can NOT fill in this gap, I think we may be able to construct a similar inductive chain of thought by using as the ngc carpet region not the one all the way out to 2p(k-1), but maybe only out to 2p(k-2).
What thinkst?
I should really clean up this mess now instead of leaving this pile of crap in your hands, but I'm tired and going to go to sleep, or at least to bed, which means I am quitting. Please help if you can!
Thanks 40,000 quintillion, aloha, Alan
From: Fred Strohm
Sent: Saturday, January 16, 2021 9:11 PM
To: Alan DiCenzo
Subject: Re: CORRUPTION HUMAN RIGHTS FREEDOM
Hi Alan,
A very sad situation.
Power corrupting the corrupt.
All the best,
Fred
On Sat, Jan 16, 2021 at 8:42 PM Alan DiCenzo wrote:
Hello everyone who will read this,
THE LEVEL OF CORRUPTION IN AMERICA HAS REACHED A THRESHOLD WHERE THE PRESERVATION OF FREEDOM AND THE PROTECTION OF HUMAN RIGHTS ARE NO LONGER POSSIBLE. Meanwhile, as always when this happens, the great mass of the people are sleeping.
We are now no longer different from Communist China, the old Soviet Union, North Korea, Cuba and Venezuela.
How long will THIS VERY POST last on THIS social media? This is a TEST.
The courts no longer do their sacred duty.
We have a one-party system.
Big Tech and the Mass Media now run an INFORMATION MONOPOLY as tight and irresistible as anything in the world’s great despotic dictatorships.
It matters not that now the choke-hold on information is run by Big Tech and the Mass Media, supposedly private enterprise entities, as opposed to the government. Their stranglehold is every bit as murderous as anything our Founders could have ever envisioned, every bit as choking as anything in Nazi Germany or Communist Russia or Communist Korea.
America, America, if you don’t wake up right now and see this, and, despite party affiliation, IMMEDIATELY AND URGENTLY make laws and run challenges in the courts to overthrow these tech monopolies and bring some true COMPETITION INTO THE MARKET so that WE THE PEOPLE AND OUR DIVERSE LEADERS TOO can continue to communicate in openness and freedom, then nothing else will matter.
NOTHING ELSE WILL MATTER. THIS IS NO LONGER ABOUT TRUMP, BELIEVE ME.
Cyber Anguish
THE ETERNAL NOW
Intuitively there is a “now” that is sliding along across all that is, i.e., across the all of the opposite of nothing. But Einstein wanted us to believe that such a universal “now” must be impossible, owing to the impossibility of objectively defining even the simultaneity of events.
So then, suppose we have two entangled particles, photons, or whatever. One is in my hand, or at least in my lab. The other is all the way on the other side of the known universe, or even in the most distant possible other member of the multiverse, or one of the most distant possible members, in case the multiverse is infinite.
I now “measure” my entangled particle, or do whatever the hell the quantum physicists say you can do to an entangled particle that will make the same (or the symmetric opposite) thing happen to the other particle at the same instant. That “other” particle thus indeed does exactly the same/opposite thing at exactly the same instant. Klyznaaabdittzuugo is standing in her lab over there and watches this happen. “Wow!” she says, “I wonder how far away the other particle that did that is.”
So, did we just have a definition of the UNIVERSAL NOW?
OK, after ALL OF THAT ABOVE, I’ve now made corrections today, 25 Jan 2021, which leave me in the silly position again of not knowing what is wrong with my “PROOF” of the Goldbach Conjecture. For all I know, this morning, I have finally proved the Goldbach Conjecture. But, of course, that seems about as probably as my winning the $1 billion lotto. Nevertheless, here it is:
Jan 25, 2021 1 AM UNDERSTANDING OF AD INDUCTION “PROOF”
Alan DiCenzo
Mon 1/25/2021 1:54 AM
Hi Fred,
We begin with this sparkling sentence, which is a typical example of STROHMIAN OVERFLOWING GENEROSITY: “Your explanation is a model of clarity. And I think I understand.”
HA! I WISH my explanation was a model of clarity. But, I see now that despite all my trying in the last iteration to “get clear” (sorry for the Scientology overtones), I STILL did NOT do so. But, I think, oddly (no pun intended) that that is both BAD NEWS and GOOD NEWS. FIRST, at least one part of the bad news is that ALL of your confusion and misunderstanding in what you wrote just below was due to ME and my ridiculous tripping over my own toes in trying to figure out both what I wanted to do, and HOW to express it. So, none of that is your fault.
The “good news” is that, after incrementally correcting my understanding, use of indices, and clarity of expression (still far from perfect), I now find myself in the ridiculous position of thinking the “proof” below actually does PROVE GC. Thus, once again we are STUCK with the challenge of figuring out what the hell is wrong, n’est-ce pas?
OK. SO: First, let me recopy and paste right here some of the most CRUCIAL, CRITICAL TEXT OF ALL from the “proof” below, and begin to FINALLY CORRECT IT, at least as per this latest iteration of my “wobbly” understanding of my own work, as I keep getting confused by my own indices and my own inductive assumptions. The text involves words right around eq (2). In this pasted text, I’ll attempt to insert in RED: the clarifications and corrections needed (STILL needed, damn it!):
“ACTUAL “PROOF”: We now are allowed to use p(k-1) for the first time. By that we mean, we consider for the first time all sums of pairs of primes where p(k-1) is now one OR BOTH of the summands in the pair of primes. EMPHASIS, AGAIN, BECAUSE VERY IMPORTANT: Either or BOTH of the two “post induction” summands are now always to be potentially larger than before, i.e., now less than or equal to p(k-1), and NOT merely less than or equal to p(k-2). That is because, in accordance with the rules for correct mathematical induction, we replace the entire structure in the inductive assumption with the analogous structure, but with the inductive (k-2) incremented up by 1 to (k-1), everywhere it is used EXCEPT FOR that part of the statement which is what is itself to be proved.
But we see right away that every such sum using p(k-1) is also expressible thusly, in this “other way,” so that the sum now “looks exactly like” a sum involving only a lesser prime, and p(k-2), rather than p(k-1):
p(j) + p(k-1) = p(j) + p(k-2) + delta[(k-2),(k-1)] (2)
where j as an index now ranges or roves over all primes <=p(k-1), and NOT merely
From: Fred Strohm
Sent: Sunday, January 24, 2021 6:39 PM
To: Alan DiCenzo
Subject: Re: GC: SELF CONFUSION? Re: GC INDUCTION AGAIN? WAIT A MINUTE! Re: FLAWS: Re: GOLDBACH CONJECTURE: A STRANGE, UNEXPECTED LINE OF REASONING
Hi Alan,
Your explanation is a model of clarity. And I think I understand.
But if we prove that all even integers up to
2p[k-1]-p[k-2]
are sums of two primes
(and I may not be stating this exactly right)
then it seems we have only extended the carpet ahead from p[k-1] the same distance as the distance from p[k-2] to p[k-1].
But what if
p[k] – p[k-1]
is a much greater distance
than
p[k-1] – p[k-2]
What can we do then?
But I may still be missing something critical.
All the best,
Fred
On Sun, Jan 24, 2021 at 4:34 PM Alan DiCenzo wrote:
Hi Fred,
Ha! You said “Your steps along your path are clear to you. And my steps along my path are clear to me. But each of us finds it hard to follow our friend along his path.”
But in truth, my steps along my own path are not necessarily clear even to ME. So, in that case, it would be no wonder that they would not be clear to YOU!
Well, meanwhile, it’s now Sun, Jan 24, 2021 at 227 pm and I am just now this instant finding and looking at this email of yours. I may be more awake than previously. That remains to be determined. But I’m hoping at this moment, and possibly feeling, that I may be able to summarize and phrase my understanding in a very BRIEF few sentences, thereby aiding both my own ability to understand my own path, and yours. Let me try that. After doing that, I will try to start understanding the last email of yours where you tried to clarify your path to me. I’m feeling good energy now, which might mean I have a slim chance of actually understanding something more, and that now is a good time to try reading and following your email!
But first, here is my VERY brief “summary” of my own understanding of my path, at least at THIS moment. I’ll begin by phrasing our inductive assumption in several different, but hopefully equivalent ways, to hopefully facilitate understanding and comprehension for both of us:
Inductively, we are assuming, and are JUSTIFIED in assuming (since we have MANY “first cases” which we can choose to use if we wish, all the way up to 4 quintillion, apparently), that at the kth odd prime, pk, ALL the even integers prior to p(k-1) can be expressed as the sum of just two primes, and that, in every case, these two primes are ACTUALLY LESS THAN p(k-2).
I’ll say this in a different, second way: We are currently “phrasing this” in these words: “Using ONLY primes < p(k-2) we can COVER (Where the word "cover" always for our purposes means "completely cover with no gaps," so that we can be lazy and type just a bit less) ALL THE WAY UP TO p(k-1)."
Phrased in still other, hopefully suggestive, words, we may say "Inductively, we are saying and assuming that all the primes up to ONLY p(k-2) have a "reaching power so great " that they actually cover all the way to p(k-1)."
Trying to communicate this same idea in still another way, we might say, phrased visually or pictorially, we are saying and assuming inductively:
3…(Lots of primes in here)…p(k-2)—————-p(k-1)- – – – – – – – – – -p(k)
___________________________________________________________
Fig (1)
(I’m hoping and praying that Fig (1) appears in your received email as it appears to me right now….)
where, in fig (1), we are “showing” or “illustrating” that all the primes up through ONLY p(k-2) produce enough sum pairs to cover all the way further across the dashed-line “gap” to p(k-1). In the same figure, we are also showing with a “less-dashed line” to p(k), that we are NOT YET covered all the way to p(k), or at least that WE DO NOT KNOW AT THIS POINT IN OUR “PROOF” THAT WE ARE COVERED ALL THE WAY TO p(k).
(So, did all that, admittedly a violation of “briefness,” facilitate any understanding for either one of us? I’ll be interested to hear….)
****************
OK, enough on all these different ways of stating our inductive assumption, which by the way ACTUALLY IS INTENDED TO MEAN THAT THE COVERAGE EXTENDS TO THE FIRST EVEN NUMBER BEYOND p(k-1). I waited to add that “wrinkle” until now, for easier comprehension, I hope. Now, for what we have to prove:
What we now WANT TO SHOW VIA OUR INDUCTIVE “PROOF” IS THAT OUR INDUCTIVE ASSUMPTION ALLOWS US TO PROVE THAT COVERAGE ACTUALLY EXTENDS ALL THE WAY TO p(k), I.E., THAT THE “LESS DASHED LINE” FROM p(k-1) TO p(k) IS ACTUALLY A “FULLY DASHED LINE,” JUST LIKE THE SEGMENT BETWEEN p(k-2) AND p(k-1).
**************
ACTUAL “PROOF”: We now are allowed to use p(k-1) for the first time. By that we mean, we consider for the first time all sums of pairs of primes where p(k-1) is now one of the summands in the pair of primes. The other summand is always one of the primes less than or equal to p(k-1), and NOT merely less than or equal to p(k-2).
But we see right away that every such sum using p(k-1) is also expressible thusly, in this “other way,” so that the sum now “looks exactly like” a sum involving only a lesser prime, and p(k-2), rather than p(k-1):
p(j) + p(k-1) = p(j) + p(k-2) + delta[(k-2),(k-1)] (2)
***************
—>Restart here using the “reddened” corrections from above, taking advantage now of the recopied eq (2a):
p(k-1)+1 + p(k-1)+1 = [p(k-1)+2] + p(k-2) + delta[(k-2),(k-1)] (2b)
where we are now using in (2b) the fact that j as an index now ranges or roves over all primes <=p(k-1), and NOT merely <=p(k-2), and that additional strengthening wrinkle in the inductive assumption, of coverage one even integer beyond the new prime. Eq (2b) shows us an "exact copy" of a perfect picture of the situation with our inductive assumption all over again, i.e., sums only using primes up to p(k-2), except for now TWO KEY CHANGES IN THE PICTURE.
THE FIRST KEY CHANGE IN THE PICTURE IN FIG (1) (And this honestly may be the first time I am noticing it) is what we just said, namely that one or both of the possible choices for pj are in fact p(k-1), and not only p(k-2).
(By the way: Right here might be (or might NOT be; sorry) the perfect place to notice that we have these specific, possibly isolated, even integers covered as pairwise prime sums:
Specific pair sums: p(k-2)+p(k-2) and p(k-1)+p(k-2) and p(k-1)+p(k-1)
(3) )
THE SECOND KEY CHANGE IN THE PICTURE in (2a) is of course the appearance of the "delta" which we can now see clearly serves the function only of "shifting" the magnitude or value of all of these "old sums" produced from the inductive assumption, to NEW VALUES to the right of p(k-1). The "AMOUNT" or "MAGNITUDE" of the right-shift, or right increase-in-value, is of course delta[(k-2),(k-1)].
But, prior to the "shifting" or "increasing in value" of the even integers we are covering, or producing by our pairwise sums, we know that ALL these "inductive-assumption sums" ALREADY covered the region ALL THE WAY FROM 3 THROUGH p(k-1)+1. THUS, AFTER THE SHIFTING OR INCREASING IN VALUE, we know that we have NEWLY COVERED all the even integers, i.e., with NO gaps, hence as NEW CARPET, from
[p(k-1)+1] thru [p(k-1)+1]+[p(k-2)+1]
+ delta[(k-2),(k-1)]
or
[p(k-1)+1] thru [p(k-1)+1]+[p(k-2)+1]
+ {p(k-1) – p(k-2)}
where the part {} is from the definition of our delta,
or
[p(k-1)+1] THRU [2p(k-1)] + 2 + p(k-2)-p(k-2)
and so finally:
p(k-1)+1] THRU [2p(k-1)] + 2 (4)
So, from (3) and (4) we see that in total (i.e., including "old carpet"), we have now covered all the way from
3 through [2p(k-1)] + 2 (5)
QED
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Hmm…..producing the "brief" (?) summary has perhaps allowed us to see clearly for the first time what we have truly covered at this point in the proof (Which is pretty much the end of the proof.)
What we have truly covered, I believe now, (at 150 am on Mon Jan 25, 2021) is what is shown in (5) AND those 3 possibly isolated even integers shown in (3). In other words, we have 3 more even integers covered, shown in (3), than we might have suspected at first.
Now, by Bertrand-Strohm, or Strohm-Bertrand (SB), we have DEFINITELY COVERED all the way to pk, AND in fact FURTHER, BY MUCH MORE THAN 2 and THAT MIGHT BE really KEWL.
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Hmm….this "summarizing" has been quite useful to me. I see that, even IF the "proof" is flawed, it STILL produces something potentially interesting and useful, maybe QUITE interesting and useful. What thinkst?
This is probably a good place to stop for now, according to my own laziness.
Maybe now I can take a look at the interesting work of FS in previous emails!
Oh—but wait; before I do that, let me first comment in RED on your email just below….
Thanks, aloha, Alan