It seems a queer relationship offhand and I have to investigate it.Quote:
Originally Posted by YesNo
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It seems a queer relationship offhand and I have to investigate it.Quote:
Originally Posted by YesNo
I have some of it already. It is tying in with what I am doing and what I was almost able to see last time I quit the problem.Quote:
Originally Posted by desiresjab
But still, why on earth is there a connection between factorials and the product of these oddly labled triangular numbers? I have not seen that part yet.
You have been so attentive and perspicuous I have to give you something. The hypothetical very large factorial in Brocard's problem has to be equal to 8 times some single triangular number. That fact should not have eluded me for so long, for I have long had everything I needed to realize it, but I am rather slow at this business. I miss things a math teacher would see routinely.
Can you figure out why the factorial must be equal to 8 times some single triangular number for there to be another pair of Brown numbers? I think you can. I believe it is within your range, from what you have shown me.
This might work as a way to show that a solution of Brocard's problem is 8 times a triangular number.
Let n and m be integers such that n! + 1 = m2. Then n! = m2 - 1 = (m - 1)(m + 1).
Brocard's problem does not work for n = 1 and so n > 1. Notice that m must be odd because otherwise for n > 1, n! and m have a common factor. That means m - 1 is even. Let 2r = m - 1. Then 2r + 2 = m + 1.
Then (m - 1)(m + 1) = 2r (2r + 2) = 4(r (r + 1)). Multiply this by 2/2 = 1 to get 8(r(r + 1)/2).
A triangular number, T, is the sum of all positive integers less than or equal to that number. That sum can be written algebraically as T(T + 1)/2. This shows that r is the desired triangular number since it has the algebraic form of a triangular number.
You took your own way, which is to be expected from different minds. I arrived at this idea from another factorization of the factorial.
Observe: Incidentally (or perhaps not), both 120 and 720 are factorials and triangular numbers. Triangular numbers can be square numbers sometimes, as well. But they can never occur together because a factorial can never be a square, which has an easy proof but is intuitively clear to anyone who thinks about it hard enough. I have been around these exact issues for a while now and I have to an extent humanized them into normal language as far as I have understood them. Some algebra is necessary, as this is not the 12th century.
Now to the heart of it.
8[(n)(n+1)]/2= 4[n(n+1)]=
4n2+4n=2n(2n+2).
Consider: Every whole number is a square root. Every odd integral square root lies between two numbers of the form 2n and (2n+2) for any n.
When two numbers on either side of a number x are multipled together the product is always one less than the square of x. (x-1)(x+1)=x2-1.
The Secret: It does not have to be proven that such and such a square can or cannot exist. The kernel of the problem is whether a very large factorial can ever be factored precisely into the form 2n(2n+2). If it cannot be, then such a breed of square can never repeat itself beyond the three known examples.
The simple power of this factoring approach may be missed. Our factorial is huge because n itself is very large, so factorialized n is really large.
Because they are only two apart on the number line, this makes 2n and 2n+2 next door neighbors in the ordered set of even numbers. It also makes one highly even and the other barely even. These two factors that are expected to produce a factorial, can share only a single factor, i.e. one factor of 2. They are not far enough apart to share anything else.
Can two large next door neighbors in the world of evenness ever contain between them precisely and only all the factors of a factorial? Can two such neighbors exist? If a factorial cannot be factored this way it cannot meet the qualifications of Brocard's problem.
There is quite a bit more. We cannot determine which of the factors is highly even and which is barely even. I call them super factors and SF for short.
* * * *
The possibilities for the last two digits of the SFs are listed below vertically. The middle number might be paired with either factor in its group. Remember, the SFs are very large numbers themselves and these are only the last two digits of the possible SFs. Observe that the digits of the long factors in each group are identical copies of one another except for their last two digits, the one's and ten's positions. The same is true for the SFs in group B.
A
xpnuh000...02
xpnuh000...00
xpnuh000...98
B
jbvfg...52
jbvfg...50
jbvfg...48
In the case of B all the factors of 5 are with the middle factor, but it has only one factor of 2 to make zeros with, and both its possible mates 52 and 48 are highly even without any 5.
In the case of A all the factors of 5 are with the abundance of 2's in the middle SF, and the traditional tail of zeros is observed. The highly even number is in the middle this time and both its possible mates 98 and 02 are barely even.
We can also observe now that our hypothetical integral square root of (n!+1) which lies exactly between and contiguous to the SFs, must have last digits of 49, 51, 99 or 01.
It is hard to say which of the numerous collected facts and observations might next help to further understanding. The proof does not have to be about squares at all. It could be about triangles.
That is an interesting way of looking at the problem, considering one factor being highly even and the other barely even. Together those two factors should equal n! which has zeros in the units and tens positions after n is greater than or equal to 10 so a factor of 100 is in n!. I can see how one of those two numbers should have many zeros in them. I suspect there should be n/5 number of zeros in n!.
There are trivial cases where a factorial could be a square, such as 0! and 1!, but as soon as one gets larger than 1 there are primes involved. The largest prime less than or equal to n would not have another prime like it in n! to pair up with to make a square out of that number.
You mentioned the factorials that are also triangular numbers like 120 and 720. I wonder when would there be an r such that for some n, n! = r(r+1)/2. That would be like saying the product of positive integers less than n is equal to the sum of positive integers less than r.
While bicycling in the neighborhood, it occurred to me that you could do something similar to the highly even and barely even with the Brown numbers. As you mentioned, assuming n! = (m - 1)(m + 1) for n > 1, then both m - 1 and m + 1 are even, but since they have a difference of 2, one of them has only one factor of 2 and the other has the rest of the factors of 2 that are in n!. Now take any other prime p > 2 in n!. All powers of that prime will be in either m - 1 or m + 1. That prime cannot be shared across those two factors since they have a difference of 2.
For the three known solutions,
4! = (4) (6) = (22) ((2)(3))
5! = (10) (12) = ((2)(5)) ((22)(3)
7! = (70) (72) = ((2)(5)(7)) ( (23)(32))
One way to show that the solutions are finite is to try to see if this additional constraint forces there to be no solutions after a certain point.
On the right track, but you have to remember to include those factors in the numbers less than or equal to 100 which can be divided by 5 more than once, 25, 50, 75, 100. There is a formula for this using sigma and the floor function.Quote:
Originally Posted by YesNo
This is precisely the reasonQuote:
Originally Posted by YesNo
I made a computational mistake regarding 720. It is not a triangular number. In fact, it is conjectured there are no more triangualr factorials after 1, 6 and 120, but this conjecture remains unproven and in about the same state as Brocard, generating lots of phd dissertations in mathematics as brains gnaw away on various corners of it.Quote:
Originally Posted by YesNo
Apparently, the question of are there anymore triangular factorials is related to a more general and deeper problem which, if solved, would suddenly lead to the solution of this problem and perhaps Brocard's too. That is, a general solution to this:
n!=a!b!c!...
Some examples are known, and brains are gnawing away.
* * * * *
Triangulars were likely a dead end for Brocard's problem, unless one could show Brocard's factorial has to be triangular to fulfill the problem. Then there would be a connection.
What is still of major interest is the connection between triangulars and factorials in this identity:
(2n)!=2n k=1∏n T2k-1.
Major interest. What does it say? I think it says that every even factorial is the product of some powers of 2 and some odd labled triangular numbers, which is just a jewel.
(2n)! can be manipulated as follows.
Let me get back. I have to recheck my notes for errors...
(A.) (2n)!=2n k=1∏n T2k-1.
(2n)!=2(n)· (2n-1)· 2(n-1)·(2n-3)·2(n-2)...·(2n-2n+1)
Note that 2 can be factored out of every other term. Since there are 2n terms there are n terms from which 2 can be factored. We now withdraw n factors of 2 and place them in front of cap pi, just as in the identity (A.) above, leaving:
(2n)!= 2n[n·(2n-1)·(n-1)·(2n-3)·(n-2)...·(2n-2n+1)]=
Looking now exclusively at the factors above in brackets, all we have to do is rearrange the terms to see what is going on.
[(2n-1)·(2n-3)...·(2n-2n+1)...·(n)·(n-1)·(n-2)...·(n-n+1)]
Simply amazing. What we now have is this:
(2n)!=2n[(n!)·(2n-1)!!].
Caution: Double factorials are not the same as nested factorials. The double factorial means to multiply together all the odd numbers from a certain point down to 1.
I am not yet seeing in the above bold forumla the connection to triangular numbers I was hoping would jump out at me, but the double factorial coming into play is a bit fascinating, as now the factorial and the double factorial multiplied together by definition must have the same prime factorization as a string of odd labled triangular numbers multiplied together, since the two products are the same. This fact needs to be drawn out of the algebra, if I can find the manipulations. I need the manipulations that break what is inside the brackets down to a visible (recognizable) triangular connection that we know is there, again, by definition, from the identity itself.
Note: It is always possible my manipulations, though correct, went the wrong way to uncover the triangular connection, leaving us with a curiosity that is merely interesting.
This is the first I've heard of double factorials, but they make sense: https://en.wikipedia.org/wiki/Double_factorial
I knew about double factorials, but this is the first time I ever ran into one out in the wild. I had been wondering if they were mere toys or relevant to anything. I should have known. In math everything is always relevant on some level because everything is connected. The road that leads ever on is sometimes hard to find.Quote:
Originally Posted by YesNo
All these new connections tell me something else--I am nowhere near the last view of Ramnujan on Brocard's problem. Wherever he stopped and gave up, I have not arrived to yet, if ever I will. To do so was a stated goal when I started the problem.
* * * * *
A friend of mine wonders if the world did not in the long run lose out by Ramanujan's journey to England. I had always assumed that he picked up enough mathematical formality from Hardy and associates at Cambridge to untangle his genius from unnecessary quests and his few incorrect notions to make it a great blessing that he undertook his voyage. There was great value to us living now in Ramanujan's being shepherded toward some of the most important mathematcial problems of his time and of all time and brought right up to specs on the frontiers of research by several of the world's preeminent number theorists who recognized the Injun's awesome abstract power.
The greatest mathematicians always best everyone of their time, doing here and there what was impossible for other great talents. They come up with formulas long thought to be impossible, solve a problem from antiquity or invent new tools. Most of Ramanujan's tools were a personal thing he could not understand himself, so his forte was producing amazing formulas that also looked amazing. When you see them, you know you were not made to go there. You will consider yourself fortunate if years of study garners a half decent understanding of just a few of the identities he pulled from nowhere, directed, he said, by a household goddess.
Ramanujan's formulas are too long and difficult to try to set up on the house word processor. People will have to take a look for themselves. There was a shorter one that was amazing, but I cannot find it.
I guess it is still an open question whether we got lucky or unlucky by Ramanujan's trip to England.
I have a small improvement to announce. Improvement sounds much better than correction. In the formula
(2n)!=2n[(n!)·(2n-1)!!d, n+(1 or 2)],
We had to add that subscript of "d" at the end because the double factorial is not a complete one, I realized after something kept nagging me. Going downwards, it begins on (2n-1) and descends to the next odd number after n. So it begins on n+1 or n+2, depending on whether n is even or odd. A full written expression of the function would have to account for this with something like the piecemeal notation of the Legendre symbol using the oversized bracket. That is hard on the house word processor. So is defining the range of cap pi since the superscript and subscript cannot be put up properly at the same time. One or the other, but not both.
On the function above I could have used a "u" subscript for upward. Either way you write it down, the double factorial will start and end in the same place, just the notation encodes it differently.
What can be done with the beast above, I do not know. I cannot see that triangular relationship the way we have it expressed so far.
That goddess, Namajiri, was a form of Lakshmi. https://en.wikipedia.org/wiki/Namagiri_Thayar
In the good old days, poets would credit muses with what they produced and I think they meant it. Ramanujan still meant it.
Here is something on triangular factorials that gives an argument that only 1 and 3 such: http://math.stackexchange.com/questi...angular-number
In their example the triangle has to equal the factorial, which is very easy to prove. The actual problem we are looking at and the more interesting question is whether any factorials equal any triangle, beyond the three known solutions.
It may be just eerie coincidence, but there are only three known solutions of Brocard too. With all the connections between mathematical objects, it makes one wonder. Brocard's three solutions lead back to three different numbers, though, than the three solutions to the factorial/triangular question. Whew!
That gets me to wondering further if there are numerous such unsolved problems in which there are exactly three known easy solutions and the existence of more can neither be proved or disproved plus computer computations have shown there are no more solutions of the function out to a vast input? In some famous problems does it stop at four solutions? How about five solutions? How are small numbers (or any numbers) distributed over a large number of unsolved problems of this type?
Wouldn't it be nice if all you had to do was plug in the number of known solutions, then plug in the extent to which you had searched for more solutions, and as output you would receive back an answer as to whether any more solutions existed? Someday I believe there might actually be a formula that works on similar principles that can decide which unsolved problems should not be worked on any longer. I know this goes against Godel. But Godel himself went against what were considered unassailable notions.
I see what you mean by the factorial/triangular problem. I was assuming the n used in the factorial had to be the n used in the triangular number. But they could be different.
If Brocard's problem and the factorial/triangular problem are related there should be some underlying explanation for the relationship. However, the existence of only three solutions suggests that maybe there is such a relationship yet to be discovered.
There were three features about mathematics that Hilbert wanted to show: (1) completeness, (2) consistency and (3) decidabliity. Godel showed that (1) and (2) cannot be achieved, but Church and Turing showed that (3) was not attainable either. https://en.wikipedia.org/wiki/Entscheidungsproblem However, the idea of giving a computer examples of input and correct output and then asking it to create a model based on that training data and then use that to predict what the correct output would be for arbitrary input underlies "machine learning" or "artificial intelligence".
1, 3, 5...
Add this string up and you get a square. You always get a square no matter how long or short you make the consecutive string of odd numbers.
Also, all squares are the sum of two consecutive triangular numbers.
Also, cube each consecutive integer and add them up. Then add each consecutive integer again and square the result. In other words:
13+23+33...=(1+2+3+...)2
Also, the difference between the squares of two consecutive triangular numbers is a cube.
Also, since every square number is the sum of consecutive odd integers, so is the square of a triangular number.
This last one could be very important to us, since we have the sum of some odd integers in our derived equation.
This one makes geometric sense. Start with 1 dot and then to get the next square add a dot to the left and the bottom plus one in the corner. In general if one has a square of side n dots, then one needs n dots on the left side and n dots on the bottom and then one dot in the corner to get the next larger square. That would be 2n + 1 extra dots added to the n2 dots already there. The previous square would have needed n - 1 dots on the side and the bottom and one on the edge or 2n - 1 dots. So the sequence of squares is the sum of the odd integers.Quote:
Originally Posted by desiresjab
For this one use the closed form of the triangular number, T, as T(T+1)/2. Then algebraically add the closed form of T + 1 to it. One should get a square form.Quote:
Originally Posted by desiresjab
I don't see an explanation for this one. I didn't look for counterexamples.Quote:
Originally Posted by desiresjab
I don't see an explanation for this one either, but I did check it for n < 16.Quote:
Originally Posted by desiresjab
This last one doesn't seem to be correct. So, I will look for a counterexample. Try 4 = 1 + 3. That one works. Try 9 = 3 + 5? That doesn't work, so 9 = 32 is a counterexample.Quote:
Originally Posted by desiresjab
Edit: I see it now. It is not the sum of two consecutive odd integers but the sum of the odd integers starting with 1 up to some point.
Sorry, I did not word that one very well.Quote:
Originally Posted by YesNo
The part that you add to get the next and the next and the next figurate number, the Greeks called the gnomon. It is actually a useful word.
With an unending proliferation of relationships, it is no wonder one runs into sudden connections when dealing with figurates.
Every other triangular number is a hexagonal number.
Every pentagonal number is 1/3 of a triangular number.
All even perfect numbers are triangular Tp with prime p.
666 is the largest repdigit triangular number (Bellew and Weger, 1975).
The latter would have been a great problem to work on, if it were not already solved. I envy the people who got to solve it. That is my kind of problem.
The sum of all the triangular numbers up to the nth triangular number is the nth terahedral number.
The sixth heptagonal number minus the sixth hexgonal number is the fifth triangular number.
It seems triangular numbers are generators for any figurate number where gnomons are applicable. If you know the right triangular numbers you can calculate all the other figurates, it does seem.
Well, I guess I am not banned.
And finally:
1=13, 3+5=23, 7+9+11=33, 13+15+17+19=43,...
This about takes the cake, or is the frosting on the cake.
The additive properties of numbers and their multiplicative properties being friendly but not related by family is part of what keeps numbers so mysterious. There is still a lot of work left to do in the additive properties. Unfortunately, none of it will be accesible to civilians the way the properties of triangles and squares are. I doubt if elliptic equations will become common to people. That is about as likely as eighth graders of the future comfortably reading Finnegan's Wake.
Most of the properties of figurate numbers seem to be additive. There is the exception that the product of sums of squares is also a sum of squares.
In the factorial problem one of the factors of (2n)! is a product of triangular numbers. This just hit me in the head that I have set that problem up wrong. We do not have an upward double factorial. That is the labeling! What we have is an upward sequence of triangualr numbers multiplied together. Excuse me again.
I have not made a mistake at all, except to think I made one in the first place. Double excuse me, and the double factorial is still on!! Yes, I derived it that way. I am happy again, but too tired to think math tonight, perhaps..Quote:
Originally Posted by desiresjab
What can I say? I am making mistakes left and right. Mistakes can eventually lead to the truth. Careless algebra on a word processor not meant for it, instead of doing the algebra on paper first, is the main reason for the mistakes.Quote:
Originally Posted by desiresjab
Calculation instead of algebra has shown that within the brackets of the formula
(2n)!=2n[(n!)·(2n-1)!!]
we have a full double factorial instead of a partial one. I do not know if this will make a difference, but it is certainly more neat with the notation and more pure of form. I do not like mistakes, but I like this result. This time it is byond dispute, for indeed
10!=25(5!9!!).
Awfully neat and suggestive, but I still have no suggestions.
10!=25(5!9!!) and
10!=25(T1·T3·T5·T7·T9)
Really cool.
Not only eighth graders. When it comes to something like Finnegans Wake I ask myself would I rather spend my time trying to understand that or trying to understand quantum physics or gravitational theory or elliptic curves or Brocard's problem or Sierpinski's problem? If one wants an impossible task one might as well choose an interesting one. It might turn out not to be so impossible after all.Quote:
Originally Posted by desiresjab
I checked it in a Google sheet. It worked.Quote:
Originally Posted by desiresjab
I keep asking myself, "How did they know how many factors of 2 to draw out front of the cap pi? We can reverse engineer the forumla pretty easily for 10 simply by taking a factor of 2 from each even number 10 or below. Then we notice two factors of 2 remain above 5 and between 10 , and tqo factors are missing below five. Easy replacement. But how did they know to do it? They would have no reason to do that unless they had a clue from somewhere else.
They derived their formula from something.. "Our" formula came from somewhere too. I want to get to their formula for the above one.
* * * * *
Trying to solve unsolved problems, one lives for such moments-- side trips through wonderland. One understands one will not solve the problem. In trying anyway one runs into questions asked by one's self which open up new vistas and allow learning to go on in the approximate area of the problem when progress has stalled. It happens every time. Now I have this new question to answer, and I will not stop until it is answered. Where is the connection between triangles, factorials and double factorials? Even when I know, I may not be any closer to solving Brocard, but I will have better connections among numbers.
I learned a little more. Apparently, double factorials were not even around as a function until roughly mid 20th century. Bringing those two out front of cap pi must have been someone's idea when defining the function. The function is pretty natural, if you look at
n!!=2kk! for even numbers, and
n!!=(2n)!..=..(n!)....
.......2kk!......(n-1)!!
for the odds.
Very reminiscent of what I derived from (2n)! These are attractive to me. I think they look good.
Even and odd double factorials have to be defined separately. Odd ones can weirdly be extended to negative values.
It doesn't seem obvious to me either why someone would factor out those 2s. However, the exponent makes sense. There will be at least n/2 factors of 2 in n!.
However, if one is multiplying triangular numbers together to see what relationships pop up, someone might have seen that the difference between them and the n! is a curious factor of 2 to the n/2 power and guessed the relationship. The Greeks would have been thinking along geometric lines rather than algebraic ones.
This formula makes intuitive sense. The n!! is n(n-2)(n-4)...2. Factor a 2 out of each of these n/2 factors and you get the 2n/2(n/2)!. Let k = n/2.Quote:
Originally Posted by desiresjab
It does make intuitive sense. It is the way we ended up doing it by simply removing a factor of 2 from each even number. But we knew what we were after. We were simply trying to manipulate the formula for (2n)! by manipulating (2n)! to see where we would arrive, i.e. if we could arrive back at the forumla given for it including triangular numbers. Of course we didn't, we ended up at our factorial times double factorial thing, which obviously is that product of triangular numbers in disguise.Quote:
Originally Posted by YesNo
Here is the difficulty right now. For odd numbers, when I remove n factors of 2 from the numbers above and below n, I need a proof that the number of such factors remaining above n is exactly the amount needed to replace those taken from below n. This part is not intuitively clear to me and I believe a proof must have been provided at some point.
The key to making it intuitively clear may lie simply in studying the function on even numbers, which I have not done yet. I need a little more time with all these formulas. I am glad there is something to sort out--we cannot grow unless there is. Do I believe this digression will be helpful with Brocard? Probably not, but it is increasing our understanding of numbers, and that will.
I have to be away for a few days again. I see the way to settle the question perhaps. If the total 2's in 2n minus the trotal 2's in(2n-1) equaled n and was proven algebraically, I suppose that might do it. Go ahead and prove that while I am gone if you have a mind to.
Here is what I know for sure. The higher powers of 2 between n and 2n have to equal the lowest power of 2 between 1 and n. In other words, for n=5, after removing that first layer of 2's there is still 22 left where 8 was. 22 is exactly what was skimmed off between 1 and n. I am asking how they know this to be true in general. I am probably missing something quite basic. Maybe I will find it before I leave for a few days off.
Specifically, it needs to be shown that
i=2Σ2n↓2n/2i↓=↓n/2↓
where the arrows indicate the floor function, which is always rounded to the lowest whole value. This may prove to be a lousy way to set the problem up to find the answer. I just know it is correct. The value on the left (for 2n) is without its lowest and most numerous power of two, the value on the right (for n) is calculated only for the value of its lowest power of 2. The two should be equal.
It also says that the factors of 2 in 2n! is equal to 2n, doesn't it? In other words there are n such factors. Which is what we are trying to show. It is a nifty equality. One has to internalize that.
Well, one should prove it first, even though it is known to be true. I cannot see it intuitively, but I believe it is seeable that way.
I guess I do not see that intuitively--yet, at least. It is what I said was intuitively clear a few posts ago. But I do not see that the remaining number of 2's between n and 2n after the first layer is peeled off should always be equal to what was peeled off between 1 and n. But maybe that fact is just a consequence of the law. No, it is the law in slightly different expression.
A consequence of this would be that whenever n is a power of 2, (n)! will have (n-1) factors of 2. A power of 2 factorialized always has a value of one less than 2k. 4! is guaranteed to have three factors of 2, 8! is guaranteed to have 7, etc. People who have worked in binary know this.
For even n, n!! skips every other factor one would normally see in n!. That is, it looks like this product: 2*4*6*...*n. There are n/2 factors. Now remove 2 from each of those factors. You get 2n/2(1*2*3*...*n/2) = 2n/2(n/2)!. Let k = n/2 to simplify the notation and you get 2kk!.
For odd n, n!! skips just like for the even n. It looks like this product: 1*3*5*...*n. This is the same thing as multiplying all the numbers together less than or equal to n and then dividing out the even ones: (1*2*3*...*n)/(2*4*6*...*(n-1)). But that is n!/(n-1)!!. Note that n-1 is even since n is odd. We already have a way to write n!! when it is even and so we get the following for n odd: n!! = n!/(n-1)!! = n!/(2(n-1)/2((n-1)/2)!). Let k = (n-1)/2 and this simplifies to n!/(2kk!).
I was thinking about those double factorials. They are a way to split a factorial into the even factors and the odd factors in this manner:
n! = n!! (n-1)!!
For the even n, we can factor out the n/2 powers of 2 and continue the process. For example, if n is even then n!! = 2n/2(n/2)! Now take the (n/2)! and split it into even and odd factors. Repeat the process until all the factors of 2 are removed from n! leaving 2 to some power and the odd factors.
I suppose one could also do this for other primes but this notation is specific to 2. What one gets is the prime factorization of n!.
On my trip it became crystal clear where the oversight lay that was causing my disagreement with the formula. Math requires lots of solitude and good marijuana. I had overlooked the detail in my reckoning that the lower half of the double factorial was being filled in with odd small numbers. Here is the formula again:
(2n)!=2n[n!(2n-1)!!].
Only n factors of 2 are needed to produce the full double factorial with its complete lower end because that is exactly how many factors of 2 are contained in the upper interval. I have not seen it proved anywhere that the upper interval will always contain exactly n factors of 2, but having now examined the mechanics of the details, I fully accept that it does and has been proven elsewhere, probably by induction. It is one of those facts of life of numbers I did not know. There are plenty of them I did not know before but came to understand. I like this formula and I am glad we stopped to consider it. Whether it can help in the quest for Brocard awaits more solitude and marijuana. I believe there is a connection and I believe I see a hint of it through the smoke, sir. It is really a matter of connecting notations with the problem.
Maybe more vision will enable me to see why the part of the formula in brackets ( the factorial times the double factorial) is equivalent to multiplying the oddly labled traingular numbers together. I look forward to seeing that, and I suppose it is what I must stick with for the moment.
I was trying to make sense out of the abc conjecture. I think I have a basic understanding of what it is trying to say primarily from this wikipedia article: https://en.wikipedia.org/wiki/Abc_conjecture
What I don't see at the moment is why it implies that Brocard's problem has only finitely many solutions, but this reference is supposed to provide the key: http://www.mat.univ.szczecin.pl/file...dramanujan.pdf
I do not see it clearly yet, either. One first needs to study Diophantine equations to get the basics of their solution and become familiarized with the usual methods in the field.Quote:
Originally Posted by YesNo
I am still trying to figure out the triangular connection to the other problem, but I am looking at this problem too. I would like to understand the abc conjecture better, so I suppose I will. You never know when the insight will come, except that it will be when you are concentrating your best.