Cosmology

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.

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Originally Posted by desiresjab

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.

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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.

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Originally Posted by desiresjab

Also, all squares are the sum of two consecutive triangular numbers.

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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.

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Originally Posted by desiresjab

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

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I don't see an explanation for this one. I didn't look for counterexamples.

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Originally Posted by desiresjab

Also, the difference between the squares of two consecutive triangular numbers is a cube.

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I don't see an explanation for this one either, but I did check it for n < 16.

Quote:

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Originally Posted by desiresjab

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.

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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.

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.

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Originally Posted by YesNo

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.

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.

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Sorry, I did not word that one very well.

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 T_p 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.

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Originally Posted by desiresjab

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.

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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..

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Originally Posted by desiresjab

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..

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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.

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.

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Originally Posted by desiresjab

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.

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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.

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Originally Posted by desiresjab

10!=2⁵(5!9!!) and

10!=2⁵(T₁·T₃·T₅·T₇·T₉)

Really cool.

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I checked it in a Google sheet. It worked.

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.

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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!!=2^kk! for even numbers, and

n!!=(2n)!..=..(n!)....
.......2^kk!......(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.

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Originally Posted by desiresjab

n!!=2^kk! for even numbers

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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 2^n/2(n/2)!. Let k = n/2.

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Originally Posted by YesNo

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 2^n/2(n/2)!. Let k = n/2.

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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.

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.