...in under a week.
Take one dot. From there, you can make a 3-dot equilateral triangle, a four-dot square, a five dot regular pentagon, etc. From those, you can build a 6-dot trianlge, a 9-dot square, a 12-dot pentagon, etc. So any number that is the total number of dots in a regular r-gon so built is a r-gonal number. For convenience, say 0 is an r-gonal number for any r. Can someone prove that EVERY positive whole number can be represented in at least one way as the sum of r r-gonal numbers, for every r?
Hints as needed, although I bet someone can come up with a prettier proof than mine...
Originally posted by royalchickenThis isn't anywhere near the problem, but first things first, I want to make sure we are thinking of the same r-gonal numbers before I start racking my brain. So, if we are thinking of the same number, then the general formula for the number of points in a given r-gonal number is:
...in under a week.
Take one dot. From there, you can make a 3-dot equilateral triangle, a four-dot square, a five dot regular pentagon, etc. From those, you can build a 6-dot trianlge, a 9-dot square, a 12-dot pentagon, etc. So any number that is the total number of dots in a regular r-gon so built is a r-gonal number. For convenience, say 0 is ...[text shortened]... ery r?
Hints as needed, although I bet someone can come up with a prettier proof than mine...
N[r,w] = (4 + r(w - 1) - 2w)w/2
where r is the number of sides and w is the number of dots for a given side, vertices included. r>=3, w >=2.
Am I ok so far? Hate to start thinking about the proof if I'm not thinking of the same thing as you!
Originally posted by maggoteerI still don't see how pentagonal and higher numbers work graphically 😳
This isn't anywhere near the problem, but first things first, I want to make sure we are thinking of the same r-gonal numbers before I start racking my brain. So, if we are thinking of the same number, then the general formula for the number of points in a given r-gonal number is:
N[r,w] = (4 + r(w - 1) - 2w)w/2
where r is the number of sides and w is t ...[text shortened]... ok so far? Hate to start thinking about the proof if I'm not thinking of the same thing as you!
Originally posted by AcolyteAssume I'm thinking of the same shape as royalchicken. Then for a given polygonal shape, say a pentagram, imagine as follows. A next bigger pentagram, extends the edge length of the next smaller pentagram by one dot; but they always share the two edges adjacent to the single "common" vertice. Notice that along those two shared edges, the dots that were vertices of the smaller pentagram are now "edge" vertices of the bigger pentagram.
I still don't see how pentagonal and higher numbers work graphically 😳
Um, bejeese,I just read that. I doubt it helped! Yeah, this is a time when a picture would be worth about a mole of words....
And I probably should have recast my formula; as I've written it, I consider any r-gonal number of length 1 to be a single dot. The first square, ie the 4 dot one, is in my formula P[4,2]. P[4,1] is the "degenerate" square with one dot. 😕
Acolyte & Maggoteer (quite possibly the two coolest usernames)-
I'll tell you my general formula for clarification. It barely helps at all for the proof. Let A(r,n) be the nth r-gonal number. Then:
A(r,n) = n + (r-2)(n-1)n/2 = n + (r-2)A(3,n-1)
So some of the the 2-gons, 3-gons, 4-gons, 5-gons are:
0, 1, 2, 3, 4, 5....
0, 1, 3, 6, 10, 15...
0, 1, 4, 9, 16, 25...
0, 1, 5, 12, 22, 35...
Basically, you will recognise that each term in one of these sequences is formed by adding the next term in an arithmetic series with common diffrence r-2 and initial term 1 (except the 0's). I want you to prove that every natural number is the sum of 3 3-gons, 4 4-gons, etc.
My proof took two weeks of evenings and is basically analytic in character. Something combinatorial might be more natural. Good luck.
~mark (Je ne suis pas un poulet, mais je pense que je me conduis comme un poulet quand je joue un match de la chesse.)
Sorry, maggoteer, I think I explained it poorly the first time. When I discussed these numbers in terms of polygons, this was mostly because I was following Acolyte's advice and not making this look like a math(s) exam. Think of it, if you like, in the way I expressed above.
Originally posted by royalchickenAt least our formula expand out to the same! So I'm at leas thinking what your thinking. Um..not sure I'll get much further....😕 Acoylte, the field is yours!😵🙄
Let A(r,n) be the nth r-gonal number. Then:
A(r,n) = n + (r-2)(n-1)n/2 = n + (r-2)A(3,n-1)
Originally posted by royalchickenSo did you construct a proof by mathematical induction, using the assumption that 0 is an r-gonal number for any r as its base?
Acolyte & Maggoteer (quite possibly the two coolest usernames)-
I'll tell you my general formula for clarification. It barely helps at all for the proof. Let A(r,n) be the nth r-gonal number. Then:
A(r,n) = n + (r-2)(n-1)n/2 = n + (r-2)A(3,n-1)
So some of the the 2-gons, 3-gons, 4-gons, 5-gons are:
0, 1, 2, 3, 4, 5....
0, 1, 3, 6, 10, ...[text shortened]... t making this look like a math(s) exam. Think of it, if you like, in the way I expressed above.
Originally posted by bbarrHey! Slyly trying to get a hint! But it is Valentine's day...Tried that. That was, it seems, the most reasonable thing to do. But I was unsuccessful. So I let k(r,n) be the nth number that satisfies the conjecture. I then let X(r,x) be the number of k(r,n) between 0 and x inclusive for some real number x. All I showed was that x = X(r,x) + O(1) for all x in R. Obviously, this implies that since X is a step function, the set of k(r,n) is equal to N for all r. But I won't go into any more detail, except under torture.
So did you construct a proof by mathematical induction, using the assumption that 0 is an r-gonal number for any r as its base?
The use of 0 was just so that exactly r r-gons are used. For example, 6 = 3+3+0 or 6+0+0, but it cannot be formed as the sum of three 3-gons in any other way.