# I'll be very impressed if anyone solves this...

royalchicken
Posers and Puzzles 11 Feb '03 22:08
1. royalchicken
CHAOS GHOST!!!
11 Feb '03 22:08
...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...
2. zach918
Banned
11 Feb '03 22:21
MARK-
interesting puzzle....hate to change the subject but, why did you delete your profile pic ðŸ˜• it was a good picture ðŸ™‚
3. royalchicken
CHAOS GHOST!!!
11 Feb '03 22:25
I never had a profile pic-I haven't got a star.
4. thire
Xebite
11 Feb '03 23:141 edit
Originally posted by royalchicken
I never had a profile pic-I haven't got a star.
I guess he knows... this ðŸ˜‰
5. 11 Feb '03 23:54
Originally posted by zach918
MARK-
interesting puzzle....hate to change the subject but, why did you delete your profile pic ðŸ˜• it was a good picture ðŸ™‚
are you talking about the T1000 Mark maby?
6. royalchicken
CHAOS GHOST!!!
12 Feb '03 18:51
Probably. But enough of this! Anyone solve it yet? Or should we wait for Acolyte?
7. maggoteer
The MAKIA
13 Feb '03 07:54
Originally posted by royalchicken
...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...
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 the number of dots for a given side, vertices included. r&gt;=3, w &gt;=2.
Am I ok so far? Hate to start thinking about the proof if I'm not thinking of the same thing as you!
8. Acolyte
Now With Added BA
13 Feb '03 09:41
Originally posted by maggoteer
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!
I still don't see how pentagonal and higher numbers work graphically ðŸ˜³
9. maggoteer
The MAKIA
13 Feb '03 19:101 edit
Originally posted by Acolyte
I still don't see how pentagonal and higher numbers work graphically ðŸ˜³
Assume 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 &quot;common&quot; vertice. Notice that along those two shared edges, the dots that were vertices of the smaller pentagram are now &quot;edge&quot; vertices of the bigger pentagram.

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 &quot;degenerate&quot; square with one dot. ðŸ˜•
10. royalchicken
CHAOS GHOST!!!
14 Feb '03 02:262 edits
Acolyte &amp; 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.
11. royalchicken
CHAOS GHOST!!!
14 Feb '03 02:29
Originally posted by Acolyte
I still don't see how pentagonal and higher numbers work graphically ðŸ˜³
It's okay. Visualisation can be a handicap.

Just build one more layer on the outside. THe total number for pentagons is n(3n-1)/2 =&gt; 0,1,5,12...
12. maggoteer
The MAKIA
14 Feb '03 04:341 edit
Originally posted by royalchicken
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)
At 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....ðŸ˜• Acoylte, the field is yours!ðŸ˜µðŸ™„
13. bbarr
Chief Justice
14 Feb '03 05:21
Originally posted by royalchicken
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.
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?
14. royalchicken
CHAOS GHOST!!!
14 Feb '03 16:441 edit
Originally posted by bbarr
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?
Hey! 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.

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.
15. Acolyte
Now With Added BA
14 Feb '03 17:58
Originally posted by royalchicken
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.
Or equivalently, any number can be expressed as the sum of at most r r-gonal numbers for all r.