| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
Hexagonal numbers are the counts of objects arranged in a hexagonal lattice. For example, * * * * * * * * * * * * * * * * * * * * * * * * * * * So h(1) = 1, h(2) = 7, h(3) = 19, etc. Problem [1]: (Easy) Find a formula for h(n) as a polynomial in n. Problem [2]: (Hard) Arrange the numbers 1..h(n), for some n, to form a 'magic' hexagon in which all the rows - horizontal, 60 degrees, or 120 degrees - add to the same sum. Besides the trivial case for h(1), THERE IS ONLY ONE OTHER SOLUTION (except for rotations and reflections). Hints for both problems follow. Hint[1]: There are (at least) two geometrical approaches, each based on the formula for triangular numbers t(n)=n(n+1)/2. A LARGE hint for problem 2 follows: Hint[2]: Only h(3) works.
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 1064.1 | re Problem [1]: (Easy) Find a formula for h(n) as a polynomial in n. | AITG::DERAMO | Daniel V. {AITG,ZFC}:: D'Eramo | Fri Apr 21 1989 22:49 | 27 |
I would expect a quadratic relationship, so the first three values are sufficient to set up a table of differences, but I threw in a fourth anyway: n: 1 2 3 4 h(n): 1 7 19 37 1st diff: 6 12 18 2nd diff: 6 6 The function n has first difference 1, remaining differences 0. The function n(n-1) has first difference 2n, second difference 2, remaining differences 0. The function n(n-1)(n-2) has first difference 3n(n-1), etc. Actually we won't need the third and higher degree terms here: h(n) = an(n-1) + bn + c = [1, 7, 19, 37, ...] 1st diff = 2an + b = [6, 12, 18, ...] 2nd diff = 2a = [6, 6, 6, ...] Substituting 1 for n where it occurs, these give 2a = 6 so a = 3. Then 2an + b = 6 + b = 0 so b = 0. Finally, 3n(n-1) + c = c = 1. So h(n) = 3n(n-1) + 1 = 3n^2 - 3n + 1. Check: n: 1 2 3 4 ... 3n^2 - 3n + 1: 1 7 19 37 ... Dan | |||||
| 1064.2 | but how to arrange them? | HERON::BUCHANAN | Andrew @vbo DTN 828-5805 | Mon Apr 24 1989 09:22 | 19 |
| 1064.3 | same difference | VINO::JMUNZER | Mon Apr 24 1989 16:50 | 12 | |
* @ @ @ @ * If each side is k long, then there 6 triangles at
@ * @ @ @ * @
@ @ * @ @ * @ @ 1 + 2 + ... + (k-2) = .5 * (k-2) * (k-1)
@ @ @ * @ * @ @ @
@ @ @ @ * * @ @ @ @ at-signs each, and 6 lines at (k-1) stars each,
* * * * * # * * * * * and one center sharp. That's
@ @ @ @ * * @ @ @ @
@ @ @ * @ * @ @ @ 3 * (k-2) * (k-1) + 6 * (k-1) + 1
@ @ * @ @ * @ @
@ * @ @ @ * @ which agrees, of course, with .1.
* @ @ @ @ *
John
| |||||
| 1064.4 | Lots of approaches to problem 1 | AKQJ10::YARBROUGH | I prefer Pi | Mon Apr 24 1989 18:17 | 6 |
You can get the result by summing six little triangles; by taking the difference of one large triangle and three little ones; by taking the difference between a large rhombus and two little triangles; or algebraically by differences; or ... Lynn | |||||
| 1064.5 | the answer | UTRUST::DEHARTOG | set vms /posix | Mon Apr 24 1989 20:57 | 12 |
Just entering this conference (looks interesting!). Here are my 1st
2cts:
3 19 16
17 7 2 12
18 1 5 4 10
11 6 8 13
9 14 15
Hans.
| |||||
| 1064.6 | Fresh blood | NIZIAK::YARBROUGH | I PREFER PI | Wed Apr 26 1989 15:45 | 1 |
Well done, and welcome to the conference! - Lynn Yarbrough | |||||