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Thus named because it looks like arranging and rearranging huge numbers of draughtsmen in a giant parking lot.

Let's consider three types of numbers:
• triangles: a(a+1)/2
• squares: b2
• centered hexagons: 3c(c+1) + 1
where a, b and c are natural numbers.

And two questions:

• Which numbers are both triangular and square, triangular and hexagonal, square and hexagonal?
• Which triangles divide into smaller triangles?
Solutions consist of one or more recurrent series of the type U(a,b,c)F, all with the same factor.

The first question
can be generalized to diophantine equations of the type p1a2+q1a + r1 = p2b2+q2b + r2, where pi, qi and ri are constants.
Here are the above mentioned examples:

Triangles through squares: a(a+1)/2 = b2
 a: 1 8 49 288 1681 9800 ... U(0,1,2)6. b: 1 6 35 204 1189 6930 ... U(0,1,0)6, or the primus of 6.

Triangles through hexagons: a(a+1)/2 = 3c(c+1)+1
 a: 1 13 133 1321 13081 129493 ... U(1,1,4)10. c: 0 5 54 539 5340 52865 ... U(-1,0,4)10.

Squares through hexagons: b2 = 3c(c+1)+1
 b: 1 13 181 2521 35113 489061 ... U(1,1,0)14, or the tertius of 14. c: 0 7 104 1455 20272 282359 ... U(-1,0,6)14.

The second question
is about triangular division and thus about equations of the type a2+a = k(b2+b).
There are no solutions if k is a square. For non-square 'k', solutions approach the square root of 'k'.
Here are some examples:

a2+a = 2(b2+b)
 a: 0 3 20 119 696 4059 23660 ... U(0,3,2)6. b: 0 2 14 84 492 2870 16730 ... U(0,2,2)6.

a2+a = 3(b2+b)
 a: 0 2 9 35 132 494 1845 ... U(0,2,1)4. b: 0 1 5 20 76 285 1065 ... U(0,1,1)4.

a2+a = 5(b2+b)
 a: 0 5 14 99 260 1785 4674 32039 83880 574925 1505174 ... U(0,14,8)18 combined with U(-1,5,8)18. b: 0 2 6 44 116 798 2090 14328 37512 257114 673134 ... U(0,6,8)18 combined with U(0,2,8)18.

a2+a = 6(b2+b)
 a: 0 3 8 35 84 351 836 3479 8280 34443 81968 340955 ... U(0,8,4)10 combined with U(-1,3,4)10. b: 0 1 3 14 34 143 341 1420 3380 14061 33463 139194 ... U(0,3,4)10 combined with U(0,1,4)10.