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%I A248864 #37 Mar 25 2015 01:30:22
%S A248864 4,5,6,5,6,7,6,7,8,7,8,7,6,9,8,7,8,8,8,9,8,9
%N A248864 Minimal perimeter of an n-dollar construction consisting of 3-dollar triangles and 4-dollar squares.
%C A248864 The squares and equilateral triangles have edge lengths of 1.
%C A248864 What ratios of triangle cost to square cost make large minimal perimeter shapes dominated by squares?  What ratios make them dominated by triangles? - _Gordon Hamilton_, Mar 17 2015
%C A248864 a(19) - a(27) have not been proved to be optimal. - _Gordon Hamilton_, Mar 17 2015
%e A248864 a(6) = 4 is created by gluing two equilateral triangles along an edge to make a rhombus with perimeter 4:
%e A248864 .
%e A248864                           /\
%e A248864                          /  \
%e A248864                         /____\
%e A248864                         \    /
%e A248864                          \  /
%e A248864                           \/
%e A248864 .
%e A248864 a(16) = 8 because 4($4) = $16 and the four squares can be arranged so the shape has perimeter 8:
%e A248864 .
%e A248864                     +------+------+
%e A248864                     |      |      |
%e A248864                     |      |      |
%e A248864                     |      |      |
%e A248864                     +------+------+
%e A248864                     |      |      |
%e A248864                     |      |      |
%e A248864                     |      |      |
%e A248864                     +------+------+
%e A248864 .
%e A248864 a(17) = 7 because 3($3) + 2($4) = $17 and three triangles can be built on top of two squares to create a shape with perimeter 7:
%e A248864                         _______
%e A248864                        /\     /\
%e A248864                       /  \   /  \
%e A248864                      /    \ /    \
%e A248864                     +------+------+
%e A248864                     |      |      |
%e A248864                     |      |      |
%e A248864                     |______|______|
%e A248864 .
%e A248864 a(18) = 6 because 6($3) = $18 and the six triangles can be built into a hexagon of perimeter 6.
%e A248864                          ______
%e A248864                         /\    /\
%e A248864                        /  \  /  \
%e A248864                       /____\/____\
%e A248864                       \    /\    /
%e A248864                        \  /  \  /
%e A248864                         \/____\/
%e A248864 .
%e A248864 a(19) = 9 because 5($3) + 1($4) = $19 and this is one of the minimal perimeter shapes:
%e A248864                          ______
%e A248864                         /\    /
%e A248864                        /  \  /
%e A248864                       /____\/____
%e A248864                       \    /\    /
%e A248864                        \  /  \  /
%e A248864                         \/____\/
%e A248864                          |    |
%e A248864                          |    |
%e A248864                          |____|
%e A248864 .
%e A248864 a(20) = 8 because 4($3) + 2($4) = $20 and four triangles can be built on top of two squares to create this minimal perimeter shape:
%e A248864                            +
%e A248864                           / \
%e A248864                          /   \
%e A248864                         /_____\
%e A248864                        /\     /\
%e A248864                       /  \   /  \
%e A248864                      /    \ /    \
%e A248864                     +------+------+
%e A248864                     |      |      |
%e A248864                     |      |      |
%e A248864                     |______|______|
%e A248864 .
%e A248864 a(21) = 7 because 7($3) = $21 and the seven triangles can be built into this minimal perimeter shape.
%e A248864                          ______
%e A248864                         /\    /\
%e A248864                        /  \  /  \
%e A248864                       /____\/____\
%e A248864                       \    /\    /\
%e A248864                        \  /  \  /  \
%e A248864                         \/____\/____\
%e A248864 .
%e A248864 a(26) = 8 because 6($3) + 2($4) = $26 and the following shape minimizes the perimeter:
%e A248864                         _______
%e A248864                        /\     /\
%e A248864                       /  \   /  \
%e A248864                      /    \ /    \
%e A248864                     +------+------+
%e A248864                     |      |      |
%e A248864                     |      |      |
%e A248864                     |      |      |
%e A248864                     +------+------+
%e A248864                      \    / \    /
%e A248864                       \  /   \  /
%e A248864                        \/_____\/
%e A248864 .
%e A248864 a(33) = 9 because 7($3) + 3($4) = $33 and the following construction works: Take a triangle. Glue the three squares to its three edges. Use the remaining 6 triangles to make the convex shape of perimeter 9.
%K A248864 nonn,more
%O A248864 6,1
%A A248864 _Gordon Hamilton_, Mar 03 2015

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