Number of convex quadrilaterals constructed from polyabolos. ------------------------------------------------------------ o 1 o 8 2 o o 7 3 o o 6 4 o 5 o Number the sides of an octahedron 1..8 clockwise. Any quadrilateral constructed from polyabolos will have 4 of these as sides which can be used to identify the basic shape. Up to rotation and reflection there are 8 possible shapes. Number of solutions to n = a * b where 1 <= a <= b: = ceiling(numdiv(n)/2) where numdiv(n) is the number of divisors of n. = A038548(n) Number of solutions to n = a * b where 1 <= a < b: = floor(numdiv(n)/2)) = A056924(n) Shapes 1357 (rectangle) area in cells: 2 * s1 * s2 must be even; require 1 <= s1, 1 <= s2. for uniqueness up to rotation also require s1 <= s2 for n even, number of solutions is then ceiling(numdiv(n/2)/2) Shape 2468 (rectangle at 45 degrees) area in cells: 4 * s1 * s2 must be multiple of 4; required 1 <= s1, 1 <= s2. for uniqeness up to rotation also require s1 <= s2 for n a multiple of 4, number of solutions is then ceiling(numdiv(n/4)/2) Shape 1256 (rhombus) area in cells: 2 * s1 * s2 must be even; require 1 <= s1, 1 <= s2. for n even, number of solutions is then numdiv(n/2) Let b(n) = number of solutions to n = a*a + 2*a*b where 1 <= a and 1 <= b: Now a*a + 2*a*b = a*(a + 2*b) = ((a+b)-b)*(a+b)+b) = (a+b)^2 - b^2 is then the number of representations of n as the difference of two positive squares = A100073(n) Can be computed with PARI: b(n) = if(n%2, floor(numdiv(n)/2), if(n%4, 0, floor(numdiv(n/4)/2))); Shape 1257 (right trapezoid) area in cells 2*s1*s2 + s2^2; require 1 <= s1, 1 <= s2. number of solutions is then b(n) Shape 1246 (right trapezoid at 45 degrees) area in cells 4*s2*s4 + 2*s4^2 must be even; require 1 <= s2, 1 <= s4. for n even, number of solutions is then b(n/2) Shape 1258 (trapezoid) area in cells 2*s1*s2 + 2*s2^2 = 2*s2*(s1+s2) must be even; require 1 <= s1, 1 <= s2. for n even, number of solutions is then floor(numdiv(n/2)/2) since s1+s2 > s2 Shape 1236 (trapezoid at 45 degrees) area in cells 2*s1*s2 + s2^2; require 1 <= s1, 1 <= s2. number of solutions is then b(n) Let d(n) = number of solutions to n = b^2 - 2*a^2 where 1 <= a and 2*a < b: Can be computed with PARI: d(n) = my(t); sum(k=1, floor(sqrt((n-1)/2)), issquare(n+2*k^2,&t) && t>2*k); Shape 1247 (kite) area in cells (s1 + 2*s2)^2 - 2*s2^2; require 1 <= s1, 1 <= s2. number of solitions is then d(n) Total ----- Terms for all n: 2*b(n) + d(n) Terms for n is a multiple of 2: numdiv(n/2) + ceiling(numdiv(n/2)/2) + floor(numdiv(n/2)/2) + b(n/2) = 2 * numdiv(n/2) + b(n/2) Terms for n is a multiple of 4: ceiling(numdiv(n/4)/2) As a PARI script: a(n) = 2*b(n) + d(n) + if(n%2, 0, 2*numdiv(n/2) + b(n/2)) + if(n%4, 0, ceil(numdiv(n/4)/2));