One might hope to generalize the factorial reflection formula, z! (-z)! = pi z/(sin pi z) to include things like z! (wz)! (z/w)!, with w^3 = 1, but these have unilaterally infinite rows of singularities, whereas finite combinations of elementary functions have their singularities bilaterally infinite. So the only oddgons we can do are the special cases where pairs of vertices sum to integers. E.g. for triangles of integer radius centered at the origin, 2 even! (even w)! (even w )! even + 1 %pi even 2 even! = -----------------------------------------------------, even ---- 2 /===\ sqrt(3) %pi even | | 2 2 sqrt(3) sinh(----------------) | | (4 k + 3 even ) 2 | | k = 1 with a similar expression for odd!. We can also slide the triangle left or right by an integer. For a 2n-gon of any phase and radius, we can just combine opposite vertices in pairs: %i %pi n 2 n - 1 %i %pi k -------- /===\ -------- n 2 n | | n %i %pi %e z | | (%e z)! = - ---------------------, | | f (%pi z) k = 0 n n - 1 %i %pi k /===\ -------- | | n f (z) = | | sin(%e z). n | | k = 0 f_n(z) is fairly interesting. For the decagon, f_5(z) = i pi 2 i pi 3 i pi 4 i pi ---- ------ ------ ------ 5 5 5 5 sin(z) sin(e z) sin(e z) sin(e z) sin(e z) 15 25 5 z 521 z = z - ----- + ---------------- + . . . 93555 4593988395871875 inf ==== j 10 j + 5 \ (- 1) (fib(10 j + 6) + fib(10 j + 4) + 1) (2 z) 5 > -------------------------------------------------------- / (10 j + 5)! ==== j = 0 = ---------------------------------------------------------------- 16 (after strenuous simplification) where the fibby factor can be written as Lucas(10j+5) + 1, since Lucas(n) = fib(n-1) + fib(n+1) = fib(2n)/fib(n). The powers of 2 in the coefficients exactly cancel in terms 5,15,25,45,65,75,105,... (What is the largest n for which the numerator of 2^n/n! is 2? 4? 8? ...) For the hexagon, f_3(z) = %i sqrt(3) 1 %i sqrt(3) 1 sin(z) sin((---------- - -) z) sin((---------- + -) z) 2 2 2 2 inf ==== i 6 i + 3 \ 3 (- 1) (2 z) = > (- ---------------------) / 4 (6 i + 3)! ==== i = 0 9 15 21 3 z 4 z 2 z = - z + --- - --------- + -------------- + . . ., 945 212837625 64965492466875 and for the octagon, f_4(z) = (%i - 1) z (%i + 1) z %i sin(z) sinh(z) sin(----------) sin(----------) = sqrt(2) sqrt(2) inf ==== 4 i + 2 4 i + 2 i 8 i + 4 \ %i ((2 sqrt(2) + 4) + (4 - 2 sqrt(2)) ) (- 1) z

(- --------------------------------------------------------------------) / 2 (8 i + 4)! ==== i = 0

12 20 4 %i z (59 %i) z = - %i z + ------ - ------------- + . . . 9450 1302566265000 --rwg PS, "square" tree rings: Replace one side of a square with a semicircle to make a figure with two corners and three straight sides. Near me is a large, sawn-off tree stump whose outer rings are routinely round, but about halfway in gradually deform to the aforementioned shape, with sides quite straight and corners quite sharp. Explain.