A047776 Number of chiral pairs of asymmetric dissectable polyhedra with n tetrahedral cells (type A).
0, 0, 0, 0, 2, 11, 71, 370, 2005, 10682, 58167, 320116, 1789210, 10121965, 57933469, 334919626, 1953800059, 11489466014, 68053583772, 405713887061, 2433000197471, 14668527134167, 88869448492895, 540834097467624, 3304961431043989, 20273201718862728, 124798671079300720, 770762029389852807
Offset: 1
Links
- L. W. Beineke and R. E. Pippert, Enumerating dissectable polyhedra by their automorphism groups, Canad. J. Math., 26 (1974), 50-67.
- Robert A. Russell, Mathematica Graphics3D program for A047776 examples
Crossrefs
Cf. A007173 (oriented), A027610 (unoriented), A371350 (chiral), A001764 (rooted), A047775 (type B), A047774 (type C). A047773 (type D), A047762 (type E), A047760 (type F), A047758 (type G), A047754 (type H), A047753 (type I), A047752 (type J), A047751 (type K), A047771 (type L), A047769 (type M), A047766 (type N|O), A047765 (type P), A047764 (type Q).
Programs
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Mathematica
Table[If[n < 5, 0, Binomial[3 n, 2 n + 2]/(3 n (n - 1)) - If[OddQ[n], Binomial[3 n/2 - 1/2, n + 1] 3/(n - 1), 7 Binomial[3 n/2, n + 1]/(3 n)] - Switch[Mod[n, 3], 1, Binomial[n - 1, 2 n/3 + 1/3]/(n - 1), 2, Binomial[n - 1, 2 n/3 + 2/3]/(n - 2), _, 0] + Switch[Mod[n, 4], 1, Binomial[3 n/4 - 3/4, n/2 + 1/2] 2/(3 (n - 1)) + Binomial[3 n/4 + 1/4, n/2 + 3/2] 4/(n - 1) + Binomial[3 n/4 - 3/4, n/2 + 1/2] 4/(n + 3), 2, Binomial[3 n/4 - 1/2, n/2 + 1] 8/(n - 2), 3, Binomial[3 n/4 - 1/4, n/2 + 3/2] 12/(n - 3), 0, Binomial[3 n/4 - 1, n/2 + 1] 12/(n - 4)] + Switch[Mod[n, 6], 1, Binomial[n/2 - 1/2, n/3 + 2/3] 6/(n - 1), 2, Binomial[n/2 - 1, n/3 + 1/3] 4/(n - 2) + Binomial[n/2, n/3 + 4/3] 6/(n - 2) + Binomial[n/2 - 1, n/3 + 1/3] 6/(n + 4), 4, Binomial[n/2 - 1, n/3 + 2/3] 12/(n - 4), 5, Binomial[n/2 - 1/2, n/3 + 1/3] 9/(n + 4), _, 0] + Switch[Mod[n, 12], 2, -Binomial[n/4 - 1/2, n/6 + 2/3] 12/(n - 2), 5, Binomial[n/4 - 5/4, n/6 - 5/6] 2/(n + 1), 8, -Binomial[n/4 - 1, n/6 - 1/3] 12/(n + 4), _, 0] - Switch[Mod[n, 24], 5, Binomial[n/8 - 5/8, n/12 - 5/12] 12/(n + 7), 17, Binomial[n/8 - 9/8, n/12 - 5/12] 24/(n + 7), , 0]]/2, {n, 1, 60}] (* _Robert A. Russell, Apr 09 2012 *)
Formula
From Robert A. Russell, Mar 31 2024: (Start)
a(n) = A001764(n)/(12(n+1)) - A047775(n)/2 - A047774(n)/3 - A047773(n)/6 - A047762(n)/2 - A047760(n)/4 - A047758(n)/4 - A047754(n)/4 - A047753(n)/8 - A047752(n)/12 - A047751(n)/24 - A047771(n)/2 - A047769(n)/2 - A047766(n)/6 - A047766(n)/6 - A047765(n)/4 - A047764(n)/12.
G.f.: (G(z^4) + G(z^6) - 2)/(2z) - z/3 + G(z)/6 - G(z)^2/12 + z*G(z)^4/24 - 7*G(z^2)/12 - 3z*G(z^2)^2/8 - z*G(z^3)/6 - z^2*G(z^3)^2/12 + G(z^4)/2 - z*G(z^4)/6 + (z*G(z^4)^2 + z^2*G(z^4)^2 + z*G(z^6))/2 + z^2*G(z^6)/12 + (z^2*G(z^6)^2 + z^4*G(z^6)^2 - z^2*G(z^12))/2 + z^5*G(z^12)/6 - (z^8*G(z^12)^2 + z^5*G(z^24) + z^17*G(z^24)^2)/2, where G(z) = 1 + z*G(z)^3 is the g.f. for A001764. (End)
Comments