A047775 Number of dissectable polyhedra with n tetrahedral cells and symmetry of type B.

0, 0, 0, 0, 2, 5, 11, 25, 66, 131, 349, 708, 1911, 3856, 10604, 21597, 59961, 123266, 345060, 715198, 2015416, 4206926, 11919257, 25032840, 71246129, 150413234, 429750208, 911379241, 2612614298, 5562367173, 15991792731, 34164355260

Offset: 1

Author

N. J. A. Sloane

Keywords

• nonn

Comments

One of 17 different symmetry types comprising A007173 and A027610 and one of 10 for A371351. Also the number of tetrahedral clusters or polyominoes of the regular tiling with Schläfli symbol {3,3,oo}, both having type B achiral symmetry and n tetrahedral cells. The plane of symmetry bisects a tetrahedral cell (321); the order of the symmetry group is 2. An achiral polyomino is identical to its reflection. - Robert A. Russell, Mar 29 2024

Links

• L. W. Beineke and R. E. Pippert, Enumerating dissectable polyhedra by their automorphism groups, Canad. J. Math., 26 (1974), 50-67.
• S. J. Cyvin, Jianji Wang, J. Brunvoll, Shiming Cao, Ying Li, B. N. Cyvin, and Yugang Wang, Staggered conformers of alkanes: complete solution of the enumeration problem, J. Molec. Struct. 413-414 (1997), 227-239.
• Robert A. Russell, Mathematica Graphics3D program for A047775 examples

Crossrefs

Cf. A047772.

Cf. A007173 (oriented), A027610 (unoriented), A371351 (achiral), A001764 (rooted), A047749 (type U), A047751 (type K), A047753 (type I), A047760 (type F), A047764 (type Q), A047765 (type P), A047773 (type D).

Programs

Mathematica

Table[If[n<5,0,If[OddQ[n],2Binomial[(3n-1)/2,(n-1)/2]/(n+1)+If[1==Mod[n,4],2Binomial[3(n-1)/4,(n-1)/4]/(n+1),0],Binomial[3n/2,n/2]/(n+1)-If[OddQ[n/2],4Binomial[(3n-2)/4,(n-2)/4],2Binomial[3n/4,n/4]]/(n+2)]/2+If[2==Mod[n,6],3Binomial[(n-2)/2,(n-2)/6]/(n+1)+If[2==Mod[n,12],6Binomial[(n-2)/4,(n-2)/12],12Binomial[n/4-1,(n-8)/12]]/(n+4),0]/2-Switch[Mod[n,4],1,4Binomial[(3n+1)/4,(n-1)/4],3,2Binomial[(3n+3)/4,(n+1)/4],,0]/(n+3)-Switch[Mod[n,6],1,3Binomial[(n-1)/2,(n-1)/6]/(n+2),2,6Binomial[n/2,(n-2)/6]/(n+4),4,6Binomial[(n-2)/2,(n-4)/6]/(n+2),,0]-If[5==Mod[n,6],3Binomial[(n+1)/2,(n+1)/6]/(n+4)-Switch[Mod[n,24],5,12Binomial[(n-5)/8,(n-5)/24],17,24Binomial[(n-9)/8,(n-17)/24],,0]/(n+7),0]],{n,50}] (* _Robert A. Russell, Mar 29 2024 *)

Formula

a(n) = (1/2)*(A047749(n) - 2*A047773(n) - 2*A047760(n) - A047753(n) - A047751(n) - A047764(n) - A047765(n)).

G.f.: (2 - G(z^4) - G(z^6))/z + (G(z^2) + z*G(z^2)^2 - G(z^4) + z*G(z^4) - z^2*G(z^4)^2 + z^2*G(z^6) + z^2*G(z^12) + z^8*G(z^12)^2) / 2 + z - z*G(z^4)^2 - z*G(z^6) - z^2*G(z^6)^2 - z^4*G(z^6)^2 + z^5*G(z^24) + z^17*G(z^24)^2, where G(z) = 1 + z*G(z)^3 is the g.f. for A001764. - Robert A. Russell, Mar 29 2024