cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A047776 Number of chiral pairs of asymmetric dissectable polyhedra with n tetrahedral cells (type A).

Original entry on oeis.org

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

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Author

Keywords

Comments

One of 17 different symmetry types comprising A007173 and A027610 and one of 7 for A371350. Also the number of tetrahedral clusters or polyominoes of the regular tiling with Schläfli symbol {3,3,oo}, both asymmetric (type A) with n tetrahedral cells. The order of the symmetry group is 1. Each member of a chiral pair is a reflection but not a rotation of the other. - Robert A. Russell, Mar 31 2024

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

  • 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)

A001762 Number of labeled n-vertex dissections of a ball.

Original entry on oeis.org

1, 1, 10, 180, 4620, 152880, 6168960, 293025600, 15990004800, 984647664000, 67493121696000, 5094263446272000, 419688934689024000, 37465564582397952000, 3601861863990534144000, 370962724717928318976000, 40744403224500159055872000
Offset: 3

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Author

Keywords

Comments

This is the number of labeled Apollonian networks (planar 3-trees). - Allan Bickle, Feb 20 2024

Examples

			There is one maximal planar graph with 4 vertices, and one way to label it, so a(4) = 1.
		

References

  • L. W. Beineke and R. E. Pippert, Enumerating labeled k-dimensional trees and ball dissections, pp. 12-26 of Proceedings of Second Chapel Hill Conference on Combinatorial Mathematics and Its Applications, University of North Carolina, Chapel Hill, 1970. Reprinted in Math. Annalen, 191 (1971), 87-98.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Julia
    using Combinatorics
    a(n) = n < 4 ? 1 : binomial(BigInt(n),3)*factorial(BigInt(3*n-9))÷factorial(BigInt(2*n-4))
    print([a(n) for n in 3:28]) # Paul Muljadi, Mar 27 2024
  • Mathematica
    Join[{1}, Table[Binomial[n, 3]*(3*n - 9)!/(2*n - 4)!, {n, 4, 25}]] (* T. D. Noe, Aug 10 2012 *)
  • Python
    from math import factorial
    from sympy import binomial
    def a(n):
        if n < 4:
            return 1
        else:
            return binomial(n, 3) * factorial(3*n-9) // factorial(2*n-4)
    print([a(n) for n in range(3, 21)]) # Paul Muljadi, Mar 05 2024
    

Formula

a(n) = binomial(n,3)*(3*n-9)!/(2*n-4)!, n >= 4; a(3) = 1.
a(n) ~ 3^(3*n - 19/2) * n^(n-2) / (2^(2*n - 5/2) * exp(n)). - Vaclav Kotesovec, Mar 14 2024

Extensions

More terms from Wolfdieter Lang
Name clarified by Andrey Zabolotskiy, Mar 15 2024

A121180 Alkane systems (see Cyvin reference for precise definition).

Original entry on oeis.org

0, 0, 0, 1, 1, 5, 6, 26, 32, 133, 176, 708, 952, 3861, 5302, 21604, 29960
Offset: 1

Views

Author

N. J. A. Sloane, Aug 17 2006

Keywords

Comments

Appears to be A047774 without every third term (all omitted terms are zeros). - Andrey Zabolotskiy, Jul 29 2023

Crossrefs

Cf. other columns of Cyvin et al.'s Table 5: A027610 (spectral isomers), A007173 (stereoisomers), A047775 (C_s), A047772 (C_i), A047774 (C_3, apparently), A047767 (C_{2h}), A047761 (C_{2v}), A047773 (C_{3v}, apparently).
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