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.

Showing 1-3 of 3 results.

A090372 Number of unrooted planar 3-constellations with n triangles.

Original entry on oeis.org

1, 6, 22, 174, 1479, 16808, 201834, 2631594, 35965555, 512062566, 7528425420, 113708935808, 1756853846316, 27676951028496, 443411345677658, 7209139541742750, 118738765611199983, 1978360119497335826
Offset: 1

Views

Author

Valery A. Liskovets, Dec 01 2003

Keywords

Comments

These are planar maps with bicolored faces having n black triangular faces and an arbitrary number of white faces of degrees multiple to 3. The vertices can be and are colored so that any black triangle is colored counterclockwise 1,2,3. Isomorphisms are required to respect the colorings. Also unrooted bi-Eulerian maps with bicolored both vertices and faces and with 2n edges; the maps are considered up to color-preserve isomorphism.

Crossrefs

Programs

  • Maple
    with(numtheory): C_3 := proc(n) local s,d; if n=0 then RETURN(1) else s := -3^n*binomial(3*n,n); for d in divisors(n) do s := s+phi(n/d)*3^d*binomial(3*d,d) od; RETURN((4/(3*n))*(3^n*binomial(3*n,n)/((2*n+1)*(2*n+2))+s/2)); fi; end;
  • Mathematica
    a[0] = 1; a[n_] := Module[{s, d}, s = -3^n Binomial[3n, n]; Do[s = s + EulerPhi[n/d] 3^d Binomial[3d, d], {d, Divisors[n]}]; (4/(3n)) (3^n Binomial[3n, n]/((2n+1)(2n+2)) + s/2)];
    Array[a, 18] (* Jean-François Alcover, Jul 24 2018, from Maple *)

A090374 Number of rooted planar 4-constellations with n quadrangles: rooted planar maps with bicolored faces having n black quadrangular faces and an arbitrary number of white faces of degrees multiple to 4.

Original entry on oeis.org

1, 10, 160, 3200, 72960, 1813504, 47923200, 1325629440, 37991219200, 1120005652480, 33789432561664, 1039157228994560, 32480974549811200, 1029463445864448000, 33023079530417356800, 1070513886720329515008, 35026358912891580579840, 1155516042520241436098560
Offset: 1

Views

Author

Valery A. Liskovets, Dec 01 2003

Keywords

Crossrefs

Programs

  • Maple
    A090374 := proc(n)
        5*4^(n-1)*binomial(4*n, n)/((3*n+1)*(3*n+2))
    end proc:
    seq(A090374(n),n=1..40) ; # R. J. Mathar, Mar 29 2023
  • Mathematica
    a[n_] := 5 2^(2n) (4n-1)! / ((n-1)! (3n+2)!);
    Array[a, 18] (* Jean-François Alcover, Aug 28 2019 *)
  • PARI
    vector(20, n, 5*4^(n-1)*binomial(4*n, n)/((3*n+1)*(3*n+2))) \\ Michel Marcus, Dec 11 2014

Formula

a(n) = 5*4^(n-1)*binomial(4*n, n)/((3*n+1)*(3*n+2)). - corrected by Michel Marcus, Dec 11 2014
D-finite with recurrence 3*n*(3*n+2)*(3*n+1)*a(n) -32*(4*n-3)*(2*n-1)*(4*n-1)*a(n-1)=0. - R. J. Mathar, Mar 29 2023

Extensions

More terms from Michel Marcus, Dec 11 2014

A298358 a(n) is the number of rooted 3-connected bicubic planar maps with 2n vertices.

Original entry on oeis.org

1, 0, 0, 1, 0, 3, 7, 15, 63, 168, 561, 1881, 6110, 21087, 72174, 250775, 883116, 3125910, 11174280, 40209852, 145590720, 530358095, 1941862860, 7144623447, 26403493545, 97971775008, 364903633215, 1363847131450, 5113975285788, 19233646581282
Offset: 1

Views

Author

Michael D. Weiner, Jan 17 2018

Keywords

Examples

			A(x) = x + x^4 + 3*x^6 + 7*x^7 + 15*x^8 + 63*x^9 + 168*x^10 + 561*x^11 + ...
		

Crossrefs

Programs

  • Mathematica
    kmax = 30; b[0] = 1; b[n_] := 3*2^(n - 1)*CatalanNumber[n]/(n + 2);
    G[x_] = Sum[b[k] x^k, {k, 0, kmax}];
    A[_] = 1;
    Do[A[x_] = G[x/(1 + A[x] + O[x]^k)^3] - 1 // Normal, {k, 1, kmax + 1}];
    CoefficientList[A[x], x][[2 ;; -2]] (* Jean-François Alcover, Jun 19 2018 *)
  • PARI
    seq(N) = {
      my(x='x+O('x^(N+2)), y=(1-sqrt(1-8*x))/(4*x), g=(1+4*y-y^2)/4);
      Vec(subst(g-1, 'x, serreverse(x*g^3)));
    };
    seq(30) \\ Gheorghe Coserea, Apr 11 2018

Formula

G.f.: A(x) = G(x/(1+A(x))^3)-1 where A(x*(G(x))^3) = G(x)-1 and G(x) = g.f. for A000257.
From Gheorghe Coserea, Apr 16 2018: (Start)
y = A(x)/x satisfies: 0 = x^6*y^7 + 6*x^5*y^6 + 15*x^4*y^5 + 4*x^3*(5 - 3*x)*y^4 + x^2*(15 - 37*x)*y^3 + x*(16*x^2 - 39*x + 6)*y^2 + (24*x^2 - 15*x + 1)*y + (9*x - 1).
A(x) = serreverse((1+x)^3*(1 + 12*x - (1-4*x)^(3/2))/(2*(4*x+3)^2)); equivalently, it can be rewritten as A(x) = serreverse((y-1)*(y^2+y-1)^3/(y^5*(3*y-2)^2)), where y = A000108(x). (End)
a(n) ~ 3 * 2^(9*n-1) / (sqrt(Pi) * 17^(5/2) * 5^(3*n - 5/2) * n^(5/2)). - Gheorghe Coserea and Vaclav Kotesovec, Apr 16 2018
Showing 1-3 of 3 results.