A266240 Triangle read by rows: T(n,g) is the number of rooted 2n-face triangulations in an orientable surface of genus g.

1, 4, 1, 32, 28, 336, 664, 105, 4096, 14912, 8112, 54912, 326496, 396792, 50050, 786432, 7048192, 15663360, 6722816, 11824384, 150820608, 544475232, 518329776, 56581525, 184549376, 3208396800, 17388675072, 30117189632, 11100235520, 2966845440

Offset: 0

Author

Gheorghe Coserea, Dec 25 2015

Keywords

• nonn
• tabf

Comments

Row n contains floor((n+3)/2) terms.

Examples

			Triangle starts:
n\g    [0]          [1]          [2]          [3]          [4]
[0]    1;
[1]    4,           1;
[2]    32,          28;
[3]    336,         664,         105;
[4]    4096,        14912,       8112;
[5]    54912,       326496,      396792,      50050;
[6]    786432,      7048192,     15663360,    6722816;
[7]    11824384,    150820608,   544475232,   518329776,   56581525;
[8]    184549376,   3208396800,  17388675072, 30117189632, 11100235520;
[9]    ...
		

Links

• Gheorghe Coserea, Rows n = 0..200, flattened
• Edward A. Bender, Zhicheng Gao, L. Bruce Richmond, The map asymptotics constant tg, The Electronic Journal of Combinatorics, Volume 15 (2008), Research Paper #R51.
• Zhicheng Gao, A Formula for the Bivariate Map Asymptotics Constants in terms of the Univariate Map Asymptotics Constants, The Electronic Journal of Combinatorics, Volume 17 (2010), Research Paper #R155.
• I. P. Goulden and D. M. Jackson, The KP hierarchy, branched covers, and triangulations, Advances in Mathematics, Volume 219, Issue 3, 20 October 2008, Pages 932-951.

Crossrefs

Columns k=0-4 give: A002005, A269473, A269474, A269475, A269476.

Row sums give A062980.

Cf. A269418, A269419.

Programs

Mathematica

T[n_ /; n >= 0, g_] /; 0 <= g <= (n+1)/2 := f[n, g]/(3n+2); T[, ] = 0; f[n_ /; n >= 1, g_ /; g >= 0] := f[n, g] = 4*(3*n+2)/(n+1)*(n*(3*n-2)*f[n - 2, g-1] + Sum[f[i, h]*f[n-2-i, g-h], {i, -1, n-1}, {h, 0, g}]); f[-1, 0] = 1/2; f[0, 0] = 2; f[, ] = 0; Table[Table[T[n, g], {g, 0, Floor[(n + 1)/2]}], {n, 0, 9}] // Flatten (* Jean-François Alcover, Feb 27 2016 *)

PARI

N = 10;
m = matrix(N+2, N+2);
mget(n,g) = {
  if (g < 0 || g > (n+1)/2, return(0));
  return(m[n+2,g+1]);
}
mset(n,g,v) = {
  m[n+2,g+1] = v;
}
Cubic() = {
  mset(-1,0,1/2);
  mset(0,0,2);
  for (n = 1, N,
  for (g = 0, (n+1)\2,
    my(t1 = n * (3*n-2) * mget(n-2, g-1),
       t2 = sum(i = -1, n-1, sum(h = 0, g,
                mget(i,h) * mget(n-2-i, g-h))));
    mset(n, g, 4*(3*n+2)/(n+1) * (t1 + t2))));
  my(a = vector(N+1));
  for (n = 0, N,
    a[n+1] = vector(1 + (n+1)\2);
    for (g = 0, (n+1)\2,
         a[n+1][g+1] = mget(n, g));
    a[n+1] = a[n+1]/(3*n+2));
  return(a);
}
concat(Cubic())

Formula

T(n,g) = f(n,g)/(3*n+2) for all n >= 0 and 0 <= g <= (n+1)/2, where f(n,g) satisfies the quadratic recurrence equation f(n,g) = 4*(3*n+2)/(n+1)*(n*(3*n-2)*f(n-2,g-1) + Sum_{i=-1..n-1} Sum_{h=0..g} f(i,h)*f(n-2-i, g-h)) for n >= 1 and g >= 0 with the initial conditions f(-1,0)=1/2, f(0,0)=2 and f(n,g)=0 for g < 0 or g > (n+1)/2.

For column g, as n goes to infinity we have T(n,g) ~ 3*6^((g-1)/2) * t(g) * n^(5*(g-1)/2) * (12*sqrt(3))^n, where t(g) = (A269418(g)/A269419(g)) / (2^(g-2) * gamma((5*g-1)/2)) and gamma is the Gamma function. - Gheorghe Coserea, Feb 26 2016