A214569 Irregular triangle read by rows: T(n,k) is the number of rooted trees having n vertices and isomorphic (as rooted trees) to k ordered trees (n>=1, k>=1).

1, 1, 2, 3, 1, 5, 3, 1, 6, 8, 4, 2, 10, 17, 7, 8, 1, 5, 11, 34, 16, 25, 3, 18, 0, 3, 1, 1, 0, 3, 16, 63, 27, 65, 6, 56, 1, 16, 5, 4, 0, 22, 0, 0, 1, 0, 0, 1, 0, 2, 0, 0, 0, 1, 19, 111, 47, 154, 12, 138, 3, 65, 13, 13, 0, 95, 0, 0, 3, 5, 0, 13, 0, 8, 1, 0, 0, 13, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 26, 186, 73, 348, 18, 319, 6, 208, 35, 32, 0, 308, 0, 2, 13, 34, 0, 58, 0, 29, 1, 0, 0, 88, 0, 0, 1, 1, 0, 16, 0, 0, 0, 0, 1, 18, 0, 0, 0, 8, 0, 2, 0, 0, 0, 0, 0, 5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 6

Offset

1,3

Comments

Row n contains A214570(n) entries.

T(n,1) = A003238(n).

Sum(T(n,k), k=1..n) = A000081(n) = number of rooted trees with n vertices.

Sum(k*T(n,k), k=1..n) = A000108(n-1) (the Catalan numbers).

T(n,k) is also the number of size k equivalence classes of function representations as x^x^...^x with n x's and parentheses inserted in all possible ways. T(4,2) = 1: (x^x)^(x^x) == (x^(x^x))^x; T(5,3) = 1: ((x^x)^x)^(x^x) == ((x^x)^(x^x))^x == ((x^(x^x))^x)^x. - Alois P. Heinz, Aug 31 2012

Links

Alois P. Heinz, Rows n = 1..16, flattened

Formula

No formula available. Entries have been obtained by counting (using Maple) the rooted trees (identified by their Matula-Goebel numbers) with the required properties (using A061775 and A206487).

Example

Row 4 is 3,1: among the four rooted trees with 4 vertices the path tree P_4, the star tree K_{1,3}, and the tree in the shape of Y are isomorphic only to themselves, while A - B - C - D with root at B is isomorphic to itself and to A - B - C - D with root at C.
Triangle starts:
   1;
   1;
   2;
   3,  1;
   5,  3,  1;
   6,  8,  4,  2;
  10, 17,  7,  8, 1,  5;
  11, 34, 16, 25, 3, 18, 0, 3, 1, 1, 0, 3;
  ...

Maple

F:= proc(n) option remember; `if`(n=1, [x+1],
      [seq(seq(seq(f^g, g=F(n-i)), f=F(i)), i=1..n-1)])
    end:
T:= proc(n) option remember; local i, l, p;
      l:= map(f->coeff(series(f, x, n+1), x, n), F(n)):
      p:= proc() 0 end: forget(p);
      for i in l do p(i):= p(i)+1 od:
      l:= map(p, l); forget(p);
      for i in l do p(i):= p(i)+1 od:
      seq(p(i)/i, i=1..max(l[]))
    end:
seq(T(n), n=1..10);  # Alois P. Heinz, Aug 31 2012

Mathematica

F[n_] := F[n] = If[n == 1, {x+1}, Flatten[Table[Table[Table[f^g, {g, F[n-i]}], {f, F[i]}], {i, 1, n-1}]]]; T[n_] := T[n] = Module[{i, l, p}, l = Map[Function[ {f}, Coefficient[Series[f, {x, 0, n+1}], x, n]], F[n]]; Clear[p]; p[_] = 0; Do[ p[i] = p[i]+1 , {i, l}]; l = Map[p, l]; Clear[p]; p[_] = 0; Do[p[i] = p[i]+1, {i, l}]; Table[p[i]/i, {i, 1, Max[l]}]]; Table[T[n], {n, 1, 10}] // Flatten (* Jean-François Alcover, May 28 2015, after Alois P. Heinz *)

Crossrefs

Cf. A000108, A000081, A003238, A214570, A214571, A061775, A206487, A215703.

Sequence in context: A345990 A261555 A134734 * A263409 A047706 A300518

Adjacent sequences: A214566 A214567 A214568 * A214570 A214571 A214572

Keyword

nonn,tabf,hard

Author

Emeric Deutsch, Jul 28 2012

Status

approved