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A214571
Irregular triangle read by rows: T(n,k) is the number of ordered trees having n vertices and isomorphic (as rooted trees) to k ordered trees (n >= 1, k >= 1).
5
1, 1, 2, 3, 2, 5, 6, 3, 6, 16, 12, 8, 10, 34, 21, 32, 5, 30, 11, 68, 48, 100, 15, 108, 0, 24, 9, 10, 0, 36, 16, 126, 81, 260, 30, 336, 7, 128, 45, 40, 0, 264, 0, 0, 15, 0, 0, 18, 0, 40, 0, 0, 0, 24, 19, 222, 141, 616, 60, 828, 21, 520, 117, 130, 0, 1140, 0, 0, 45, 80, 0, 234, 0, 160, 21, 0, 0, 312, 0, 0, 0, 0, 0, 120, 0, 0, 0, 0, 0, 36, 0, 0, 0, 40, 26, 372, 219, 1392, 90, 1914, 42, 1664, 315, 320, 0, 3696, 0, 28, 195, 544, 0, 1044, 0, 580, 21, 0, 0, 2112, 0, 0, 27, 28, 0, 480, 0, 0, 0, 0, 35, 648, 0, 0, 0, 320, 0, 84, 0, 0, 0, 0, 0, 240, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 360
OFFSET
1,3
COMMENTS
Row n contains A214570(n) entries.
T(n,1) = A003238(n).
Sum_{k=1..n} T(n,k) = A000108(n) (the Catalan numbers).
Sum_{k=1..n} T(n,k)/k = A000081(n) (the number of rooted trees with n vertices).
T(n,k) = k*A214569(n,k).
T(n,k) is also the number of function representations as x^x^...^x with n x's and parentheses inserted in all possible ways that are equivalent to (describe the same function as) k-1 other representations. T(4,2) = 2: (x^x)^(x^x), (x^(x^x))^x; T(5,3) = 3: ((x^x)^x)^(x^x), ((x^x)^(x^x))^x, ((x^(x^x))^x)^x. - Alois P. Heinz, Aug 31 2012
LINKS
FORMULA
No formula available. Entries have been obtained from T(n,k)= k*A214569(n,k).
EXAMPLE
Row 4 is 3,2: among the five ordered 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 and A - B - C - D with root at C are isomorphic among themselves.
Triangle starts:
1;
1;
2;
3, 2;
5, 6, 3;
6, 16, 12, 8;
10, 34, 21, 32, 5, 30;
11, 68, 48, 100, 15, 108, 0, 24, 9, 10, 0, 36;
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=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, 1, Max[l]}]]; Table[T[n], {n, 1, 10}] // Flatten (* Jean-François Alcover, May 28 2015, after Alois P. Heinz *)
KEYWORD
nonn,tabf,hard
AUTHOR
Emeric Deutsch, Jul 28 2012
STATUS
approved