cp's OEIS Frontend

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).

%I A214569 #34 Mar 06 2019 18:44:27
%S A214569 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,
%T A214569 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,
%U A214569 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
%N 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).
%C A214569 Row n contains A214570(n) entries.
%C A214569 T(n,1) = A003238(n).
%C A214569 Sum(T(n,k), k=1..n) = A000081(n) = number of rooted trees with n vertices.
%C A214569 Sum(k*T(n,k), k=1..n) = A000108(n-1) (the Catalan numbers).
%C A214569 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
%H A214569 Alois P. Heinz, <a href="/A214569/b214569.txt">Rows n = 1..16, flattened</a>
%F A214569 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).
%e A214569 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.
%e A214569 Triangle starts:
%e A214569    1;
%e A214569    1;
%e A214569    2;
%e A214569    3,  1;
%e A214569    5,  3,  1;
%e A214569    6,  8,  4,  2;
%e A214569   10, 17,  7,  8, 1,  5;
%e A214569   11, 34, 16, 25, 3, 18, 0, 3, 1, 1, 0, 3;
%e A214569   ...
%p A214569 F:= proc(n) option remember; `if`(n=1, [x+1],
%p A214569       [seq(seq(seq(f^g, g=F(n-i)), f=F(i)), i=1..n-1)])
%p A214569     end:
%p A214569 T:= proc(n) option remember; local i, l, p;
%p A214569       l:= map(f->coeff(series(f, x, n+1), x, n), F(n)):
%p A214569       p:= proc() 0 end: forget(p);
%p A214569       for i in l do p(i):= p(i)+1 od:
%p A214569       l:= map(p, l); forget(p);
%p A214569       for i in l do p(i):= p(i)+1 od:
%p A214569       seq(p(i)/i, i=1..max(l[]))
%p A214569     end:
%p A214569 seq(T(n), n=1..10);  # _Alois P. Heinz_, Aug 31 2012
%t A214569 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_ *)
%Y A214569 Cf. A000108, A000081, A003238, A214570, A214571, A061775, A206487, A215703.
%K A214569 nonn,tabf,hard
%O A214569 1,3
%A A214569 _Emeric Deutsch_, Jul 28 2012