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Triangular array read by rows: T(n,k) is the number of unlabeled rooted trees with n nodes in which the largest subtree (of the root) has exactly k nodes; n>=2, 1<=k<=n-1.
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%I #35 Mar 28 2013 11:10:13

%S 1,1,1,1,1,2,1,2,2,4,1,2,4,4,9,1,3,7,8,9,20,1,3,9,16,18,20,48,1,4,12,

%T 30,36,40,48,115,1,4,18,38,81,80,96,115,286,1,5,21,64,144,180,192,230,

%U 286,719,1,5,27,92,216,400,432,460,572,719,1842

%N Triangular array read by rows: T(n,k) is the number of unlabeled rooted trees with n nodes in which the largest subtree (of the root) has exactly k nodes; n>=2, 1<=k<=n-1.

%C Row sums are A000081.

%C Diagonal terms (i.e., T(n,n-1)) are A000081(n-1).

%H Alois P. Heinz, <a href="/A218666/b218666.txt">Rows n = 2..142, flattened</a>

%F O.g.f. for column k: A_k(x) - A_{k-1}(x) with

%F A_k(x) = x*Product_{j=1..k} 1/(1 - x^j)^A000081(j).

%e ....o..........o..........o..........o....

%e ....|..........|........./ \......../|\...

%e ....o..........o........o o......o o o..

%e ....|........./ \.......| .............

%e ....o........o o......o .............

%e ....|.....................................

%e ....o.....................................

%e T(4,3) = 2 because the 2 leftmost graphs have 3 nodes in the largest subtree. T(4,2) = T(4,1) = 1 as seen in the two rightmost graphs.

%e 1,

%e 1, 1;

%e 1, 1, 2;

%e 1, 2, 2, 4;

%e 1, 2, 4, 4, 9;

%e 1, 3, 7, 8, 9, 20;

%e 1, 3, 9, 16, 18, 20, 48;

%e 1, 4, 12, 30, 36, 40, 48, 115;

%e 1, 4, 18, 38, 81, 80, 96, 115, 286;

%p g:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0, add(

%p binomial(g((i-1)$2)+j-1, j)*g(n-i*j, i-1), j=0..n/i))) end:

%p h:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0, add(

%p binomial(g((i-1)$2)+j-1, j)*g(n-i*j, i-1), j=1..n/i))) end:

%p T:= (n, k)-> h(n-1, k):

%p seq(seq(T(n, k), k=1..n-1), n=2..16); # _Alois P. Heinz_, Mar 27 2013

%t nn=10;f[list_]:=Select[list,#>0&];t[x_]:=Sum[a[n]x^n,{n,0,nn}];b=Level[Table[sol=SolveAlways[0==Series[t[x]-x Product[1/(1-x^i)^a[i],{i,1,n}],{x,0,nn}],x];Table[a[n],{n,0,nn}]/.sol,{n,0,nn}],{2}];Map[f,Drop[Transpose[Table[b[[k+1]]-b[[k]],{k,1,nn}]],2]]//Grid

%K nonn,tabl

%O 2,6

%A _Geoffrey Critzer_, Mar 27 2013