login
Number A(n,k) of rooted trees with n nodes such that no more than k subtrees extending from the same node have the same number of nodes; square array A(n,k), n>=0, k>=0, read by antidiagonals.
13

%I #22 May 27 2019 09:06:08

%S 0,0,1,0,1,0,0,1,1,0,0,1,1,1,0,0,1,1,2,2,0,0,1,1,2,3,3,0,0,1,1,2,4,7,

%T 6,0,0,1,1,2,4,8,15,12,0,0,1,1,2,4,9,18,34,25,0,0,1,1,2,4,9,19,43,79,

%U 51,0,0,1,1,2,4,9,20,46,102,190,111,0,0,1,1,2,4,9,20,47,110,250,457,240,0

%N Number A(n,k) of rooted trees with n nodes such that no more than k subtrees extending from the same node have the same number of nodes; square array A(n,k), n>=0, k>=0, read by antidiagonals.

%H Alois P. Heinz, <a href="/A318753/b318753.txt">Antidiagonals n = 0..200, flattened</a>

%F A(n,k) = Sum_{j=0..k} A318754(n,j) for n > 0.

%F A(n,n+j) = A000081(n) for j >= -1.

%e Square array A(n,k) begins:

%e 0, 0, 0, 0, 0, 0, 0, 0, 0, ...

%e 1, 1, 1, 1, 1, 1, 1, 1, 1, ...

%e 0, 1, 1, 1, 1, 1, 1, 1, 1, ...

%e 0, 1, 2, 2, 2, 2, 2, 2, 2, ...

%e 0, 2, 3, 4, 4, 4, 4, 4, 4, ...

%e 0, 3, 7, 8, 9, 9, 9, 9, 9, ...

%e 0, 6, 15, 18, 19, 20, 20, 20, 20, ...

%e 0, 12, 34, 43, 46, 47, 48, 48, 48, ...

%e 0, 25, 79, 102, 110, 113, 114, 115, 115, ...

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

%p binomial(A(i, k)+j-1, j)*g(n-i*j, i-1, k), j=0..min(k, n/i))))

%p end:

%p A:= (n, k)-> g(n-1$2, k):

%p seq(seq(A(n, d-n), n=0..d), d=0..14);

%t g[n_, i_, k_] := g[n, i, k] = If[n == 0, 1, If[i < 1, 0, Sum[Binomial[A[i, k] + j - 1, j]*g[n - i*j, i - 1, k], {j, 0, Min[k, n/i]}]]];

%t A[n_, k_] := g[n - 1, n - 1, k];

%t Table[A[n, d - n], {d, 0, 14}, {n, 0, d}] // Flatten (* _Jean-François Alcover_, May 27 2019, after _Alois P. Heinz_ *)

%Y Columns k=0-10 give: A063524, A032305, A213920, A318797, A318798, A318799, A318800, A318801, A318802, A318803, A318804.

%Y Rows n=0-2 give: A000004, A000012, A057427.

%Y Main diagonal gives A000081.

%Y Cf. A318754.

%K nonn,tabl

%O 0,19

%A _Alois P. Heinz_, Sep 02 2018