|
%I #16 Jun 01 2021 15:43:07
%S 1,0,1,0,1,1,0,1,1,1,0,1,2,1,1,0,1,2,2,1,1,0,1,4,3,2,1,1,0,1,4,5,3,2,
%T 1,1,0,1,7,7,6,3,2,1,1,0,1,8,12,8,6,3,2,1,1,0,1,12,18,15,9,6,3,2,1,1,
%U 0,1,14,27,23,16,9,6,3,2,1,1,0,1,21,42,39,26,17,9,6,3,2,1,1
%N Number T(n,k) of n-node unlabeled rooted trees with every leaf at height k; triangle T(n,k), n>=1, 0<=k<=n-1, read by rows.
%H Alois P. Heinz, <a href="/A244925/b244925.txt">Rows n = 1..141, flattened</a>
%e The A048816(5) = 5 rooted trees with 5 nodes with every leaf at the same height sorted by height are:
%e : o : o o : o : o :
%e : /( )\ : / \ | : | : | :
%e : o o o o : o o o : o : o :
%e : : | | /|\ : | : | :
%e : : o o o o o : o : o :
%e : : : / \ : | :
%e : : : o o : o :
%e : : : : | :
%e : : : : o :
%e : : : : :
%e : ---1--- : -----2----- : --3-- : -4- :
%e Thus row 5 = [0, 1, 2, 1, 1].
%e Triangle T(n,k) begins:
%e 1;
%e 0, 1;
%e 0, 1, 1;
%e 0, 1, 1, 1;
%e 0, 1, 2, 1, 1;
%e 0, 1, 2, 2, 1, 1;
%e 0, 1, 4, 3, 2, 1, 1;
%e 0, 1, 4, 5, 3, 2, 1, 1;
%e 0, 1, 7, 7, 6, 3, 2, 1, 1;
%e 0, 1, 8, 12, 8, 6, 3, 2, 1, 1;
%e 0, 1, 12, 18, 15, 9, 6, 3, 2, 1, 1;
%e 0, 1, 14, 27, 23, 16, 9, 6, 3, 2, 1, 1;
%e ...
%p with(numtheory):
%p T:= proc(n, k) option remember; `if`(n=1, 1, `if`(k=0, 0,
%p add(add(`if`(d<k, 0, T(d, k-1)*d), d=divisors(j))*
%p T(n-j, k), j=1..n-1)/(n-1)))
%p end:
%p seq(seq(T(n, k), k=0..n-1), n=1..14);
%t T[n_, k_] := T[n, k] = If[n == 1, 1, If[k == 0, 0, Sum[ Sum[ If[d<k, 0, T[d, k-1]*d], {d, Divisors[j]}] * T[n-j, k], {j, 1, n-1}]/(n-1)]]; Table[Table[T[n, k], {k, 0, n-1}], {n, 1, 14}] // Flatten (* _Jean-François Alcover_, Jan 28 2015, after _Alois P. Heinz_ *)
%Y Columns k=0-10 give: A000007(n-1), A000012 (for n>0), A002865(n-1) (for n>2), A048808, A048809, A048810, A048811, A048812, A048813, A048814, A048815.
%Y T(2n+1,n) gives A074045.
%Y Row sums give A048816.
%K nonn,tabl
%O 1,13
%A _Alois P. Heinz_, Jul 08 2014
|
