%I #34 Oct 24 2018 08:20:53
%S 1,1,1,1,1,2,1,0,3,5,1,0,1,10,15,1,0,0,3,41,52,1,0,0,1,9,196,203,1,0,
%T 0,0,4,40,1057,877,1,0,0,0,1,10,210,6322,4140,1,0,0,0,0,5,30,1176,
%U 41393,21147,1,0,0,0,0,1,15,175,7273,293608,115975,1,0,0,0,0,0,6,35,1176,49932,2237921,678570
%N Square array A(n,k), n>=0, k>=0, read by antidiagonals, where sequence a_k of column k is the exponential transform of C(n,k).
%C A(n,k) is also the number of ways of placing n labeled balls into indistinguishable boxes, where in each filled box k balls are seen at the top. E.g. A(3,1)=10:
%C |1.| |2.| |3.| |1|2| |1|2| |1|3| |1|3| |2|3| |2|3| |1|2|3|
%C |23| |13| |12| |3|.| |.|3| |2|.| |.|2| |1|.| |.|1| |.|.|.|
%C +--+ +--+ +--+ +-+-+ +-+-+ +-+-+ +-+-+ +-+-+ +-+-+ +-+-+-+
%H Alois P. Heinz, <a href="/A145460/b145460.txt">Antidiagonals n = 0..140, flattened</a>
%H N. J. A. Sloane, <a href="/transforms.txt">Transforms</a>
%F A(0,k) = 1 and A(n,k) = Sum_{i=0..n-1} binomial(n-1,i) * binomial(i+1,k) * A(n-1-i,k) for n > 0. - _Seiichi Manyama_, Sep 28 2017
%e Square array A(n,k) begins:
%e 1, 1, 1, 1, 1, 1, ...
%e 1, 1, 0, 0, 0, 0, ...
%e 2, 3, 1, 0, 0, 0, ...
%e 5, 10, 3, 1, 0, 0, ...
%e 15, 41, 9, 4, 1, 0, ...
%e 52, 196, 40, 10, 5, 1, ...
%p exptr:= proc(p) local g; g:=
%p proc(n) option remember; `if`(n=0, 1,
%p add(binomial(n-1, j-1) *p(j) *g(n-j), j=1..n))
%p end: end:
%p A:= (n,k)-> exptr(i-> binomial(i, k))(n):
%p seq(seq(A(n, d-n), n=0..d), d=0..12);
%t Exptr[p_] := Module[{g}, g[n_] := g[n] = If[n == 0, 1, Sum[Binomial[n-1, j-1] *p[j]*g[n-j], {j, 1, n}]]; g]; A[n_, k_] := Exptr[Function[i, Binomial[i, k]]][n]; Table[Table[A[n, d-n], {n, 0, d}], {d, 0, 12}] // Flatten (* _Jean-François Alcover_, Jan 15 2014, translated from Maple *)
%o (Ruby)
%o def ncr(n, r)
%o return 1 if r == 0
%o (n - r + 1..n).inject(:*) / (1..r).inject(:*)
%o end
%o def A(k, n)
%o ary = [1]
%o (1..n).each{|i| ary << (0..i - 1).inject(0){|s, j| s + ncr(i - 1, j) * ncr(j + 1, k) * ary[i - 1 - j]}}
%o ary
%o end
%o def A145460(n)
%o a = []
%o (0..n).each{|i| a << A(i, n - i)}
%o ary = []
%o (0..n).each{|i|
%o (0..i).each{|j|
%o ary << a[i - j][j]
%o }
%o }
%o ary
%o end
%o p A145460(20) # _Seiichi Manyama_, Sep 28 2017
%Y Columns k=0-9 give: A000110, A000248, A133189, A145453, A145454, A145455, A145456, A145457, A145458, A145459.
%Y A(2n,n) gives A029651.
%Y Cf.: A007318, A143398, A292948.
%K nonn,tabl
%O 0,6
%A _Alois P. Heinz_, Oct 10 2008