%I #33 Sep 06 2018 12:49:06
%S 1,1,1,2,1,1,3,2,0,1,5,3,2,0,1,7,6,2,1,0,1,11,7,3,2,1,0,1,15,13,5,3,1,
%T 1,0,1,22,16,9,3,3,1,1,0,1,30,25,10,6,3,2,1,1,0,1,42,33,16,8,5,3,2,1,
%U 1,0,1,56,49,23,13,6,5,2,2,1,1,0,1,77,61,31,15,10,5,5,2,2,1,1,0,1
%N Triangle T(n,k), n>=0, 0<=k<=n, read by rows: T(n,k) is the number of partitions of n containing a clique of size k.
%C All parts of a number partition with the same value form a clique. The size of a clique is the number of elements in the clique. Each partition has a clique of size 0.
%H Alois P. Heinz, <a href="/A183568/b183568.txt">Rows n = 0..140, flattened</a>
%F G.f. of column k: (1-Product_{j>0} (1-x^(k*j)+x^((k+1)*j))) / (Product_{j>0} (1-x^j)).
%e T(5,2) = 2, because 2 (of 7) partitions of 5 contain (at least) one clique of size 2: [1,2,2], [1,1,3].
%e Triangle T(n,k) begins:
%e 1;
%e 1, 1;
%e 2, 1, 1;
%e 3, 2, 0, 1;
%e 5, 3, 2, 0, 1;
%e 7, 6, 2, 1, 0, 1;
%e 11, 7, 3, 2, 1, 0, 1;
%e 15, 13, 5, 3, 1, 1, 0, 1;
%p b:= proc(n, i, k) option remember; `if`(n=0, [1, 0], `if`(i<1, [0, 0],
%p add((l->`if`(j=k, [l[1]$2], l))(b(n-i*j, i-1, k)), j=0..n/i)))
%p end:
%p T:= (n, k)-> (l-> l[`if`(k=0, 1, 2)])(b(n, n, k)):
%p seq(seq(T(n, k), k=0..n), n=0..12);
%t b[n_, i_, k_] := b[n, i, k] = If[n == 0, {1, 0}, If[i < 1, {0, 0}, Sum[Function[l, If[j == k, {l[[1]], l[[1]]}, l]][b[n - i*j, i-1, k]], {j, 0, n/i}]] ]; t[n_, k_] := Function[l, l[[If[k == 0, 1, 2]]]][b[n, n, k]]; Table[Table[t[n, k], {k, 0, n}], {n, 0, 12}] // Flatten (* _Jean-François Alcover_, Dec 16 2013, translated from Maple *)
%Y Columns k=0-10 give: A000041, A183558, A183559, A183560, A183561, A183562, A183563, A183564, A183565, A183566, A183567.
%Y Differences between columns 0 and k (0<k<=10) give: A007690, A116645, A118807, A184639, A184640, A184641, A184642, A184643, A184644, A184645.
%Y T(2*k+1,k+1) gives A002865.
%K nonn,tabl
%O 0,4
%A _Alois P. Heinz_, Jan 05 2011