%I #20 Mar 27 2022 22:24:23
%S 1,1,1,1,1,3,1,7,1,1,15,6,1,31,25,1,1,63,90,10,1,127,301,65,1,1,255,
%T 966,350,15,1,511,3025,1701,140,1,1,1023,9330,7770,1050,21,1,2047,
%U 28501,34105,6951,266,1,1,4095,86526,145750,42525,2646,28
%N Irregular triangle read by rows, Stirling numbers of the second kind: columns shifted to allow (1, 1, 2, 2, 3, 3, ...) terms per row.
%C Row sums = A024427: (1, 1, 2, 4, 9, 22, 58, 164, 495, 1587, ...).
%C T(n,k) is the number of ways to partition {1,2,...,n+1} into exactly k blocks such that each block contains at least 2 elements and the smallest 2 elements in each block are consecutive integers. - _Geoffrey Critzer_, Dec 02 2013
%H Alois P. Heinz, <a href="/A136011/b136011.txt">Rows n = 1..200, flattened</a>
%F Given A008277, the Stirling number of the second kind triangle, left column = (1, 1, 1, ...); all other columns start at 3rd term of previous column.
%F O.g.f. for column k: Product_{i=1..k} x^2/(1 - i*x). - _Geoffrey Critzer_, Dec 02 2013
%F T(n,k) = Stirling2(n+1-k,k). - _Alois P. Heinz_, Dec 04 2013
%e First few rows of the triangle:
%e 1;
%e 1;
%e 1, 1;
%e 1, 3;
%e 1, 7, 1;
%e 1, 15, 6;
%e 1, 31, 25, 1;
%e 1, 63, 90, 10;
%e 1, 127, 301, 65, 1;
%e 1, 255, 966, 350, 15;
%e ...
%e T(5,3) = 1 because we have {1,2},{3,4},{5,6}.
%e T(6,3) = 6 because we have {1,2,7},{3,4},{5,6}; {1,2},{3,4,7},{5,6}; {1,2},{3,4},{5,6,7}; {1,2},{3,4,5},{6,7}; {1,2,3},{4,5},{6,7}; {1,2,5},{3,4},{6,7}. - _Geoffrey Critzer_, Dec 02 2013
%p T:= (n, k)-> Stirling2(n+1-k, k):
%p seq(seq(T(n, k), k=1..(n+1)/2), n=1..20); # _Alois P. Heinz_, Dec 04 2013
%t nn=15;Range[0,nn]!;Map[Select[#,#>0&]&,Drop[Transpose[Table[CoefficientList[Series[Product[x^2/(1-i x),{i,1,k}],{x,0,nn}],x],{k,1,nn/2}]],2]]//Grid (* _Geoffrey Critzer_, Dec 02 2013 *)
%Y Cf. A008277, A024427.
%K nonn,tabf
%O 1,6
%A _Gary W. Adamson_, Dec 09 2007