%I #33 May 17 2018 08:30:20
%S 1,1,1,1,2,1,1,4,4,1,1,6,11,6,1,1,10,28,26,9,1,1,14,61,86,50,12,1,1,
%T 22,136,276,236,92,16,1,1,30,275,770,927,530,150,20,1,1,46,580,2200,
%U 3551,2782,1130,240,25,1,1,62,1141,5710,12160,12632,6987,2130,355,30,1
%N Triangle read by rows: T(n,k) = number of parity alternating partitions of [n] into k blocks (1 <= k <= m).
%C The first element of any block may be odd or even and then the parity of terms alternates within each block. - _Alois P. Heinz_, Jun 28 2016
%C Let a(n,k,i) be the number of parity alternating partitions of n into k blocks, i of which have even maximal elements. Dzhumadil'daev and Yeliussizov, Proposition 5.3, give recurrences for a(n,k,i), which depend on the parity of n. It is easy to verify that the solution to these recurrences is given by a(2*n,k,i) = Stirling2(n,i)*Stirling2(n+1,k+1-i) and a(2*n+1,k,i) = Stirling2(n+1,i+1) * Stirling2(n+1,k-i). The formula below for the table entries T(n,k) follows from this observation. - _Peter Bala_, Apr 09 2018
%H Alois P. Heinz, <a href="/A274310/b274310.txt">Rows n = 1..141, flattened</a>
%H Askar Dzhumadil'daev and Damir Yeliussizov, <a href="http://www.combinatorics.org/ojs/index.php/eljc/article/view/v22i4p10">Walks, partitions, and normal ordering</a>, Electronic Journal of Combinatorics, 22(4) (2015), #P4.10.
%F T(n,k) = Sum_{i = 0..k-1} Stirling2(floor((n+2)/2), i+1) * Stirling2(floor((n+1)/2), k-i). - _Peter Bala_, Apr 09 2018
%e Triangle begins:
%e 1;
%e 1, 1;
%e 1, 2, 1;
%e 1, 4, 4, 1;
%e 1, 6, 11, 6, 1;
%e 1, 10, 28, 26, 9, 1;
%e 1, 14, 61, 86, 50, 12, 1;
%e 1, 22, 136, 276, 236, 92, 16, 1;
%e ...
%e From _Alois P. Heinz_, Jun 28 2016: (Start)
%e T(5,1) = 1: 12345.
%e T(5,2) = 6: 1234|5, 123|45, 125|34, 12|345, 145|23, 1|2345.
%e T(5,3) = 11: 123|4|5, 12|34|5, 125|3|4, 12|3|45, 14|23|5, 1|234|5, 1|23|45, 145|2|3, 14|25|3, 1|25|34, 1|2|345.
%e T(5,4) = 6: 12|3|4|5, 1|23|4|5, 14|2|3|5, 1|2|34|5, 1|25|3|4, 1|2|3|45.
%e T(5,5) = 1: 1|2|3|4|5. (End)
%p A274310 := proc (n, k) local i;
%p with(combinat):
%p add(Stirling2(floor((1/2)*n+1), i+1)*Stirling2(floor((1/2)*n+1/2), k-i), i = 0..k-1);
%p end proc:
%p for n from 1 to 10 do
%p seq(A274310(n, k), k = 1..n);
%p end do; # _Peter Bala_, Apr 09 2018
%t T[n_, k_] = Sum[StirlingS2[Floor[(n + 2)/2], i + 1] * StirlingS2[Floor[(n + 1)/2], k - i], {i, 0, k - 1}];
%t Table[T[n, k], {n, 1, 10}, {k, 1, n}] // Flatten (* _Jean-François Alcover_, May 17 2018, after _Peter Bala_ *)
%Y Row sums give A124419(n+1).
%Y Cf. A274547, A274581.
%K nonn,tabl,easy
%O 1,5
%A _N. J. A. Sloane_, Jun 23 2016
%E More terms from _Alois P. Heinz_, Jun 26 2016