A274310 Triangle read by rows: T(n,k) = number of parity alternating partitions of [n] into k blocks (1 <= k <= m).

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, 22, 136, 276, 236, 92, 16, 1, 1, 30, 275, 770, 927, 530, 150, 20, 1, 1, 46, 580, 2200, 3551, 2782, 1130, 240, 25, 1, 1, 62, 1141, 5710, 12160, 12632, 6987, 2130, 355, 30, 1

Offset: 1

Author

N. J. A. Sloane, Jun 23 2016

Keywords

• nonn
• tabl
• easy

Comments

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

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

Examples

			Triangle begins:
  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,  22, 136, 276, 236,  92,  16,   1;
  ...
From _Alois P. Heinz_, Jun 28 2016: (Start)
T(5,1) = 1: 12345.
T(5,2) = 6: 1234|5, 123|45, 125|34, 12|345, 145|23, 1|2345.
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.
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.
T(5,5) = 1: 1|2|3|4|5. (End)
		

Links

• Alois P. Heinz, Rows n = 1..141, flattened
• Askar Dzhumadil'daev and Damir Yeliussizov, Walks, partitions, and normal ordering, Electronic Journal of Combinatorics, 22(4) (2015), #P4.10.

Crossrefs

Row sums give A124419(n+1).

Cf. A274547, A274581.

Programs

Maple

A274310 := proc (n, k) local i;
with(combinat):
   add(Stirling2(floor((1/2)*n+1), i+1)*Stirling2(floor((1/2)*n+1/2), k-i), i = 0..k-1);
end proc:
for n from 1 to 10 do
   seq(A274310(n, k), k = 1..n);
end do; # Peter Bala, Apr 09 2018

Mathematica

T[n_, k_] = Sum[StirlingS2[Floor[(n + 2)/2], i + 1] * StirlingS2[Floor[(n + 1)/2], k - i], {i, 0, k - 1}];
Table[T[n, k], {n, 1, 10}, {k, 1, n}] // Flatten (* Jean-François Alcover, May 17 2018, after Peter Bala *)

Formula

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

Extensions

More terms from Alois P. Heinz, Jun 26 2016