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Number T(n,k) of partitions of [n] having k blocks containing their own index when blocks are ordered with decreasing largest elements; triangle T(n,k), n>=0, 0<=k<=ceiling(n/2), read by rows.
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%I #25 Jan 11 2022 18:31:44

%S 1,0,1,1,1,1,3,1,6,7,2,16,25,10,1,73,91,35,4,298,390,163,25,1,1453,

%T 1797,755,128,7,7366,9069,3919,737,55,1,40689,49106,21485,4304,380,11,

%U 238258,284537,126273,26695,2696,110,1,1483306,1751554,785435,173038,19272,976,16

%N Number T(n,k) of partitions of [n] having k blocks containing their own index when blocks are ordered with decreasing largest elements; triangle T(n,k), n>=0, 0<=k<=ceiling(n/2), read by rows.

%H Alois P. Heinz, <a href="/A350647/b350647.txt">Rows n = 0..200, flattened</a>

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Partition_of_a_set">Partition of a set</a>

%F Sum_{k=1..ceiling(n/2)} k * T(n,k) = A350648(n).

%e T(4,0) = 6: 432|1, 42|31, 42|3|1, 4|31|2, 4|3|21, 4|3|2|1.

%e T(4,1) = 7: 4321, 43|21, 43|2|1, 421|3, 4|321, 4|32|1, 41|3|2.

%e T(4,2) = 2: 431|2, 41|32.

%e T(5,2) = 10: 5431|2, 541|32, 531|42, 51|432, 521|4|3, 5|421|3, 5|42|31, 5|42|3|1, 51|4|32, 51|4|3|2.

%e T(5,3) = 1: 51|42|3.

%e Triangle T(n,k) begins:

%e 1;

%e 0, 1;

%e 1, 1;

%e 1, 3, 1;

%e 6, 7, 2;

%e 16, 25, 10, 1;

%e 73, 91, 35, 4;

%e 298, 390, 163, 25, 1;

%e 1453, 1797, 755, 128, 7;

%e 7366, 9069, 3919, 737, 55, 1;

%e 40689, 49106, 21485, 4304, 380, 11;

%e 238258, 284537, 126273, 26695, 2696, 110, 1;

%e ...

%p b:= proc(n, m) option remember; expand(`if`(n=0, 1, add(

%p `if`(j=n, x, 1)*b(n-1, max(m, j)), j=1..m+1)))

%p end:

%p T:= n-> (p-> seq(coeff(p, x, i), i=0..ceil(n/2)))(b(n, 0)):

%p seq(T(n), n=0..14);

%t b[n_, m_] := b[n, m] = Expand[If[n == 0, 1, Sum[

%t If[j == n, x, 1]*b[n-1, Max[m, j]], {j, 1, m+1}]]];

%t T[n_] := With[{p = b[n, 0]},

%t Table[Coefficient[p, x, i], {i, 0, Ceiling[n/2]}]];

%t Table[T[n], {n, 0, 14}] // Flatten (* _Jean-François Alcover_, Jan 11 2022, after _Alois P. Heinz_ *)

%Y Columns k=0-1 give: A350649, A350650.

%Y Row sums give A000110.

%Y T(2n,n) gives A000124(n-1) for n>=1.

%Y T(2n+1,n+1) gives A000012.

%Y Cf. A259691, A350648.

%K nonn,tabf

%O 0,7

%A _Alois P. Heinz_, Jan 09 2022