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%I #26 Mar 12 2024 07:46:53
%S 1,0,1,1,0,0,1,1,1,1,1,0,0,1,4,1,1,2,2,2,1,1,0,0,1,11,4,4,5,5,6,3,3,3,
%T 3,2,1,1,0,0,1,41,11,11,15,15,19,20,13,10,11,8,8,5,4,4,3,2,1,1,0,0,1,
%U 162,41,41,52,52,63,67,78,41,45,39,39,33,30,20,17,14,10,10,6,5,4,3,2,1,1,0,0,1
%N Number T(n,k) of partitions of [n] whose singletons sum to k; triangle T(n,k), n>=0, 0<=k<=A000217(n), read by rows.
%H Alois P. Heinz, <a href="/A370945/b370945.txt">Rows n = 0..50, flattened</a>
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Partition_of_a_set">Partition of a set</a>
%F Sum_{k=0..A000217(n)} k * T(n,k) = A105479(n+1).
%F T(n,A161680(n)) = A370946(n).
%F T(n,A000217(n)) = 1.
%e T(4,0) = 4: 1234, 12|34, 13|24, 14|23.
%e T(4,1) = 1: 1|234.
%e T(4,2) = 1: 134|2.
%e T(4,3) = 2: 124|3, 1|2|34.
%e T(4,4) = 2: 123|4, 1|24|3.
%e T(4,5) = 2: 1|23|4, 14|2|3.
%e T(4,6) = 1: 13|2|4.
%e T(4,7) = 1: 12|3|4.
%e T(4,10) = 1: 1|2|3|4.
%e Triangle T(n,k) begins:
%e 1;
%e 0, 1;
%e 1, 0, 0, 1;
%e 1, 1, 1, 1, 0, 0, 1;
%e 4, 1, 1, 2, 2, 2, 1, 1, 0, 0, 1;
%e 11, 4, 4, 5, 5, 6, 3, 3, 3, 3, 2, 1, 1, 0, 0, 1;
%e ...
%p h:= proc(n) option remember; `if`(n=0, 1,
%p add(h(n-j)*binomial(n-1, j-1), j=2..n))
%p end:
%p b:= proc(n, i, m) option remember; `if`(n>i*(i+1)/2, 0,
%p `if`(n=0, h(m), b(n, i-1, m)+b(n-i, min(n-i, i-1), m-1)))
%p end:
%p T:= (n, k)-> b(k, min(n, k), n):
%p seq(seq(T(n, k), k=0..n*(n+1)/2), n=0..7);
%t h[n_] := h[n] = If[n == 0, 1,
%t Sum[h[n-j]*Binomial[n-1, j-1], {j, 2, n}]];
%t b[n_, i_, m_] := b[n, i, m] = If[n > i*(i + 1)/2, 0,
%t If[n == 0, h[m], b[n, i - 1, m] + b[n - i, Min[n - i, i - 1], m - 1]]];
%t T[n_, k_] := b[k, Min[n, k], n];
%t Table[Table[T[n, k], { k, 0, n*(n + 1)/2}], {n, 0, 7}] // Flatten (* _Jean-François Alcover_, Mar 12 2024, after _Alois P. Heinz_ *)
%Y Column k=0 gives A000296.
%Y Row sums give A000110.
%Y Row lengths give A000124.
%Y Reversed rows converge to A370946.
%Y T(n,n) gives A370947.
%Y Cf. A000217, A053632, A105479, A369596.
%K nonn,tabf
%O 0,15
%A _Alois P. Heinz_, Mar 06 2024