

A132312


Triangle read by rows: T(n,k) = number of partitions of binomial(n,k) into distinct parts of the first n rows of Pascal's triangle, 0<=k<=n.


4



0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 3, 2, 1, 1, 4, 7, 6, 7, 4, 1, 1, 4, 11, 14, 14, 11, 4, 1, 1, 5, 28, 57, 56, 57, 28, 5, 1, 1, 7, 73, 273, 434, 434, 273, 73, 7, 1, 1, 10, 189, 1411, 3479, 3980, 3479, 1411, 189, 10, 1, 1, 11, 300, 4138, 16293, 26555, 26555, 16293, 4138, 300, 11, 1
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OFFSET

0,17


COMMENTS

T(n,k) = T(n,nk);
T(n,0) = 1 for n>0;
A000009(n)  1 <= T(n,1) <= A000009(n) for n>1;


LINKS

Alois P. Heinz, Rows n = 0..28, flattened
Index entries for triangles and arrays related to Pascal's triangle


EXAMPLE

T(9,1) = A000009(9)1 = 7;
A007318(5,2) = A007318(10,1) = 10:
T(5,2) = #{6+4, 6+3+1, 4+3+2+1} = 3,
but T(10,1) = A000009(10) = 10.


MATHEMATICA

T[n_] := T[n] = Table[Binomial[m, k], {m, 0, n1}, {k, 0, m}] // Flatten // Union;
T[n_, k_] /; k <= n/2 := T[n, k] = Select[ IntegerPartitions[ Binomial[n, k], Length[T[n]], T[n]], Length[#] == Length[Union[#]]&] // Length;
T[n_, k_] /; k > n/2 := T[n, k] = T[n, nk];
Table[Print["T[", n, ", ", k, "] = ", T[n, k]]; T[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* JeanFrançois Alcover, Oct 02 2020 *)


CROSSREFS

Cf. A132311, A007318, A126257, A014631.
Sequence in context: A270921 A038529 A176259 * A090431 A107336 A282013
Adjacent sequences: A132309 A132310 A132311 * A132313 A132314 A132315


KEYWORD

nonn,tabl


AUTHOR

Reinhard Zumkeller, Aug 18 2007


STATUS

approved



