A182114 Number of partitions of n with largest inscribed rectangle having area <= k; triangle T(n,k), 0<=n, 0<=k<=n, read by rows.

1, 0, 1, 0, 0, 2, 0, 0, 1, 3, 0, 0, 0, 2, 5, 0, 0, 0, 1, 5, 7, 0, 0, 0, 0, 5, 7, 11, 0, 0, 0, 0, 3, 7, 13, 15, 0, 0, 0, 0, 1, 5, 16, 18, 22, 0, 0, 0, 0, 0, 3, 17, 21, 27, 30, 0, 0, 0, 0, 0, 1, 16, 22, 34, 38, 42, 0, 0, 0, 0, 0, 0, 13, 21, 39, 48, 54, 56

Offset: 0

Author

Alois P. Heinz, Apr 12 2012

Keywords

• nonn
• look
• tabl

Comments

T(n,k) = A000041(k) for n

Sum_{n>=0} T(n,k) = A115725(k).

Examples

			T(5,4) = 5 because there are 5 partitions of 5 with largest inscribed rectangle having area <= 4: [1,1,1,2], [1,2,2], [1,1,3], [2,3], [1,4].
T(9,5) = 3: [1,1,1,2,4], [1,1,1,1,5], [1,1,2,5].
Triangle T(n,k) begins:
  1;
  0, 1;
  0, 0, 2;
  0, 0, 1, 3;
  0, 0, 0, 2, 5;
  0, 0, 0, 1, 5, 7;
  0, 0, 0, 0, 5, 7, 11;
  0, 0, 0, 0, 3, 7, 13, 15;
  0, 0, 0, 0, 1, 5, 16, 18, 22;
  0, 0, 0, 0, 0, 3, 17, 21, 27, 30;
  0, 0, 0, 0, 0, 1, 16, 22, 34, 38, 42;
  ...
		

Links

• Alois P. Heinz, Rows n = 0..140, flattened

Crossrefs

Diagonal gives: A000041.

T(n,n-1) = A144300(n) = A000041(n) - A000005(n).

T(n+d,n) for d=2-10 give: A218623, A218624, A218625, A218626, A218627, A218628, A218629, A218630, A218631.

Cf. A115723, A115725.

Programs

Maple

b:= proc(n, i, t, k) option remember; `if`(n=0, 1,
      `if`(i=1, `if`(t+n<=k, 1, 0), `if`(i<1, 0, b(n, i-1, t, k)+
       add(`if`(t+j<=k/i, b(n-i*j, i-1, t+j, k), 0), j=1..n/i))))
    end:
T:= (n, k)-> b(n, n, 0, k):
seq(seq(T(n, k), k=0..n), n=0..15);

Mathematica

b[n_, i_, t_, k_] := b[n, i, t, k] = If[n == 0, 1, If[i == 1, If[t + n <= k, 1, 0], If[i < 1, 0, b[n, i - 1, t, k] + Sum[If[t + j <= k/i, b[n - i*j, i - 1, t + j, k], 0], {j, 1, n/i}]]]] ; T[n_, k_] := b[n, n, 0, k]; Table[Table[T[n, k], {k, 0, n}], {n, 0, 15}] // Flatten (* Jean-François Alcover, Dec 27 2013, translated from Maple *)

Formula

T(n,k) = Sum_{i=1..k} A115723(n,i) for n>0, T(0,0) = 1.

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