%I #27 Feb 17 2022 07:34:07

%S 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,

%T 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,

%U 34,38,42,0,0,0,0,0,0,13,21,39,48,54,56

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

%C T(n,k) = A000041(k) for n<k is omitted from the triangle.

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

%H Alois P. Heinz, <a href="/A182114/b182114.txt">Rows n = 0..140, flattened</a>

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

%e 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].

%e T(9,5) = 3: [1,1,1,2,4], [1,1,1,1,5], [1,1,2,5].

%e Triangle T(n,k) begins:

%e 1;

%e 0, 1;

%e 0, 0, 2;

%e 0, 0, 1, 3;

%e 0, 0, 0, 2, 5;

%e 0, 0, 0, 1, 5, 7;

%e 0, 0, 0, 0, 5, 7, 11;

%e 0, 0, 0, 0, 3, 7, 13, 15;

%e 0, 0, 0, 0, 1, 5, 16, 18, 22;

%e 0, 0, 0, 0, 0, 3, 17, 21, 27, 30;

%e 0, 0, 0, 0, 0, 1, 16, 22, 34, 38, 42;

%e ...

%p b:= proc(n, i, t, k) option remember; `if`(n=0, 1,

%p `if`(i=1, `if`(t+n<=k, 1, 0), `if`(i<1, 0, b(n, i-1, t, k)+

%p add(`if`(t+j<=k/i, b(n-i*j, i-1, t+j, k), 0), j=1..n/i))))

%p end:

%p T:= (n, k)-> b(n, n, 0, k):

%p seq(seq(T(n, k), k=0..n), n=0..15);

%t 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 *)

%Y Diagonal gives: A000041.

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

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

%Y Cf. A115723, A115725.

%K nonn,look,tabl

%O 0,6

%A _Alois P. Heinz_, Apr 12 2012