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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.

%I A182114 #27 Feb 17 2022 07:34:07
%S A182114 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 A182114 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 A182114 34,38,42,0,0,0,0,0,0,13,21,39,48,54,56
%N 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.
%C A182114 T(n,k) = A000041(k) for n<k is omitted from the triangle.
%C A182114 Sum_{n>=0} T(n,k) = A115725(k).
%H A182114 Alois P. Heinz, <a href="/A182114/b182114.txt">Rows n = 0..140, flattened</a>
%F A182114 T(n,k) = Sum_{i=1..k} A115723(n,i) for n>0, T(0,0) = 1.
%e A182114 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 A182114 T(9,5) = 3: [1,1,1,2,4], [1,1,1,1,5], [1,1,2,5].
%e A182114 Triangle T(n,k) begins:
%e A182114   1;
%e A182114   0, 1;
%e A182114   0, 0, 2;
%e A182114   0, 0, 1, 3;
%e A182114   0, 0, 0, 2, 5;
%e A182114   0, 0, 0, 1, 5, 7;
%e A182114   0, 0, 0, 0, 5, 7, 11;
%e A182114   0, 0, 0, 0, 3, 7, 13, 15;
%e A182114   0, 0, 0, 0, 1, 5, 16, 18, 22;
%e A182114   0, 0, 0, 0, 0, 3, 17, 21, 27, 30;
%e A182114   0, 0, 0, 0, 0, 1, 16, 22, 34, 38, 42;
%e A182114   ...
%p A182114 b:= proc(n, i, t, k) option remember; `if`(n=0, 1,
%p A182114       `if`(i=1, `if`(t+n<=k, 1, 0), `if`(i<1, 0, b(n, i-1, t, k)+
%p A182114        add(`if`(t+j<=k/i, b(n-i*j, i-1, t+j, k), 0), j=1..n/i))))
%p A182114     end:
%p A182114 T:= (n, k)-> b(n, n, 0, k):
%p A182114 seq(seq(T(n, k), k=0..n), n=0..15);
%t A182114 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 A182114 Diagonal gives: A000041.
%Y A182114 T(n,n-1) = A144300(n) = A000041(n) - A000005(n).
%Y A182114 T(n+d,n) for d=2-10 give: A218623, A218624, A218625, A218626, A218627, A218628, A218629, A218630, A218631.
%Y A182114 Cf. A115723, A115725.
%K A182114 nonn,look,tabl
%O A182114 0,6
%A A182114 _Alois P. Heinz_, Apr 12 2012