A259478
Partition containment triangle.
Original entry on oeis.org
1, 2, 2, 3, 4, 3, 5, 8, 7, 5, 7, 12, 13, 12, 7, 11, 20, 23, 25, 19, 11, 15, 28, 35, 42, 39, 30, 15, 22, 42, 54, 70, 70, 66, 45, 22, 30, 58, 78, 105, 114, 119, 99, 67, 30, 42, 82, 112, 158, 178, 202, 186, 155, 97, 42, 56, 110, 154, 223, 262, 313, 314, 292, 226, 139, 56, 77, 152, 215, 319, 383, 479, 503, 511, 442, 336, 195, 77
Offset: 1
T(3,2) = 4, the pairs of partitions are ((3)/(2)), ((2,1)/(2)), ((2,1)/(1,1)), ((1,1,1)/(1,1))
and the diagrams are:
x x 0 , x x , x 0 , x
0 x x
0
Triangle begins:
n=1; 1
n=2; 2 2
n=3; 3 4 3
n=4; 5 8 7 5
n=5; 7 12 13 12 7
n=6; 11 20 23 25 19 11
- I. G. MacDonald: "Symmetric functions and Hall polynomials", Oxford University Press, 1979. Page 4.
Cf.
A000070,
A259479,
A259480,
A259481,
A161492,
A227309,
A006958,
A297388,
A303851,
A303852,
A303861,
A303862,
A303863.
-
b:= proc(n, i, t) option remember; expand(`if`(n=0 or i=1,
`if`(t=0, 1, add(x^j, j=0..n)), b(n, i-1, min(i-1, t))+
add(b(n-i, min(i, n-i), min(j, n-i))*x^j, j=0..t)))
end:
T:= n-> (p-> seq(coeff(p, x, i), i=1..n))(b(n$3)):
seq(T(n), n=1..15); # Alois P. Heinz, Jul 05 2015, revised Jan 10 2018
-
majorsweak[left_List,right_List]:=Block[{le1=Length[left],le2=Length[right]},If[le2>le1||Min[Sign[left-PadRight[right,le1]]]<0,False,True]];
Table[Sum[ If[! majorsweak[\[Lambda], \[Mu]], 0, 1] , {\[Lambda], IntegerPartitions[n] }, {\[Mu], IntegerPartitions[m]}], {n, 7}, {m, n}]
(* Second program: *)
b[n_, m_, i_, j_, t_] := b[n, m, i, j, t] = If[m > n, 0, If[n == 0, 1, If[i < 1, 0, If[t && j > 0, b[n, m, i, j - 1, t], 0] + If[i > j, b[n, m, i - 1, j, False], 0] + If[i > n || j > m, 0, b[n - i, m - j, i, j, True]]]]]; T[n_, m_] := b[n, m, n, m, True]; Table[Table[T[n, m], {m, 1, n}], {n, 1, 14}] // Flatten (* Jean-François Alcover, Aug 27 2016, after Alois P. Heinz *)
A259480
T(n,m) counts connected skew Ferrers diagrams of shape lambda/mu with lambda and mu partitions of n and m respectively (0<=m<=n).
Original entry on oeis.org
0, 1, 0, 2, 0, 0, 3, 0, 0, 0, 5, 1, 0, 0, 0, 7, 2, 0, 0, 0, 0, 11, 5, 2, 0, 0, 0, 0, 15, 8, 4, 0, 0, 0, 0, 0, 22, 14, 10, 3, 0, 0, 0, 0, 0, 30, 21, 18, 7, 1, 0, 0, 0, 0, 0, 42, 32, 32, 17, 6, 0, 0, 0, 0, 0, 0, 56, 45, 50, 31, 15, 2, 0, 0, 0, 0, 0, 0, 77, 65, 80, 58, 36, 11, 2, 0, 0, 0, 0, 0, 0
Offset: 0
T(7,2) = 4, the pairs of partitions are ((4,3)/(2)), ((3,3,1)/(2)), ((3,2,2)/(1,1)) and ((2,2,2,1)/(1,1));
The diagrams are:
x x 0 0 , x x 0 , x 0 0 , x 0
0 0 0 0 0 0 x 0 x 0
0 0 0 0 0
0
Triangle begins:
k=0 1 2 3 4 5 6 7
n=0; 0
n=1; 1 0
n=2; 2 0 0
n=3; 3 0 0 0
n=4; 5 1 0 0 0
n=5; 7 2 0 0 0 0
n=6; 11 5 2 0 0 0 0
n=7; 15 8 4 0 0 0 0 0
- I. G. MacDonald: "Symmetric functions and Hall polynomials"; Oxford University Press, 1979. Page 4.
-
(* see A259479 *) factor[\[Lambda],\[Mu]]/;majorsweak[\[Lambda],\[Mu]]:=Block[{a1,a2,a3},a1=Apply[Join,Table[{i,j},{i,Length[\[Lambda]]},{j,\[Lambda][[i]],\[Lambda][[Min[i+1,Length[\[Lambda]]]]],-1}]];
a2=Map[{First[#],First[#]>Length[\[Mu]]||\[Mu][[First[#]]]<#[[2]]}&,a1];a3=Map[First,DeleteCases[SplitBy[a2,MatchQ[#,{,False}]&],{{,False}}],{2}];
Flatten[redu[Part[\[Lambda],#], Part[PadRight[\[Mu],Length[\[Lambda]],0],#]/. 0->Sequence[]]&/@Map[Union,a3],1]];
Table[Sum[Boole[majorsweak[\[Lambda],\[Mu]]&&redu[\[Lambda],\[Mu]]==factor[\[Lambda],\[Mu]]=={\[Lambda],\[Mu]}],{\[Lambda],Partitions[n]},{\[Mu],Partitions[k]}],{n,0,12},{k,0,n}]
A259479
Skew diagrams, both connected or not.
Original entry on oeis.org
1, 1, 0, 2, 0, 0, 3, 1, 0, 0, 5, 3, 0, 0, 0, 7, 5, 2, 0, 0, 0, 11, 9, 6, 1, 0, 0, 0, 15, 13, 12, 6, 0, 0, 0, 0, 22, 20, 22, 14, 3, 0, 0, 0, 0, 30, 28, 36, 27, 13, 2, 0, 0, 0, 0, 42, 40, 56, 48, 31, 11, 1, 0, 0, 0, 0, 56, 54, 82, 77, 59, 33, 9, 0, 0, 0, 0, 0, 77, 75, 120, 121, 106, 72, 30, 6, 0, 0, 0, 0, 0
Offset: 0
T(6,2) = 6, the pairs of partitions are ((4,2)/(2)), ((3,3)/(2)), ((3,2,1)/(2)), ((3,2,1)/(1,1)), ((2,2,2)/(1,1)) and ((2,2,1,1)/(1,1))
and the diagrams are:
x x 0 0 , x x 0 , x x 0 , x 0 0 , x 0 , x 0
0 0 0 0 0 0 0 x 0 x 0 x 0
0 0 0 0 0
0
Triangle begins:
k=0 1 2 3 4 5 6
n=0; 1
n=1; 1 0
n=2; 2 0 0
n=3; 3 1 0 0
n=4; 5 3 0 0 0
n=5; 7 5 2 0 0 0
n=6; 11 9 6 1 0 0 0
- I. G. MacDonald: "Symmetric functions and Hall polynomials", Oxford University Press, 1979. Page 4.
-
majorsweak[left_List, right_List]:=Block[{le1=Length[left], le2=Length[right]}, If[le2>le1||Min[Sign[left-PadRight[right, le1]]]<0, False, True]];
redu1[\[Lambda],\[Mu]]/;majorsweak[\[Lambda],\[Mu]]:=Delete[#,List/@DeleteCases[Table[i Boole[\[Lambda][[i]]==\[Mu][[i]]],{i,Length[\[Mu]]}],0]]&/@{\[Lambda],\[Mu]};
redu[\[Lambda],\[Mu]]/;majorsweak[\[Lambda],\[Mu]]:=TransposePartition/@Apply[redu1,TransposePartition/@redu1[\[Lambda],\[Mu]]];
Table[Sum[Boole[majorsweak[\[Lambda],\[Mu]]&&redu[\[Lambda],\[Mu]]=={\[Lambda],\[Mu]}],{\[Lambda],Partitions[n]},{\[Mu],Partitions[k]}],{n,0,12},{k,0,n}];
A330369
Triangle read by rows: T(n,k) (1 <= k <= n) is the total number of right angles of size k in all partitions of n.
Original entry on oeis.org
1, 0, 2, 0, 0, 3, 1, 0, 1, 4, 2, 0, 0, 2, 5, 3, 2, 0, 2, 3, 6, 4, 4, 0, 0, 4, 4, 7, 5, 6, 3, 0, 3, 6, 5, 8, 7, 8, 7, 0, 1, 6, 8, 6, 9, 9, 10, 11, 4, 0, 6, 9, 10, 7, 10, 13, 12, 15, 10, 0, 2, 11, 12, 12, 8, 11
Offset: 1
Triangle begins:
1;
0, 2;
0, 0, 3;
1, 0, 1, 4;
2, 0, 0, 2, 5;
3, 2, 0, 2, 3, 6;
4, 4, 0, 0, 4, 4, 7;
5, 6, 3, 0, 3, 6, 5, 8;
7, 8, 7, 0, 1, 6, 8, 6, 9;
9, 10, 11, 4, 0, 6, 9, 10, 7, 10;
13, 12, 15, 10, 0, 2, 11, 12, 12, 8, 11;
Figure 1 below shows the Ferrers diagram of the partition of 24: [7, 6, 3, 3, 2, 1, 1, 1]. Figure 2 shows the right-angles diagram of the same partition. Note that in this last diagram we can see the size of the three right angles as follows: the first right angle has size 14 because it contains 14 square cells, the second right angle has size 8 and the third right angle has size 2.
.
. Right-angles Right
Part Ferrers diagram Part diagram angle
_ _ _ _ _ _ _
7 * * * * * * * 7 | _ _ _ _ _ _| 14
6 * * * * * * 6 | | _ _ _ _| 8
3 * * * 3 | | | | 2
3 * * * 3 | | |_|
2 * * 2 | |_|
1 * 1 | |
1 * 1 | |
1 * 1 |_|
.
Figure 1. Figure 2.
.
For n = 8 the partitions of 8 and their respective right-angles diagrams are as follows:
.
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _
1| |8 2| _|8 3| _ _|8 4| _ _ _|8 5| _ _ _ _|8
1| | 1| | 1| | 1| | 1| |
1| | 1| | 1| | 1| | 1| |
1| | 1| | 1| | 1| | 1|_|
1| | 1| | 1| | 1|_|
1| | 1| | 1|_|
1| | 1|_|
1|_|
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
6| _ _ _ _ _|8 7| _ _ _ _ _ _|8 8|_ _ _ _ _ _ _ _|8
1| | 1|_|
1|_|
.
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
2| _|7 3| _ _|7 4| _ _ _|7 5| _ _ _ _|7 6| _ _ _ _ _|7
2| |_|1 2| |_| 1 2| |_| 1 2| |_| 1 2|_|_| 1
1| | 1| | 1| | 1|_|
1| | 1| | 1|_|
1| | 1|_|
1|_|
.
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
2| _|6 3| _ _|6 3| _ _|6 4| _ _ _|6 4| _ _ _|6 5| _ _ _ _|6
2| | |2 2| | | 2 3| |_ _|2 2| | | 2 3| |_ _| 2 3|_|_ _| 2
2| |_| 2| |_| 1| | 2|_|_| 1|_|
1| | 1|_| 1|_|
1|_|
.
_ _ _ _ _ _ _ _ _
2| _|5 3| _ _|5 4| _ _ _|5
2| | |3 3| | _|3 4|_|_ _ _|3
2| | | 2|_|_|
2|_|_|
.
There are 5 right angles of size 1, so T(8,1) = 5.
There are 6 right angles of size 2, so T(8,2) = 6.
There are 3 right angles of size 3, so T(8,3) = 3.
There are no right angle of size 4, so T(8,4) = 0.
There are 3 right angles of size 5, so T(8,5) = 3.
There are 6 right angles of size 6, so T(8,6) = 6.
There are 5 right angles of size 7, so T(8,7) = 5.
There are 8 right angles of size 8, so T(8,8) = 8.
Hence the 8th row of triangle is [5, 6, 3, 0, 3, 6, 5, 8].
Note that the sum of the terms after the last zero is 3 + 6 + 5 + 8 = 22, equaling A000041(8) = 22, the number of partitions of 8.
- G. E. Andrews, Theory of Partitions, Cambridge University Press, 1984, page 143 [Defines the right angles in the Ferrers graph of a partition. - N. J. A. Sloane, Nov 20 2020]
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