A264753
Irregular triangle read by rows: T(n,k) = A127671(n,k)/A036040(n,k), n >= 1 and 1 <= k <= A000041(n).
Original entry on oeis.org
1, 1, -1, 1, -1, 2, 1, -1, -1, 2, -6, 1, -1, -1, 2, 2, -6, 24, 1, -1, -1, -1, 2, 2, 2, -6, -6, 24, -120, 1, -1, -1, -1, 2, 2, 2, 2, -6, -6, -6, 24, 24, -120, 720, 1, -1, -1, -1, -1, 2, 2, 2, 2, 2, -6, -6, -6, -6, -6, 24, 24, 24, -120, -120, 720, -5040
Offset: 1
The first few rows of the T(n,k) triangle:
n = 1: 1
n = 2: 1, -1
n = 3: 1, -1, 2
n = 4: 1, -1, -1, 2, -6
n = 5: 1, -1, -1, 2, 2, -6, 24
n = 6: 1, -1, -1, -1, 2, 2, 2, -6, -6, 24, -120
n = 7: 1, -1, -1, -1, 2, 2, 2, 2, -6, -6, -6, 24, 24, -120, 720
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nmax:=8: with(combinat): A008284 := proc(n, k) option remember; if k < 0 or n < 0 then 0 elif k = 0 then if n = 0 then 1 else 0 fi else A008284(n-1, k-1) + A008284(n-k, k) fi end: for n from 1 to nmax do p:=0: k:=1: while k < numbpart(n)+1 do p := p+1: k1 := A008284(n, p): while k1 > 0 do A264753(n, k) := (-1)^(p+1)*(p-1)!: k := k+1: k1 := k1-1: od: od: od: seq(seq(A264753(n, k), k = 1..numbpart(n)), n = 1..nmax);
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nMax = 8; A008284[n_, k_] := A008284[n, k] = If[k<0 || n<0, 0, If[k == 0, If[n == 0, 1, 0], A008284[n-1, k-1] + A008284[n-k, k]]]; For[n = 1, n <= nMax, n++, p = 0; k = 1; While[k < PartitionsP[n]+1, p = p+1; k1 = A008284[n, p]; While[k1>0, A264753[n, k] = (-1)^(p+1)*(p-1)!; k = k+1; k1 = k1-1]]]; Table[Table[A264753[n, k], {k, 1, PartitionsP[n]}], {n, 1, nMax}] // Flatten (* Jean-François Alcover, Oct 01 2016, translated from Maple *)
A134284
A certain partition array in Abramowitz-Stegun order (A-St order), called M_0(3)/M_0.
Original entry on oeis.org
1, 3, 1, 10, 3, 1, 35, 10, 9, 3, 1, 126, 35, 30, 10, 9, 3, 1, 462, 126, 105, 100, 35, 30, 27, 10, 9, 3, 1, 1716, 462, 378, 350, 126, 105, 100, 90, 35, 30, 27, 10, 9, 3, 1, 6435, 1716, 1386, 1260, 1225, 462, 378, 350, 315, 300, 126, 105, 100, 90, 81, 35, 30, 27, 10, 9, 3, 1
Offset: 1
[1]; [3,1]; [10,3,1]; [35,10,9,3,1]; [126,35,30,10,9,3,1]; ...
a(4,3) = 9 = 3^2 because (2^2) is the k=4 partition of n=4 in A-St order and s2(3,2,1)=3.
A262030
Partition array in Abramowitz-Stegun order: Schur functions evaluated at 1.
Original entry on oeis.org
1, 3, 1, 10, 8, 1, 35, 45, 20, 15, 1, 126, 224, 175, 126, 75, 24, 1, 462, 1050, 1134, 490, 840, 896, 175, 280, 189, 35, 1, 1716, 4752, 6468, 4704, 4950, 7350, 3528, 2646, 2400, 2940, 784, 540, 392, 48, 1, 6435, 21021, 34320, 33264, 13860, 27027, 50688, 41580, 25872, 15876, 17325, 29700, 15120, 14700, 1764, 5775, 7680, 2352, 945, 720, 63, 1
Offset: 1
The irregular triangle begins (commas separate entries for partitions of like numbers of parts in A-St order):
n\k 1 2 3 4 5 6 7 8 9 10 11
1: 1
2: 3, 1
3: 10, 8, 1
4: 35, 45 20, 15, 1
5: 126, 224 175, 126 75, 24, 1
6: 462, 1050 1134 490, 840 896 175, 280 189,35, 1
...
Row 7: 1716, 4752 6468 4704, 4950 7350 3528 2646, 2400 2940 784, 540 392, 48, 1;
Row 8: 6435, 21021 34320 33264 13860, 27027 50688 41580 25872 15876, 17325 29700 15120 14700 1764, 5775 7680 2352, 945 720, 63, 1.
...
n = 4, k = 4: lambda(4, 4) = (2,1,1,0) (m=3), SSYT (we use semicolons to separate the three rows): [1,1;2;3], [1,1;2;4], [1,1;3;4],
[1,2;2;3], [1,2;2;4], [1,2;3;4],
[1,3;2;3], [1,3;2;4], [1,3;3;4],
[1,4;2;3], [1,4;2;4], [1,4;3;4],
[2,2;3;4], [2,3;3;4], [2,4;3;4], hence a(4, 4) = 15. The three tableaux with distinct numbers are standard Young tableaux and give A117506(4, 4) = 3.
- Macdonald, I. G., Symmetric functions and Hall polynomials, Oxford University Press, 1979.
- Stanley, R. P., Enumerative Combinatorics, Vol. 2, Cambridge University Press 1999, sect. 7.30, pp. 308-316.
A226651
Multidimensional Young numbers: Given a d-dimensional partition of n, this is the number of ways to fill the associated d-dimensional Young diagram with the integers 1 to n such that the entries are increasing in each positive (orthogonal) direction.
Original entry on oeis.org
1, 1, 1, 2, 1, 3, 2, 6, 1, 4, 5, 6, 12, 8, 24, 1, 5, 9, 10, 5, 16, 20, 25, 30, 20, 16, 60, 40, 120, 1, 6, 14, 15, 14, 25, 20, 21, 30, 54, 60, 30, 96, 40, 66, 61, 75, 90, 48, 120, 150, 180, 120, 96, 80, 360, 240, 720, 1, 7, 20, 21, 28, 64, 35, 14, 70, 56, 90, 42, 42, 98, 105, 98, 245, 140, 147
Offset: 1
The ordering of "conjugacy classes" of partitions begins:
(1), (2), (3), (2+1), (4), (3+1), (2+2), ((2+1)+(1)), (5), (4+1), (3+2), (3+1+1), ((3+1)+(1)), ((2+2)+(1)),
(((2+1)+(1))+((1))), ...
The 14th partition, ((2+2)+(1)), is associated to the Young diagram with cubes centered at p_1=(0,0,0), p_2=(1,0,0), p_3=(0,1,0), p_4=(1,1,0), and p_5=(0,0,1). The possible ways to fill the cubes centered on these points so that the numbers are increasing in all directions are;
(For each i=1:5, the i-th integer in a sequence below is placed on p_i.)
1-2-3-4-5
1-3-2-4-5
1-2-3-5-4
1-3-2-5-4
1-2-4-5-3
1-4-2-5-3
1-3-4-5-2
1-4-3-5-2
Hence the 14th term is 8.
The 48th partition, ((2+2)+(2+2)), can be represented as a cube divided into octants. The integers 1 and 8 must lie in opposite octants. Of the three octants adjacent to the one which contains 1, one must contain 2 and one must contain 3. This gives 6 possibilities. For each of these possibilities there are 4 numbers (4, 5, 6, and 7) to choose from for the number placed in the remaining cube in the plane that contains 1, 2, and 3. Regardless of this choice, there are 2 ways to fill in the remaining three octants. Thus there are 6*4*2=48 ways to fill the octants all together--that is, the 48th multidimensional Young number is 48.
Example of recursion:
The partition: p|--6=((3+2)+(1)) covers the following partitions of 5:
q_1|--5=(3+2)
q_2|--5=((3+1)+(1))
q_3|--5=((2+2)+(1)) Thus Y(p)=Y(q_1)+Y(q_2)+Y(q_3)=5+12+8=25
Original entry on oeis.org
0, 0, 0, 2, 10, 44, 168, 636, 2364, 8984, 34672, 138104, 564408, 2382288, 10333152, 46173968, 211733776, 997182752, 4809439296, 23758139808, 119951644320, 618882541760, 3257839688320, 17492182188992, 95680426983360
Offset: 1
Row five of A117506 is 1 5 9 5 10 16 5 10 9 5 1.
Row five of A007318 is 1 5 10 10 5 1.
So included values are 9 5 16 5 9;
therefore a(5) = 44 = 76 - 32.
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Table[Sum[(2k)!/k!/2^k Binomial[n, 2k], {k, 0, n/2}] - 2^(n - 1) // FullSimplify, {n, 26}] (* Robert G. Wilson v, Sep 27 2006 *) (* or *)
Table[HypergeometricU[ -(n/2), 1/2, -(1/2)]/(-(1/2))^(-(-n/2)) - 2^(n - 1), {n, 26}] (* Robert G. Wilson v, Sep 27 2006 *) (* or *)
(* first do *) Needs["DiscreteMath`Combinatorica`"] (* then *) Table[NumberOfTableaux[M[Star[n+1]]] - 2^(n - 1), {n, 26}] (* Robert G. Wilson v, Sep 27 2006 *)
A167203
Number of Young tableaux with n cells and k inversions.
Original entry on oeis.org
1, 2, 4, 8, 2, 16, 10, 32, 34, 5, 5, 64, 98, 35, 35, 128, 258, 154, 154, 14, 42, 14
Offset: 1
The irregular triangle begins
..1
..2
..4
..8...2
.16..10
.32..34...5...5
.64..98..35..35
128.258.154.154..14..42..14
etc.
A340398
Number of spanning trees in the Bruhat graph of the symmetric group.
Original entry on oeis.org
1, 1, 81, 22799473113563136
Offset: 1
For n=3 the number of spanning trees is 81 since the graph is the complete bipartite graph K_{3,3}.
For n=4, the following calculation demonstrates the formula:
31: 6 - (1-1) - (2-1) - (3-1) - (1-2) = 4;
22: 6 - (1-1) - (2-1) - (1-2) - (2-2) = 6;
211: 6 - (1-1) - (2-1) - (1-2) - (1-3) = 8;
1111: 6 - (1-1) - (1-2) - (1-3) - (1-4) = 12.
Hence the number of spanning trees is given by (1/4!) * 4^9 * 6^4 * 8^9 * 12^1 = 2^48 * 3^4 = 22799473113563136.
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