A321331 -id:A321331 - cp's OEIS frontend

A339030 T(n, k) = Sum_{p in P(n, k)} card(p), where P(n, k) is the set of set partitions of {1,2,...,n} where the largest block has size k and card(p) is the number of blocks of p. Triangle T(n, k) for 0 <= k <= n, read by rows.

1, 0, 1, 0, 2, 1, 0, 3, 6, 1, 0, 4, 24, 8, 1, 0, 5, 85, 50, 10, 1, 0, 6, 300, 280, 75, 12, 1, 0, 7, 1071, 1540, 525, 105, 14, 1, 0, 8, 3976, 8456, 3570, 840, 140, 16, 1, 0, 9, 15219, 47208, 24381, 6552, 1260, 180, 18, 1

Offset: 0

Peter Luschny, Nov 22 2020

			Triangle starts:
0: [1]
1: [0, 1]
2: [0, 2, 1]
3: [0, 3, 6,     1]
4: [0, 4, 24,    8,     1]
5: [0, 5, 85,    50,    10,    1]
6: [0, 6, 300,   280,   75,    12,   1]
7: [0, 7, 1071,  1540,  525,   105,  14,   1]
8: [0, 8, 3976,  8456,  3570,  840,  140,  16,  1]
9: [0, 9, 15219, 47208, 24381, 6552, 1260, 180, 18, 1]
.
T(4,0) = 0  = 0*card({})
T(4,1) = 4  = 4*card({1|2|3|4}).
T(4,2) = 24 = 3*card({12|3|4, 13|2|4, 1|23|4, 14|2|3, 1|24|3, 1|2|34})
            + 2*card({12|34, 13|24, 14|23}).
T(4,3) = 8  = 2*card({123|4, 124|3, 134|2, 1|234}).
T(4,4) = 1  = 1*card({1234}).
.
Seen as the projection of a 2-dimensional statistic this is, for n = 6:
[  0   0    0     0     0    0   0]
[  0   0    0     0     0    0   6]
[  0   0    0    45   180   75   0]
[  0   0   20   180    80    0   0]
[  0   0   30    45     0    0   0]
[  0   0   12     0     0    0   0]
[  0   1    0     0     0    0   0]
The row sum projection gives row 6 of this triangle, and the column sum projection gives [0, 1, 62, 270, 260, 75, 6], which appears in a decapitated version as row 5 in A321331.

Cf. A005493 with 1 prepended are the row sums.

Cf. A080510, A262071, A321331.

def A339030Row(n):
    if n == 0: return [1]
    M = matrix(n + 1)
    for k in (1..n):
        for p in SetPartitions(n):
            if p.max_block_size() == k:
                M[k, len(p)] += p.cardinality()
    return [sum(M[k, j] for j in (0..n)) for k in (0..n)]
for n in (0..9): print(A339030Row(n))

A367198 T(n, k) = Sum_{m = 0..n-1} Stirling1(m+1, k)binomial(n, m)(-1)^(n + k), where "Stirling1" are the signed Stirling numbers of the first kind.

Original entry on oeis.org

1, 1, 2, 4, 6, 3, 15, 30, 18, 4, 76, 165, 125, 40, 5, 455, 1075, 930, 380, 75, 6, 3186, 8015, 7679, 3675, 945, 126, 7, 25487, 67536, 70042, 37688, 11550, 2044, 196, 8, 229384, 634935, 702372, 414078, 144417, 30870, 3990, 288, 9, 2293839, 6591943, 7696245, 4886390, 1885065, 463092, 73080, 7200, 405, 10

Offset: 1

Thomas Scheuerle, Nov 10 2023

To use the unsigned Stirling numbers rewrite the formula as: T(n, k) = Sum_{m = 0..n-1} abs(Stirling1(m+1, k))*binomial(n, m)*(-1)^(1+m+n). Replacing in this formula Stirling1 (A008275) by Stirling2 (A048993) one obtains a shifted version of A321331.

			Triangle begins:
    1;
    1,    2;
    4,    6,   3;
   15,   30,  18,   4;
   76,  165, 125,  40,  5;
  455, 1075, 930, 380, 75, 6;

Cf. A002411, A002467 (first column), A000027 (main diagonal), A008275.

Cf. A180191(n+1) (row sums), A321331 (variant with Stirling2).

T := (n, k) -> local m; add(Stirling1(m+1, k)*binomial(n, m)*(-1)^(n + k), m = 0..n-1): seq(seq(T(n, k), k = 1..n), n = 1..9);  # Peter Luschny, Nov 10 2023

T(n,k) = sum(m=0, n-1, stirling(m+1, k)*binomial(n, m)*(-1)^(n+k))

T(n+1, n) = n^2*(n+1)/2 = A002411(n).
T(n, n-2) = 6*T(n-1, n-3) - 15*T(n-2, n-4) + 20*T(n-3, n-5) - 15*T(n-4, n-6) + 6*T(n-5, n-7) - T(n-6, n-8), for n > 8.
T(n, n-k) = (-1)^k*Sum_{m=0..n-1} Stirling1(m+1, n-k)*binomial(n, m).

cp's OEIS Frontend

A339030 T(n, k) = Sum_{p in P(n, k)} card(p), where P(n, k) is the set of set partitions of {1,2,...,n} where the largest block has size k and card(p) is the number of blocks of p. Triangle T(n, k) for 0 <= k <= n, read by rows.

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SageMath

A367198 T(n, k) = Sum_{m = 0..n-1} Stirling1(m+1, k)binomial(n, m)(-1)^(n + k), where "Stirling1" are the signed Stirling numbers of the first kind.

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cp's OEIS Frontend

A339030 T(n, k) = Sum_{p in P(n, k)} card(p), where P(n, k) is the set of set partitions of {1,2,...,n} where the largest block has size k and card(p) is the number of blocks of p. Triangle T(n, k) for 0 <= k <= n, read by rows.

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Author

Keywords

Examples

Crossrefs

Programs

SageMath

A367198 T(n, k) = Sum_{m = 0..n-1} Stirling1(m+1, k)*binomial(n, m)*(-1)^(n + k), where "Stirling1" are the signed Stirling numbers of the first kind.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

PARI

Formula

A367198 T(n, k) = Sum_{m = 0..n-1} Stirling1(m+1, k)binomial(n, m)(-1)^(n + k), where "Stirling1" are the signed Stirling numbers of the first kind.