A339031 - OEIS

A339031

T(n, k) = k*P(n, k), where P(n, k) is the number of partitions of an n-set with k blocks, the largest of which has the size n - k + 1. Triangle T(n, k) for 0 <= k <= n, read by rows.

1, 0, 1, 0, 1, 2, 0, 1, 6, 3, 0, 1, 8, 18, 4, 0, 1, 10, 30, 40, 5, 0, 1, 12, 45, 80, 75, 6, 0, 1, 14, 63, 140, 175, 126, 7, 0, 1, 16, 84, 224, 350, 336, 196, 8, 0, 1, 18, 108, 336, 630, 756, 588, 288, 9

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OFFSET

0,6

LINKS

Table of n, a(n) for n=0..54.

FORMULA

T(n, k) = k*binomial(n, k - 1) for n >= 1 and 0 < k < n, and T(n, 0) = 0^n, T(n, n) = n.

EXAMPLE

Triangle starts:

0: [1]

1: [0, 1]

2: [0, 1, 2]

3: [0, 1, 6, 3]

4: [0, 1, 8, 18, 4]

5: [0, 1, 10, 30, 40, 5]

6: [0, 1, 12, 45, 80, 75, 6]

7: [0, 1, 14, 63, 140, 175, 126, 7]

8: [0, 1, 16, 84, 224, 350, 336, 196, 8]

9: [0, 1, 18, 108, 336, 630, 756, 588, 288, 9]

T(4, 1) = 1 = 1*card({1234})

T(4, 2) = 8 = 2*card({123|4, 124|3, 134|2, 1|234})

T(4, 3) = 18 = 3*card({12|3|4, 13|2|4, 1|23|4, 14|2|3, 1|24|3, 1|2|34})

T(4, 4) = 4 = 4*card({1|2|3|4})

MAPLE

A339031 := proc(n, k) if k = 0 then 0^n elif k = n then n

else k*binomial(n, k-1) fi end:

seq(seq(A339031(n, k), k=0..n), n=0..9);

PROG

(SageMath) # Shows the combinatorial interpretation.

def A339031Row(n):

if n == 0: return [1]

M = matrix(n + 2)

for k in (1..n):

for p in SetPartitions(n):

if p.max_block_size() == k:

M[len(p), k] += p.cardinality()

return [M[k, n-k+1] for k in (0..n)]

for n in (0..9): print(A339031Row(n))

CROSSREFS

Cf. A339032 (row sums), A339030.

Sequence in context: A376983 A367795 A343825 * A367270 A365770 A059299

Adjacent sequences: A339028 A339029 A339030 * A339032 A339033 A339034

KEYWORD

nonn,tabl

AUTHOR

Peter Luschny, Nov 22 2020

STATUS

approved