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A339031
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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.
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2
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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
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0,6
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LINKS
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FORMULA
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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.
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EXAMPLE
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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]
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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})
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MAPLE
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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);
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PROG
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(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))
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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