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A182700
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Triangle T(n,k) = n*A000041(n-k), 0<=k<=n, read by rows.
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8
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0, 1, 1, 4, 2, 2, 9, 6, 3, 3, 20, 12, 8, 4, 4, 35, 25, 15, 10, 5, 5, 66, 42, 30, 18, 12, 6, 6, 105, 77, 49, 35, 21, 14, 7, 7, 176, 120, 88, 56, 40, 24, 16, 8, 8, 270, 198, 135, 99, 63, 45, 27, 18, 9, 9, 420, 300, 220, 150, 110, 70, 50, 30, 20, 10, 10, 616, 462, 330, 242, 165, 121, 77, 55, 33
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OFFSET
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0,4
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COMMENTS
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T(n,k) is the sum of the parts of all partitions of n that contain k as a part, assuming that all partitions of n have 0 as a part: Thus, column 0 gives the sum of the parts of all partitions of n.
By definition all entries in row n>0 are divisible by n.
Row sums are 0, 2, 8, 21, 48, 95, 180, 315, 536, 873, 1390, 2145,...
The partitions of n+k that contain k as a part can be obtained by adding k to every partition of n assuming that all partitions of n have 0 as a part.
For example, the partitions of 6+k that contain k as a part are
k + 6
k + 3 + 3
k + 4 + 2
k + 2 + 2 + 2
k + 5 + 1
k + 3 + 2 + 1
k + 4 + 1 + 1
k + 2 + 2 + 1 + 1
k + 3 + 1 + 1 + 1
k + 2 + 1 + 1 + 1 + 1
k + 1 + 1 + 1 + 1 + 1 + 1
The partition number A000041(n) is also the number of partitions of m*(n+k) into parts divisible by m and that contain m*k as a part, with k>=0, m>=1, n>=0 and assuming that all partitions of n have 0 as a part.
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LINKS
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FORMULA
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T(n,k) = A182701(n,k), n>=1 and k>=1.
T(n,n) = n = min { T(n,k); 0<=k<=n }.
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EXAMPLE
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For n=7 and k=4 there are 3 partitions of 7 that contain 4 as a part. These partitions are (4+3)=7, (4+2+1)=7 and (4+1+1+1)=7. The sum is 7+7+7 = 7*3 = 21. By other way, the partition number of 7-4 is A000041(3) = p(3)=3, then 7*3 = 21, so T(7,4) = 21.
Triangle begins with row n=0 and columns 0<=k<=n :
0,
1, 1,
4, 2, 2,
9, 6, 3, 3,
20,12,8, 4, 4,
35,25,15,10,5, 5,
66,42,30,18,12,6, 6
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MAPLE
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A182700 := proc(n, k) n*combinat[numbpart](n-k) ; end proc:
seq(seq(A182700(n, k), k=0..n), n=0..15) ;
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MATHEMATICA
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Table[n*PartitionsP[n-k], {n, 0, 11}, {k, 0, n}] // Flatten (* Robert Price, Jun 23 2020 *)
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PROG
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(PARI) A182700(n, k) = n*numbpart(n-k)
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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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