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A300185
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Irregular triangle read by rows: T(n, {j,k}) is the number of partitions of n that have exactly j parts equal to k; 1 <= j <= n, 1 <= k <= n.
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1
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1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 2, 1, 1, 0, 1, 2, 1, 0, 0, 0, 2, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 2, 2, 1, 1, 0, 1, 3, 1, 1, 0, 0, 0, 2, 1, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 4, 2, 2, 1, 1, 0, 1, 4, 2, 1, 0, 0, 0, 0, 4, 1, 0, 0, 0, 0, 0, 3, 0
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OFFSET
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1,31
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COMMENTS
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If superfluous zeros are removed from the right side of each row, the row lengths = 1,2,1,3,1,1,4,2,... = A010766.
Sum of each N X N block of rows = 1,2,4,7,12,19,... = A000070.
The sum of the partitions of n that are over-counted in each block of N x N rows = A000070(n) - A000041(n) = A058884(n), n >= 1.
Concatenation of first row from each N X N block = A116598.
As noted by Joerg Arndt in A116598, the first row from each N X N block in reverse converges to A002865. Two sequences emerge from alternating second rows in reverse: for 2n, converges to even-indexed terms in A027336, and for 2n+1, converges to odd-indexed terms in A027336.
Counting the rows in each N X N block where columns j=2 > 0 and j=3 through j=n are all zeros produces A008615(n), n > 0.
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LINKS
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EXAMPLE
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\ j 1 2 3 4 5
k
n
1: 1 1
2: 1 0 1
2 1 0
3: 1 1 0 1
2 1 0 0
3 1 0 0
4: 1 1 1 0 1
2 1 1 0 0
3 1 0 0 0
4 1 0 0 0
5: 1 2 1 1 0 1
2 2 1 0 0 0
3 2 0 0 0 0
4 1 0 0 0 0
5 1 0 0 0 0
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MATHEMATICA
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Array[With[{s = IntegerPartitions[#]}, Table[Count[Map[Count[#, k] &, s], j], {k, #}, {j, #}]] &, 7] // Flatten (* Michael De Vlieger, Feb 28 2018 *)
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PROG
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(Python) # See Stauduhar link.
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CROSSREFS
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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