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A210950
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Triangle read by rows: T(n,k) = number of parts in the k-th column of the partitions of n but with the partitions aligned to the right margin.
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3
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1, 1, 2, 1, 2, 3, 1, 2, 4, 5, 1, 2, 4, 6, 7, 1, 2, 4, 7, 10, 11, 1, 2, 4, 7, 11, 14, 15, 1, 2, 4, 7, 12, 17, 21, 22, 1, 2, 4, 7, 12, 18, 25, 29, 30, 1, 2, 4, 7, 12, 19, 28, 36, 41, 42, 1, 2, 4, 7, 12, 19, 29, 40, 50, 55, 56, 1, 2, 4, 7, 12, 19, 30, 43
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OFFSET
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1,3
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COMMENTS
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Index of the first partition of n that has k parts, when the partitions of n are listed in reverse lexicographic order, as in Mathematica's IntegerPartitions[n]. - Clark Kimberling, Oct 16 2023
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LINKS
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FORMULA
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T(n,k) = Sum_{j=1..n} A210951(j,k).
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EXAMPLE
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For n = 6 the partitions of 6 aligned to the right margin look like this:
.
. 6
. 3 + 3
. 4 + 2
. 2 + 2 + 2
. 5 + 1
. 3 + 2 + 1
. 4 + 1 + 1
. 2 + 2 + 1 + 1
. 3 + 1 + 1 + 1
. 2 + 1 + 1 + 1 + 1
. 1 + 1 + 1 + 1 + 1 + 1
.
The number of parts in columns 1-6 are
. 1, 2, 4, 7, 10, 11, the same as the 6th row of triangle.
Triangle begins:
1;
1, 2;
1, 2, 3;
1, 2, 4, 5;
1, 2, 4, 6, 7;
1, 2, 4, 7, 10, 11;
1, 2, 4, 7, 11, 14, 15;
1, 2, 4, 7, 12, 17, 21, 22;
1, 2, 4, 7, 12, 18, 25, 29, 30;
1, 2, 4, 7, 12, 19, 28, 36, 41, 42;
1, 2, 4, 7, 12, 19, 29, 40, 50, 55, 56;
1, 2, 4, 7, 12, 19, 30, 43, 58, 70, 76, 77;
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MATHEMATICA
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m[n_, k_] := Length[IntegerPartitions[n][[k]]]; c[n_] := PartitionsP[n];
t[n_, h_] := Select[Range[c[n]], m[n, #] == h &, 1];
Column[Table[t[n, h], {n, 1, 20}, {h, 1, 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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