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A276468
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Irregular triangular array: T(n,i) = number of partitions of n having crossover index k; see Comments.
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3
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1, 2, 2, 1, 4, 1, 4, 2, 1, 7, 3, 1, 7, 6, 1, 1, 12, 8, 1, 1, 12, 12, 4, 1, 1, 19, 16, 5, 1, 1, 19, 25, 8, 2, 1, 1, 30, 34, 9, 2, 1, 1, 30, 44, 17, 6, 2, 1, 1, 45, 59, 20, 7, 2, 1, 1, 45, 81, 31, 12, 3, 2, 1, 1, 67, 108, 36, 13, 3, 2, 1, 1, 67, 132, 64, 18, 9
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OFFSET
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1,2
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COMMENTS
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Suppose that P = [p(1),p(2),...,p(k)] is a partition of n, where p(1) >= p(2) >= ... >= p(k). The crossover index of P is the least h such that p(1)+...+p(h) > = n/2. Equivalently for k > 1, p(1)+...+p(h) >= p(h+1)+...+p(k). The n-th row sum is the number of partitions of n, A000041. The bisections of column 1 are given by A000070. The limit of the reversal of row n is given by A000041.
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LINKS
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EXAMPLE
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First 15 rows (indexed by column 1):
1... 1
2... 2
3... 2 1
4... 4 1
5... 4 2 1
6... 7 3 1
7... 7 6 1 1
8... 12 8 1 1
9... 12 12 4 1 1
10.. 19 16 5 1 1
11... 19 25 8 2 1 1
12.. 30 34 9 2 1 1
13.. 30 44 17 6 2 1 1
14.. 45 59 20 7 2 1 1
15.. 45 81 31 12 3 2 1 1
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MATHEMATICA
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p[n_] := p[n] = IntegerPartitions[n]; t[n_, k_] := t[n, k] = p[n][[k]];
q[n_, k_] := q[n, k] = Select[Range[50], Sum[t[n, k][[i]], {i, 1, #}] >= n/2 &, 1];
u[n_] := u[n] = Flatten[Table[q[n, k], {k, 1, Length[p[n]]}]];
c[n_, k_] := c[n, k] = Count[u[n], k];
v = Table[c[n, k], {n, 1, 25}, {k, 1, Ceiling[n/2]}];
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CROSSREFS
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KEYWORD
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nonn,easy,tabf
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AUTHOR
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STATUS
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approved
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