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A274310
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Triangle read by rows: T(n,k) = number of parity alternating partitions of [n] into k blocks (1 <= k <= m).
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5
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1, 1, 1, 1, 2, 1, 1, 4, 4, 1, 1, 6, 11, 6, 1, 1, 10, 28, 26, 9, 1, 1, 14, 61, 86, 50, 12, 1, 1, 22, 136, 276, 236, 92, 16, 1, 1, 30, 275, 770, 927, 530, 150, 20, 1, 1, 46, 580, 2200, 3551, 2782, 1130, 240, 25, 1, 1, 62, 1141, 5710, 12160, 12632, 6987, 2130, 355, 30, 1
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OFFSET
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1,5
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COMMENTS
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The first element of any block may be odd or even and then the parity of terms alternates within each block. - Alois P. Heinz, Jun 28 2016
Let a(n,k,i) be the number of parity alternating partitions of n into k blocks, i of which have even maximal elements. Dzhumadil'daev and Yeliussizov, Proposition 5.3, give recurrences for a(n,k,i), which depend on the parity of n. It is easy to verify that the solution to these recurrences is given by a(2*n,k,i) = Stirling2(n,i)*Stirling2(n+1,k+1-i) and a(2*n+1,k,i) = Stirling2(n+1,i+1) * Stirling2(n+1,k-i). The formula below for the table entries T(n,k) follows from this observation. - Peter Bala, Apr 09 2018
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LINKS
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FORMULA
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T(n,k) = Sum_{i = 0..k-1} Stirling2(floor((n+2)/2), i+1) * Stirling2(floor((n+1)/2), k-i). - Peter Bala, Apr 09 2018
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EXAMPLE
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Triangle begins:
1;
1, 1;
1, 2, 1;
1, 4, 4, 1;
1, 6, 11, 6, 1;
1, 10, 28, 26, 9, 1;
1, 14, 61, 86, 50, 12, 1;
1, 22, 136, 276, 236, 92, 16, 1;
...
T(5,1) = 1: 12345.
T(5,2) = 6: 1234|5, 123|45, 125|34, 12|345, 145|23, 1|2345.
T(5,3) = 11: 123|4|5, 12|34|5, 125|3|4, 12|3|45, 14|23|5, 1|234|5, 1|23|45, 145|2|3, 14|25|3, 1|25|34, 1|2|345.
T(5,4) = 6: 12|3|4|5, 1|23|4|5, 14|2|3|5, 1|2|34|5, 1|25|3|4, 1|2|3|45.
T(5,5) = 1: 1|2|3|4|5. (End)
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MAPLE
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with(combinat):
add(Stirling2(floor((1/2)*n+1), i+1)*Stirling2(floor((1/2)*n+1/2), k-i), i = 0..k-1);
end proc:
for n from 1 to 10 do
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MATHEMATICA
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T[n_, k_] = Sum[StirlingS2[Floor[(n + 2)/2], i + 1] * StirlingS2[Floor[(n + 1)/2], k - i], {i, 0, k - 1}];
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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