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A363041
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Triangle read by rows: T(n,k) = Stirling2(n+1,k)/binomial(k+1,2) if n-k is even, else 0 (1 <= k <= n).
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2
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1, 0, 1, 1, 0, 1, 0, 5, 0, 1, 1, 0, 15, 0, 1, 0, 21, 0, 35, 0, 1, 1, 0, 161, 0, 70, 0, 1, 0, 85, 0, 777, 0, 126, 0, 1, 1, 0, 1555, 0, 2835, 0, 210, 0, 1, 0, 341, 0, 14575, 0, 8547, 0, 330, 0, 1, 1, 0, 14421, 0, 91960, 0, 22407, 0, 495, 0, 1
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OFFSET
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1,8
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COMMENTS
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A companion triangle to the triangle of Hultman numbers A164652.
The triangle of Hultman numbers can be constructed from the triangle of Stirling cycle numbers ( |A008275(n,k)| )n,k>=1 by removing the triangular number factor n*(n-1)/2 from every other entry in the n-th row (n >= 2) and setting the remaining entries to 0.
Here we carry out the analogous construction starting with the triangle of Stirling numbers of the second kind A008277, but now removing the triangular number factor k*(k+1)/2 from every other entry in the k-th column and setting the remaining entries to 0.
Do these numbers have a combinatorial interpretation?
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LINKS
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FORMULA
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Let P(n,x) = (1 - x)*(1 - 2*x)*...*(1 - n*x). The g.f. for the k-th column of the triangle is (1/(k*(k + 1)))*x^(k-1)*(1/P(k,x) - 1/P(k,-x)) = (x^k)*(x^k*R(k-1,1/x))/((1 - x^2)*(1 - 4*x^2)*...*(1 - k^2*x^2)), where R(n,x) denotes the n-th row polynomial of A164652. (Since the entries of triangle A164652 are integers, it follows that the entries of the present triangle are also integers.)
It appears that the matrix product (|A008275|)^-1 * A164652 * A008277 = I_1 + A363041 (direct sum, where I_1 is the 1 X 1 identity matrix). See the Example section.
The sequence of row sums of the inverse array begins [1, 1, 0, -4, 0, 120, 0, -12096, 0, 3024000, 0, -1576143360, 0, 1525620096000, 0, -2522591034163200, 0, 6686974460694528000, 0, -27033456071346536448000, ...], and appears to be essentially A129825.
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EXAMPLE
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Triangle begins
k = 1 2 3 4 5 6 7 8 9 10
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
n = 1: 1
2: 0 1
3: 1 0 1
4: 0 5 0 1
5: 1 0 15 0 1
6: 0 21 0 35 0 1
7: 1 0 161 0 70 0 1
8: 0 85 0 777 0 126 0 1
9: 1 0 1555 0 2835 0 210 0 1
10: 0 341 0 14575 0 8547 0 330 0 1
...
/ 1 \ /1 \ /1 \ /1 \
|-1 1 | |0 1 | |1 1 | |0 1 |
| 1 -3 1 | |1 0 1 | |1 3 1 | = |0 0 1 |
|-1 7 -6 1 | |0 5 0 1 | |1 7 6 1 | |0 1 0 1 |
| 1 -15 25 -10 1| |8 0 15 0 1| |1 15 25 10 1| |0 0 5 0 1 |
| ... | |... | |... | |0 1 0 15 0 1|
| | | | | | |... |
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MAPLE
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A362041:= (n, k)-> `if`(n-k mod 2 = 0, Stirling2(n+1, k)/binomial(k+1, 2), 0):
for n from 1 to 10 do seq(A362041(n, k), k = 1..n) od;
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PROG
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(PARI) T(n, k) = if ((n-k) % 2, 0, stirling(n+1, k, 2)/binomial(k+1, 2)); \\ Michel Marcus, May 23 2023
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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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