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A237996
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Triangular array read by rows. T(n,k) is the number of even permutations of {1,2,...,n} that have exactly k cycles, n >= 0, 0 <= k <= n.
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1
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1, 0, 1, 0, 0, 1, 0, 2, 0, 1, 0, 0, 11, 0, 1, 0, 24, 0, 35, 0, 1, 0, 0, 274, 0, 85, 0, 1, 0, 720, 0, 1624, 0, 175, 0, 1, 0, 0, 13068, 0, 6769, 0, 322, 0, 1, 0, 40320, 0, 118124, 0, 22449, 0, 546, 0, 1, 0, 0, 1026576, 0, 723680, 0, 63273, 0, 870, 0, 1
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OFFSET
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0,8
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COMMENTS
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If a permutation of {1, 2, ..., n} is written as a product of m disjoint cycles (where the fixed points of the permutation are viewed as 1-cycles) then the parity of the permutation is (-1)^(n-m). It is an even permutation if the number of cycles of even length is even (possibly zero), and it is an odd permutation if the number of cycles of even length is odd. - Peter Bala, Jun 25 2024
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REFERENCES
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J. Riordan, Introduction to Combinatorial Analysis, Wiley, 1958, page 87, problem # 20.
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LINKS
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FORMULA
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E.g.f.: exp(y*A(x))*cosh(y*B(x)) where A(x)= log((1 + x)/(1 - x))^(1/2) and B(x)=log(1/(1-x^2)^(1/2).
If n and k are both even or both odd, then T(n, k) is equal to the Stirling cycle number |s(n, k)| = A132393(n, k), and 0 otherwise.
n-th row polynomial R(n, x) = (1/2)*( x*(x + 1)*...*(x + n + 1) + x*(x - 1)*...*(x - n - 1) ).
For n >= 1, the zeros of R(n, x) are purely imaginary. (End)
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EXAMPLE
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1,
0, 1,
0, 0, 1,
0, 2, 0, 1,
0, 0, 11, 0, 1,
0, 24, 0, 35, 0, 1,
0, 0, 274, 0, 85, 0, 1,
0, 720, 0, 1624, 0, 175, 0, 1,
0, 0, 13068, 0, 6769, 0, 322, 0, 1,
0, 40320, 0, 118124, 0, 22449, 0, 546, 0, 1,
0, 0, 1026576, 0, 723680, 0, 63273, 0, 870, 0, 1
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MAPLE
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with(combinat):
b:= proc(n, i, t) option remember; expand(`if`(n=0, t, `if`(i<1,
0, add(x^j*multinomial(n, n-i*j, i$j)*(i-1)!^j/j!*b(n-i*j,
i-1, irem(t+`if`(irem(i, 2)=0, j, 0), 2)), j=0..n/i))))
end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..n))(b(n$2, 1)):
# Alternative:
A132393 := (n, k) -> abs(Stirling1(n, k)):
T := (n, k) -> ifelse((n::even and k::even) or (n::odd and k::odd), A132393(n, k),
0): seq(seq(T(n, k), k = 0..n), n = 0..9); # Peter Luschny, Jun 26 2024
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MATHEMATICA
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nn=11; a=Log[((1+x)/(1-x))^(1/2)]; b=Log[1/(1-x^2)^(1/2)]; Table[Take[(Range[0, nn]!CoefficientList[Series[Exp[y a]Cosh[y b] , {x, 0, nn}], {x, y}])[[n]], n], {n, 1, nn}]//Grid
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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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