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Search: id:A145894
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| A145894 |
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Triangle read by rows: T(n,k) is the number of permutations p of {1,2,...,n} such that j and p(j) are of the same parity for k values of j (0<=k<=n). |
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+0
2
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| 1, 0, 1, 1, 0, 1, 0, 4, 0, 2, 4, 0, 16, 0, 4, 0, 36, 0, 72, 0, 12, 36, 0, 324, 0, 324, 0, 36, 0, 576, 0, 2592, 0, 1728, 0, 144, 576, 0, 9216, 0, 20736, 0, 9216, 0, 576, 0, 14400, 0, 115200, 0, 172800, 0, 57600, 0, 2880, 14400, 0, 360000, 0, 1440000, 0, 1440000, 0
(list; table; graph; listen)
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OFFSET
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0,8
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COMMENT
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Mirror image of A145893.
Without the 0's, it is the triangle of A134434.
Sum of entries in row n = n! =A000142(n).
T(n,n)=A010551(n).
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FORMULA
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T(2n,2k)=[n!*binom(n,k)]^2; T(2n+1,2k+1)=[(n+1)!*binom(n,k)]^2/(k+1); elsewhere T(n,k)=0.
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EXAMPLE
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T(3,1)=4 because we have 132, 312, 213 and 231.
Triangle starts:
1;
0,1;
1,0,1;
0,4,0,2;
4,0,16,0,4;
0,36,0,72,0,12;
36,0,324,0,324,0,36;
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MAPLE
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T:=proc(n, k) if `mod`(n, 2) = 0 and `mod`(k, 2) = 0 then factorial((1/2)*n)^2*binomial((1/2)*n, (1/2)*k)^2 elif `mod`(n, 2) = 1 and `mod`(k, 2) = 1 then 2*factorial((1/2)*n+1/2)^2*binomial((1/2)*n-1/2, (1/2)*k-1/2)^2/(k+1) else 0 end if end proc: for n from 0 to 10 do seq(T(n, k), k=0..n) end do; # yields sequence in triangular form
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CROSSREFS
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A000142, A134434, A145893, A010551.
Sequence in context: A109157 A123314 A058997 this_sequence A021881 A021717 A016679
Adjacent sequences: A145891 A145892 A145893 this_sequence A145895 A145896 A145897
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KEYWORD
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nonn,tabl
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AUTHOR
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Emeric Deutsch (deutsch(AT)duke.poly.edu), Nov 30 2008
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