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Search: id:A152666
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| A152666 |
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Triangle read by rows: T(n,k) is the number of permutations of {1,2,...,n} kaving k runs of odd entries (1<=k<=ceil(n/2)). For example, the permutation 321756498 has 3 runs of odd entries: 3, 175 and 9. |
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+0
4
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| 1, 2, 4, 2, 12, 12, 36, 72, 12, 144, 432, 144, 576, 2592, 1728, 144, 2880, 17280, 17280, 2880, 14400, 115200, 172800, 57600, 2880, 86400, 864000, 1728000, 864000, 86400, 518400, 6480000, 17280000, 12960000, 2592000, 86400, 3628800, 54432000
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OFFSET
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1,2
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COMMENT
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Sum of entries in row n is n! (=A000142(n)).
Row n contains ceil(n/2) entries.
T(n,1)=A010551(n+1).
Sum(k*T(n,k),k>=1) = A052618(n-1).
Mirror image of A134435.
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FORMULA
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T(2n,k) = (n!)^2*binom(n+1,k)binom(n-1,k-1);
T(2n+1,k) = n!(n+1)!binom(n,k-1)binom(n+1,k).
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EXAMPLE
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T(3,2)=2 because we have 123 and 321.
T(4,2)=12 because we have 1234, 1432, 3214, 3412, 1243, 3241 and their reverses.
Triangle starts:
1;
2;
4,2;
12,12;
36,72,12;
144,432,144;
576,2592,1728,144.
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MAPLE
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ae := proc (n, k) options operator, arrow: factorial(n)^2*binomial(n+1, k)*binomial(n-1, k-1) end proc: ao := proc (n, k) options operator, arrow: factorial(n)*factorial(n+1)*binomial(n, k-1)*binomial(n+1, k) end proc: T := proc (n, k) if `mod`(n, 2) = 0 then ae((1/2)*n, k) else ao((1/2)*n-1/2, k) end if end proc: for n to 12 do seq(T(n, k), k = 1 .. ceil((1/2)*n)) end do; # yields sequence in triangular form
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CROSSREFS
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A000142, A010551, A052618, A152667, A134435
Sequence in context: A138770 A137777 A006018 this_sequence A153801 A062867 A113539
Adjacent sequences: A152663 A152664 A152665 this_sequence A152667 A152668 A152669
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KEYWORD
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nonn,tabf
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AUTHOR
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Emeric Deutsch (deutsch(AT)duke.poly.edu), Dec 14 2008
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