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Search: id:A145892
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| A145892 |
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Triangle read by rows: T(n,k) is the number of permutations of {1,2,...,n} having k adjacent pairs of the form (even,even) (0<=k<=floor(n/2)-1). |
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+0
3
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| 1, 1, 2, 6, 12, 12, 72, 48, 144, 432, 144, 1440, 2880, 720, 2880, 17280, 17280, 2880, 43200, 172800, 129600, 17280, 86400, 864000, 1728000, 864000, 86400, 1814400, 12096000, 18144000, 7257600, 604800, 3628800, 54432000, 181440000, 181440000
(list; graph; listen)
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OFFSET
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0,3
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COMMENT
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Row n contains floor(n/2) entries (n>=2).
Sum of entries in row n = n! =A000142(n).
Sum(k*T(n,k),k>=0) = A077612(n).
T(2n,k)=A134435(2n,k).
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FORMULA
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T(2n,k)=(n!)^2*binom(n-1,k)*binomial(n+1,k+1); T(2n+1,k)=n!(n+1)!binom(n-1,k)binom(n+2,k+2).
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EXAMPLE
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T(4,1)=12 because we have 1243, 1423, 1324, 1342, 3124, 3142, 2413, 4213, 2431, 4231, 3241 and 3421.
Triangle starts:
1;
1;
2;
6;
12,12;
72,48;
144,432,144;
1440,2880,720.
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MAPLE
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T:=proc(n, k) if `mod`(n, 2) = 0 then factorial((1/2)*n)^2*binomial((1/2)*n-1, k)*binomial((1/2)*n+1, k+1) else factorial((1/2)*n-1/2)*factorial((1/2)*n+1/2)*binomial((1/2)*n-3/2, k)*binomial((1/2)*n+3/2, k+2) end if end proc: 1; 1; for n from 2 to 12 do seq(T(n, k), k = 0 .. floor((1/2)*n)-1) end do; # yields sequence in triangular form
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CROSSREFS
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A000142, A077612, A134434, A134435, A145891
Sequence in context: A066791 A062723 A152667 this_sequence A154712 A079849 A057895
Adjacent sequences: A145889 A145890 A145891 this_sequence A145893 A145894 A145895
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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), Nov 30 2008
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