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A152878
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Triangle read by rows: T(n,k) is the number of permutations of {1,2,...,n} with maximal number of initial entries of the same parity equal to k (1<=k<=ceil(n/2)).
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0
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1, 2, 4, 2, 16, 8, 72, 36, 12, 432, 216, 72, 2880, 1440, 576, 144, 23040, 11520, 4608, 1152, 201600, 100800, 43200, 14400, 2880, 2016000, 1008000, 432000, 144000, 28800, 21772800, 10886400, 4838400, 1814400, 518400, 86400, 261273600
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| Sum of entries in row n is n! =A000142(n).
Row n contains ceil(n/2) entries.
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REFERENCES
| E. Deutsch and J. H. Nieto, Mathematics Magazine, Problem 1823, Vol. 83, No. 3, 2010, pp. 230-231. [From Emeric Deutsch (deutsch(AT)duke.poly.edu), Aug 12 2010]
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FORMULA
| a(2n,k)=2nk!(2n-k-1)!binom(n,k);
a(2n+1,k)=n!(n+1)!*binom(2n-k+1,n).
Contribution from Emeric Deutsch (deutsch(AT)duke.poly.edu), Aug 12 2010: (Start)
T(n,k)= (ceil(n/2)*binom(floor(n/2),k) + floor(n/2)*binom(ceil(n/2),k))*k!*(n-k-1)! (from J. H. Nieto's solution).
(End)
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EXAMPLE
| T(4,2)=8 because we have 1324, 1342, 3124, 3142, 2413, 2431, 4213 and 4231.
T(5,3)=12 because the first 3 entries form a permutation of (1,3,5) (6 choices) and the last 2 entries form a permutation of {2,4} (2 choices).
Triangle starts:
1;
2;
4,2;
16,8;
72,36,12;
432,216,72;
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MAPLE
| ae := proc (n, k) options operator, arrow: 2*n*factorial(k)*factorial(2*n-k-1)*binomial(n, k) end proc: ao := proc (n, k) options operator, arrow: factorial(n)*factorial(n+1)*binomial(2*n-k+1, n) end proc: a := proc (n, k) if `mod`(n, 2) = 0 and k <= (1/2)*n then ae((1/2)*n, k) elif `mod`(n, 2) = 1 and k <= ceil((1/2)*n) then ao((1/2)*n-1/2, k) else 0 end if end proc: for n to 12 do seq(a(n, k), k = 1 .. ceil((1/2)*n)) end do; # yields sequence in triangular form
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CROSSREFS
| A000142
Sequence in context: A152877 A071353 A134763 * A186526 A100944 A059890
Adjacent sequences: A152875 A152876 A152877 * A152879 A152880 A152881
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KEYWORD
| nonn,tabf
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AUTHOR
| Emeric Deutsch (deutsch(AT)duke.poly.edu), Dec 26 2008
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