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%I #11 Nov 26 2019 20:07:07
%S 1,2,2,4,12,12,12,72,36,144,432,144,144,1728,2592,576,2880,17280,
%T 17280,2880,2880,57600,172800,115200,14400,86400,864000,1728000,
%U 864000,86400,86400,2592000,12960000,17280000,6480000,518400
%N Triangle read by rows: T(n,k) is the number of permutations of {1,2,...,n} having k odd entries that are followed by a smaller entry (n >= 1, k >= 0).
%C Row n has ceiling(n/2) entries. T(2n,0) = T(2n+1,0) = n!(n+1)! = A010790(n).
%C T(n,k) is also the number of permutations of {1,2,...,n} having k adjacent pairs of the form (odd, odd) (0 <= k <= ceiling(n,2)-1). Example: T(3,1)=4 because we have 132, 213, 312 and 231. - _Emeric Deutsch_, Dec 14 2008
%H S. Kitaev and J. Remmel, <a href="http://dx.doi.org/10.1007/s00026-007-0313-2">Classifying descents according to parity</a>, Annals of Combinatorics, 11, 2007, 173-193.
%F T(2n,k) = (n!)^2*C(n-1,k) C(n+1,k+1); T(2n+1,k) = n!(n+1)! * C(n,k) * C(n+1,k).
%e T(3,1) = 4 because we have 132, 312, 231 and 321.
%e Triangle starts:
%e 1;
%e 2;
%e 2, 4;
%e 12, 12;
%e 12, 72, 36;
%e 144, 432, 144;
%p T:=proc(n,k) if `mod`(n,2)=0 then binomial((1/2)*n-1, k)*binomial((1/2)* n+1, k+1)*factorial((1/2)*n)^2 elif `mod`(n,2)=1 then factorial((1/2)*n-1/2)*factorial((1/2)*n+1/2)*binomial((1/2)*n-1/2, k)*binomial((1/2)* n+1/2, k) else 0 end if end proc: for n to 11 do seq(T(n,k), k=0..ceil((1/2)*n)-1) end do; # yields sequence in triangular form
%Y Cf. A010790, A134434.
%K nonn,tabf
%O 1,2
%A _Emeric Deutsch_, Nov 22 2007
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