|
%I #13 Aug 12 2019 23:15:12
%S 2,6,12,12,48,72,144,432,144,720,2880,1440,2880,17280,17280,2880,
%T 17280,129600,172800,43200,86400,864000,1728000,864000,86400,604800,
%U 7257600,18144000,12096000,1814400,3628800,54432000,181440000,181440000
%N Triangle read by rows: T(n,k) is the number of permutations of {1,2,...,n} having k runs of even entries (n >= 2, 1 <= k <= floor(n/2)). For example, the permutation 321756498 has 3 runs of even entries: 2, 64 and 8.
%C Sum of entries in row n is n! (=A000142(n)). Row n contains floor(n/2) entries.
%C T(n,1) = A152873(n).
%C Sum_{k>=1} k*T(n,k) = A152668(n).
%C Mirror image of A145892.
%F T(2n,k) = (n!)^2 * binomial(n+1,k) binomial(n-1,k-1);
%F T(2n+1,k) = n!*(n+1)!*binomial(n-1,k-1)*binomial(n+2,k) (n >= 1).
%e T(4,2) = 12 because we have 1234, 3214, 1432, 3412, 2134, 2314 and their reverses.
%e Triangle starts:
%e 2;
%e 6;
%e 12, 12;
%e 48, 72;
%e 144, 432, 144;
%e 720, 2880, 1440;
%p 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-1, k-1)*binomial(n+2, 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 .. floor((1/2)*n)) end do; # yields sequence in triangular form
%Y Cf. A000142, A152666, A152668, A145892, A152873.
%K nonn,tabf
%O 2,1
%A _Emeric Deutsch_, Dec 14 2008
|
