%I #10 Oct 12 2023 11:22:44
%S 2,4,2,12,8,4,48,36,24,12,240,192,144,96,48,1440,1200,960,720,480,240,
%T 10080,8640,7200,5760,4320,2880,1440,80640,70560,60480,50400,40320,
%U 30240,20160,10080,725760,645120,564480,483840,403200,322560,241920,161280,80640
%N Triangle read by rows: T(n,k) is the number of permutations of {1,2,...,n} such that there are exactly k entries between the entries 1 and 2 (n>=2, 0<=k<=n-2).
%C Sum of row n = n! = A000142(n).
%C The expected value of k is (n-2)/3. [_Geoffrey Critzer_, Dec 19 2009]
%F T(n,k) = 2*(n-k-1)*(n-2)!.
%F T(n,0) = 2(n-1)! = A052849(n-1).
%F T(n,1) = A052582(n-2).
%F T(n,2) = A052609(n-2).
%F T(n,3) = 12*A005990(n-3).
%F T(n,4) = 48*A061206(n-5).
%F T(n,n-2) = 2(n-2)! (A052849).
%F Sum_{k=0..n-2} k*T(n,k) = n!*(n-2)/3 = A090672(n-1).
%e T(4,2)=4 because we have 1342, 1432, 2341 and 2431.
%e Triangle starts:
%e 2;
%e 4,2;
%e 12,8,4;
%e 48,36,24,12;
%e 240,192,144,96,48;
%e ...
%p T:=proc(n,k) if n-2 < k then 0 else (2*n-2*k-2)*factorial(n-2) end if end proc; for n from 2 to 10 do seq(T(n, k),k=0..n-2) end do; # yields sequence in triangular form
%t Table[Table[2 (n - r) (n - 2)!, {r, 1, n - 1}], {n, 1, 10}] // Grid (* _Geoffrey Critzer_, Dec 19 2009 *)
%Y Cf. A000142, A052489, A052582, A052609, A005990, A061206, A052849, A090672.
%K nonn,tabl
%O 2,1
%A _Emeric Deutsch_, Apr 06 2008