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%I #17 Dec 09 2019 10:37:20
%S 1,1,1,3,1,2,12,4,2,6,60,20,10,6,24,360,120,60,36,24,120,2520,840,420,
%T 252,168,120,720,20160,6720,3360,2016,1344,960,720,5040,181440,60480,
%U 30240,18144,12096,8640,6480,5040,40320
%N Triangle read by rows: T(n,k) is the number of permutations of {1,2,...,n} in which the size of the last cycle is k (the cycles are ordered by increasing smallest elements; 1 <= k <=n).
%C Row sums are the factorials (A000142). T(n,1)=n!/2 for n>=2; Sum(k*T(n,k),k=1..n)=s(n,2)=A000254(n) (Stirling numbers of the first kind).
%H Alois P. Heinz, <a href="/A136125/b136125.txt">Rows n = 1..141, flattened</a>
%F T(n,k) = n!/[k(k+1)] if k<n; T(n,n)=(n-1)!.
%F Rec. rel.: T(n,k) = (n-1-k)*T(n-1,k) + (k-1)T(n-1,k-1) for 1 < k < n.
%e T(4,2) = 4 because we have (1)(2)(34), (13)(24), (12)(34) and (14)(23).
%e Triangle starts:
%e 1;
%e 1, 1;
%e 3, 1, 2;
%e 12, 4, 2, 6;
%e 60, 20, 10, 6, 24;
%p T:=proc(n,k) if k < n then factorial(n)/(k*(k+1)) elif k = n then factorial(n-1) else 0 end if end proc: for n to 9 do seq(T(n,k),k=1..n) end do; # yields sequence in triangular form
%p # second Maple program:
%p b:= proc(n, l) option remember; `if`(n=0, x^l, add(
%p binomial(n-1, j-1)*b(n-j, j)*(j-1)!, j=1..n))
%p end:
%p T:= n-> (p-> (seq(coeff(p, x, i), i=1..n)))(b(n, 0)):
%p seq(T(n), n=1..12); # _Alois P. Heinz_, Dec 08 2018
%t T[n_, n_] := (n-1)!;
%t T[n_, k_] := n!/(k(k+1));
%t Table[T[n, k], {n, 1, 12}, {k, 1, n}] // Flatten (* _Jean-François Alcover_, Dec 09 2019 *)
%Y Cf. A000142, A000254.
%Y T(2n,n) gives A322450.
%K nonn,tabl
%O 1,4
%A _Emeric Deutsch_, Jan 10 2008
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