%I #12 Jul 21 2017 10:49:16
%S 1,1,1,2,1,3,6,2,4,12,24,6,10,20,60,120,24,36,60,120,360,720,120,168,
%T 252,420,840,2520,5040,720,960,1344,2016,3360,6720,20160,40320,5040,
%U 6480,8640,12096,18144,30240,60480,181440,362880,40320,50400,64800
%N Triangle read by rows: T(n,k) is the number of permutations of {1,2,...,n} in which k is the largest entry in the cycle containing 1 (1 <= k <= n).
%C Row sums are the factorials (A000142).
%C T(n,n) = n!/2 = A001710(n) (n>=2).
%C Sum_{k=1..n} k*T(n,k) = A121586(n).
%D Solution to Problem 1831 by J. W. Grossman. Mathematics Magazine, 83, No. 5, 2010, pp. 392-393.
%H J. W. Grossman, <a href="http://www.jstor.org/stable/10.4169/mathmaga.83.5.0392a">Solution to Problem 1831 proposed by Emeric Deutsch</a>, Mathematics Magazine, 83, No. 5, 2010, pp. 392-393.
%F T(n,1)=(n-1)!; T(n,k)=n!/((n-k+1)(n-k+2)) for 2 <= k <= n.
%e T(4,3)=4 because we have (132)(4), (13)(24), (123)(4), (13)(2)(4).
%e Triangle starts:
%e 1;
%e 1, 1;
%e 2, 1, 3;
%e 6, 2, 4, 12;
%e 24, 6, 10, 20, 60;
%e 120, 24, 36, 60, 120, 360;
%p T:=proc(n,k) if k=1 then factorial(n-1) elif k <= n then factorial(n)/((n-k+1)*(n-k+2)) else 0 end if end proc: for n to 10 do seq(T(n,k),k=1..n) end do; # yields sequence in triangular form
%Y Cf. A000142, A001710, A121586.
%K nonn,tabl
%O 1,4
%A _Emeric Deutsch_, Nov 10 2008