%I #13 Jul 01 2023 08:27:10
%S 1,1,0,1,0,1,2,1,1,2,6,4,3,2,9,24,18,14,11,9,44,120,96,78,64,53,44,
%T 265,720,600,504,426,362,309,265,1854,5040,4320,3720,3216,2790,2428,
%U 2119,1854,14833,40320,35280,30960,27240,24024,21234,18806,16687,14833
%N Triangle of rank k of permutations of {1,2,...,n}.
%C The rank k of a permutation of n elements is the first position of a fixed point. If there is no fixed point then k=n+1 and R(n,n+1)=A000166(n), the derangements numbers (subfactorials).
%D Ch. A. Charalambides, Enumerative Combinatorics, Chapman & Hall/CRC, Boca Raton, Florida, 2002, p. 176, Table 5.3 (without row n=0 and column k=1), p. 185.
%H Wolfdieter Lang, <a href="/A134830/a134830.txt">First 10 rows and more</a>.
%F R(n,k)=0 if n+1<k, R(n,n+1)=D(n),n>=0, with D(n):=A000166(n) the derangements numbers (subfactorials), R(n,k)=sum((-1)^j*binomial(k-1,j)*(n-j-1)!,j=0..k-1), k from 1,..,n.
%F Subtriangle without diagonal k=n+1: R(n,k)=sum(binomial(n-k,j)*D(k+j-1),j=0..n-k), k=1,...,n, n>=1, with D(n):=A000166(n).
%F R(n,k) = R(n,k-1) - R(n-1,k-1), R(0,0)=1, R(n,1)=(n-1)!.
%e Triangle begins:
%e [1];
%e [1,0];
%e [1,0,1];
%e [2,1,1,2];
%e [6,4,3,2,9];
%e [24,18,14,11,9,44];
%e ...
%e R(4,2)=4 from the four rank k=2 partitions of 4 elements (3,2,1,4), (3,2,4,1), (4,2,1,3) and (4,2,3,1).
%Y Row sums n!=A000142(n). Alternating row sums A134831.
%K nonn,easy,tabl
%O 0,7
%A _Wolfdieter Lang_, Jan 21 2008