A144090 - OEIS

%I #22 Apr 27 2016 16:29:38

%S 1,2,0,3,6,3,4,24,36,8,5,60,210,220,45,6,120,780,1920,1590,264,7,210,

%T 2205,9940,19005,12978,1855,8,336,5208,37520,130200,203952,118664,

%U 14832,9,504,10836,114408,630630,1783656,2369556,1201464,133497

%N Triangle read by rows: T(n,k) is the number of partial bijections (or subpermutations) of an n-element set of height k (height(alpha) = |Im(alpha)|) and with exactly 1 fixed point.

%H A. Laradji and A. Umar, <a href="http://dx.doi.org/10.1007/s00233-007-0732-8">Combinatorial results for the symmetric inverse semigroup</a>, Semigroup Forum 75, (2007), 221-236.

%F T(n,k) = (n!/(n-k)!)*Sum_{m=0..k-1} ((-1)^m/m!)*C(n-1-m,k-1-m).

%e T(3,2) = 6 because there are exactly 6 partial bijections (on a 3-element set) with exactly 1 fixed point and of height 2, namely: (1,2)->(1,3), (1,2)->(3,2), (1,3)->(1,2), (1,3)->(2,3), (2,3)->(2,1), (2,3)->(1,3)- the mappings are coordinate-wise.

%e First six rows:

%e 1

%e 2      0

%e 3      6      3

%e 4     24     36       8

%e 5     60    210     220      45

%e 6    120    780    1920    1590    264

%t Table[(n!/(n - k)!) Sum[ ((-1)^m/m!) Binomial[n - 1 - m, k - 1 - m], {m, 0, k - 1}], {n, 9}, {k, n}] // Flatten (* _Michael De Vlieger_, Apr 27 2016 *)

%o (PARI) T(n,k) = (n!/(n-k)!)*sum(m=0,k-1,((-1)^m/m!)*binomial(n-1-m,k-1-m));

%o for (n=1, 10, for (k=1, n, print1(T(n,k), ", "))) \\ _Michel Marcus_, Apr 27 2016

%Y Rows sums are A144086.

%Y Main diagonal gives A000240.

%K nonn,tabl

%O 1,2

%A _Abdullahi Umar_, Sep 11 2008