A183153 - OEIS

%I #15 Mar 29 2013 19:04:31

%S 1,1,1,1,4,1,1,9,5,1,1,16,14,6,1,1,25,30,20,7,1,1,36,55,50,27,8,1,1,

%T 49,91,105,77,35,9,1,1,64,140,196,182,112,44,10,1,1,81,204,336,378,

%U 294,156,54,11,1,1,100,285,540,714,672,450,210,65,12,1,1,121,385,825,1254,1386,1122,660

%N T(n,k) is the number of order-preserving partial isometries of an n-chain of height k (height of alpha = |Im(alpha)|).

%C The matrix inverse starts

%C 1;

%C -1,1;

%C 3,-4,1;

%C -7,11,-5,1;

%C 15,-26,16,-6,1;

%C -31,57,-42,22,-7,1;

%C 63,-120,99,-64,29,-8,1;

%C -127,247,-219,163,-93,37,-9,1;

%C 255,-502,466,-382,256,-130,46,-10,1;

%C ...perhaps related to A054143. - _R. J. Mathar_, Mar 29 2013

%H R. Kehinde, A. Umar, <a href="http://arxiv.org/abs/1101.0049">On the semigroup of partial isometries of a finite chain</a>, arXiv:1101.0049

%F T(n,0)=1. T(n,k)=(2*n-k+1)*C(n,k)/(k+1) if k>0.

%e T(3,2)=5 because there are exactly 5 order-preserving partial isometries (on a 3-chain) of height 2, namely: (1,2)-->(1,2); (1,2)-->(2,3); (2,3)-->(1,2); (2,3)-->(2,3); (1,3)-->(1,3), the mappings are coordinate-wise.

%e Triangle begins as:

%e 1;

%e 1, 1;

%e 1, 4, 1;

%e 1, 9, 5, 1;

%e 1, 16, 14, 6, 1;

%e 1, 25, 30, 20, 7, 1;

%e 1, 36, 55, 50, 27, 8, 1;

%e 1, 49, 91, 105, 77, 35, 9, 1;

%o (PARI) T(n,k)=if(k==0,1, (2*n-k+1)*binomial(n,k)/(k+1));

%o for(n=0,17,for(k=0,n,print1(T(n,k),", ")))

%Y Cf. A097813 (row sums), A125165, A121306, A029653.

%K nonn,tabl,easy

%O 0,5

%A _Abdullahi Umar_, Dec 27 2010