a recursion left-right or right-left

Wouter Meeussen w.meeussen.vdmcc at vandemoortele.be
Fri Jun 15 15:21:16 CEST 2001



Clear[T];
T[0,0]:=0;            T[n_,m_]/;(m>n):=0;
T[n_,0]:= x^(n-1) ;
T[n_,m_]:=T[n,m]=  T[n,m-1]-  T[n-1,m];


xit=Table[T[n,m],{n,0,5},{m,0,n}]

{{0},
 {1, 1}, 
 {x, -1 + x, -1 + x},
 {x^2, 1 - x + x^2, 2 - 2*x + x^2,  2 - 2*x + x^2},
 {x^3, -1 + x - x^2 + x^3, -3 + 3*x - 2*x^2 + x^3, 
             -5 + 5*x - 3*x^2 + x^3, -5 + 5*x - 3*x^2 + x^3}, 
 {x^4, 1 - x + x^2 - x^3 + x^4, 4 - 4*x + 3*x^2 - 2*x^3 + x^4, 
              9 - 9*x + 6*x^2 - 3*x^3 + x^4, 14 - 14*x + 9*x^2 - 4*x^3 + x^4, 
             14 - 14*x + 9*x^2 - 4*x^3 + x^4}}

the same polynome table can be generated right to left (rare?):
with  definition  (using A009766)
catx[n_]:=Sum[(n+k)!/(n+1)!/k! (n-k+1)(-1)^k x^(n-k),{k,0,n}]

Clear[T];
T[0,0]:=0;            T[n_,m_]/;(m>n):=0 ;
T[n_,n_]:= catx[n-1] ;
T[n_,m_]:=T[n,m]= Expand[ T[n,m+1] + T[n-1,m+1]  ];

The polynomes at x=1 give the value of the recursion:

T[n_,0]:= 1;          T[n_,m_]/;(m>n):=0 ;
T[n_,m_]:=T[n,m]=  T[n,m-1]-  T[n-1,m];

this gives table
1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, -1, -2, -2, 1, 1, 2, 4, 6, 6, 1, 0, -2, 
-6, -12, -18, -18, 1, 1, 3, 9, 21, 39, 57, 57, 1, 0, -3, -12, -33, -72, -129, 
-186, -186, 1, 1, 4, 16, 49, 121, 250, 436, 622, 622, 1, 0, -4, -20, -69, 
-190, -440, -876, -1498, -2120, -2120, 1, 1, 5, 25, 94, 284, 724, 1600, 3098, 
5218, 7338, 7338, 1, 0, -5, -30, -124, -408, -1132, -2732, -5830, -11048, 
-18386, -25724, -25724

and row sums:
1, 2, 1, 4, -4, 20, -55, 188, -620, 2122, -7336, 25726, -91142, 325880, 
-1174279, 4260284, -15548692, 57048050, -210295324, 778483934, -2892818242, 
10786724390, -40347919624, 151355847014, -569274150154, 2146336125650, 
-8110508473250, 30711521221378, -116518215264490, 442862000693440, 
-1686062250699431, 6429286894263740, -24552388991392228, 93891870710425442, 
-359526085719652660, 1378379704593824302, -5290709340633314594

could anyone check this? I've been known to blunder

Wouter 


wouter.meeussen at vandemoortele.com








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