right-half-sum triangle - Matrix form of Linear Recurrence

[Paul D Hanna]

Dear Seqfans,
     Here is a matrix formulation of the linear recurrence.
First, to state the recurrence formula better:
a(0)=1
 
a(n) = C(2^n+n-1,n) - Sum_{k=0..n-1} a(k)*C(2^n-2^k+n-k-1,n-k)
 
where  C(n,k) = binomial(n,k) .
 
PARI CODE (improved): 
{a(n)=if(n==0,1,binomial(2^n+n-1,n)-sum(k=0,n-1,a(k)*binomial(2^n-2^k+n-k-1,n-k)))}

 
MATFRIX FORM.
Define matrix T by:
 
T(n,k) = C(2^n-2^k+n-k-1, n-k)
 
Triangular matrix T begins:
1;
1,1;
6,2,1;
84,21,4,1;
3060,560,78,8,1;
324632,40920,4060,300,16,1;
109453344,8936928,595665,30856,1176,32,1;
124397910208,6249655776,264566400,9078630,240464,4656,64,1;
  
The Inverse matrix starts:
1;
-1,1;
-4,-2,1;
-47,-13,-4,1;
-1812,-300,-46,-8,1;
-224380,-24100,-2124,-172,-16,1;
-87372452,-6220470,-350177,-15944,-664,-32,1;
-109591022246,-5020041906,-184889864,-5333670,-123472,-2608,-64,1;
 
Define 'diagonal' vector D by:
 
D[n] = C(2^n+n-1,n)
 
D=[1, 2, 10, 120, 3876, 376992, 119877472, 131254487936, 
509850594887712, 7145544812472168960, 
364974894538906616240640, 
68409601066028072105113098240, 47312269462735023248040155132636160, ...]
 
 
Then Max's nice sequence is given by: 
 
[T^-1]*D = A
 
= [1, 1, 2, 7, 44, 516, 11622, 512022, 44588536, 7718806044]~



I will submit to the OEIS the above matrix, its inverse, and the vector D tonight.

Thanks,
      Paul