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Matrix inverse of triangle A098568, where A098568(n, k) = C( (k+1)*(k+2)/2 + n-k-1, n-k) for n>=k>=0.
3

%I #5 Jun 13 2015 11:07:36

%S 1,0,1,0,-1,1,0,2,-3,1,0,-7,12,-6,1,0,37,-67,39,-10,1,0,-268,498,-311,

%T 95,-15,1,0,2496,-4701,3045,-1015,195,-21,1,0,-28612,54298,-35901,

%U 12560,-2675,357,-28,1,0,391189,-745734,499157,-179717,40635,-6097,602,-36,1,0,-6230646,11911221,-8034267,2945010

%N Matrix inverse of triangle A098568, where A098568(n, k) = C( (k+1)*(k+2)/2 + n-k-1, n-k) for n>=k>=0.

%F (1) T(n,k) = (-1)^(n-k)*[A107876^(k*(k+1)/2)](n,k); i.e., column k equals signed column k of A107876^(k*(k+1)/2). G.f.s for column k: (2) 1 = Sum_{j>=0} T(j+k,k)*x^j/(1-x)^( j*(j+1)/2) + j*k + k*(k+1)/2); (3) 1 = Sum_{j>=0} T(j+k,k)*x^j*(1+x)^( j*(j-1)/2) + j*k + k*(k+1)/2).

%e Triangle begins:

%e 1;

%e 0, 1;

%e 0, -1, 1;

%e 0, 2, -3, 1;

%e 0, -7, 12, -6, 1;

%e 0, 37, -67, 39, -10, 1;

%e 0, -268, 498, -311, 95, -15, 1;

%e 0, 2496, -4701, 3045, -1015, 195, -21, 1;

%e 0, -28612, 54298, -35901, 12560, -2675, 357, -28, 1;

%e 0, 391189, -745734, 499157, -179717, 40635, -6097, 602, -36, 1; ...

%o (PARI) /* Matrix Inverse of A098568 */ T(n,k)=local(M=matrix(n+1,n+1,r,c,if(r>=c,binomial((c-1)*(c-2)/2+r-2,r-c)))); return((M^-1)[n+1,k+1])

%o (PARI) /* Obtain by G.F. */ T(n,k)=polcoeff(1-sum(j=0, n-k-1, T(j+k,k)*x^j/(1-x+x*O(x^n))^(j*(j+1)/2+j*k+k*(k+1)/2)), n-k)

%Y Cf. A098568, A107876; unsigned columns: A107877, A107887.

%K sign,tabl

%O 0,8

%A _Paul D. Hanna_, Aug 27 2006