A121437 - OEIS

%I #14 Nov 28 2016 13:39:48

%S 1,-3,1,6,-4,1,-16,14,-6,1,63,-62,33,-9,1,-351,365,-215,72,-13,1,2609,

%T -2790,1731,-642,143,-18,1,-24636,26749,-17076,6696,-1664,261,-24,1,

%U 284631,-311769,202356,-81963,21684,-3831,444,-31,1,-3909926,4305579,-2822991,1166310,-320515,60768,-8012,713,-39,1

%N Matrix inverse of triangle A122177, where A122177(n,k) = C( k*(k+1)/2 + n-k + 2, n-k) for n>=k>=0.

%F (1) T(n,k) = A121436(n-1,k) - A121436(n-1,k+1). (2) T(n,k) = (-1)^(n-k)*[A107876^(k*(k+1)/2 + 3)](n,k); i.e., column k equals signed column k of A107876^(k*(k+1)/2 + 3). G.f.s for column k: (3) 1 = Sum_{j>=0} T(j+k,k)*x^j/(1-x)^( j*(j+1)/2) + j*k + k*(k+1)/2 + 3); (4) 1 = Sum_{j>=0} T(j+k,k)*x^j*(1+x)^( j*(j-1)/2) + j*k + k*(k+1)/2 + 3).

%F From _Benedict W. J. Irwin_, Nov 26 2016: (Start)

%F Conjecture: The sequence (column 2 of triangle) 14, -62, 365, -2790, 26749, ... is described by a series of nested sums:

%F 14    =  Sum_{i=1..4} (i+1),

%F -62   = -Sum_{i=1..4} (Sum_{j=1..i+1} (j+2)),

%F 365   =  Sum_{i=1..4} (Sum_{j=1..i+1} (Sum_{k=1..j+2} (k+3))),

%F -2790 = -Sum_{i=1..4} (Sum_{j=1..i+1} (Sum_{k=1..j+2} (Sum_{l=1..k+3} (l+4)))). (End)

%e Triangle begins:

%e 1;

%e -3, 1;

%e 6, -4, 1;

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

%e 63, -62, 33, -9, 1;

%e -351, 365, -215, 72, -13, 1;

%e 2609, -2790, 1731, -642, 143, -18, 1;

%e -24636, 26749, -17076, 6696, -1664, 261, -24, 1;

%e 284631, -311769, 202356, -81963, 21684, -3831, 444, -31, 1; ...

%o (PARI) /* Matrix Inverse of A122177 */ T(n,k)=local(M=matrix(n+1,n+1,r,c,if(r>=c,binomial((c-1)*(c-2)/2+r+1,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+3)), n-k)

%Y Cf. A098568, A107876, A122177, A107885.

%K sign,tabl

%O 0,2

%A _Paul D. Hanna_, Aug 27 2006