%I #28 Jun 30 2022 08:40:12
%S 1,1,1,1,3,1,1,9,9,1,1,29,72,29,1,1,99,626,626,99,1,1,351,6084,13869,
%T 6084,351,1,1,1275,64974,347020,347020,64974,1275,1,1,4707,744193,
%U 9952274,21537270,9952274,744193,4707,1,1,17577,8965323,321541977,1545936516,1545936516,321541977,8965323,17577,1
%N Triangle of coefficients of characteristic polynomial of negative Pascal matrix with (i,j)-th entry -C(i+j-2,i-1).
%H P. Di Francesco, P. Zinn-Justin and J.-B. Zuber, <a href="https://arxiv.org/abs/math-ph/0410002">Determinant Formulae for some Tiling Problems and Application to Fully Packed Loops</a>, arXiv:math-ph/0410002, 2004.
%H W. F. Lunnon, <a href="http://www.fq.math.ca/Scanned/15-3/lunnon.pdf">The Pascal matrix</a>, Fib. Quart. vol. 15 (1977) pp. 201-204.
%H Luca Guido Molinari, <a href="https://arxiv.org/abs/2206.14428">Graphene nanocones and Pascal matrices</a>, arXiv:2206.14428 [math.CO], 2022.
%e 1;
%e 1,1;
%e 1,3,1;
%e 1,9,9,1;
%e 1,29,72,29,1;
%e ...
%t P[n_] := Table[Binomial[i + j - 2, i - 1], {i, 1, n}, {j, 1, n}];
%t row[0] = {1};
%t row[n_] := CoefficientList[ CharacteristicPolynomial[P[n], x], x] // Abs;
%t Table[row[n], {n, 0, 9}] // Flatten (* _Jean-François Alcover_, Aug 09 2018 *)
%o (PARI) T(n,k)=if(n<0,0,(-1)^(n+k)*polcoeff(charpoly(matrix(n,n,i,j,binomial(i+j-2,i-1))),k))
%o (PARI) T(n,k)=if(n<0,0,polcoeff(charpoly(-matrix(n,n,i,j,binomial(i+j-2,i-1))),k))
%Y Sum of k-th row is A006366(n). Columns give A006134, A006135, A006136.
%K nonn,tabl
%O 0,5
%A _Fred Lunnon_, Dec 11 1999