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%I #18 Sep 20 2019 07:24:49
%S 1,-1,1,-2,1,1,-4,4,-1,1,-6,11,-6,1,1,-8,22,-24,9,-1,1,-10,37,-62,46,
%T -12,1,1,-12,56,-128,148,-80,16,-1,1,-14,79,-230,367,-314,130,-20,1,1,
%U -16,106,-376,771,-920,610,-200,25,-1,1,-18,137
%N Array: row n shows the coefficients of the characteristic polynomial of the n-th principal submatrix of A203905.
%C Let p(n)=p(n,x) be the characteristic polynomial of the n-th principal submatrix. The zeros of p(n) are positive, and they interlace the zeros of p(n+1). See A202605 for a guide to related sequences.
%C If we omit the main diagonal of this array and ignore the signs of the entries then the resulting array, reading the rows in reverse order, appears to equal the Riordan array (1/((1 + x)*(1 - x)^3), x/(1 - x)^2), whose generating function begins 1 + (2 + t)*x + (4 + 4*t + t^2)*x^2 + (6 + 11*t + 6*t^2 + t^3)*x^3 + (9 + 24*t + 22*t^2 + 8*t^3 + t^4)*x^4 + .... - _Peter Bala_, Sep 17 2019
%D (For references regarding interlacing roots, see A202605.)
%e Top of the array:
%e 1...-1
%e 1...-2....1
%e 1...-4....4...-1
%e 1...-6...11...-6....1
%e 1...-8...22...-24...9...-1
%t t = {1, 0}; t1 = Flatten[{t, t, t, t, t, t, t, t, t, t}];
%t f[k_] := t1[[k]];
%t U[n_] := NestList[Most[Prepend[#, 0]] &, #,
%t Length[#] - 1] &[Table[f[k], {k, 1, n}]];
%t L[n_] := Transpose[U[n]];
%t p[n_] := CharacteristicPolynomial[L[n].U[n], x];
%t c[n_] := CoefficientList[p[n], x]
%t TableForm[Flatten[Table[p[n], {n, 1, 10}]]]
%t Table[c[n], {n, 1, 12}]
%t Flatten[%] (* A203906 *)
%t TableForm[Table[c[n], {n, 1, 10}]]
%t Table[p[n] /. x -> -1, {n, 1, 16}] (* A166516 *)
%Y Cf. A202605, A166516.
%K tabf,sign
%O 1,4
%A _Clark Kimberling_, Jan 08 2012
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