A204021 Triangle read by rows: row n shows the coefficients of the characteristic polynomial of the n-th principal submatrix of min(2i-1,2j-1) (A157454).

1, 1, -1, 2, -4, 1, 4, -12, 9, -1, 8, -32, 40, -16, 1, 16, -80, 140, -100, 25, -1, 32, -192, 432, -448, 210, -36, 1, 64, -448, 1232, -1680, 1176, -392, 49, -1, 128, -1024, 3328, -5632, 5280, -2688, 672, -64, 1, 256, -2304, 8640, -17472, 20592

Offset: 0

Author

Clark Kimberling, Jan 11 2012

Keywords

• tabl
• sign

Comments

Let p(n)=p(n,x) be the characteristic polynomial of the n-th principal submatrix. The zeros of p(n) are real, and they interlace the zeros of p(n+1). See A202605 and A204016 for guides to related sequences.

a(0)=1 by convention. - Philippe Deléham, Nov 17 2013

The n roots of the n-th polynomial are 1/(1+cos((2*k-1)*Pi/(2*n))) for k = 1..n. See my pdf in the link section for the proof. - Jianing Song, Dec 01 2023

Examples

			Top of the triangle:
  1
  1....-1
  2....-4.....1
  4....-12....9....-1
  8....-32....40...-16....1
  16...-80....140..-100...25....-1
  32...-192...432..-448...210...-36....1
  ...
-448=2*(-100)-2*140-(-32). - _Philippe Deléham_, Nov 17 2013
		

References

• (For references regarding interlacing roots, see A202605.)

Links

• Eric Weisstein's World of Mathematics, Morgan-Voyce polynomials
• Jianing Song, Roots of the characteristic polynomials

Crossrefs

Cf. A157454, A202605, A204016. A085478, A211957.

Programs

Mathematica

f[i_, j_] := Min[2 i - 1, 2 j - 1];
m[n_] := Table[f[i, j], {i, 1, n}, {j, 1, n}]
TableForm[m[6]] (* 6x6 principal submatrix *)
Flatten[Table[f[i, n + 1 - i],
  {n, 1, 15}, {i, 1, n}]]   (* A157454 *)
p[n_] := CharacteristicPolynomial[m[n], x];
c[n_] := CoefficientList[p[n], x]
TableForm[Flatten[Table[p[n], {n, 1, 10}]]]
Table[c[n], {n, 1, 12}]
Flatten[%]                  (* A204021 *)
TableForm[Table[c[n], {n, 1, 10}]]

Formula

From Peter Bala, May 01 2012: (Start)

The triangle appears to be a signed version of the row reverse of A211957.

If true, then for 0 <= k <= n-1, T(n,k) = (-1)^k*n/(n-k)*2^(n-k-1)*binomial(2*n-k-1,k) and Sum_{k = 0..n} T(n,k)*x^(n-k) = 1/2*(-1)^n*(b(2*n,-2*x) + 1)/b(n,-2*x), where b(n,x) := Sum_{k = 0..n} binomial(n+k,2*k)*x^k are the Morgan-Voyce polynomials of A085478.

Conjectural o.g.f.: t*(1-x-x^2*t)/(1-2*t*(1-x)+t^2*x^2) = (1-x)*t + (2-4*x+x^2)*t^2 + .... (End)

T(n,k)=2*T(n-1,k)-2*T(n-1,k-1)-T(n-2,k-2), T(0,0)=T(1,0)=1, T(1,1)=-1, T(n,k)=0 of k<0 or if k>n. - Philippe Deléham, Nov 17 2013