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A202673
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Array: row n shows the coefficients of the characteristic polynomial of the n-th principal submatrix of the symmetric matrix A115263 based on (1,2,3,4,...); by antidiagonals.
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3
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1, -1, 1, -6, 1, 1, -12, 20, -1, 1, -18, 75, -50, 1, 1, -24, 166, -328, 105, -1, 1, -30, 293, -1050, 1134, -196, 1, 1, -36, 456, -2432, 5140, -3312, 336, -1, 1, -42, 655, -4690, 15471, -20814, 8514, -540, 1, 1, -48, 890, -8040, 36771, -80584
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OFFSET
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1,4
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COMMENTS
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Let p(n)=p(n,x) be the characteristic polynomial of the n-th principal submatrix of A115262 (when A115262 is formatted as a square matrix). The zeros of p(n) are positive, and they interlace the zeros of p(n+1).
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LINKS
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EXAMPLE
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The 1st principal submatrix (ps) of A115263 is {{1}} (using Mathematica matrix notation), with p(1)=1-x and zero-set {1}.
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The 2nd ps is {{1,2},{2,5}}, with p(2)=1-6x+x^2 and zero-set {0.171..., 5.828...}.
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The 3rd ps is {{1,2,3},{2,5,8},{3,8,14}}, with p(3)=1-12x+20x^2-x^3 and zero-set {0.099..., 0.516..., 19.383...}.
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Top of the array:
1....-1
1....-6.....1
1...-12....20.....-1
1...-18....75....-50....1
1...-24...166...-328..105..-1
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MATHEMATICA
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U[n_] := NestList[Most[Prepend[#, 0]] &, #, Length[#] - 1] &[Table[k, {k, 1, n}]];
L[n_] := Transpose[U[n]];
F[n_] := CharacteristicPolynomial[L[n].U[n], x];
c[n_] := CoefficientList[F[n], x]
TableForm[Flatten[Table[F[n], {n, 1, 10}]]]
Table[c[n], {n, 1, 12}]
Flatten[%]
TableForm[Table[c[n], {n, 1, 10}]]
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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