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A267082
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Coefficient of x^4 in the minimal polynomial of the continued fraction [1^n,2^(1/3),1,1,...], where 1^n means n ones.
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7
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0, 6, 456, 6240, 131238, 2238780, 41011296, 730283034, 13143304440, 235581102912, 4229156006790, 75876624195564, 1361636473680576, 24432987781993530, 438436202143461288, 7867390833380267040, 141174789462751501926, 2533277512666920359964
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OFFSET
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0,2
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COMMENTS
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See A265762 for a guide to related sequences.
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LINKS
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FORMULA
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a(n) = 13*a(n-1) + 104*a(n-2) - 260*a(n-3) - 260*a(n-4) + 104*a(n-5) + 13*a(n-6) - a(n-7) for n > 8.
G.f.: (6 (x + 62 x^2 - 114 x^3 + 823 x^4 - 182 x^5 - 28 x^6 + 2 x^7))/(1 - 14 x - 90 x^2 + 350 x^3 - 90 x^4 - 14 x^5 + x^6).
a(n) = 14*a(n-1) + 90*a(n-2) - 350*a(n-3) + 90*a(n-4) + 14*a(n-5) - a(n-6) for n > 7.
G.f.: 6*x*(1 + 62*x - 114*x^2 + 823*x^3 - 182*x^4 - 28*x^5 + 2*x^6)/((1 - 3*x + x^2)*(1 + 7*x + x^2)*(1 - 18*x + x^2)).
(End)
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EXAMPLE
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Let u = 2^(1/3), and let p(n,x) be the minimal polynomial of the number given by the n-th continued fraction:
[u,1,1,1,...] has p(0,x) = -5 - 15 x - 6 x^2 - 9 x^3 + 3 x^5 + x^6, so that a(0) = 0.
[1,u,1,1,1,...] has p(1,x) = -11 + 45 x - 66 x^2 + 35 x^3 + 6 x^4 - 15 x^5 + 5 x^6, so that a(1) = 6;
[1,1,u,1,1,1...] has p(2,x) = 131 - 633 x + 1110 x^2 - 969 x^3 + 456 x^4 - 111 x^5 + 11 x^6, so that a(2) = 456.
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MATHEMATICA
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u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {2^(1/3)}, {{1}}];
f[n_] := FromContinuedFraction[t[n]];
t = Table[MinimalPolynomial[f[n], x], {n, 0, 30}]
Coefficient[t, x, 0]; (* A267078 *)
Coefficient[t, x, 1]; (* A267079 *)
Coefficient[t, x, 2]; (* A267080 *)
Coefficient[t, x, 3]; (* A267081 *)
Coefficient[t, x, 4]; (* A267082 *)
Coefficient[t, x, 5]; (* A267083 *)
Coefficient[t, x, 6]; (* A266527 *)
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PROG
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(PARI) concat([0], 6*Vec((1 + 62*x - 114*x^2 + 823*x^3 - 182*x^4 - 28*x^5 + 2*x^6)/((1 - 3*x + x^2)*(1 + 7*x + x^2)*(1 - 18*x + x^2)) + O(x^30))) \\ Andrew Howroyd, Mar 07 2018
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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