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A068203
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Chebyshev T-polynomials T(n,15) with Diophantine property.
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7
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1, 15, 449, 13455, 403201, 12082575, 362074049, 10850138895, 325142092801, 9743412645135, 291977237261249, 8749573705192335, 262195233918508801, 7857107443850071695, 235451028081583642049, 7055673735003659189775, 211434761022028192051201
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OFFSET
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0,2
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COMMENTS
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Let (x_n, y_n) be n-th solution to the Pell equation x^2 = 14*y^2 + 1. Sequence gives {x_n}.
Except for the first term, positive values of x (or y) satisfying x^2 - 30xy + y^2 + 224 = 0. - Colin Barker, Feb 24 2014
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LINKS
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FORMULA
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x_n + y_n*sqrt(14) = (x_1 + y_1*sqrt(14))^n.
a(n) = (-15/2-2*sqrt(14))*(-1/(-15-4*sqrt(14)))^n/(-15-4*sqrt(14))+(2*sqrt(14)-15/2)*(-1/(-15+4*sqrt(14)))^n/(-15+4*sqrt(14)). Recurrence: a(n) = 30*a(n-1)-a(n-2). G.f.: (1-15*x)/(1-30*x+x^2). - Vladeta Jovovic, Mar 25 2002
a(n) = T(n, 15)= (S(n, 30)-S(n-2, 30))/2 = S(n, 30)-15*S(n-1, 30) with T(n, x), resp. S(n, x), Chebyshev's polynomials of the first, resp.second, kind. See A053120 and A049310. S(n, 30)=A097313(n). - Wolfdieter Lang, Aug 31 2004
a(n) = sum(((-1)^k)*(n/(2*(n-k)))*binomial(n-k, k)*(2*15)^(n-2*k), k=0..floor(n/2)), n>=1. - Wolfdieter Lang, Aug 31 2004
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MAPLE
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Digits := 1000: q := seq(floor(evalf(((15+4*sqrt(14))^n+(15-4*sqrt(14))^n)/2)+0.1), n=1..30);
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MATHEMATICA
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a[0] = 1; a[1] = 15; a[n_] := 30a[n-1] - a[n-2]; Table[a[n], {n, 0, 16}] (* or *) LinearRecurrence[{30, -1}, {1, 15}, 17] (* Indranil Ghosh, Feb 18 2017 *)
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PROG
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(Sage) [lucas_number2(n, 30, 1)/2 for n in range(0, 15)] # Zerinvary Lajos, Jun 27 2008
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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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EXTENSIONS
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STATUS
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approved
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