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A200216
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Danilov's sequence of x satisfying 0 < |x^3-y^2| < sqrt(x) with integer (x,y).
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12
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OFFSET
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1,1
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COMMENTS
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The values (a(n)+1)/5 are perfect squares: for sqrt((a(n)+1)/5) see A200335. - Jasinski
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REFERENCES
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LINKS
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FORMULA
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Conjecture: a(n) = 3461450947505*a(n-1) + 6440022564931157490*a(n-2) - 6440022564931157490*a(n-3) - 3461450947505*a(n-4) + a(n-5). - R. J. Mathar, Nov 15 2011
Conjecture: g.f. 4092*(1-z)/(5*(1+1860498*z+z^2)) - 7/(10*(z-1)) + 930249*(1-z)/(10*(1-3461452808002*z+z^2)). - R. J. Mathar, Nov 15 2011
125*y^2 *(54 + 50*x^3 - 25*y^2)=(9 - 6*x + 5*x^2)*(-9 + 15*x + 25*x^2)^2. - Artur Jasinski, Nov 16 2011
7/10 - (6/5)*(-1)^n * (1/2)*(f^(15*(2n-1))-(1/f)^(15(2n-1)))+(1/20)*(f^(30(2n-1))+(1/f)^(30(2n-1)))), where f is golden ratio constant = (1+sqrt(5)/2). - Artur Jasinski, Nov 17 2011
7/10+(3/5)*L(15*(2 n - 1))*(-1)^(n+1)+(1/20)*L(30*(2 n - 1)) where L(k) is the k-th Lucas number: A000204(n) or A000032(n+1). - Artur Jasinski, Nov 18 2011
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EXAMPLE
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93844^3 -(round(sqrt(93844^3))^2 < sqrt(93844).
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MATHEMATICA
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aa = {}; uu = 682 + 61*Sqrt[125]; Do[vv = Expand[uu^(2*n - 1)]; tt = ((-1)^n vv[[1]] + 57)/125; xx = (5^5*tt^2 - 3000*tt + 719); yy = Round[N[Sqrt[xx^3], 1000]]; dd = xx^3 - yy^2; AppendTo[aa, xx], {n, 1, 6}]; aa
(* second program follows the generator formula *)
data = Table[(7/10 - (6/5)*(-1)^n*(1/2)*(f^(15*(2 n - 1)) - (1/f)^(15 (2 n - 1))) + (1/20)*(f^(30 (2 n - 1)) + (1/f)^(30 (2 n - 1)))) /. f -> GoldenRatio, {n, 1, 6}]; data // FunctionExpand // ExpandAll // Simplify (* Bob Hanlon (hanlonr(AT)cox.net) *)
(* third program uses the Lucas numbers formula *)
Table[7/10 + (-1)^(n + 1) 3/5 LucasL[15*(2 n - 1)] +
1/20 LucasL[30*(2 n - 1)] , {n, 1, 10}] (* Artur Jasinski, Nov 18 2011 *)
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PROG
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(PARI) u = quadunit(20)^5
for(i=1, 6, v = u^(2*i-1); t = ((-1)^i * real(v) + 57) / 125; print(5^5*t^2 - 3000*t + 719) ) \\ Noam D. Elkies
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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