cp's OEIS Frontend

A200216 Danilov's sequence of x satisfying 0 < |x^3-y^2| < sqrt(x) with integer (x,y).

%I A200216 #98 May 13 2025 08:25:27
%S A200216 93844,322001299796379844,1114592308630995805123571151844,
%T A200216 3858108676488182444301031186675778188809844,
%U A200216 13354661111806898918013326915229994453818137920195953844
%N A200216 Danilov's sequence of x satisfying 0 < |x^3-y^2| < sqrt(x) with integer (x,y).
%C A200216 For y values see A200217.
%C A200216 For x^3-y^2 values see A200218.
%C A200216 The values (a(n)+1)/5 are perfect squares: for sqrt((a(n)+1)/5) see A200335.
%C A200216 This sequence is an infinite subset of A078933. - _Artur Jasinski_, Nov 27 2011
%D A200216 For references see A078933 and A179386.
%H A200216 L. V. Danilov, <a href="http://mi.mathnet.ru/mz5944">Letter to the Editor</a>, Mat. Zametki, 1984, Volume 36, Issue 3, Pages 457-458.
%H A200216 L. V. Danilov, <a href="http://dx.doi.org/10.1007/BF01141949">Letter to the editor</a>, Math. Notes 36 (3) (1984) 726.
%H A200216 R. D'Mello, <a href="http://arxiv.org/abs/1410.0078">Marshall Hall's Conjecture and Gaps Between Integer Points on Mordell Elliptic Curves</a>, arXiv preprint arXiv:1410.0078 [math.NT], 2014.
%F A200216 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
%F A200216 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
%F A200216 3125*A200218(n)^2 + 6750*A200218(n) + 729 = 2916*a(n). - _Artur Jasinski_, Nov 15 2011
%F A200216 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
%F A200216 a(n) = 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))), where f is golden ratio constant = (1+sqrt(5)/2). - _Artur Jasinski_, Nov 17 2011
%F A200216 a(n) = 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
%e A200216 |93844^3 - (round(sqrt(93844^3)))^2| < sqrt(93844).
%t A200216 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
%t A200216 (* second program follows the generator formula *)
%t A200216 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) *)
%t A200216 (* third program uses the Lucas numbers formula *)
%t A200216 Table[7/10 + (-1)^(n + 1) 3/5 LucasL[15*(2 n - 1)] +
%t A200216   1/20 LucasL[30*(2 n - 1)] , {n, 1, 10}] (* _Artur Jasinski_, Nov 18 2011 *)
%o A200216 (PARI) u = quadunit(20)^5
%o A200216 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_
%o A200216 (Python)
%o A200216 from sympy import lucas
%o A200216 def A200216(n): return (14+12*(lucas(k:=30*n-15) if n&1 else -lucas(k:=30*n-15))+lucas(k<<1))//20 # _Chai Wah Wu_, Jun 19 2024
%Y A200216 Cf. A078933, A179107, A179108, A179109, A179387, A179388, A199496.
%Y A200216 Cf. A200217, A200218, A200335.
%K A200216 nonn
%O A200216 1,1
%A A200216 _Artur Jasinski_, Nov 14 2011