A200216 Danilov's sequence of x satisfying 0 < |x^3-y^2| < sqrt(x) with integer (x,y).
93844, 322001299796379844, 1114592308630995805123571151844, 3858108676488182444301031186675778188809844, 13354661111806898918013326915229994453818137920195953844
Offset: 1
Keywords
Examples
|93844^3 - (round(sqrt(93844^3)))^2| < sqrt(93844).
Links
- L. V. Danilov, Letter to the Editor, Mat. Zametki, 1984, Volume 36, Issue 3, Pages 457-458.
- L. V. Danilov, Letter to the editor, Math. Notes 36 (3) (1984) 726.
- R. D'Mello, Marshall Hall's Conjecture and Gaps Between Integer Points on Mordell Elliptic Curves, arXiv preprint arXiv:1410.0078 [math.NT], 2014.
Crossrefs
Programs
-
Mathematica
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 *)
-
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
-
Python
from sympy import lucas 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
Formula
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
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
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
Comments