A179813 Values x for record minima of the positive distance d between the fifteenth power of a positive integer x and the square of an integer y such that d = x^15 - y^2 (x <> k^2 and y <> k^15).

2, 3, 5, 6, 7, 8, 10, 11, 17, 18, 23, 24, 27, 35, 45, 55, 56, 76, 78, 84, 111, 114, 115, 117, 118, 139, 164, 172, 175, 176, 179, 183, 188, 190, 193, 305, 316, 377, 395, 461, 466, 483, 485, 654, 747, 868, 877, 931, 1045, 1434, 1822, 2199, 2645, 2754, 3171, 3961

Offset: 1

Author

Artur Jasinski, Jul 28 2010

Keywords

• nonn

Comments

Distance d is equal to 0 when x = k^2 and y = k^15.

For x values see A179813.

For y values see A179814.

Conjecture: For any positive number x >= A179813(n), the distance d between the fifteenth power of x and the square of any y (such that x <> k^2 and y <> k^15) can't be less than A179812(n).

Crossrefs

Cf. A179107, A179108, A179109, A179386, A179387, A179388, A179407, A179408, A179784, A179785, A179786, A179790, A179791, A179792, A179793, A179794, A179795, A179798, A179799, A179800, A179812, A179813, A179814.

Programs

Mathematica

d = 15; max = 1000; vecd = Table[10^100, {n, 1, max}]; vecx = Table[10^100, {n, 1, max}]; vecy = Table[10^100, {n, 1, max}]; len = 1; Do[m = Floor[(n^d)^(1/2)]; k = n^d - m^2; If[k != 0, ile = 0; Do[If[vecd[[z]] < k, ile = ile + 1], {z, 1, len}]; len = ile + 1; vecd[[len]] = k; vecx[[len]] = n; vecy[[len]] = m], {n, 1, 10000000}]; dd = {}; xx = {}; yy = {}; Do[AppendTo[dd, vecd[[n]]]; AppendTo[xx, vecx[[n]]]; AppendTo[yy, vecy[[n]]], {n, 1, len}]; xx