A075259 Number of solutions (x,y,z) to 3/(2n+1) = 1/x + 1/y + 1/z satisfying 0 < x < y < z and odd x, y, z.

0, 1, 2, 1, 1, 5, 2, 1, 3, 5, 1, 12, 8, 3, 3, 5, 14, 8, 6, 4, 7, 20, 1, 9, 6, 3, 22, 11, 3, 11, 31, 24, 5, 10, 3, 11, 16, 20, 6, 23, 2, 35, 7, 3, 35, 15, 25, 16, 47, 8, 12, 54, 3, 9, 8, 4, 42, 41, 22, 11, 8, 25, 8, 15, 5, 61, 92, 3, 7, 16, 28, 47, 37, 7, 10, 40, 23, 13, 11, 29, 11, 75, 3

Offset: 1

Author

T. D. Noe, Sep 10 2002

Keywords

• nice
• nonn

Comments

N. J. A. Sloane and R. H. Hardin conjecture a(n) > 0 for n > 1. All of the solutions can be printed by removing the comment symbols from the Mathematica program. For the solution (x,y,z) having the largest z value, see (A075260, A075261, A075262). See A073101 for the 4/n conjecture due to Erdős and Straus.

The conjecture was proved by Thomas Hagedorn and Gary Mulkey. - T. D. Noe, Jan 03 2005

Examples

			a(3)=2 because there are two solutions: 3/7 = 1/3+1/11+1/231 and 3/7 = 1/3+1/15+1/35.
		

References

• R. K. Guy, Unsolved Problems in Theory of Numbers, Springer-Verlag, Third Edition, 2004, D11.

Links

• T. D. Noe, Table of n, a(n) for n=1..499
• Thomas R. Hagedorn, A proof of a conjecture on Egyptian fractions, Amer. Math Monthly, 107 (2000), 62-63.
• Stan Wagon, Problem of the Week 848: An Odd Egyptian Puzzle
• Eric Weisstein's World of Mathematics, Egyptian Fraction

Crossrefs

Cf. A073101, A075260, A075261, A075262.

Programs

Mathematica

m = 3; For[lst = {}; n = 3, n <= 200, n = n + 2, cnt = 0; xr = n/m; If[IntegerQ[xr], xMin = xr + 1, xMin = Ceiling[xr]]; If[IntegerQ[3xr], xMax = 3xr - 1, xMax = Floor[3xr]]; For[x = xMin, x <= xMax, x++, yr = 1/(m/n - 1/x); If[IntegerQ[yr], yMin = yr + 1, yMin = Ceiling[yr]]; If[IntegerQ[2yr], yMax = 2yr + 1, yMax = Ceiling[2yr]]; For[y = yMin, y <= yMax, y++, zr = 1/(m/n - 1/x - 1/y); If[y > x && zr > y && IntegerQ[zr], z = zr; If[OddQ[x y z], cnt++;(*Print[n, " ", x, " ", y, " ", z]*)]]]]; AppendTo[lst, cnt]]; lst

Extensions

More terms from T. D. Noe, Oct 15 2002