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A075259
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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.
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4
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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
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OFFSET
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1,3
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COMMENTS
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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
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REFERENCES
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R. K. Guy, Unsolved Problems in Theory of Numbers, Springer-Verlag, Third Edition, 2004, D11.
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LINKS
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EXAMPLE
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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.
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MATHEMATICA
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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
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CROSSREFS
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KEYWORD
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nice,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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