

A075248


Number of solutions (x,y,z) to 5/n = 1/x + 1/y + 1/z satisfying 0 < x < y < z.


11



0, 1, 2, 1, 1, 3, 5, 9, 6, 3, 12, 5, 18, 15, 10, 5, 21, 11, 22, 18, 15, 8, 55, 30, 15, 20, 43, 20, 45, 5, 24, 35, 23, 36, 53, 10, 21, 52, 62, 6, 62, 12, 73, 69, 16, 11, 92, 38, 84, 34, 50, 11, 77, 56, 80, 45, 38, 34, 142, 6, 23, 96, 53, 53, 67, 15, 66, 70, 124, 12, 148, 21, 57
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OFFSET

2,3


COMMENTS

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 (A075249, A075250, A075251). See A073101 for the 4/n conjecture due to Erdos and Straus.


LINKS

T. D. Noe, Table of n, a(n) for n=2..1000
Ron Knott Egyptian Fractions
Eric Weisstein's World of Mathematics, Egyptian Fraction


EXAMPLE

a(4)=2 because there are two solutions: 5/4 = 1/1+1/5+1/20 and 5/4 = 1/1+1/6+1/12.


MATHEMATICA

m = 5; For[lst = {}; n = 2, n <= 100, n++, 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; cnt++; (*Print[n, " ", x, " ", y, " ", z]*)]]]; AppendTo[lst, cnt]]; lst
f[n_] := Length@ Solve[5/n == 1/x + 1/y + 1/z && 0 < x < y < z, {x, y, z}, Integers]; Array[f, 74] (* Robert G. Wilson v, Jul 17 2013 *)


CROSSREFS

Cf. A073101, A075249, A075250, A075251.
Sequence in context: A241188 A145236 A162206 * A128325 A111528 A144303
Adjacent sequences: A075245 A075246 A075247 * A075249 A075250 A075251


KEYWORD

nice,nonn


AUTHOR

T. D. Noe, Sep 10 2002


STATUS

approved



