cp's OEIS Frontend

A075248 Number of solutions (x,y,z) to 5/n = 1/x + 1/y + 1/z satisfying 0 < x < y < z.

%I A075248 #21 Feb 16 2025 08:32:47
%S A075248 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,
%T A075248 20,45,5,24,35,23,36,53,10,21,52,62,6,62,12,73,69,16,11,92,38,84,34,
%U A075248 50,11,77,56,80,45,38,34,142,6,23,96,53,53,67,15,66,70,124,12,148,21,57
%N A075248 Number of solutions (x,y,z) to 5/n = 1/x + 1/y + 1/z satisfying 0 < x < y < z.
%C A075248 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 Erdős and Straus.
%H A075248 T. D. Noe, <a href="/A075248/b075248.txt">Table of n, a(n) for n = 2..1000</a>
%H A075248 Ron Knott, <a href="http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fractions/egyptian.html">Egyptian Fractions</a>.
%H A075248 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/EgyptianFraction.html">Egyptian Fraction</a>.
%e A075248 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.
%t A075248 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
%t A075248 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 *)
%Y A075248 Cf. A073101, A075249, A075250, A075251.
%K A075248 nice,nonn
%O A075248 2,3
%A A075248 _T. D. Noe_, Sep 10 2002