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Search: id:A075248
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| A075248 |
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Number of solutions (x,y,z) to 5/n = 1/x + 1/y + 1/z satisfying 0 < x < y < z. |
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+0
5
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| 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
(list; graph; listen)
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OFFSET
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2,3
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COMMENT
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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.
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LINKS
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T. D. Noe, Table of n, a(n) for n=2..1000
Eric Weisstein's World of Mathematics, Egyptian Fraction
Ron Knott Egyptian Fractions
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EXAMPLE
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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.
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MATHEMATICA
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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
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CROSSREFS
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Cf. A073101, A075249, A075250, A075251.
Sequence in context: A060082 A102225 A145236 this_sequence A128325 A111528 A144303
Adjacent sequences: A075245 A075246 A075247 this_sequence A075249 A075250 A075251
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KEYWORD
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nice,nonn
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AUTHOR
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T. D. Noe (noe(AT)sspectra.com), Sep 10 2002
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