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 A073101 Number of integer solutions (x,y,z) to 4/n = 1/x + 1/y + 1/z satisfying 0 < x < y < z. 31
 0, 0, 1, 1, 2, 5, 5, 6, 4, 9, 7, 15, 4, 14, 33, 22, 4, 21, 9, 30, 25, 22, 19, 45, 10, 17, 25, 36, 7, 72, 17, 62, 27, 22, 59, 69, 9, 29, 67, 84, 7, 77, 12, 56, 87, 39, 32, 142, 16, 48, 46, 53, 13, 82, 92, 124, 37, 30, 25, 178, 11, 34, 147, 118, 49, 94, 15, 67, 51, 176, 38, 191, 7 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,5 COMMENTS In 1948 Erdős and Straus conjectured that for any positive integer n >= 2 the equation 4/n = 1/x + 1/y + 1/z has a solution with positive integers x, y and z (without the additional requirement 0 < x < y < z). 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 (A075245, A075246, A075247). See A075248 for Sierpiński's conjecture for 5/n. See (A257839, A257840, A257841) for the lexicographically smallest solutions, and A257843 for the differences between these and those with largest z-value. - M. F. Hasler, May 16 2015 LINKS T. D. Noe, Table of n, a(n) for n = 1..1000, (corrected by Peter Luschny, Jan 19 2019) Christian Elsholtz, Sums Of k Unit Fractions, Trans. Amer. Math. Soc. 353 (2001), 3209-3227. David Eppstein, Algorithms for Egyptian Fractions Paul Erdős, Az 1/z_1 + 1/z_2 + ... + 1/z_n = a/b egyenlet egész számú megoldásairól, (On a Diophantine equation), Mat. Lapok, 1:192-210, 1050. Math. Rev. 13:208b. Ron Knott, Egyptian Fractions Eric Weisstein's World of Mathematics, Egyptian Fraction EXAMPLE a(5)=2 because there are two solutions: 4/5 = 1/2 + 1/4 + 1/20 and 4/5 = 1/2 + 1/5 + 1/10. MAPLE A:= proc(n)    local x, t, p, q, ds, zs, ys, js, tot, j; tot:= 0; for x from 1+floor(n/4) to ceil(3*n/4)-1 do     t:= 4/n - 1/x;     p:= numer(t);     q:= denom(t);     ds:= convert(select(d -> (d < q) and d + q mod p = 0,           numtheory:-divisors(q^2)), list);     ys:= map(d -> (d+q)/p, ds);     zs:= map(d -> (q^2/d+q)/p, ds);     js:= select(j -> ys[j] > x, [\$1..nops(ds)]);     tot:= tot + nops(js); od; tot; end proc: seq(A(n), n=2..100); # Robert Israel, Aug 22 2014 MATHEMATICA (* download Egypt.m from D. Eppstein's site and put it into MyOwn directory underneath Mathematica\AddOns\StandardPackages *) Needs["MyOwn`Egypt`"]; Table[ Length[ EgyptianFraction[4/n, Method -> Lexicographic, MaxTerms -> 3, MinTerms -> 3, Duplicates -> Disallow, OutputFormat -> Plain]], {n, 5, 80}] m = 4; 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[4/n == 1/x + 1/y + 1/z && 0 < x < y < z, {x, y, z}, Integers]; Array[f, 72, 2] (* Robert G. Wilson v, Jul 17 2013 *) PROG (Haskell) import Data.Ratio ((%), numerator, denominator) a073101 n = length [(x, y) |    x <- [n `div` 4 + 1 .. 3 * n `div` 4],   let y' = recip \$ 4%n - 1%x,    y <- [floor y' + 1 .. floor (2*y') + 1], let z' = recip \$ 4%n - 1%x - 1%y,    denominator z' == 1 && numerator z' > y && y > x] -- Reinhard Zumkeller, Jan 03 2011 (PARI) A073101(n)=sum(c=n\4+1, n*3\4, sum(b=c+1, ceil(2/(t=4/n-1/c))-1, numerator(t-1/b)==1)) \\ M. F. Hasler, May 15 2015 CROSSREFS Cf. A075245, A075246, A075247, A075248. Sequence in context: A094236 A205444 A270705 * A235526 A130851 A130856 Adjacent sequences:  A073098 A073099 A073100 * A073102 A073103 A073104 KEYWORD nonn AUTHOR Robert G. Wilson v, Aug 18 2002 EXTENSIONS Edited by T. D. Noe, Sep 10 2002 Extended to offset 1 with a(1)=0 by M. F. Hasler, May 16 2015 STATUS approved

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