A075248 -id:A075248 - cp's OEIS frontend

A073101 Number of integer solutions (x,y,z) to 4/n = 1/x + 1/y + 1/z satisfying 0 < x < y < z.

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

Offset: 1

Robert G. Wilson v, Aug 18 2002

nonn

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

			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.

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
Wikipedia, Erdős-Straus conjecture

Cf. A075245, A075246, A075247, A075248.

Cf. A192787 (# distinct solutions with x <= y <= z).

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

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

(* 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 *)

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

Edited by T. D. Noe, Sep 10 2002

Extended to offset 1 with a(1) = 0 by M. F. Hasler, May 16 2015

A192787 Number of distinct solutions of 4/n = 1/a + 1/b + 1/c in positive integers satisfying 1 <= a <= b <= c.

Original entry on oeis.org

0, 1, 3, 3, 2, 8, 7, 10, 6, 12, 9, 21, 4, 17, 39, 28, 4, 26, 11, 36, 29, 25, 21, 57, 10, 20, 29, 42, 7, 81, 19, 70, 31, 25, 65, 79, 9, 32, 73, 96, 7, 86, 14, 62, 93, 42, 34, 160, 18, 53, 52, 59, 13, 89, 98, 136, 41, 33, 27, 196, 11, 37, 155, 128, 49, 103, 17, 73, 55, 185, 40, 211, 7, 32, 129, 80, 97, 160, 37, 292

Offset: 1

Charles R Greathouse IV, Jul 10 2011

nonn

The Erdős-Straus conjecture is that a(n) > 0 for n > 1. Swett verified the conjecture for n < 10^14.
Vaughan shows that the number of n < x with a(n) = 0 is at most x exp(-c * (log x)^(2/3)) for some c > 0.
See A073101 for the 4/n conjecture due to Erdős and Straus.

			a(1) = 0, since 4/1 = 4 cannot be expressed as the sum of three reciprocals.
a(2) = 1 because 4/2 = 1/1 + 1/2 + 1/2, and there are no other solutions.
a(3) = 3 since 4/3 = 1 + 1/4 + 1/12 = 1 + 1/6 + 1/6 = 1/2 + 1/2 + 1/3.
a(4) = 3 = A002966(3).

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
Christian Elsholtz, Terence Tao, Counting the number of solutions to the Erdős-Straus equation in unit fractions, arXiv:1107.1010 [math.NT], 2010-2015.
Maria Monks, Ameya Velingker, On the Erdős-Straus conjecture: Properties of solutions to its underlying diophantine equation. See also.
Alan Swett, The Erdos-Strauss Conjecture, 1999.
R. C. Vaughan, On a problem of Erdős, Straus and Schinzel, Mathematika 17 (1970), pp. 193-198.
Wikipedia, Erdős-Straus conjecture.
Konstantine Zelator, An ancient Egyptian problem: the diophantine equation 4/n=1/x+1/y+1/z, n>or=2, arXiv:0912.2458 [math.GM], 2009.

A292581 counts the solutions with multiplicity. A073101 counts solutions with a, b, and c distinct.
Cf. A292581, A292624, A192789, A075245, A075246, A075247, A075248.
Cf. A004194, A226641, A226642, A226644, A226645, A226646.
Cf. A337432 (solutions with minimal c).

A192787 := proc(n) local t,a,b,t1,count; t:= 4/n; count:= 0; for a from floor(1/t)+1 to floor(3/t) do t1:= t - 1/a; for b from max(a,floor(1/t1)+1) to floor(2/t1) do if type( 1/(t1 - 1/b),integer) then count:= count+1; end if end do end do; count; end proc; # Robert Israel, Feb 19 2013

f[n_] := Length@ Solve[ 4/n == 1/x + 1/y + 1/z && 1 <= x <= y <= z, {x, y, z}, Integers]; Array[f, 70] (* Allan C. Wechsler and Robert G. Wilson v, Jul 19 2013 *)

a(n, show=0)=my(t=4/n, t1, s, c); for(a=1\t+1, 3\t, t1=t-1/a; for(b=max(1\t1+1, a), 2\t1, c=1/(t1-1/b); if(denominator(c)==1&&c>=b, s++; show&&print("4/",n," = 1/",a," + 1/",b," + 1/",c)))); s \\ variant with print(...) added by Robert Munafo, Feb 19 2013, both combined through option "show" by M. F. Hasler, Jul 02 2022

Corrected at the suggestion of Allan C. Wechsler by Charles R Greathouse IV, Feb 19 2013

Examples and cross-references added by M. F. Hasler, Feb 19 2013

A227610 Number of ways 1/n can be expressed as the sum of three distinct unit fractions: 1/n = 1/x + 1/y + 1/z satisfying 0 < x < y < z.

Original entry on oeis.org

1, 6, 15, 22, 30, 45, 36, 62, 69, 84, 56, 142, 53, 124, 178, 118, 67, 191, 74, 274, 227, 145, 87, 342, 146, 162, 216, 322, 100, 461, 84, 257, 304, 199, 435, 508, 79, 204, 360, 580, 115, 587, 98, 455, 618, 192, 129, 676, 217, 417, 369, 449, 119, 573, 543, 759, 367, 240, 166, 1236, 102, 261, 857, 428, 568, 717, 115, 537, 460, 1018, 155, 1126, 112, 276, 839

Offset: 1

Robert G. Wilson v, Jul 17 2013

nonn

See A073101 for the 4/n conjecture due to Erdős and Straus.

			a(1)=1 because 1 = 1/2 + 1/3 + 1/6;
a(2)=6 because 1/2 = 1/3 + 1/7 + 1/42 = 1/3 + 1/8 + 1/24 = 1/3 + 1/9 + 1/18 = 1/3 + 1/10 + 1/15 = 1/4 + 1/5 + 1/20 = 1/4 + 1/6 + 1/12;
a(3)=15 because 1/3 = 1/x + 1/y + 1/z presented as {x,y,z}: {4,13,156}, {4,14,84}, {4,15,60}, {4,16,48}, {4,18,36}, {4,20,30}, {4,21,28}, {5,8,120}, {5,9,45}, {5,10,30}, {5,12,20}, {6,7,42}, {6,8,24}, {6,9,18}, {6,10,15}; etc.

Jud McCranie, Table of n, a(n) for n = 1..500
Christian Elsholtz, Sums Of k Unit Fractions, Trans. Amer. Math. Soc. 353 (2001), 3209-3227.
David Eppstein, Algorithms for Egyptian Fractions
David Eppstein, Ten Algorithms for Egyptian Fractions, Wolfram Library Archive.
Ron Knott, Egyptian Fractions
Eric Weisstein's World of Mathematics, Egyptian Fraction
Index entries for sequences related to Egyptian fractions

Cf. A002966, A073546.
Cf. A227611 (2/n), A075785 (3/n), A073101 (4/n), A075248 (5/n), A227612.
Cf. A347566, A347569.

f[n_] := Length@ Solve[1/n == 1/x + 1/y + 1/z && 0 < x < y < z, {x, y, z}, Integers]; Array[f, 70]

A226644 Number of ways to express 5/n as Egyptian fractions in just three terms; i.e., 5/n = 1/x + 1/y + 1/z satisfying 1<=x<=y<=z.

Original entry on oeis.org

0, 1, 2, 4, 3, 4, 4, 7, 12, 10, 3, 17, 6, 21, 21, 12, 6, 26, 13, 28, 22, 18, 9, 61, 36, 18, 24, 48, 22, 57, 5, 27, 38, 26, 42, 60, 11, 24, 56, 70, 6, 71, 13, 79, 79, 19, 12, 99, 41, 96, 38, 55, 12, 84, 62, 86, 50, 41, 36, 160, 6, 26, 104, 57, 59, 76, 16, 71, 74, 136, 12, 158, 22, 60, 196, 52, 65, 103, 25, 128, 46, 30, 15, 224, 73, 32, 58, 141, 38, 211, 71, 67, 59, 41, 80, 151, 24, 97, 222, 292

Offset: 1

Allan C. Wechsler and Robert G. Wilson v, Aug 17 2013

nonn

See A073101 for the 4/n conjecture due to Erdős and Straus.

Cf. A075248, A004194, A226641, A226642, A192787, A226645, A226646.

f[n_] := Length@ Solve[ 5/n == 1/x + 1/y + 1/z && 1 <= x <= y <= z, {x, y, z}, Integers]; Array[f, 70]

A349083 The number of three-term Egyptian fractions of rational numbers x/y, 0 < x/y < 1, ordered as below. The sequence is the number of (p,q,r) such that x/y = 1/p + 1/q + 1/r where p, q, and r are integers with p < q < r.

Original entry on oeis.org

6, 15, 5, 22, 6, 3, 30, 9, 7, 2, 45, 15, 6, 5, 1, 36, 14, 6, 5, 3, 1, 62, 22, 16, 6, 5, 3, 2, 69, 21, 15, 4, 9, 5, 2, 1, 84, 30, 15, 9, 6, 7, 2, 2, 1, 56, 22, 13, 7, 3, 5, 2, 0, 0, 0, 142, 45, 22, 15, 12, 6, 9, 5, 3, 1, 2, 53, 17, 8, 4, 5, 1, 6, 3, 1, 1, 1, 0, 124, 36, 27, 14, 18, 6, 6, 5, 2, 3, 1, 1, 0

Offset: 1

Jud McCranie, Nov 09 2021

The sequence are the terms in a triangle, where the rows correspond to the denominator of the rational number (starting with row 2, column 1) and the columns correspond to the numerators:
            x = 1   2   3   4   5   Rationals x/y:
Row 1: (y=2)    6                   1/2
Row 2: (y=3)   15,  5               1/3, 2/3
Row 3: (y=4)   22,  6,  3           1/4, 2/4, 3/4
Row 4: (y=5)   30,  9,  7,  2       1/5, 2/5, 3/5, 4/5
Row 5: (y=6)   45, 15,  6,  5,  1   1/6, 2/6, 3/6, 4/6, 5/6
Alternatively, order the rational numbers, x/y, 0 < x/y < 1, in this order: 1/2, 1/3, 2/3, 1/4, 2/4, 3/4, 1/5, 2/5, ... The numerators of the n-th rational number are A002260(n) and the denominators are A003057(n).

			The sixth rational number is 3/4;
  3/4 = 1/2 + 1/5 + 1/20
      = 1/2 + 1/6 + 1/12
      = 1/3 + 1/4 + 1/5,
so a(6)=3.

Jud McCranie, Table of n, a(n) for n = 1..990

Cf. A002260, A003057, A349082.

Columns x=1..5: A227610, A227611, A075785, A073101, A075248.

Efrac3(x,y)=sum(p=if(y%x,y\x,y\x+1),3*y\x, my(N=x/y-1/p); sum(q=max(if(numerator(N)==1,1\N+1,1\N),p+1),2\N, my(M=N-1/q,r=1/M); type(r)=="t_INT" && qCharles R Greathouse IV, Nov 09 2021

A227611 Number of ways 2/n can be expressed as the sum of three distinct unit fractions: 2/n = 1/x + 1/y + 1/z with 0 < x < y < z.

Original entry on oeis.org

0, 1, 5, 6, 9, 15, 14, 22, 21, 30, 22, 45, 17, 36, 72, 62, 22, 69, 29, 84, 77, 56, 39, 142, 48, 53, 82, 124, 30, 178, 34, 118, 94, 67, 176, 191, 29, 74, 151, 274, 37, 227, 37, 145, 220, 87, 57, 342, 80, 146, 138, 162, 39, 216, 214, 322, 134, 100, 73, 461, 31, 84, 316, 257, 197, 304, 47, 199, 166, 435, 69, 508, 34, 79, 317

Offset: 1

Robert G. Wilson v, Jul 17 2013

nonn

See A073101 for the 4/n conjecture due to Erdős and Straus.

Cf. A002966, A073546.

Cf. A227610 (1/n), A075785 (3/n), A073101 (4/n), A075248 (5/n), A227612.

f[n_] := Length@ Solve[2/n == 1/x + 1/y + 1/z && 0 < x < y < z, {x, y, z}, Integers]; Array[f, 75]

A227612 Table read by antidiagonals: Number of ways m/n can be expressed as the sum of three distinct unit fractions, i.e., m/n = 1/x + 1/y + 1/z satisfying 0 < x < y < z and read by antidiagonals.

Original entry on oeis.org

1, 0, 6, 0, 1, 15, 0, 1, 5, 22, 0, 0, 1, 6, 30, 0, 0, 1, 3, 9, 45, 0, 0, 1, 1, 7, 15, 36, 0, 0, 0, 2, 2, 6, 14, 62, 0, 0, 0, 1, 1, 5, 6, 22, 69, 0, 0, 0, 1, 1, 1, 5, 16, 21, 84, 0, 0, 0, 0, 1, 1, 3, 6, 15, 30, 56, 0, 0, 0, 0, 1, 4, 1, 5, 4, 15, 22, 142, 0, 0, 0, 0, 0, 1, 1, 3, 9, 9, 13, 45, 53

Offset: 1

Robert G. Wilson v, Jul 17 2013

The main diagonal is 1, 1, 1, 1, 1, 1, 1, ..., ; i.e., 1 = 1/2 + 1/3 + 1/6.

			  m\n| 1  2   3   4   5   6   7   8   9  10  11   12  13   14   15
  ---+------------------------------------------------------------
   1 | 1  6  15  22  30  45  36  62  69  84  56  142  53  124  178  A227610
   2 | 0  1   5   6   9  15  14  22  21  30  22   45  17   36   72  A227611
   3 | 0  1   1   3   7   6   6  16  15  15  13   22   8   27   30  A075785
   4 | 0  0   1   1   2   5   5   6   4   9   7   15   4   14   33  A073101
   5 | 0  0   1   2   1   1   3   5   9   6   3   12   5   18   15  A075248
   6 | 0  0   0   1   1   1   1   3   5   7   5    6   1    6    9  n/a
   7 | 0  0   0   1   1   4   1   2   2   2   2    9   6    6    7  n/a
   8 | 0  0   0   0   1   1   1   1   1   2   0    5   3    5   15  n/a
   9 | 0  0   0   0   0   1   1   3   1   1   0    3   1    2    7  n/a
  10 | 0  0   0   0   0   1   0   2   2   1   0    1   1    3    5  n/a
.
Antidiagonals are {1}, {0, 6}, {0, 1, 15}, {0, 1, 5, 22}, {0, 0, 1, 6, 30}, {0, 0, 1, 3, 9, 45}, ...

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
Christian Elsholtz, Sums Of k Unit Fractions, Trans. Amer. Math. Soc. 353 (2001), 3209-3227.
David Eppstein, Algorithms for Egyptian Fractions
David Eppstein, Ten Algorithms for Egyptian Fractions, Wolfram Library Archive.
Ron Knott Egyptian Fractions
Oakland University, The Erdős Number Project
Eric Weisstein's World of Mathematics, Egyptian Fraction
Index entries for sequences related to Egyptian fractions

Cf. A002966, A073546, A227610 (1/n), A227611 (2/n), A075785 (3/n), A073101 (4/n), A075248 (5/n).

f[m_, n_] := Length@ Solve[m/n == 1/x + 1/y + 1/z && 0 < x < y < z, {x, y, z}, Integers]; Table[ f[n, m - n + 1], {m, 12}, {n, m, 1, -1}]

A075249 x-value of the solution (x,y,z) to 5/n = 1/x + 1/y + 1/z satisfying 0 < x < y < z and having the largest z-value. The y and z components are in A075250 and A075251.

Original entry on oeis.org

1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 13, 14, 13, 13, 13, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 16, 16, 16, 16, 16, 17, 17, 17, 17, 17, 18

Offset: 3

T. D. Noe, Sep 10 2002

See A075248 for more details.

Cf. A075248, A075250, A075251.

For[xLst={}; yLst={}; zLst={}; n=3, n<=100, n++, cnt=0; xr=n/5; If[IntegerQ[xr], x=xr+1, x=Ceiling[xr]]; While[yr=1/(5/n-1/x); If[IntegerQ[yr], y=yr+1, y=Ceiling[yr]]; cnt==0&&y>x, While[zr=1/(5/n-1/x-1/y); cnt==0&&zr>y, If[IntegerQ[zr], z=zr; cnt++; AppendTo[xLst, x]; AppendTo[yLst, y]; AppendTo[zLst, z]]; y++ ]; x++ ]]; xLst

Is a(n) = A047252(n-3)-n+4 ? - Ralf Stephan, Feb 24 2004

A075250 y-value of the solution (x,y,z) to 5/n = 1/x + 1/y + 1/z satisfying 0 < x < y < z and having the largest z-value. The x and z components are in A075249 and A075251.

Original entry on oeis.org

2, 5, 3, 4, 5, 9, 19, 7, 9, 13, 20, 43, 13, 17, 23, 37, 77, 21, 27, 37, 58, 121, 31, 40, 55, 85, 175, 43, 56, 75, 116, 239, 57, 73, 99, 153, 313, 73, 93, 127, 194, 397, 91, 116, 157, 241, 491, 111, 141, 191, 292, 595, 133, 169, 229, 349, 709, 157, 95, 269, 410, 833

Offset: 3

T. D. Noe, Sep 10 2002

See A075248 for more details.

Cf. A075248, A075249, A075251.

For[xLst={}; yLst={}; zLst={}; n=3, n<=100, n++, cnt=0; xr=n/5; If[IntegerQ[xr], x=xr+1, x=Ceiling[xr]]; While[yr=1/(5/n-1/x); If[IntegerQ[yr], y=yr+1, y=Ceiling[yr]]; cnt==0&&y>x, While[zr=1/(5/n-1/x-1/y); cnt==0&&zr>y, If[IntegerQ[zr], z=zr; cnt++; AppendTo[xLst, x]; AppendTo[yLst, y]; AppendTo[zLst, z]]; y++ ]; x++ ]]; yLst

A075251 z-value of the solution (x,y,z) to 5/n = 1/x + 1/y + 1/z satisfying 0 < x < y < z and having the largest z-value. The x and y components are in A075249 and A075250.

Original entry on oeis.org

6, 20, 6, 12, 70, 72, 342, 42, 99, 156, 780, 1806, 156, 272, 1564, 1332, 5852, 420, 945, 4070, 6670, 14520, 930, 1560, 2970, 7140, 30450, 1806, 1736, 16800, 26796, 56882, 3192, 5256, 29304, 23256, 97656, 5256, 11439, 16002, 75078, 157212, 8190, 13340

Offset: 3

T. D. Noe, Sep 10 2002

See A075248 for more details.

Cf. A075248, A075249, A075250.

For[xLst={}; yLst={}; zLst={}; n=3, n<=100, n++, cnt=0; xr=n/5; If[IntegerQ[xr], x=xr+1, x=Ceiling[xr]]; While[yr=1/(5/n-1/x); If[IntegerQ[yr], y=yr+1, y=Ceiling[yr]]; cnt==0&&y>x, While[zr=1/(5/n-1/x-1/y); cnt==0&&zr>y, If[IntegerQ[zr], z=zr; cnt++; AppendTo[xLst, x]; AppendTo[yLst, y]; AppendTo[zLst, z]]; y++ ]; x++ ]]; zLst

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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.

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A192787 Number of distinct solutions of 4/n = 1/a + 1/b + 1/c in positive integers satisfying 1 <= a <= b <= c.

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A227610 Number of ways 1/n can be expressed as the sum of three distinct unit fractions: 1/n = 1/x + 1/y + 1/z satisfying 0 < x < y < z.

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A226644 Number of ways to express 5/n as Egyptian fractions in just three terms; i.e., 5/n = 1/x + 1/y + 1/z satisfying 1<=x<=y<=z.

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A349083 The number of three-term Egyptian fractions of rational numbers x/y, 0 < x/y < 1, ordered as below. The sequence is the number of (p,q,r) such that x/y = 1/p + 1/q + 1/r where p, q, and r are integers with p < q < r.

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A227611 Number of ways 2/n can be expressed as the sum of three distinct unit fractions: 2/n = 1/x + 1/y + 1/z with 0 < x < y < z.

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A227612 Table read by antidiagonals: Number of ways m/n can be expressed as the sum of three distinct unit fractions, i.e., m/n = 1/x + 1/y + 1/z satisfying 0 < x < y < z and read by antidiagonals.

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A075249 x-value of the solution (x,y,z) to 5/n = 1/x + 1/y + 1/z satisfying 0 < x < y < z and having the largest z-value. The y and z components are in A075250 and A075251.

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