A192787 -id:A192787 - 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

A004194 Number of partitions of 1/n into 3 reciprocals of positive integers.

Original entry on oeis.org

3, 10, 21, 28, 36, 57, 42, 70, 79, 96, 62, 160, 59, 136, 196, 128, 73, 211, 80, 292, 245, 157, 93, 366, 156, 174, 230, 340, 106, 497, 90, 269, 322, 211, 453, 538, 85, 216, 378, 604, 121, 623, 104, 473, 648, 204, 135, 706, 227, 437, 387, 467, 125, 601, 561, 783, 385

Offset: 1

Scott Aaronson (philomath(AT)voicenet.com)

nonn

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

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

Justus Springer, Table of n, a(n) for n = 1..5000 (terms 1..100 from Robert G. Wilson)
K. S. Brown, Unit Fraction Partitions
Index entries for sequences related to Egyptian fractions

Cf. A227610, A226641, A226642, A192787, A226644, A226645, A226646.

a[n_] := Length@ Solve[ 1/n == 1/x + 1/y + 1/z && 1 <= x <= y <= z, {x, y, z}, Integers]; Array[a, 70] (* Allan C. Wechsler and Robert G. Wilson v, Aug 17 2013 *)

a(n)=my(t=1/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++))); s \\ Charles R Greathouse IV, Jun 12 2013

More terms from David W. Wilson, Aug 15 1996

A075245 x-value of the solution (x,y,z) to 4/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 A075246 and A075247.

Original entry on oeis.org

1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 9, 9, 9, 9, 10, 10, 10, 10, 11, 11, 11, 11, 12, 12, 12, 12, 13, 14, 13, 13, 14, 14, 14, 14, 15, 15, 15, 15, 16, 16, 16, 16, 17, 17, 17, 17, 18, 18, 18, 18, 19, 20, 19, 19, 20, 20, 20, 20, 21, 21, 21

Offset: 3

T. D. Noe, Sep 10 2002

See A073101 for more details.

a(n) = floor(n/4) + 1, at least up to n = 2000, except for some n = 8k+1 (k = 6, 9, 11, 14, 20, 21, 24, 29, 30, 35, 39, 41, 44, 45, 50, ...), where a(n) is one larger than a(n-1) and a(n+1). - M. F. Hasler, Jul 02 2022

			For n = 3, we have a(3) = 1 = x in 4/3 = 1/x + 1/y + 1/z with y = 4 and z = 12 which is the largest possible z: Indeed, x < y < z gives 4/3 < 3/x, so only x = 1 and 2 are possible, and then with y < z, 2/y > 4/3 - 1/x is impossible for x = 2 < y < 12/5 and for x = 1 < y < 6 only y = 4 gives a solution.

M. F. Hasler, Table of n, a(n) for n = 3..2000

Cf. A073101, A075246, A075247, A192787.

A075245:= proc () local t, n, a, b, t1, largex, largez; for n from 3 to 100 do t := 4/n; largez := 0; largex := 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 `and`(type(1/(t1-1/b), integer), a < b, b < 1/(t1-1/b)) then if largez < 1/(t1-1/b) then largez := 1/(t1-1/b); largex := a end if end if end do end do; lprint(n, largex) end do end proc; # [program derived from A192787] Patrick J. McNab, Aug 20 2014

For[xLst={}; yLst={}; zLst={}; n=3, n<=100, n++, cnt=0; xr=n/4; If[IntegerQ[xr], x=xr+1, x=Ceiling[xr]]; While[yr=1/(4/n-1/x); If[IntegerQ[yr], y=yr+1, y=Ceiling[yr]]; cnt==0&&y>x, While[zr=1/(4/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

apply( {A075245(n,c=1,a)=for(x=n\4+1, 3*n\4, my(t=4/n-1/x); for(y=max(1\t,x)+1, ceil(2/t)-1, t-1/y >= c && break; numerator(t-1/y)==1 && [c,a]=[t-1/y,x])); a}, [3..99]) \\ M. F. Hasler, Jul 02 2022

Conjecture: a(n) = floor(n/4) + d, with d = 1 except for some n = 8k+1 (k = 6, 9, 11, 14, 20, 21, 24, 29, 30, 35, 39, ...) where d = 2 . - M. F. Hasler, Jul 02 2022

A075246 y-value of the solution (x,y,z) to 4/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 A075245 and A075247.

Original entry on oeis.org

4, 3, 4, 7, 15, 7, 10, 16, 34, 13, 18, 29, 61, 21, 30, 46, 96, 31, 43, 67, 139, 43, 60, 92, 190, 57, 78, 121, 249, 73, 100, 154, 316, 91, 124, 191, 391, 111, 154, 232, 474, 133, 181, 277, 565, 157, 99, 326, 664, 183, 248, 379, 771, 211, 286, 436, 886, 241, 326

Offset: 3

T. D. Noe, Sep 10 2002

See A073101 for more details.

See A257840 for a variant that differs from a(89) on. - M. F. Hasler, Jul 03 2022

M. F. Hasler, Table of n, a(n) for n = 3..2000

Cf. A073101 (number of solutions), A075245 (x values), A075247 (z values), A192787 (number of solutions with x <= y <= z), A257840 (variant: lex earliest solution, not largest z).

A075246:=proc() local t, n,a,b,t1,largey,largez; for n from 3 to 100 do t:=4/n; largez:=0; largey:=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 and(type(1/(t1 - 1/b),integer),alargez then largez:=(1/(t1 - 1/b)); largey:=b; end if end if end do; end do; lprint(n, largey) end do; end proc; # [program derived from A192787] Patrick J. McNab, Aug 20 2014

For[xLst={}; yLst={}; zLst={}; n=3, n<=100, n++, cnt=0; xr=n/4; If[IntegerQ[xr], x=xr+1, x=Ceiling[xr]]; While[yr=1/(4/n-1/x); If[IntegerQ[yr], y=yr+1, y=Ceiling[yr]]; cnt==0&&y>x, While[zr=1/(4/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

apply( {A075246(n, c=1, a, t)=for(x=n\4+1, 3*n\4, for(y=max(1\t=4/n-1/x, x)+1, ceil(2/t)-1, t-1/y >= c && break; numerator(t-1/y)==1 && [c, a]=[t-1/y, y])); a}, [3..99]) \\ M. F. Hasler, Jul 03 2022

A075247 Largest possible z-value of an integer solution (x,y,z) to 4/n = 1/x + 1/y + 1/z satisfying 0 < x < y < z. The x and y components are in A075245 and A075246.

Original entry on oeis.org

12, 6, 20, 42, 210, 42, 90, 240, 1122, 156, 468, 812, 3660, 420, 510, 2070, 9120, 930, 1806, 4422, 19182, 1806, 2100, 8372, 35910, 3192, 9048, 14520, 61752, 5256, 9900, 23562, 99540, 8190, 22940, 36290, 152490, 12210, 6314, 53592, 224202, 17556

Offset: 3

T. D. Noe, Sep 10 2002

See A073101 for more details.

			For n = 6 we have a(n) = 42 the largest possible z in a solution of 4/n = 2/3 = 1/x + 1/y + 1/z with 0 < x < y < z in the integers. Indeed, from 1/x < 2/3 < 3/x we have 3/2 < x < 9/2. For x = 2 we get 2/y > 2/3 - 1/2 = 1/6 > 1/y <=> 6 < y < 12, and each of these y except y = 11 yields a solution, with z = 42, 24, 18, 15 and 12. There are no other possible z values: x = 3 gives 2/y > 1/3 <=> y < 6 and indeed y = 4 gives a solution with z = 12, no solution for y = 5; finally, x = 4 gives 2/y > 5/12 <=> y < 24/5, impossible with y > x.

M. F. Hasler, Table of n, a(n) for n = 3..2000

Cf. A075245 (x values), A075246 (y values), A073101 (number of solutions), A192787 (solutions with x <= y <= z).

A075247:= proc () local t, n, a, b, t1, largey, largez; for n from 3 to 100 do t := 4/n; largez := 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 `and`(type(1/(t1-1/b), integer), a < b, b < 1/(t1-1/b)) then if largez < 1/(t1-1/b) then largez := 1/(t1-1/b) end if end if end do end do; lprint(n, largez) end do end proc; # [program derived from A192787] Patrick J. McNab, Aug 20 2014

For[xLst={}; yLst={}; zLst={}; n=3, n<=100, n++, cnt=0; xr=n/4; If[IntegerQ[xr], x=xr+1, x=Ceiling[xr]]; While[yr=1/(4/n-1/x); If[IntegerQ[yr], y=yr+1, y=Ceiling[yr]]; cnt==0&&y>x, While[zr=1/(4/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

apply( {A075247(n, c=1, t)=for(x=n\4+1, 3*n\4, for(y=max(1\t=4/n-1/x, x)+1, ceil(2/t)-1, t-1/y >= c && break; numerator(t-1/y)==1 && c=t-1/y)); 1/c}, [3..99]) \\ M. F. Hasler, Jul 02 2022

A226645 Number of ways to express 6/n as Egyptian fractions in just three terms; i.e., 6/n = 1/x + 1/y + 1/z satisfying 1<=x<=y<=z.

Original entry on oeis.org

0, 1, 1, 3, 3, 3, 1, 6, 8, 10, 7, 10, 1, 9, 12, 20, 8, 21, 2, 21, 17, 16, 11, 28, 7, 11, 26, 33, 9, 36, 3, 31, 25, 24, 60, 57, 2, 11, 20, 68, 10, 42, 6, 35, 81, 23, 15, 70, 10, 37, 25, 51, 14, 79, 33, 76, 32, 30, 20, 96, 2, 17, 86, 65, 48, 62, 9, 50, 42, 138, 35, 160, 2, 18, 53, 51, 52, 59, 8, 142, 89, 34, 23, 136, 37, 24, 33, 140, 23, 196, 30, 46, 37, 32, 75, 128, 5, 43, 103, 98

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.

Daniel Leary, Table of n, a(n) for n = 1..1000

Cf. A226640, A004194, A226642, A192787, A226644, A226646.

a[n_] := Length@ Solve[ 6/n == 1/x + 1/y + 1/z && 1 <= x <= y <= z, {x, y, z}, Integers]; Array[a, 70]

A292581 Number of solutions to 4/n = 1/x + 1/y + 1/z in positive integers.

Original entry on oeis.org

0, 3, 12, 10, 12, 39, 36, 46, 30, 63, 48, 106, 24, 93, 216, 148, 24, 141, 60, 196, 162, 141, 120, 304, 60, 111, 162, 232, 42, 459, 108, 394, 174, 141, 372, 442, 54, 183, 420, 538, 42, 489, 78, 352, 540, 243, 198, 904, 102, 303, 294, 334, 78, 513

Offset: 1

Hugo Pfoertner, Sep 20 2017

nonn

Corrected version of A192786.
The Erdos-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.
After a(2) = 3, the values shown are all composite. [Jonathan Vos Post, Jul 17 2011]

			a(3)=12 because 4/3 can be expressed in 12 ways:
  4/3 =  1/1  + 1/4  + 1/12
  4/3 =  1/1  + 1/6  + 1/6
  4/3 =  1/1  + 1/12 + 1/4
  4/3 =  1/2  + 1/2  + 1/3
  4/3 =  1/2  + 1/3  + 1/2
  4/3 =  1/3  + 1/2  + 1/2
  4/3 =  1/4  + 1/1  + 1/12
  4/3 =  1/4  + 1/12 + 1/1
  4/3 =  1/6  + 1/1  + 1/6
  4/3 =  1/6  + 1/6  + 1/1
  4/3 =  1/12 + 1/1  + 1/4
  4/3 =  1/12 + 1/4  + 1/1
a(4) = 10 because 4/4 = 1 can be expressed in 10 ways:
  4/4=  1/2 + 1/3 + 1/6
  4/4=  1/2 + 1/4 + 1/4
  4/4=  1/2 + 1/6 + 1/3
  4/4=  1/3 + 1/2 + 1/6
  4/4=  1/3 + 1/3 + 1/3
  4/4=  1/3 + 1/6 + 1/2
  4/4=  1/4 + 1/2 + 1/4
  4/4=  1/4 + 1/4 + 1/2
  4/4=  1/6 + 1/2 + 1/3
  4/4=  1/6 + 1/3 + 1/2

Hugo Pfoertner, Table of n, a(n) for n = 1..1000
Christian Elsholtz and Terence Tao, Counting the number of solutions to the Erdos-Straus equation on unit fractions, arXiv:1107.1010 [math.NT], 2011-2015.
Allan Swett, The Erdos-Strauss Conjecture, 1999.
R. C. Vaughan, On a problem of Erdős, Straus and Schinzel, Mathematika 17 (1970), pp. 193-198.

For more references and links see A192787.

Cf. A073101, A192786, A292624, A337432.

checkmult[a_, b_, c_] := If[Denominator[c] == 1, If[a == b && a == c && b == c, Return[1], If[a != b && a != c && b != c, Return[6], Return[3]]], Return[0]];
a292581[n_] := Module[{t, t1, s, a, b, c, q = Quotient}, t = 4/n; s = 0; For[a = q[1, t]+1, a <= q[3, t], a++, t1 = t - 1/a; For[b = Max[q[1, t1] + 1, a], b <= q[2, t1], b++, c = 1/(t1 - 1/b); s += checkmult[a, b, c]]]; Return[s]];
Reap[For[n=1, n <= 54, n++, Print[n, " ", an = a292581[n]]; Sow[an]]][[2, 1]] (* Jean-François Alcover, Dec 02 2018, adapted from PARI *)

\\ modified version of code by Charles R Greathouse IV in A192786
checkmult (a,b,c) =
{
  if(denominator(c)==1,
     if(a==b && a==c && b==c,
        return(1),
        if(a!=b && a!=c && b!=c,
           return(6),
           return(3)
          )
       ),
     return(0)
     )
}
a292581(n) =
{
  local(t, t1, s, a, b, c);
  t = 4/n;
  s = 0;
  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);
          s+=checkmult(a,b,c);
         )
      );
     return(s);
}
for (n=1,54,print1(a292581(n),", "))

A226641 Number of ways to express 2/n as Egyptian fractions in just three terms; i.e., 2/n = 1/x + 1/y + 1/z satisfying 1<=x<=y<=z.

Original entry on oeis.org

1, 3, 8, 10, 12, 21, 17, 28, 26, 36, 25, 57, 20, 42, 81, 70, 25, 79, 32, 96, 86, 62, 42, 160, 53, 59, 89, 136, 33, 196, 37, 128, 103, 73, 185, 211, 32, 80, 160, 292, 40, 245, 40, 157, 235, 93, 60, 366, 85, 156, 147, 174, 42, 230, 223, 340, 143, 106, 76, 497, 34, 90, 331, 269, 206, 322, 50, 211, 175, 453, 72, 538, 37, 85, 332, 216, 260, 378, 69, 604, 167, 121, 79, 623, 204, 104, 203, 473, 59, 648, 253, 204, 166, 135, 318, 706, 46, 227, 427, 437

Offset: 1

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

nonn

Daniel Leary, Table of n, a(n) for n = 1..1000
Index entries for sequences related to Egyptian fractions

Cf. A227611, A004194, A226642, A192787, A226644, A226645, A226646.

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

a[n_] := Length@ Solve[ 2/n == 1/x + 1/y + 1/z && 1 <= x <= y <= z, {x, y, z}, Integers]; Array[a, 70]

A226642 Number of ways to express 3/n as Egyptian fractions in just three terms; i.e., 3/n = 1/x + 1/y + 1/z satisfying 1<=x<=y<=z.

Original entry on oeis.org

1, 3, 3, 6, 10, 10, 9, 20, 21, 21, 16, 28, 11, 33, 36, 31, 24, 57, 11, 68, 42, 35, 23, 70, 37, 51, 79, 76, 30, 96, 17, 65, 62, 50, 138, 160, 18, 51, 59, 142, 34, 136, 24, 140, 196, 46, 32, 128, 43, 98, 73, 111, 46, 211, 111, 192, 80, 63, 46, 292, 24, 81, 245

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. A075785, A004194, A226641, A192787, A226644, A226645, A226646.

f[n_] := Length@ Solve[ 3/n == 1/x + 1/y + 1/z && 1 <= 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]

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.

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A004194 Number of partitions of 1/n into 3 reciprocals of positive integers.

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A075245 x-value of the solution (x,y,z) to 4/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 A075246 and A075247.

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A075246 y-value of the solution (x,y,z) to 4/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 A075245 and A075247.

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A075247 Largest possible z-value of an integer solution (x,y,z) to 4/n = 1/x + 1/y + 1/z satisfying 0 < x < y < z. The x and y components are in A075245 and A075246.

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A226645 Number of ways to express 6/n as Egyptian fractions in just three terms; i.e., 6/n = 1/x + 1/y + 1/z satisfying 1<=x<=y<=z.

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A292581 Number of solutions to 4/n = 1/x + 1/y + 1/z in positive integers.

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