A192787 -id:A192787 - cp's OEIS frontend

A226646 Number of ways to express m/n as Egyptian fractions in just three terms, that is, m/n = 1/x + 1/y + 1/z satisfying 1 <= x <= y <= z and read by antidiagonals.

3, 1, 10, 1, 3, 21, 0, 3, 8, 28, 0, 1, 3, 10, 36, 0, 1, 3, 6, 12, 57, 0, 1, 2, 3, 10, 21, 42, 0, 0, 1, 4, 2, 10, 17, 70, 0, 0, 1, 3, 3, 8, 9, 28, 79, 0, 0, 0, 1, 3, 4, 7, 20, 26, 96, 0, 0, 1, 1, 2, 3, 4, 10, 21, 36, 62, 0, 0, 0, 1, 1, 7, 1, 7, 6, 21, 25, 160, 0, 0, 0, 1, 0, 3, 3, 6, 12, 12, 16, 57, 59

Offset: 1

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

See A073101 for the 4/n conjecture due to Erdös and Straus.
The first upper diagonal is 10, 8, 6, 2, 4, 1, 2, 1, 2, 0, 3, 0, 0, 1, 0, 0, 1, 0, 1, 0,... .
The main diagonal is: 3, 3, 3, 3, 3, 3, ... since 1 = 1/2 + 1/3 + 1/6 = 1/2 + 1/4 + 1/4 = 1/3 + 1/3 + 1/3. See A002966(3).
The first lower diagonal is 1, 3, 3, 4, 3, 7, 3, 5, 4, 6, 3, 10, 3, 6, 6, 6, 3, 9, 3, 9, ... .
The antidiagonal sum is 3, 11, 25, 39, 50, 79, 79, 104, 131, 157, 140, 229, 169, 220, 295, 282, ... .

			../n
m/ 1...2...3...4...5...6...7...8...9..10..11...12..13...14...15 =Allocation nbr.
.1 3..10..21..28..36..57..42..70..79..96..62..160..59..136..196 A004194
.2 1...3...8..10..12..21..17..28..26..36..25...57..20...42...81 A226641
.3 1...3...3...6..10..10...9..20..21..21..16...28..11...33...36 A226642
.4 0...1...3...3...2...8...7..10...6..12...9...21...4...17...39 A192787
.5 0...1...2...4...3...4...4...7..12..10...3...17...6...21...21 A226644
.6 0...1...1...3...3...3...1...6...8..10...7...10...1....9...12 A226645
.7 0...0...1...1...2...7...3...2...3...5...2...13...8...10....9 n/a
.8 0...0...0...1...1...3...3...3...1...2...0....8...3....7...19 n/a
.9 0...0...1...1...0...3...2...5...3...2...0....6...2....4...10 n/a
10 0...0...0...1...1...2...0...4...4...3...0....4...1....4....8 n/a
Triangle (by antidiagonals):
  {3},
  {1, 10},
  {1, 3, 21},
  {0, 3, 8, 28},
  {0, 1, 3, 10, 36},
  {0, 1, 3, 6, 12, 57},
  ...

Christian Elsholtz, Sums Of k Unit Fractions
David Eppstein, Algorithms for Egyptian Fractions
David Eppstein, Ten Algorithms for Egyptian Fractions
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. A227612, A226640, A226641, A226642, A192787, A226644, A226645.

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

A292624 Number of solutions to 4/p = 1/x + 1/y + 1/z in positive integers, where p is the n-th prime.

Original entry on oeis.org

3, 12, 12, 36, 48, 24, 24, 60, 120, 42, 108, 54, 42, 78, 198, 78, 156, 66, 96, 234, 42, 216, 156, 60, 48, 96, 156, 144, 90, 78, 192, 186, 102, 210, 108, 180, 144, 138, 384, 156, 276, 102, 396, 36, 138, 246, 174, 342, 216, 120, 114, 630, 48, 300

Offset: 1

Hugo Pfoertner, Sep 20 2017

nonn

Corrected version of A192788.

			a(3) = 12 because 4/(3rd prime) = 4/5 can be expressed in the following 12 ways:
  4/5 =  1/2  + 1/4  + 1/20
  4/5 =  1/2  + 1/5  + 1/10
  4/5 =  1/2  + 1/10 + 1/5
  4/5 =  1/2  + 1/20 + 1/4
  4/5 =  1/4  + 1/2  + 1/20
  4/5 =  1/4  + 1/20 + 1/2
  4/5 =  1/5  + 1/2  + 1/10
  4/5 =  1/5  + 1/10 + 1/2
  4/5 =  1/10 + 1/2  + 1/5
  4/5 =  1/10 + 1/5  + 1/2
  4/5 =  1/20 + 1/2  + 1/4
  4/5 =  1/20 + 1/4  + 1/2

For references and links see A192787.

Hugo Pfoertner, Table of n, a(n) for n = 1..1000
Christian Elsholtz, Terence Tao, Counting the number of solutions to the Erdos-Straus equation on unit fractions, arXiv:1107.1010 [math.NT], 2011-2015.
Christian Elsholtz, Terence Tao, Counting the number of solutions to the Erdos-Straus equation on unit fractions, Journal of the Australian Mathematical Society, 94(1), 50-105, 2013. doi:10.1017/S1446788712000468.

Cf. A073101, A192787, A192789, A292581.

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[Prime[n]]]; Sow[an]]][[2, 1]] (* Jean-François Alcover, Dec 02 2018, adapted from PARI *)

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)
     )
}
a292624(n) =
{
  local(t, t1, s, a, b, c);
  t = 4/prime(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(a292624(n),", "))

a(n) = A292581(A000040(n)).

A192789 Number of distinct solutions of 4/p = 1/a + 1/b + 1/c in positive integers satisfying 1<=a<=b<=c where p is the n-th prime.

Original entry on oeis.org

1, 3, 2, 7, 9, 4, 4, 11, 21, 7, 19, 9, 7, 14, 34, 13, 27, 11, 17, 40, 7, 37, 27, 10, 8, 16, 27, 25, 15, 13, 33, 32, 17, 36, 18, 31, 24, 24, 65, 26, 47, 17, 67, 6, 23, 42, 30, 58, 37, 20, 19, 106, 8, 51, 19, 71, 28, 48, 31, 17, 33, 34, 40, 79, 16, 34, 38, 21, 39, 32, 19, 110, 52, 33, 39, 86, 30, 29, 23, 15, 81, 16, 93, 19

Offset: 1

Charles R Greathouse IV, Jul 10 2011

nonn

The Erdos-Straus conjecture is equivalent to the conjecture that a(n) > 0 for all n.

			a(1) = 1 because 4/prime(1) = 1/1 + 1/2 + 1/2.

Jean-François Alcover, Table of n, a(n) for n = 1..1000
Christian Elsholtz, Terence Tao, Counting the number of solutions to the Erdos-Straus equation on unit fractions, Aug 2, 2015, arXiv:1107.1010 [math.NT], 2011-2015.
Elsholtz, C., Tao, T. Counting the number of solutions to the Erdos-Straus equation on unit fractions. Journal of the Australian Mathematical Society, 94(1), 50-105, 2013.
doi:10.1017/S1446788712000468

A292624 counts the solutions with multiplicity.

Cf. A192787, A292581, A292581.

a:= n-> A192787(ithprime(n)):
seq(a(n), n=1..70);

a[n_] := a[n] = Module[{a, b, c, r}, r = Reduce[1 <= a <= b <= c && 4/Prime[n] == 1/a + 1/b + 1/c, {a, b, c}, Integers]; Which[Head[r] === Or, Length[r], Head[r] === And, 1, r === False, 0, True, Print["error: ", r]]];
Table[an = a[n]; Print["a(", n, ") = ", an]; an, {n, 1, 84}] (* Jean-François Alcover, Nov 22 2017 *)

a(n)=my(t=4/prime(n), t1, s, c); for(a=1\t+1, 3\t, t1=t-1/a; for(b=1\t1+1, 2\t1, c=1/(t1-1/b); if(denominator(c)==1&&c>=b, s++))); s

a(n) = A192787(prime(n)). - Michel Marcus, Aug 20 2014

A337432 a(n) is the least value of z such that 4/n = 1/x + 1/y + 1/z with 0 < x <= y <= z has at least one solution.

Original entry on oeis.org

2, 3, 3, 10, 6, 14, 6, 9, 10, 33, 9, 52, 14, 12, 12, 102, 18, 57, 15, 21, 22, 138, 18, 50, 26, 27, 21, 232, 24, 248, 24, 33, 34, 30, 27, 370, 38, 39, 30, 164, 35, 258, 33, 36, 46, 329, 36, 98, 50, 51, 39, 742, 54, 44, 42, 57, 58, 885, 45, 549, 62, 56, 48, 60, 66, 603

Offset: 2

Hugo Pfoertner, Oct 13 2020

See A073101 and A192787 for the history of the problem, references, and links.

			a(6)=6 because it is the least denominator z in the A192787(6)=8 solutions
  [x, y, z]: [2, 7, 42], [2, 8, 24], [2, 9, 18], [2, 10, 15], [2, 12, 12],
  [3, 4, 12], [3, 6, 6], [4, 4, 6];
a(13)=52 because the minimum of z in the A192787(13)=4 solutions is 52:
  [4, 18, 468], [4, 20, 130], [4, 26, 52], [5, 10, 130].

Hugo Pfoertner, Table of n, a(n) for n = 2..10000 (first 576 terms from Robert Israel)

Cf. A073101, A192787, A292581.

f:= proc(n) local z,x,y;
  for z from floor(n/4)+1 do
    for x from floor(n*z/(4*z-n))+1 to z do
      y:= n*x*z/(4*x*z-n*x-n*z);
      if y::posint and y >= x and y <= z then return z fi
  od od
end proc:
map(f, [$2..100]); # Robert Israel, Oct 14 2020

a[n_] := For[z = Floor[n/4] + 1, True, z++, For[x = Floor[n(z/(4z - n))] + 1, x <= z, x++, y = n x z/(4 x z - n x - n z); If[IntegerQ[y] && x <= y <= z, Print[z]; Return [z]]]];
a /@ Range[2, 100] (* Jean-François Alcover, Oct 23 2020, after Robert Israel *)

a337432(n)={my(target=4/n,a,b,c,m=oo);for(a=1\target+1,3\target,my(t=target-1/a);for(b=max(1\t+1,a),2\t,c=1/(t-1/b);if(denominator(c)==1,m=min(m,max(a,max(b,c))))));m};
for(k=2,67,print1(a337432(k),", "))

A287116 Nonsquare integers that cannot be represented in the form 4M-d, where (a*b)|M and d|(a+b) for some positive integers a,b.

Original entry on oeis.org

288, 336, 4545

Offset: 1

Max Alekseyev, May 19 2017

If there are no more terms, the Erdos-Straus conjecture would follow.
This sequence together with the squares (A000290) and E(4) form a partition of the nonnegative integers. That is, this sequence gives nonsquare non-elements of E(4) (see Dubickas and Novikas, 2012).
There are no other terms below 2*10^9.

A. Dubickas, A. Novikas, On integers expressible by some special linear form, Acta Math. Univ. Comenianae LXXXI:2 (2012), 203-209.
Wikipedia, Erdos-Straus conjecture.

Cf. A073101, A075245, A075246, A075247, A075248, A292581, A192787, A292624, A192789.

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A226646 Number of ways to express m/n as Egyptian fractions in just three terms, that is, m/n = 1/x + 1/y + 1/z satisfying 1 <= x <= y <= z and read by antidiagonals.

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A292624 Number of solutions to 4/p = 1/x + 1/y + 1/z in positive integers, where p is the n-th prime.

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A192789 Number of distinct solutions of 4/p = 1/a + 1/b + 1/c in positive integers satisfying 1<=a<=b<=c where p is the n-th prime.

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A337432 a(n) is the least value of z such that 4/n = 1/x + 1/y + 1/z with 0 < x <= y <= z has at least one solution.

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Crossrefs

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Maple

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PARI

A287116 Nonsquare integers that cannot be represented in the form 4M-d, where (a*b)|M and d|(a+b) for some positive integers a,b.

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Keywords

Comments

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