A292624 -id:A292624 - cp's OEIS frontend

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

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

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),", "))

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

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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A192788 Erroneous version of A292624.

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

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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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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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