cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

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

Views

Author

Charles R Greathouse IV, Jul 10 2011

Keywords

Comments

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.

Examples

			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).

Links

Crossrefs

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).

Programs

  • Maple
    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
  • Mathematica
    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 *)
  • PARI
    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

Extensions

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

Views

Author

Hugo Pfoertner, Sep 20 2017

Keywords

Comments

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]

Examples

			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

Links

Crossrefs

For more references and links see A192787.
Cf. A073101, A192786, A292624, A337432.

Programs

  • Mathematica
    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 *)
  • 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),", "))

A338195 Numbers k such that 3/k = 1/x + 1/y has a solution with 0 < x < y < k.

Original entry on oeis.org

18, 20, 35, 36, 40, 45, 54, 56, 60, 70, 72, 77, 80, 84, 90, 100, 104, 105, 108, 110, 112, 120, 126, 135, 140, 143, 144, 154, 160, 162, 168, 170, 175, 176, 180, 182, 189, 198, 200, 208, 209, 210, 216, 220, 221, 224, 225, 231, 234, 240, 245, 252, 260, 264, 266, 270
Offset: 1

Views

Author

Rainer Rosenthal and Hugo Pfoertner, Oct 15 2020

Keywords

Crossrefs

Cf. A073101, A337432.
Cf. A004611 (Numbers n such that 3/n cannot be written as the sum of 2 unit fractions).

Programs

  • PARI
    for(k=3,270, forstep(y=k-1,1,-1,my(x=3/k-1/y); if(numerator(x)==1&&1/x
Showing 1-3 of 3 results.

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