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

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

Views

Author

Hugo Pfoertner, Sep 20 2017

Keywords

Comments

Corrected version of A192788.

Examples

			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

References

  • For references and links see A192787.

Links

Crossrefs

Cf. A073101, A192787, A192789, A292581.

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

Formula

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

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

Views

Author

Max Alekseyev, May 19 2017

Keywords

Comments

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.

Links

Crossrefs

Cf. A073101, A075245, A075246, A075247, A075248, A292581, A192787, A292624, A192789.
Showing 1-3 of 3 results.

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