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.

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

1,2

Comments

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

Links

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

Formula

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

Example

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

Maple

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

Mathematica

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 *)

Prog

(PARI) 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

Crossrefs

A292624 counts the solutions with multiplicity.

Cf. A192787, A292581, A292581.

Sequence in context: A194473 A091913 A212285 * A026136 A026172 A026186

Adjacent sequences: A192786 A192787 A192788 * A192790 A192791 A192792

Keyword

nonn

Author

Charles R Greathouse IV, Jul 10 2011

Status

approved