A262462 Positive integers k with pi(k^3) a square, where pi(x) denotes the number of primes not exceeding x.

1, 2, 3, 14, 1122

Offset: 1

Author

Zhi-Wei Sun, Sep 23 2015

Keywords

• nonn
• more

Comments

Conjecture: (i) The Diophantine equation pi(x^2) = y^2 with x > 0 and y > 0 has infinitely many solutions.

(ii) The only solutions to the Diophantine equation pi(x^m) = y^n with {m,n} = {2,3}, x > 0 and y > 0 are as follows:

pi(89^2) = 10^3, pi(2^3) = 2^2, pi(3^3) = 3^2, pi(14^3) = 20^2 and pi(1122^3) = 8401^2.

(iii) For m > 1 and n > 1 with m + n > 5, the equation pi(x^m) = y^n with x > 0 and y > 0 has no integral solution.

The conjecture seems reasonable in view of the heuristic arguments.

Part (ii) of the conjecture implies that the only terms of the current sequence are 1, 2, 3, 14 and 1122.

Examples

			a(1) = 1 since pi(1^3) = 0^2.
a(2) = 2 since pi(2^3) = 2^2.
a(3) = 3 since pi(3^3) = 3^2.
a(4) = 14 since pi(14^3) = pi(2744) = 400 = 20^2.
a(5) = 1122 since pi(1122^3) = pi(1412467848) = 70576801 = 8401^2.
		

Crossrefs

Cf. A000290, A000578, A000720, A064523, A262408, A262409, A262443.

Programs

Mathematica

SQ[n_]:=IntegerQ[Sqrt[n]]
f[n_]:=PrimePi[n^3]
n=0;Do[If[SQ[f[k]],n=n+1;Print[n," ",k]],{k,1,1200}]
Select[Range[1200],IntegerQ[Sqrt[PrimePi[#^3]]]&] (* Harvey P. Dale, Aug 21 2024 *)