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%I #29 Mar 25 2018 08:24:55
%S 1,2,3,14,1122
%N Positive integers k with pi(k^3) a square, where pi(x) denotes the number of primes not exceeding x.
%C Conjecture: (i) The Diophantine equation pi(x^2) = y^2 with x > 0 and y > 0 has infinitely many solutions.
%C (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:
%C 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.
%C (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.
%C The conjecture seems reasonable in view of the heuristic arguments.
%C Part (ii) of the conjecture implies that the only terms of the current sequence are 1, 2, 3, 14 and 1122.
%e a(1) = 1 since pi(1^3) = 0^2.
%e a(2) = 2 since pi(2^3) = 2^2.
%e a(3) = 3 since pi(3^3) = 3^2.
%e a(4) = 14 since pi(14^3) = pi(2744) = 400 = 20^2.
%e a(5) = 1122 since pi(1122^3) = pi(1412467848) = 70576801 = 8401^2.
%t SQ[n_]:=IntegerQ[Sqrt[n]]
%t f[n_]:=PrimePi[n^3]
%t n=0;Do[If[SQ[f[k]],n=n+1;Print[n," ",k]],{k,1,1200}]
%Y Cf. A000290, A000578, A000720, A064523, A262408, A262409, A262443.
%K nonn,more
%O 1,2
%A _Zhi-Wei Sun_, Sep 23 2015
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