A064523 -id:A064523 - cp's OEIS frontend

A115835 Square root of pi(A064523(n)).

0, 2, 3, 5, 689, 2015, 2783, 3592, 3919, 3941, 5156, 5902, 6063, 8054, 15683, 16310, 28239, 29901, 40328, 55221, 60246, 74982, 98883, 129023, 312400, 328667, 351586, 815526, 911097, 1018921, 1054125

Offset: 1

Zak Seidov, Feb 01 2006

Cf. A000720, A064517, A064523.

Sqrt@PrimePi[{ (* list of entries in A064523 *) }^2] (* Robert G. Wilson v, Feb 03 2006 *)

a(20) from Robert G. Wilson v, Feb 03 2006
Added missing terms 28239, 29901, 40328 and a(24)-a(27) from Donovan Johnson, May 30 2010
a(28)-a(31) from Chai Wah Wu, Aug 27 2019

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

Original entry on oeis.org

1, 2, 3, 14, 1122

Offset: 1

Zhi-Wei Sun, Sep 23 2015

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.

			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.

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

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 *)

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A115835 Square root of pi(A064523(n)).

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