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A262700
Primes p such that pi(p^2)*pi(q^2) is a square for some prime q < p, where pi(x) denotes the number of primes not exceeding x.
2
5, 19, 31, 151, 691, 1181, 1489, 1511, 1601, 2579, 3037, 7297, 9661, 10993, 11699, 20407, 25657, 33937, 65099, 96419, 102911, 133157, 251789, 411841, 417271, 670729, 808211, 1179907, 1671277
OFFSET
1,1
COMMENTS
Conjecture: (i) The sequence has infinitely many terms.
(ii) The Diophantine equation pi(x^n)*pi(y^n) = z^n with n > 2 and x,y,z > 0 has no solution.
REFERENCES
Zhi-Wei Sun, Problems on combinatorial properties of primes, in: M. Kaneko, S. Kanemitsu and J. Liu (eds.), Number Theory: Plowing and Starring through High Wave Forms, Proc. 7th China-Japan Seminar (Fukuoka, Oct. 28 - Nov. 1, 2013), Ser. Number Theory Appl., Vol. 11, World Sci., Singapore, 2015, pp. 169-187.
LINKS
Zhi-Wei Sun, Problems on combinatorial properties of primes, arXiv:1402.6641 [math.NT], 2014.
EXAMPLE
a(1) = 5 since pi(5^2)*pi(3^2) = 9*4 = 6^2 with 5 and 3 both prime.
a(2) = 19 since pi(19^2)*pi(2^2) = 72*2 = 12^2 with 19 and 2 both prime.
a(21) = 102911 since pi(102911^2)*pi(919^2) = pi(10590673921)*pi(844561) = 480670430*67230 = 32315473008900 = 5684670^2 with 102911 and 919 both prime.
a(22) = 133157 since pi(133157^2)*pi(19^2) = pi(17730786649)*pi(361) = 786299168*72 = 56613540096 = 237936^2 with 133157 and 19 both prime.
a(23) = 251789 since pi(251789^2)*pi(10513^2) = pi(63397700521)*pi(110523169) = 2660789341*6331444 = 16846638708338404 = 129794602^2 with 251789 and 10513 both prime.
MATHEMATICA
f[n_]:=PrimePi[Prime[n]^2]
SQ[n_]:=IntegerQ[Sqrt[n]]
n=0; Do[Do[If[SQ[f[k]*f[m]], n=n+1; Print[n, " ", Prime[m]]; Goto[aa]], {k, 1, m-1}]; Label[aa]; Continue, {m, 2, 22200}]
KEYWORD
nonn,more
AUTHOR
Zhi-Wei Sun, Sep 27 2015
EXTENSIONS
a(24)-a(29) from Chai Wah Wu, Aug 21 2019
STATUS
approved