cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A260140 Least prime p such that pi(p*n) = pi(q*n)^2 for some prime q, where pi(x) denotes the number of primes not exceeding x.

Original entry on oeis.org

2, 5, 19, 3187, 11, 2251, 12149, 19, 239, 23761, 61, 157, 8419, 10973, 1117, 9601, 58741, 37, 53359, 14533, 1063, 934811, 78487, 27647, 1249, 720221, 1616077, 30091, 5501, 131627, 2003, 67, 677, 1313843, 45413, 273943, 127241, 19661, 188317, 811, 33863, 17789, 109073, 602269, 125201, 6424897, 441647, 2512897, 2909, 836471
Offset: 1

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Author

Zhi-Wei Sun, Jul 17 2015

Keywords

Comments

Conjecture: a(n) exists for any n > 0. In general, for any integers a,b,c,n with a > 0 and n > 0, there are two elements x and y of the set {pi(p*n): p is prime} with a*x^2+b*x+c = y.
A supplement to the conjecture: For any integers b,c,n with b > 0 and n > 0, we have b*x+c = y for some elements x and y of the set {pi(p*n): p is prime}. - Zhi-Wei Sun, Aug 02 2015

Examples

			a(1) = 2 since pi(2*1) = 1^2 = pi(2*1)^2 with 2 prime.
a(4) = 3187 since pi(3187*4) = 1521 = 39^2 = pi(43*4)^2 with 43 and 3187 both prime.
a(72) = 25135867 since pi(25135867*72) = 89321401 = 9451^2 = pi(1367*72)^2 with 1367 and 25135867 both prime.
a(84) = 106788581 since pi(106788581*84) = 410224516 = 20254^2 = prime(2713*84)^2 with 2713 and 106788581 both prime.
		

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.

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=IntegerQ[Sqrt[n]]
    f[n_]:=PrimePi[n]
    Do[k=0;Label[bb];k=k+1;If[SQ[f[Prime[k]*n]]==False,Goto[bb]];Do[If[Sqrt[f[Prime[k]*n]]==f[Prime[j]*n],Goto[aa]];If[Sqrt[f[Prime[k]*n]]