OFFSET
1,1
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
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, Table of n, a(n) for n = 1..150
Zhi-Wei Sun, Problems on combinatorial properties of primes, arXiv:1402.6641 [math.NT], 2014.
EXAMPLE
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.
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]]<f[Prime[j]*n], Goto[bb]]; Continue, {j, 1, k}]; Goto[bb]; Label[aa]; Print[n, " ", Prime[k]]; Continue, {n, 1, 50}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Jul 17 2015
STATUS
approved