OFFSET
1,1
COMMENTS
Conjecture: a(n) exists for any n > 0. In general, if a,b,c are integers with a > 0 and gcd(a,b,c) = 1, and a+b or c is odd, and b^2 - 4*a*c is not a square, then there are primes p and q such that a*pi(p*n)^2 + b*pi(p*n) + c = prime(q*n).
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..72
Zhi-Wei Sun, Problems on combinatorial properties of primes, arXiv:1402.6641 [math.NT], 2014.
EXAMPLE
a(1) = 3 since pi(3*1)^2 + 1 = 5 = prime(3*1) with 3 prime.
a(8) = 24547 since pi(24547*8)^2 + 1 = 17686^2 + 1 = 312794597 = prime(2113417*8) with 24547 and 2113417 both prime.
MATHEMATICA
PQ[n_, p_] := PrimeQ[p] && PrimeQ[PrimePi[p]/n]; Do[k = 0; Label[bb]; k = k + 1; If[PQ[n, PrimePi[Prime[k] * n]^2 + 1], Goto[aa], Goto[bb]]; Label[aa]; Print[n, " ", Prime[k]]; Continue, {n, 50}]
PROG
(PARI) first(m)={my(v=vector(m), p, q, n); for(n=1, m, p=0; while(1, p++; q=1; while(primepi(prime(p)*n)^2 +1 >= prime(prime(q)*n), if(primepi(prime(p)*n)^2 +1 == prime(prime(q)*n), v[n]=prime(p); break(2), q++; )))); v; } /* Anders Hellström, Jul 19 2015 */
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Jul 19 2015
STATUS
approved