|
| |
|
|
A260219
|
|
Least prime p such that pi(p*n)^2 + 1 = prime(q*n) for some prime q.
|
|
1
|
|
|
|
3, 47, 229, 379, 11, 2687, 1181, 24547, 5, 509, 5, 16619, 877, 22543, 3, 9067, 60337, 10667, 997, 24061, 18329, 56099, 1787, 58757, 108883, 416881, 157141, 11003, 14939, 113167, 101957, 77969, 613, 27947, 52153, 158551, 6197, 36607, 25237, 27179, 330689, 203617, 77419, 708269, 87649, 340381, 267601, 67153, 123377, 21617
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
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
|
|
|
|
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
|
|
|
|
STATUS
|
approved
|
| |
|
|
