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A261281
Least positive integer k with prime(k)^2-2 and prime(prime(k))^2-2 both prime such that prime(k*n)^2-2 and prime(prime(k*n))^2-2 are all prime.
6
1, 1, 319, 134, 34, 62, 2, 536, 5215, 15, 3965, 2168, 34, 1, 1, 737, 2, 7075, 3699, 419, 132, 372, 14, 2, 34, 2, 52, 1, 668, 36561, 2, 48, 1239, 1, 401, 1613, 1646, 2472, 43, 31361, 134, 1103, 1, 5374, 6201, 466, 1, 1, 2118, 2, 1646, 1, 1343, 856, 28, 1868, 10324, 360, 2845, 6571, 65, 1, 419, 43, 1, 2, 2, 1, 889, 202
OFFSET
1,3
COMMENTS
Conjecture: a(n) exists for any n > 0. In general, any positive rational number r can be written as m/n with m and n in the set {k>0: prime(k)^2-2 and prime(prime(k))^2-2 are both prime}.
This implies that the sequence A237414 has infinitely many terms.
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.
EXAMPLE
a(2) = 1 since prime(1)^2-2 = 2^2-2 = 2, prime(prime(1))^2-2 = prime(2)^2-2 = 3^2-2 = 7, prime(1*2)^2-2 = 3^2-2 = 7, and prime(prime(1*2))^2-2 = prime(3)^2-2 = 5^2-2 = 23 are all prime.
a(3) = 319 since prime(319)^2-2 = 2113^2-2 = 4464767, prime(prime(319))^2-2 = prime(2113)^2-2 = 18443^2-2 = 340144247, prime(319*3)^2-2 = 7547^2-2 = 56957207, and prime(prime(3*319))^2-2 = prime(7547)^2-2 = 76757^2-2 = 5891637047 are all prime.
MATHEMATICA
f[n_]:=Prime[n]
q[n_]:=PrimeQ[f[n]^2-2]&&PrimeQ[f[f[n]]^2-2]
Do[k=0; Label[bb]; k=k+1; If[q[k]&&q[k*n], Goto[aa], Goto[bb]]; Label[aa]; Print[n, " ", k]; Continue, {n, 1, 70}]
PROG
(PARI) a(n) = my(k=1); while (!isprime(prime(k)^2-2) || !isprime(prime(prime(k))^2-2) || !isprime(prime(k*n)^2-2) || !isprime(prime(prime(k*n))^2-2), k++); k; \\ Michel Marcus, Aug 14 2015
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Aug 14 2015
STATUS
approved