OFFSET
1,2
COMMENTS
a(n) == 3, 15 (mod 18), for n >= 2.
It seems that a(n) = 3*p where p is a prime, for n >= 3.
a(n) is squarefree. - David A. Corneth, Jun 01 2017
From Robert Israel, Jul 14 2017: (Start)
a(149) = 13304379 = 3*11*403163 is not of the form 3*p.
The generalized Bunyakovsky conjecture implies that there are infinitely many terms of the form 3*p, and infinitely many of the form 3*11*p. - Robert Israel, Jul 14 2017
LINKS
Robert Israel, Table of n, a(n) for n = 1..300
Wikipedia, Bunyakovsky conjecture.
EXAMPLE
1 = 1*1 and 1^2 + 2*1^2 = 3, a prime.
21 = 1*21 = 3*7 = 21*1 = 7*3 => 1^2 + 2*21^2 = 883, 3^2 + 2*7^2 = 107, 21^2 + 2*1^2 = 443 and 7^2 + 2*3^2 = 67 are primes.
MAPLE
filter:= proc(m)
andmap(x -> isprime(x^2 + 2*(m/x)^2),
numtheory:-divisors(m));
end proc:
select(filter, [1, seq(seq(18*i+j, j=[3, 15]), i=0..10^5)]); # Robert Israel, Jul 14 2017
MATHEMATICA
PROG
(PARI) is(n) = d=divisors(n); for(i=1, #d, if(!isprime(d[i]^2 + 2*d[#d-i+1]^2), return(0))); n > 1 \\ David A. Corneth, Jun 01 2017
(Sage)
R = range(1, 100000)
[m for m in R if all(is_prime(d^2+2*(m//d)^2) for d in divisors(m))] # Peter Luschny, Jun 18 2017
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Lagneau, Jun 01 2017
EXTENSIONS
Name reformulated and m=1 added by Wolfdieter Lang, Jun 20 2017
STATUS
approved