OFFSET
1,1
COMMENTS
All terms are odd and squarefree.
Generalized Bunyakovsky conjecture implies for any odd prime p there are infinitely many terms of the form p*q where q is prime.
LINKS
Robert Israel, Table of n, a(n) for n = 1..10000
EXAMPLE
15 is a member because (1^2 + 15^2)/2 = 113 and (3^2 + 5^2)/2 = 17 are prime.
MAPLE
filter:= n -> andmap(d -> isprime((d^2 + (n/d)^2)/2), numtheory:-divisors(n)):
select(filter, [seq(i, i=1..3000, 2)]);
MATHEMATICA
pdnQ[n_]:=Module[{divs=Divisors[n]}, AllTrue[(#^2+(n/#)^2)/2&/@ divs, PrimeQ]]; Select[Range[1000], pdnQ] (* The program uses the AllTrue function from Mathematica version 10 *) (* Harvey P. Dale, Feb 12 2017 *)
PROG
(PARI) isp(q) = (denominator(q)==1) && isprime(q);
isok(n) = {fordiv(n, d, if (!isp((d^2 + (n/d)^2)/2), return(0)); ); return (1); } \\ Michel Marcus, Dec 11 2017
CROSSREFS
KEYWORD
nonn
AUTHOR
Robert Israel and Thomas Ordowski, Jan 24 2017
STATUS
approved