A359185 Numbers k such that for any positive integers x,y, if x*y=k then (x+y)^2+1 is a prime number.

1, 3, 5, 9, 13, 19, 23, 25, 39, 53, 55, 73, 83, 89, 109, 119, 133, 149, 155, 159, 169, 179, 203, 223, 229, 239, 263, 269, 283, 299, 305, 313, 339, 349, 383, 395, 419, 439, 443, 463, 469, 473, 543, 569, 593, 643, 653, 673, 689, 699, 703, 713, 739, 763, 859, 863, 889, 909

Offset: 1

Author

Michel Lagneau, Dec 19 2022

Keywords

• nonn

Comments

Conjecture: if a term k is a perfect square > 1, then sqrt(k) is in the sequence A236068 (Primes p such that f(f(p)) is prime, where f(z) = z^2 + 1).

The conjecture is false. A counterexample is 296147^2 = 87703045609 where 296147 = 47 * 6301. - Robert Israel, Mar 05 2024

The primes of the sequence are in A157468.

All terms except 1 are congruent to 3, 5 or 9 (mod 10). - Robert Israel, Mar 05 2024

Examples

			909 is in the sequence because 909 = 3^2*101 with 3 decompositions:
909 = 1*909 and (1+909)^2+1 = 910^2+1 = 828101 is prime;
909 = 3*303 and (3+303)^2+1 = 306^2+1 = 93637 is prime;
909 = 9*101 and (9+101)^2+1 = 110^2+1 = 12101 is prime.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A001358, A002496, A134406, A157468, A236068.

Programs

Maple

filter:= proc(n) local F;
  F:= select(t -> t^2 <= n, numtheory:-divisors(n));
  andmap(t -> isprime((t + n/t)^2+1), F)
end proc:
select(filter, [seq(i,i=1..1000,2)]); # Robert Israel, Mar 05 2024

Mathematica

t={};Do[ds=Divisors[n];k=1;While[k<=(Length[ds]+1)/2&&(ok=PrimeQ[(ds[[k]]+ds[[-k]])^2+1]),k++];If[ok,AppendTo[t,n]],{n,1,2000}];t

PARI

isok(k) = fordiv(k, d, if ((d<=k/d) && !isprime((d+k/d)^2+1), return(0));); return(1); \\ Michel Marcus, Dec 19 2022