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

1,2

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

Links

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

Example

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.

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

Prog

(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

Crossrefs

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

Sequence in context: A212530 A004132 A207187 * A065802 A245302 A118028

Adjacent sequences: A359182 A359183 A359184 * A359186 A359187 A359188

Keyword

nonn

Author

Michel Lagneau, Dec 19 2022

Status

approved