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%I #32 Mar 06 2024 04:54:50
%S 1,3,5,9,13,19,23,25,39,53,55,73,83,89,109,119,133,149,155,159,169,
%T 179,203,223,229,239,263,269,283,299,305,313,339,349,383,395,419,439,
%U 443,463,469,473,543,569,593,643,653,673,689,699,703,713,739,763,859,863,889,909
%N Numbers k such that for any positive integers x,y, if x*y=k then (x+y)^2+1 is a prime number.
%C 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).
%C The conjecture is false. A counterexample is 296147^2 = 87703045609 where 296147 = 47 * 6301. - _Robert Israel_, Mar 05 2024
%C The primes of the sequence are in A157468.
%C All terms except 1 are congruent to 3, 5 or 9 (mod 10). - _Robert Israel_, Mar 05 2024
%H Robert Israel, <a href="/A359185/b359185.txt">Table of n, a(n) for n = 1..10000</a>
%e 909 is in the sequence because 909 = 3^2*101 with 3 decompositions:
%e 909 = 1*909 and (1+909)^2+1 = 910^2+1 = 828101 is prime;
%e 909 = 3*303 and (3+303)^2+1 = 306^2+1 = 93637 is prime;
%e 909 = 9*101 and (9+101)^2+1 = 110^2+1 = 12101 is prime.
%p filter:= proc(n) local F;
%p F:= select(t -> t^2 <= n, numtheory:-divisors(n));
%p andmap(t -> isprime((t + n/t)^2+1), F)
%p end proc:
%p select(filter, [seq(i,i=1..1000,2)]); # _Robert Israel_, Mar 05 2024
%t 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
%o (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
%Y Cf. A001358, A002496, A134406, A157468, A236068.
%K nonn
%O 1,2
%A _Michel Lagneau_, Dec 19 2022