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A359185
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Numbers k such that for any positive integers x,y, if x*y=k then (x+y)^2+1 is a prime number.
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1
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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
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OFFSET
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1,2
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COMMENTS
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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
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LINKS
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EXAMPLE
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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.
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MAPLE
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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
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MATHEMATICA
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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
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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