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A281575 Numbers k such that (d^2 + (k/d)^2)/2 is prime for all divisors d of k. 1

%I #12 Dec 11 2017 04:22:20

%S 3,5,11,15,19,29,35,39,51,59,61,65,69,71,79,85,95,101,131,139,141,145,

%T 159,181,199,205,209,221,231,271,299,309,329,349,371,379,391,409,415,

%U 449,461,471,519,521,535,545,559,569,571,581,631,641,649,661,685,689,739,745,751,779,799,815,821,861

%N Numbers k such that (d^2 + (k/d)^2)/2 is prime for all divisors d of k.

%C All terms are odd and squarefree.

%C Generalized Bunyakovsky conjecture implies for any odd prime p there are infinitely many terms of the form p*q where q is prime.

%H Robert Israel, <a href="/A281575/b281575.txt">Table of n, a(n) for n = 1..10000</a>

%e 15 is a member because (1^2 + 15^2)/2 = 113 and (3^2 + 5^2)/2 = 17 are prime.

%p filter:= n -> andmap(d -> isprime((d^2 + (n/d)^2)/2), numtheory:-divisors(n)):

%p select(filter, [seq(i,i=1..3000, 2)]);

%t 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 *)

%o (PARI) isp(q) = (denominator(q)==1) && isprime(q);

%o isok(n) = {fordiv(n, d, if (!isp((d^2 + (n/d)^2)/2), return(0));); return (1);} \\ _Michel Marcus_, Dec 11 2017

%Y Contains A048161. Contained in A281505.

%K nonn

%O 1,1

%A _Robert Israel_ and _Thomas Ordowski_, Jan 24 2017

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Last modified April 24 07:06 EDT 2024. Contains 371920 sequences. (Running on oeis4.)