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A281575 Numbers k such that (d^2 + (k/d)^2)/2 is prime for all divisors d of k. 1
3, 5, 11, 15, 19, 29, 35, 39, 51, 59, 61, 65, 69, 71, 79, 85, 95, 101, 131, 139, 141, 145, 159, 181, 199, 205, 209, 221, 231, 271, 299, 309, 329, 349, 371, 379, 391, 409, 415, 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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

All terms are odd and squarefree.

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

LINKS

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

EXAMPLE

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

MAPLE

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

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

MATHEMATICA

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 *)

PROG

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

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

CROSSREFS

Contains A048161. Contained in A281505.

Sequence in context: A302590 A316151 A265121 * A105772 A244520 A034169

Adjacent sequences: A281572 A281573 A281574 * A281576 A281577 A281578

KEYWORD

nonn

AUTHOR

Robert Israel and Thomas Ordowski, Jan 24 2017

STATUS

approved

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Last modified February 5 10:05 EST 2023. Contains 360084 sequences. (Running on oeis4.)