A281575 Numbers k such that (d^2 + (k/d)^2)/2 is prime for all divisors d of k.
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
Offset: 1
Keywords
Examples
15 is a member because (1^2 + 15^2)/2 = 113 and (3^2 + 5^2)/2 = 17 are prime.
Links
- Robert Israel, Table of n, a(n) for n = 1..10000
Programs
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Maple
filter:= n -> andmap(d -> isprime((d^2 + (n/d)^2)/2), numtheory:-divisors(n)): select(filter, [seq(i,i=1..3000, 2)]);
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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 *) -
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
Comments