%I #15 Sep 01 2016 03:53:37
%S 1,2,3,4,6,7,10,13,14,16,20,24,26,36,38,40,43,54,56,66,68,74,84,90,94,
%T 110,116,117,120,124,126,130,134,146,150,156,160,170,176,180,183,184,
%U 204,206,210,224,230,236,240,250,256,260,264,270,280,284,293,300,306,314,326,327
%N Numbers k such that the prime divisors of k^2 + 1 are of the form q^2 + 1.
%H Ivan Neretin, <a href="/A181436/b181436.txt">Table of n, a(n) for n = 1..1000</a>
%e 183 is in the sequence because 183^2 + 1 = 2*5*17*197 and 2 = 1^2 + 1, 5 = 2^2+1, 17 = 4^2+1 and 197 = 14^2 + 1.
%p with(numtheory):nn:=1000:for n from 1 to nn do: x:=n^2+1:y:=factorset(x):ny:=nops(y):id:=0:for
%p q from 1 to ny do: z:=y[q]-1:zz:=sqrt(z):if zz=floor(zz) then id:=id+1:else fi:od:if id=ny then printf(`%d, `,n):else fi:od:
%t Select[Range@330, And @@ IntegerQ /@ Sqrt[FactorInteger[#^2 + 1][[All, 1]] - 1] &] (* _Ivan Neretin_, Aug 31 2016 *)
%o (PARI) isok(n) = {fn = factor(n^2+1)[,1]; for (k=1, #fn, if (!issquare(fn[k]-1), return (0));); 1;} \\ _Michel Marcus_, Sep 01 2016
%Y Cf. A180252, A002144, A002522, A005574.
%K nonn
%O 1,2
%A _Michel Lagneau_, Jan 29 2011
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