A181436 Numbers k such that the prime divisors of k^2 + 1 are of the form q^2 + 1.
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, 110, 116, 117, 120, 124, 126, 130, 134, 146, 150, 156, 160, 170, 176, 180, 183, 184, 204, 206, 210, 224, 230, 236, 240, 250, 256, 260, 264, 270, 280, 284, 293, 300, 306, 314, 326, 327
Offset: 1
Keywords
Examples
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.
Links
- Ivan Neretin, Table of n, a(n) for n = 1..1000
Programs
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Maple
with(numtheory):nn:=1000:for n from 1 to nn do: x:=n^2+1:y:=factorset(x):ny:=nops(y):id:=0:for 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:
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Mathematica
Select[Range@330, And @@ IntegerQ /@ Sqrt[FactorInteger[#^2 + 1][[All, 1]] - 1] &] (* Ivan Neretin, Aug 31 2016 *)
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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