A391137 - OEIS

%I #9 Nov 30 2025 09:52:03

%S 0,1,2,3,4,6,10,14,16,20,24,26,36,40,54,56,66,74,84,90,94,110,116,120,

%T 124,126,130,134,146,150,156,160,170,176,180,184,204,206,210,224,230,

%U 236,240,250,256,260,264,270,280,284,300,306,314,326,340,350,384,386,396,400

%N Numbers k such that the divisors of k^2 + 1 are all of the form m^2 + 1.

%C It appears to be the union of {0, 3} and A005574.

%e 3 is a term since the proper divisors of 3^2 + 1 = 10 are 1 = 0^2 + 1, 2 = 1^1 + 1, 5 = 2^2 + 1;

%e 6 is a term since 6^2 + 1 = 37 is prime and the only proper divisor 1 = 0^2 + 1;

%e 8 is not a term since the divisor 13 of 8^2 + 1 = 65 is not of the form m^2 + 1.

%t Sqrt[Select[Range[0,400]^2,AllTrue[Sqrt[Divisors[#+1]-1],IntegerQ]&]]

%Y Cf. A002496, A002522, A005574, A027750, A027751, A193432.

%K nonn,easy

%O 1,3

%A _Stefano Spezia_, Nov 30 2025