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%I #40 Sep 30 2023 11:18:54
%S 20,36,60,100,300
%N Numbers k such that tau(k) and phi(k) are the legs of a Pythagorean triple.
%C The sequence is finite since for all large enough n, we have tau(n) < n^(1/4) and phi(n) > n^(3/4) while, if x < y are the legs of a Pythagorean triangle, we always have y < x^2/2. - _Giovanni Resta_, Jul 27 2019
%C The sequence is likely complete. If a(6) exists, it satisfies tau(a(6)) > 1000. - _Max Alekseyev_, Sep 30 2023
%e 60 is in this sequence because tau(60) = 12 and phi(60) = 16, legs of the Pythagorean triple {12, 16, 20} (12^2 + 16^2 = 20^2).
%t Select[Range[300], IntegerQ@Sqrt[DivisorSigma[0, #]^2 + EulerPhi[#]^2] &] (* _Amiram Eldar_, Jul 26 2019 *)
%o (PARI) for(i = 1, 2000, a = eulerphi(i); b = numdiv(i); if(issquare(a^2 + b^2), print1(i,", ")))
%Y Cf. A000005, A000010, A020488, A062516, A063469, A063470, A112954.
%K nonn,fini,more
%O 1,1
%A _Antonio Roldán_, Jul 14 2019
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