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Conjectured numbers j such that phi(j) + phi(k) = phi(j+k) has no solution k, where phi is Euler's totient function.
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%I #41 Aug 23 2022 09:31:56

%S 3,15,21,39,45,57,69,105,147,165,177,195,213,273,285,315,345,393,399,

%T 465,489,525,585,615,633,645,651,681,717,777,807,813,843,855,879,885,

%U 903,915,933,939,1005,1035,1041,1065,1095,1149,1263,1281,1293,1317,1395

%N Conjectured numbers j such that phi(j) + phi(k) = phi(j+k) has no solution k, where phi is Euler's totient function.

%C All k < 10^8 have been checked. All of these numbers are multiples of 3.

%C The observation above is true for every term. Substituting k=j into phi(j) + phi(k) = phi(j+k) gives phi(j) + phi(j) = phi(j+j), i.e., 2*phi(j) = phi(2j), which is true for every positive even number j; thus k=j yields a solution for every positive even number j. Substituting k=2j into phi(j) + phi(k) = phi(j+k) gives phi(j) + phi(2j) = phi(j+2j), i.e., phi(j) + phi(2j) = phi(3j); since phi(j) = phi(2j) for every odd number j, this is equivalent (for odd j) to phi(j) + phi(j) = phi(3j), i.e., 2*phi(j) = phi(3j), which holds for every odd j that is not a multiple of 3; thus, k=2j yields a solution for every odd j that is not a multiple of 3. Consequently, every term of the sequence is an odd multiple of 3. - _Flávio V. Fernandes_, May 10 2022

%Y Cf. A066426 (least k such that phi(n) + phi(k) = phi(n+k)).

%Y Cf. A306771.

%K nonn

%O 1,1

%A _T. D. Noe_, Jul 15 2005