A110172 Conjectured numbers j such that phi(j) + phi(k) = phi(j+k) has no solution k, where phi is Euler's totient function.

3, 15, 21, 39, 45, 57, 69, 105, 147, 165, 177, 195, 213, 273, 285, 315, 345, 393, 399, 465, 489, 525, 585, 615, 633, 645, 651, 681, 717, 777, 807, 813, 843, 855, 879, 885, 903, 915, 933, 939, 1005, 1035, 1041, 1065, 1095, 1149, 1263, 1281, 1293, 1317, 1395

Offset: 1

Author

T. D. Noe, Jul 15 2005

Keywords

• nonn

Comments

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

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

Crossrefs

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

Cf. A306771.