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A110172
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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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1
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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
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OFFSET
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1,1
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COMMENTS
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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
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LINKS
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CROSSREFS
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Cf. A066426 (least k such that phi(n) + phi(k) = phi(n+k)).
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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