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Search: id:A055466
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| A055466 |
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Numbers n such that d(n)^2 divides phi(n) + sigma(n). |
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1
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| 1, 2, 4, 15, 39, 49, 55, 78, 81, 87, 95, 99, 110, 111, 119, 121, 125, 143, 159, 183, 184, 215, 247, 287, 295, 303, 319, 327, 335, 350, 357, 391, 407, 415, 423, 430, 447, 455, 471, 507, 511, 519, 527, 535, 543, 551, 559, 583, 591, 620, 623, 654, 655, 671, 679
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OFFSET
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1,2
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COMMENT
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Makowski proved that Phi[n]+Sigma[n] = n*d[n] iff n is a prime (see in Sivaramakrishnan, Chapter I, page 8, Theorem 3).
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REFERENCES
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Sivaramakrishnan, R. (1989), Classical Theory of Arithmetical Functions, Marcel Dekker, Inc., New York-Basel.
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FORMULA
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Integer solutions of Phi[x]+Sigma[x] = kd[x]^2 or A000203(n)+A000010(n) = k*A000005(n)^2, where k is integer.
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EXAMPLE
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true for 2 (the only prime) and some composites. n = 78: 8 divisors, Sigma = 168, Phi = 24, 168+24 = 192 = 8*8*3
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CROSSREFS
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Cf. A000005, A000010, A000203.
Sequence in context: A005219 A153945 A153942 this_sequence A148267 A148268 A148269
Adjacent sequences: A055463 A055464 A055465 this_sequence A055467 A055468 A055469
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KEYWORD
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nonn
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AUTHOR
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Labos E. (labos(AT)ana.sote.hu), Jun 27 2000
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