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A068560
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Numbers k such that phi(k) = tau(k)^2.
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5
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1, 5, 34, 63, 76, 128, 136, 170, 315, 364, 380, 444, 640, 680, 972, 1820, 1824, 1836, 2142, 2220, 4788, 4860, 6000, 8064, 8568, 8736, 9120, 9180, 10710, 23940, 40320, 42840, 43680
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OFFSET
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1,2
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COMMENTS
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This sequence is finite because phi(k) >= sqrt(k) for all k >= 6, and for any e > 0, tau(k) < k^e for k large enough. Choosing e=1/4 gives tau(k)^2 < sqrt(k) <= phi(k). It remains unknown, however, if this sequence is full. - Nathaniel Johnston, Apr 28 2011
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LINKS
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MATHEMATICA
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Do[If[EulerPhi[n] == DivisorSigma[0, n]^2, Print[n]], {n, 10^5}] (* Ryan Propper, Jun 09 2006 *)
Select[Range[10^5], EulerPhi[#] == DivisorSigma[0, #]^2 &] (* Alonso del Arte, Aug 25 2011 *)
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CROSSREFS
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KEYWORD
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easy,fini,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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