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A068559
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Numbers m such that phi(m) = tau(m)^3.
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4
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1, 85, 333, 436, 1542, 1875, 2907, 3285, 3488, 3796, 5196, 10280, 17532, 17776, 20080, 21250, 28305, 30368, 30555, 32708, 34748, 35308, 36860, 37060, 41544, 41568, 43068, 44004, 45162, 48468, 51930, 81324, 98304, 98688, 104856, 131070
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OFFSET
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1,2
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COMMENTS
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For all large enough k, we have tau(k) < k^(1/4) and phi(k) > k^(3/4). Hence, tau(k)^3 < k^(3/4) < phi(k), implying that this sequence is finite. In fact, the sequence consists of 614 terms. - Max Alekseyev, May 30 2024
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LINKS
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EXAMPLE
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MATHEMATICA
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Select[Range[132000], EulerPhi[#]==DivisorSigma[0, #]^3&] (* Harvey P. Dale, Dec 28 2022 *)
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PROG
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(PARI) isok(m) = eulerphi(m) == numdiv(m)^3; \\ Michel Marcus, Oct 18 2019
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CROSSREFS
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KEYWORD
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nonn,fini,full,changed
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AUTHOR
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STATUS
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approved
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