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A062516
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Numbers k such that 2*tau(k) = phi(k).
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12
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OFFSET
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1,1
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COMMENTS
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Sequence is finite, since for large k and suitable constants and epsilon: phi(k) - 2*tau(k) > c1*k^(2/3) - 4*c2*k^(1/2) > 0 if k > c3, so phi(k) - 2*tau(k) > 0, QED. Moreover, phi(k) = m*tau(k) has at most finitely many solutions for any constant m or even for slowly increasing functions like m(k) = k^(epsilon). - Labos Elemer, Jul 20 2001
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LINKS
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MATHEMATICA
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Select[Range[150], 2*DivisorSigma[0, #]==EulerPhi[#]&] (* Harvey P. Dale, Jun 28 2022 *)
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PROG
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(PARI) for(n=1, 1000000, if(numdiv(n)*2==eulerphi(n), print(n), ))
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CROSSREFS
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KEYWORD
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nonn,fini,full
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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