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A063470
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Numbers n such that tau(n)*4 = phi(n).
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9
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OFFSET
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1,1
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COMMENTS
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Phi(n) = k*tau(n) has at most finitely many solutions for any constant k or even for slowly increasing functions like k(n) = n^(epsilon). - Labos Elemer, Jul 20 2001
For n > 2, tau(n) > 2 and phi(n) <= n-1 so the least solution a(1) to tau(n)*k = phi(n), must be a(1) >= 2*k+1, for the case k=4, a(1) >= 2*4+1 = 9. - Enrique Pérez Herrero, May 12 2012
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LINKS
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FORMULA
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PROG
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(PARI) for(n=1, 10^6, if(numdiv(n)*4==eulerphi(n), print(n)))
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CROSSREFS
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KEYWORD
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fini,nonn,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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