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A104905
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Numbers n such that d(n)*phi(n)=sigma(n), where d(n) is number of positive divisors of n.
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3
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OFFSET
| 1,2
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COMMENTS
| d(n)*phi(n) is the product of f(p^k)=(p^k - p^(k-1))(1+k), while sigma(n) is the product of g(p^k)=(p^(k+1)-1)/(p-1) taken over all prime powers p^k in the factorization of n. We have f(p^k)<g(p^k) for p=2 and k=1 or 2; f(p^k) = g(p^k) for p=3, k=1; and f(p^k)>g(p^k) in all other cases. Furthermore, f(2)/g(2)=2/3 and f(2^2)/g(2^2)=6/7, while f(p^k)/g(p^k) > f(p)/g(p) and for p>7, f(p)/g(p) > 3/2. It easily follows that 1,3,14,42 are the only elements of this sequence. [From Max Alekseyev]
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EXAMPLE
| 42 is in the sequence because d(42)=8; phi(42)=12; sigma(42)=96 & 8*12=96.
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MATHEMATICA
| Do[If[DivisorSigma[0, n]*EulerPhi[n] == DivisorSigma[1, n], Print[n]], {n, 530000000}]
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CROSSREFS
| Cf. A063903, A104904, A104906.
Sequence in context: A050297 A117662 A196236 * A055650 A000550 A124650
Adjacent sequences: A104902 A104903 A104904 * A104906 A104907 A104908
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KEYWORD
| nonn,full,fini
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AUTHOR
| Farideh Firoozbakht (mymontain(AT)yahoo.com), Apr 13 2005
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EXTENSIONS
| full, fini from Max Alekseyev (maxale(AT)gmail.com), Feb 08 2010
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