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Search: id:A020488
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| A020488 |
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Numbers n such that tau(n) (or sigma_0(n)) = phi(n). |
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+0
17
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OFFSET
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1,2
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COMMENT
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Numbers satisfying A000005[n]=A000010[n]
Comment from Jud McCranie, Jun 17 2005: This sequence is complete because tau(n) < n^(2/3) for all n except a few small numbers, whereas phi(n) > n/(exp(gamma) * log(log(n)) + 3/(log(log(n))) for n > 2. log(log(n)) grows slowly, so phi(n) > tau(n) for all n > some relatively small constant.
Subset of A112587. - Reinhard Zumkeller (reinhard.zumkeller(AT)lhsystems.com), Sep 14 2005
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EXAMPLE
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n=1, 3, 8, 10, 18, 24, 30: tau[n]=1, 2, 4, 4, 6, 8: phi[n]=1, 2, 4, 4, 6, 8: no difference.
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MATHEMATICA
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k=1; s=Select[ Range[ 1, 100000 ], Equal[ Sign[ DivisorSigma[ k-1, # ]-EulerPhi[ # ]^k ], 0 ]& ].
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CROSSREFS
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Cf. A064374-A064377, A000005, A000010.
Cf. A112954, A062516, A063469, A063470.
Sequence in context: A022801 A128699 A104816 this_sequence A064435 A131725 A032914
Adjacent sequences: A020485 A020486 A020487 this_sequence A020489 A020490 A020491
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KEYWORD
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nonn,fini,full
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AUTHOR
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David W. Wilson (davidwwilson(AT)comcast.net)
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