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Search: id:A121850
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| A121850 |
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Numbers n such that (phi(n) + sigma(n))/(rad(n))^2 is an integer, that is (phi(n) + sigma(n)) is divisible by every prime factor of n squared. |
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| 2, 588, 864, 2430, 7776, 27000, 55296, 69984, 82134, 215622, 432000, 497664, 629856, 675000, 862488
(list; graph; listen)
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OFFSET
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1,1
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EXAMPLE
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For example, phi(588) = 168, sigma(588) = 1596, 588 = 2^2*3*7^2. The product of all prime divisors is 42, its square is 1764. Hence phi(588) + sigma(588), which is equal to 1764 is divisible by the square of each prime divisor of 588.
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MATHEMATICA
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Do[If[IntegerQ[(DivisorSigma[1, n] + EulerPhi[n])/(Times @@ Transpose[FactorInteger[n]][[1]])^2], Print[n]], {n, 2, 1000000}]
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CROSSREFS
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Cf. a(n) are numbers n such that A000010(n) + A000203(n) is divisible by A007947(n)^2. This sequence is similar to A097982.
Sequence in context: A003830 A134371 A129697 this_sequence A100011 A134796 A120830
Adjacent sequences: A121847 A121848 A121849 this_sequence A121851 A121852 A121853
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KEYWORD
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nonn
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AUTHOR
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Tanya Khovanova (tanyakh(AT)yahoo.com), Aug 30 2006
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