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A121850
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Numbers k such that (phi(k) + sigma(k))/rad(k)^2 is an integer, that is (phi(k) + sigma(k)) is divisible by every prime factor of k squared.
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2
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1, 2, 588, 864, 2430, 7776, 27000, 55296, 69984, 82134, 215622, 432000, 497664, 629856, 675000, 862488, 1499136, 1749600, 2187000, 2667168, 3449952, 3538944, 4287500, 4312440, 4478976, 4563000, 5668704, 6912000, 10800000, 13045131
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OFFSET
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1,2
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COMMENTS
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This sequence is infinite because every integer m = 32 * 3^(2r+1), r>=1 is a term: 864, 7776, 69984, ... (see De Koninck & Mercier reference). - Bernard Schott, Dec 02 2020
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REFERENCES
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J.-M. De Koninck, Those Fascinating Numbers, Amer. Math. Soc., 2009, page 71, entry 588.
J.-M. De Koninck & A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Ellipses, 2004, Problème 749, pp. 95 and 319.
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LINKS
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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, 1, 1000000}]
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PROG
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(PARI) isok(k) = (((eulerphi(k) + sigma(k)) % factorback(factorint(k)[, 1])^2) == 0); \\ Michel Marcus, Dec 03 2020
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CROSSREFS
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This sequence is similar to A097982.
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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