Original entry on oeis.org
78, 8, 709, 112, 5062, 6388, 37, 137, 1896, 46013, 57499, 2804, 670, 34, 7636, 9176, 5632, 2802, 324038, 2532, 156, 414572, 120, 517498, 30440, 47468, 26944, 11330
Offset: 1
A121850
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.
Original entry on oeis.org
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
Offset: 1
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.
- 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.
This sequence is similar to
A097982.
-
Do[If[IntegerQ[(DivisorSigma[1, n] + EulerPhi[n])/(Times @@ Transpose[FactorInteger[n]][[1]])^2], Print[n]], {n, 1, 1000000}]
-
isok(k) = (((eulerphi(k) + sigma(k)) % factorback(factorint(k)[, 1])^2) == 0); \\ Michel Marcus, Dec 03 2020
A097875
Denominator of J(n) = A000356(n)/A000309(n) (average number of 4-colorings of rank 0 in a rooted nonseparable map which is trivalent and has 2n nodes).
Original entry on oeis.org
1, 4, 24, 88, 52, 2176, 41344, 55936, 22080, 2561280, 27617280, 16570368, 34061312, 1540980736, 23512383488, 100462002176, 50231001088, 3741530783744, 4029340844032, 5798319751168, 4112761683968, 526433495547904, 12757611732533248, 129398633287122944
Offset: 1
1, 5/4, 35/24, 147/88, 99/52, 4719/2176, 102245/41344, ...
Showing 1-3 of 3 results.
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