A082473 Numbers n such that n = phi(x)*core(x) for some x <= n, where phi(x) is the Euler totient function and core(x) the squarefree part of x.
1, 2, 6, 8, 12, 20, 32, 40, 42, 48, 54, 84, 108, 110, 120, 128, 156, 160, 192, 220, 240, 252, 272, 312, 336, 342, 432, 486, 500, 504, 506, 512, 544, 640, 660, 684, 768, 812, 840, 880, 930, 936, 960, 972, 1000, 1012, 1080, 1248, 1320, 1332, 1344, 1624, 1632
Offset: 1
Keywords
References
- József Sándor and Borislav Crstici, Handbook of Number theory II, Kluwer Academic Publishers, 2004, Chapter 3, p. 224.
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Alois P. Heinz)
- Walther Janous, Problem 6588, Advanced Problems, The American Mathematical Monthly, Vol. 95, No. 10 (1988), p. 963; How Often is n*phi(n) <= x^2?, Solution to Problem 6588, ibid., Vol. 98, No. 5 (1991), pp. 446-448.
Programs
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Mathematica
With[{nn = 1700}, TakeWhile[Union@ Array[EulerPhi[#] (Sqrt@ # /. (c_: 1) a_^(b_: 0) :> (c a^b)^2) &, nn], # <= nn &]] (* Michael De Vlieger, Sep 29 2019, after Bill Gosper at A007913 *)
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PARI
isok(n) = {for (x=1, n, if (eulerphi(x)*core(x) == n, return (1));); return (0);} \\ Michel Marcus, Dec 04 2013
Formula
From Antti Karttunen, Sep 29 2019: (Start)
(End)
The number of terms not exceeding x is ~ c * sqrt(x), where c = Product_{p prime} (1 + 1/sqrt(p*(p-1)) - 1/p) = 1.3651304521... (Janous, 1988). - Amiram Eldar, Mar 10 2021
Comments