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Numbers m such that totient(m) < cototient(m).
11

%I #44 Feb 14 2021 02:23:22

%S 6,10,12,14,18,20,22,24,26,28,30,34,36,38,40,42,44,46,48,50,52,54,56,

%T 58,60,62,66,68,70,72,74,76,78,80,82,84,86,88,90,92,94,96,98,100,102,

%U 104,105,106,108,110,112,114,116,118,120,122,124,126,130,132,134,136

%N Numbers m such that totient(m) < cototient(m).

%C For powers of 2, the two function values are equal.

%C Numbers m such that m/phi(m) > 2. - _Charles R Greathouse IV_, Sep 13 2013

%C Numbers m such that A173557(m)/A007947(m) < 1/2. - _Antti Karttunen_, Jan 05 2019

%C Numbers m such that there are powers of m that are abundant. This follows from abundancy and totient being multiplicative, with the abundancy for prime p of p^k being asymptotically p/(p-1) as k -> oo; given that p/(p-1) = p^k/phi(p^k) for k >= 1. - _Peter Munn_, Nov 24 2020

%H Charles R Greathouse IV, <a href="/A054741/b054741.txt">Table of n, a(n) for n = 1..10000</a>

%H Mitsuo Kobayashi, <a href="https://doi.org/10.1142/S1793042116500445">A generalization of a series for the density of abundant numbers</a>, International Journal of Number Theory, Vol. 12, No. 3 (2016), pp. 671-677.

%F m such that A000010(m) < A051953(m).

%F a(n) seems to be asymptotic to c*n with c=1.9566...... - _Benoit Cloitre_, Oct 20 2002 [It is an old theorem that a(n) ~ cn for some c, for any sequence of the form "m/phi(m) > k". - _Charles R Greathouse IV_, May 28 2015] [c is in the interval (1.9540, 1.9562) (Kobayashi, 2016). - _Amiram Eldar_, Feb 14 2021]

%e For m = 20, phi(20) = 8, cototient(20) = 20 - phi(20) = 12, 8 = phi(20) < 20-phi(20) = 12; for m = 21, the opposite holds: phi = 12, 21-phi = 8.

%t Select[ Range[300], 2EulerPhi[ # ] < # &] (* _Robert G. Wilson v_, Jan 10 2004 *)

%o (PARI) is(n)=n>2*eulerphi(n) \\ _Charles R Greathouse IV_, Sep 13 2013

%Y A177712 is a subsequence. Complement: A115405.

%Y Positions of negative terms in A083254.

%Y Cf. A000010, A007947, A051953, A005408, A036798, A089684, A173557.

%Y Cf. A323170 (characteristic function).

%Y Complement of A000079\{1} within A119432.

%K nonn

%O 1,1

%A _Labos Elemer_, Apr 26 2000

%E Erroneous comment removed by _Antti Karttunen_, Jan 05 2019