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Numbers k >= 3 where a record value of log(phi(k))/log(lambda(k)) is reached, where phi is the Euler totient function (A000010) and lambda is the Carmichael lambda function (A002322).
1

%I #18 Dec 18 2024 09:27:06

%S 3,8,24,7280,13104,21840,32760,65520,2878785,5117840,6909084,8636355,

%T 19740240,27636336,46060560,69090840,138181680,1703601900,2271469200,

%U 3407203800,6814407600,20174525280

%N Numbers k >= 3 where a record value of log(phi(k))/log(lambda(k)) is reached, where phi is the Euler totient function (A000010) and lambda is the Carmichael lambda function (A002322).

%C Banks et al. proved that the set {log(phi(k))/log(lambda(k)) | k >= 3} is dense in [1, oo). Therefore this sequence is infinite.

%C Subsequent terms exceed 10^11. - _Lucas A. Brown_, Feb 28 2024

%H William D. Banks, Kevin Ford, Florian Luca, Francesco Pappalardi and Igor E. Shparlinski, <a href="https://doi.org/10.1007/s00605-005-0302-7">Values of the Euler Function in Various Sequences</a>, Monatshefte für Mathematik, Vol. 146, No. 1 (2005), pp 1-19, <a href="https://mospace.umsystem.edu/xmlui/bitstream/handle/10355/10673/ValuesEulerFunction.pdf">alternative link</a>.

%H Lucas A. Brown, <a href="https://github.com/lucasaugustus/oeis/blob/main/A328272.py">Python program</a>.

%e For k < 8, phi(k) = lambda(k), and log(phi(k))/log(lambda(k)) = 1. For k = 8, phi(8) = 4 and lambda(8) = 2, so log(phi(8))/log(lambda(8)) = log(4)/log(2) = 2 is a record value, and hence 8 is in this sequence.

%t r[n_] := Log[EulerPhi[n]]/Log[CarmichaelLambda[n]]; rm = 0; s = {}; Do[r1 = r[n]; If[r1 > rm, rm = r1; AppendTo[s, n]], {n, 3, 10^5}]; s

%Y Cf. A000010, A002322, A066605.

%K nonn,more,hard

%O 1,1

%A _Amiram Eldar_, Oct 10 2019

%E a(21)-a(22) from _Lucas A. Brown_, Feb 28 2024