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A279509 a(n) = largest number k such that floor(phi(k)/tau(k)) = n. 2

%I #5 Sep 08 2022 08:46:18

%S 12,60,180,240,420,480,840,462,1260,1680,1440,690,2520,2100,2160,2310,

%T 3360,2400,3780,5040,4620,3600,3300,1410,5460,4080,6300,7560,5880,

%U 4140,9240,2646,10080,6600,6480,7200,10920,8820,9360,2370,13860,8640,8160,15120

%N a(n) = largest number k such that floor(phi(k)/tau(k)) = n.

%C a(n) = largest number k such that floor(A000010(k)/A000005(k)) = A279507(k) = n.

%C Sequences b_n of numbers k such that floor(phi(k)/tau(k)) = n for n = 0..2:

%C b_0: 2, 4, 6, 12;

%C b_1: 1, 3, 8, 10, 14, 16, 18, 20, 24, 30, 36, 42, 48, 60;

%C b_2: 5, 9, 15, 22, 28, 32, 40, 54, 66, 72, 84, 90, 96, 120, 180.

%C Sequences b_n are finite for all n >= 0. See A279508 (smallest number k such that floor(phi(k)/tau(k)) = n).

%e For n = 1; a(1) = 60 because 60 is the largest number with floor(phi(60)/tau(60)) = floor(16/12) = 1.

%o (Magma) [Max([n: n in[1..100000] | Floor(EulerPhi(n) / NumberOfDivisors(n)) eq k]): k in [0..50]]

%Y Cf. A000005, A000010, A020488, A020490, A279289, A279507, A279508.

%K nonn

%O 0,1

%A _Jaroslav Krizek_, Dec 19 2016

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Last modified May 13 17:28 EDT 2024. Contains 372522 sequences. (Running on oeis4.)