A279508 -id:A279508 - cp's OEIS frontend

A279507 a(n) = floor(phi(n)/tau(n)).

1, 0, 1, 0, 2, 0, 3, 1, 2, 1, 5, 0, 6, 1, 2, 1, 8, 1, 9, 1, 3, 2, 11, 1, 6, 3, 4, 2, 14, 1, 15, 2, 5, 4, 6, 1, 18, 4, 6, 2, 20, 1, 21, 3, 4, 5, 23, 1, 14, 3, 8, 4, 26, 2, 10, 3, 9, 7, 29, 1, 30, 7, 6, 4, 12, 2, 33, 5, 11, 3, 35, 2, 36, 9, 6, 6, 15, 3, 39, 3

Offset: 1

Jaroslav Krizek, Dec 13 2016

nonn

a(n) = floor(A000010(n)/A000005(n)).
There are 11 numbers n such that phi(n) <= tau(n) and 7 numbers n such that phi(n) = tau(n); see A020490 and A020488.
Sequences b(k) of numbers n such that a(n) = k are finite for all k >=0; see A279508 (the smallest numbers n such that a(n) = k for k>=0) and A279509 (the largest numbers n such that a(n) = k for k>=0).
See A140475 (numbers n such that floor(phi(n)/tau(n)) > floor(phi(m)/tau(m)) for all m < n).

			For n=5; a(5) = floor(phi(5)/tau(5)) = floor(4/2) = 2.

Jaroslav Krizek, Table of n, a(n) for n = 1..2000

Cf. A000005, A000010, A020488, A020490, A140475, A279289, A279508, A279509.

[Floor(EulerPhi(n)/NumberOfDivisors(n)): n in[1..100]]

Table[Floor[EulerPhi[n]/DivisorSigma[0, n]], {n,1,25}] (* G. C. Greubel, Dec 13 2016 *)

for(n=1, 25, print1(floor(eulerphi(n)/numdiv(n)), ", ")) \\ G. C. Greubel, Dec 13 2016

a(n) > 1 for numbers in A279289.

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

Original entry on oeis.org

12, 60, 180, 240, 420, 480, 840, 462, 1260, 1680, 1440, 690, 2520, 2100, 2160, 2310, 3360, 2400, 3780, 5040, 4620, 3600, 3300, 1410, 5460, 4080, 6300, 7560, 5880, 4140, 9240, 2646, 10080, 6600, 6480, 7200, 10920, 8820, 9360, 2370, 13860, 8640, 8160, 15120

Offset: 0

Jaroslav Krizek, Dec 19 2016

nonn

a(n) = largest number k such that floor(A000010(k)/A000005(k)) = A279507(k) = n.
Sequences b_n of numbers k such that floor(phi(k)/tau(k)) = n for n = 0..2:
b_0: 2, 4, 6, 12;
b_1: 1, 3, 8, 10, 14, 16, 18, 20, 24, 30, 36, 42, 48, 60;
b_2: 5, 9, 15, 22, 28, 32, 40, 54, 66, 72, 84, 90, 96, 120, 180.
Sequences b_n are finite for all n >= 0. See A279508 (smallest number k such that floor(phi(k)/tau(k)) = n).

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

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

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

cp's OEIS Frontend

A279507 a(n) = floor(phi(n)/tau(n)).

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