A079814 Odd integers k such that phi(k)/k < 6/Pi^2 where phi = A000010.

15, 21, 33, 45, 63, 75, 99, 105, 135, 147, 165, 189, 195, 225, 231, 255, 273, 285, 297, 315, 345, 357, 363, 375, 399, 405, 429, 435, 441, 465, 483, 495, 525, 555, 561, 567, 585, 609, 615, 627, 645, 651, 663, 675, 693, 705, 735, 741, 759, 765, 777, 795, 819

Offset: 1

Author

Matthew Vandermast, Feb 19 2003

Keywords

• easy
• nonn
• changed

Comments

Since, as Euler proved, the random chance of two integers not having a common prime factor is 6/Pi^2, these are the odd integers that share common factors with an above average fraction of integers. Is it known, or can it be calculated, what portion of odd integers satisfy this condition? (All even numbers qualify; for all multiples of 2, phi(n)/n <= 0.5.)

The sequence is closed under multiplication by any odd number. If we include the even numbers, the sequence of primitive terms begins 2, 15, 21, 33, 663, ... . - Peter Munn, Apr 11 2021

Examples

			phi(33)/33 = 20/33 or 0.6060606...; 6/Pi^2 is 0.6079271....
		

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000010 (Euler totient function phi(n)), A280877, A280878, A280879.

Programs

Mathematica

Select[Range[1, 1000, 2], EulerPhi[#]/# < 6/Pi^2 &] (* Paolo Xausa, Aug 29 2025 *)

PARI

is(n)=n%2 && eulerphi(n)/n<6/Pi^2 \\ Charles R Greathouse IV, Sep 13 2013