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Integers k such that phi(k) = 6k/Pi^2 rounded to nearest integer.
2

%I #8 May 10 2019 16:01:58

%S 1,2,3,4,33,39,99,3237,9711,2069301,6207903,45502509,24091595067

%N Integers k such that phi(k) = 6k/Pi^2 rounded to nearest integer.

%C The average value of phi(n) can be approximated for large n by 6n/Pi^2 (Tattersall, p. 162).

%D James J. Tattersall, "Elementary Number Theory in Nine Chapters", Cambridge University Press, 2001.

%e phi(99) = 60 and 6*99/Pi^2 = 60.1848...., which rounds off to 60. Hence 99 is a term of the sequence.

%t s = Pi^2; l = {}; Do[ If[Abs[EulerPhi[n] - 6 n/ s] <= 0.5, l = Append[l, n]], {n, 1, 10^5}]; l

%Y Cf. A000010, A072355.

%Y Cf. A059956 (6/Pi^2).

%K nonn,more

%O 1,2

%A _Joseph L. Pe_, Oct 01 2002

%E a(10)-a(13) from _Amiram Eldar_, May 10 2019

%E Name edited by _Michel Marcus_, May 10 2019