%I #56 Aug 22 2021 09:25:31

%S 1,-1,-2,0,-4,2,-6,0,0,4,-10,0,-12,6,8,0,-16,0,-18,0,12,10,-22,0,0,12,

%T 0,0,-28,-8,-30,0,20,16,24,0,-36,18,24,0,-40,-12,-42,0,0,22,-46,0,0,0,

%U 32,0,-52,0,40,0,36,28,-58,0,-60,30,0,0,48,-20,-66,0,44,-24,-70,0,-72,36,0,0,60,-24,-78,0,0,40,-82,0

%N a(n) = mu(n)*phi(n) where mu(n) is the Mobius function (A008683) and phi(n) is the Euler totient function (A000010).

%C Also, a(n) = mu(n)*uphi(n) where mu(n) is the Mobius function (A008683) and uphi(n) is the unitary totient function (A047994), since phi(n) = uphi(n) when n is squarefree, while mu(n) = 0 when n is not squarefree. - _Franklin T. Adams-Watters_, May 14 2006

%C Conjecture: Sum_{n>=1} mu(n)/phi(n) = Sum_{n>=1} a(n)/phi(n)^2 = 0. It is true that Sum_{n>=1} mu(n)/phi(n)^s = 0 at least for s > 1 since: phi(2)=1, phi is multiplicative, so for n's that are squarefree, the phi(n) values can be partitioned in pairs where phi(m)=phi(2m) and mu(m) = -mu(2m). So Sum_{i=1..n} mu(i)/phi(i)^s < Sum_{j=floor(n/2)..n} 1/phi(j)^s, which approaches 0 as n increases since (1) n^(1-e) < phi(n) < n for any e > 0 and n > N(e) and (2) Sum_{i..n} 1/n^s converges for s > 1. Conjecture: Sum_{n>=1} mu(n)/phi(n)^z = 0 for Re(z) > 1.

%C Multiplicative with a(p^1) = 1-p, a(p^e) = 0, e > 1. - _Mitch Harris_, May 24 2005

%C Row sums of triangle A143153 = a signed version of the sequence such that parity = (-) iff A008683(n) = (+); 0 or (+): (1, 1, 2, 0, 4, -2, 6, 0, 0, -4, 10, 0, 12, -6, 0, 0, 0, ...). - _Gary W. Adamson_, Jul 27 2008

%C Dirichlet inverse of A003958. - _R. J. Mathar_, Jul 08 2011

%H Alois P. Heinz, <a href="/A097945/b097945.txt">Table of n, a(n) for n = 1..10000</a>

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Euler%27s_totient_function">Euler's totient function</a>

%F Dirichlet g.f.: Product_{primes p} (1-p^(1-s)+p^(-s)). - _R. J. Mathar_, Aug 29 2011

%F Sum_{d|n} abs(a(d)) = rad(n) = A007947(n). - _Rémy Sigrist_, Nov 05 2017

%F Sum_{k=1..n} abs(a(k)) ~ c * n^2, where c = A065464/2 = (1/2) * Product_{primes p} (1 - 2/p^2 + 1/p^3) = 0.21412475283854722... Equivalently, c = A065463 * 3 / Pi^2. - _Vaclav Kotesovec_, Jun 14 2020

%F From _Antti Karttunen_, Aug 20 2021: (Start)

%F a(n) = mu(n)*A000010(n) = mu(n)*A003958(n) = mu(n)*A047994(n) = mu(n)*A173557(n), where mu is Möbius mu function (A008683).

%F a(n) = A008966(n) * A023900(n) = abs(mu(n)) * A023900(n).

%F a(n) = A322581(n) - A003958(n).

%F (End)

%p with(numtheory):

%p a:= n-> mobius(n)*phi(n):

%p seq(a(n), n=1..100); # _Alois P. Heinz_, Aug 06 2012

%t Table[ MoebiusMu[n]EulerPhi[n], {n, 85}] (* _Robert G. Wilson v_, Sep 06 2004 *)

%o (PARI) a(n)=moebius(n)*eulerphi(n) \\ _Charles R Greathouse IV_, Feb 21 2013

%o (PARI) for(n=1, 100, print1(direuler(p=2, n, (1 - p*X + X))[n], ", ")) \\ _Vaclav Kotesovec_, Jun 14 2020

%Y Cf. A000010, A003958, A007947, A008683, A008966, A023900, A047994, A143153, A173557, A322581.

%K sign,mult

%O 1,3

%A _Gerald McGarvey_, Sep 04 2004

%E More terms from _Robert G. Wilson v_, Sep 06 2004

%E Edited by _N. J. A. Sloane_, May 20 2006