phi(n) = n*Product_{distinct primes p dividing n} (1 - 1/p).
Sum_{d divides n} phi(d) = n.
phi(n) = Sum_{d divides n} mu(d)*n/d, i.e., the Moebius transform of the natural numbers; mu() = Moebius function
A008683().
Dirichlet generating function Sum_{n>=1} phi(n)/n^s = zeta(s-1)/zeta(s). Also Sum_{n >= 1} phi(n)*x^n/(1 - x^n) = x/(1 - x)^2.
Multiplicative with a(p^e) = (p - 1)*p^(e-1). -
David W. Wilson, Aug 01 2001
a(n) = binomial(n+1, 2) - Sum_{i=1..n-1} a(i)*floor(n/i) (see
A000217 for inverse). -
Jon Perry, Mar 02 2004
It is a classical result (certainly known to Landau, 1909) that lim inf n/phi(n) = 1 (taking n to be primes), lim sup n/(phi(n)*log(log(n))) = e^gamma, with gamma = Euler's constant (taking n to be products of consecutive primes starting from 2 and applying Mertens' theorem). See e.g. Ribenboim, pp. 319-320. - Pieter Moree, Sep 10 2004
a(n) = Sum_{i=1..n} |k(n, i)| where k(n, i) is the Kronecker symbol. Also a(n) = n - #{1 <= i <= n : k(n, i) = 0}. -
Benoit Cloitre, Aug 06 2004 [Corrected by
Jianing Song, Sep 25 2018]
Conjecture: Sum_{i>=2} (-1)^i/(i*phi(i)) exists and is approximately 0.558 (
A335319). - Orges Leka (oleka(AT)students.uni-mainz.de), Dec 23 2004
a(n) = Sum_{i=1..n} floor(sigma_k(i*n)/sigma_k(i)*sigma_k(n)), where sigma_2 is
A001157.
a(n) = Sum_{i=1..n} floor(tau_k(i*n)/tau_k(i)*tau_k(n)), where tau_3 is
A007425.
a(n) = Sum_{i=1..n} floor(rad(i*n)/rad(i)*rad(n)), where rad is
A007947. (End)
phi(p*n) = phi(n)*(floor(((n + p - 1) mod p)/(p - 1)) + p - 1), for primes p. -
Gary Detlefs, Apr 21 2012
a(n) = lim_{s->1} n*zeta(s)*(Sum_{d divides n}
A008683(d)/(e^(1/d))^(s-1)), for n > 1. -
Mats Granvik, Jan 26 2017
Conjecture: a(n) = Sum_{a=1..n} Sum_{b=1..n} Sum_{c=1..n} 1 for n > 1. The sum is over a,b,c such that n*c - a*b = 1. -
Benedict W. J. Irwin, Apr 03 2017
a(n) = Sum_{j=1..n} gcd(j, n) cos(2*Pi*j/n) = Sum_{j=1..n} gcd(j, n) exp(2*Pi*i*j/n) where i is the imaginary unit. Notice that the Ramanujan's sum c_n(k) := Sum_{j=1..n, gcd(j, n) = 1} exp(2*Pi*i*j*k/n) gives a(n) = Sum_{k|n} k*c_(n/k)(1) = Sum_{k|n} k*mu(n/k). -
Michael Somos, May 13 2018
G.f.: x*d/dx(x*d/dx(log(Product_{k>=1} (1 - x^k)^(-mu(k)/k^2)))), where mu(n) =
A008683(n). -
Mamuka Jibladze, Sep 20 2018
G.f. A(x) satisfies: A(x) = x/(1 - x)^2 - Sum_{k>=2} A(x^k). -
Ilya Gutkovskiy, Sep 06 2019
a(n) > n/(exp(gamma)*log(log(n)) + 5/(2*log(log(n)))), except for n=223092870 (Rosser, Schoenfeld). -
Hugo Pfoertner, Jun 02 2020
Sum_{n >= 1} 1/a(n)^k is convergent iff k > 1.
a(2n) = a(n) iff n is odd, and, a(2n) > a(n) iff n is even. (End) [Actually, a(2n) = 2*a(n) for even n. -
Jianing Song, Sep 18 2022]
For n > 1, Sum_{k=1..n} phi^{(-1)}(n/gcd(n,k))*a(gcd(n,k))/a(n/gcd(n,k)) = 0, where phi^{(-1)} =
A023900.
For n > 1, Sum_{k=1..n} a(gcd(n,k))*mu(rad(gcd(n,k)))*rad(gcd(n,k))/gcd(n,k) = 0.
For n > 1, Sum_{k=1..n} a(gcd(n,k))*mu(rad(n/gcd(n,k)))*rad(n/gcd(n,k))*gcd(n,k) = 0.
Sum_{k=1..n} a(gcd(n,k))/a(n/gcd(n,k)) = n. (End)
a(n) = Sum_{d|n, e|n} gcd(d, e)*mobius(n/d)*mobius(n/e) (the sum is a multiplicative function of n by Tóth, and takes the value p^e - p^(e-1) for n = p^e, a prime power). -
Peter Bala, Jan 22 2024
Sum_{n >= 1} phi(n)*x^n/(1 + x^n) = x + 3*x^3 + 5*x^5 + 7*x^7 + ... = Sum_{n >= 1} phi(2*n-1)*x^(2*n-1)/(1 - x^(4*n-2)). For the first equality see Pólya and Szegő, problem 71, p. 126. -
Peter Bala, Feb 29 2024
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