OFFSET
0,2
COMMENTS
From Petros Hadjicostas, Dec 06 2017: (Start)
The g.f. is clear from J. Arndt's PARI program below.
The CIK transform of sequence (a(n): n>=1} with g.f. A(x) = Sum_{n>=1} a(n)*x^n has g.f. CIK(A(x)) = 1 - Sum_{n>=1} (phi(n)/n)*log(1-A(x^n)). Sometimes, the constant 1 is dropped from the formula. Here, A(x) = 2*x/(1-2*x).
To find the auxiliary sequence (c(n): n>=1} used in the formula a(n) = (1/n)*Sum_{d|n} phi(n/d)*c(d), we use the formula C(x) = Sum_{n>=1} c(n)*x^n = x*(dA(x)/dx)/(1-A(x)). Here, C(x) = 2*x/((1-4*x)*(1-2*x)), from which we can prove that c(n) = 2^n*(2^n-1) = A020522(n).
(End)
LINKS
C. G. Bower, Transforms (2)
P. Flajolet and M. Soria, The Cycle Construction, SIAM J. Discr. Math., vol. 4 (1), 1991, pp. 58-60.
P. Flajolet and M. Soria, The Cycle Construction. [pdf file]
FORMULA
From Petros Hadjicostas, Dec 06 2017: (Start)
a(n) = (1/n)*Sum_{d|n} phi(n/d)*2^d*(2^d-1) = (1/n)*Sum_{d|n} phi(n/d)*A020522(d) for n >= 1.
G.f.: 1 - Sum_{n>=1} (phi(n)/n)*log((1-4*x^n)/(1-2*x^n)).
(End)
MATHEMATICA
{1}~Join~Table[(1/n) DivisorSum[n, EulerPhi[n/#] *2^#*(2^# - 1) &], {n, 24}] (* Michael De Vlieger, Dec 06 2017 *)
PROG
(PARI)
N = 66; x = 'x + O('x^N);
f(x)=sum(n=1, N, 2^n*x^n );
gf = 1 + sum(n=1, N, eulerphi(n)/n*log(1/(1-f(x^n))) );
v = Vec(gf)
/* Joerg Arndt, Jan 21 2013 */
CROSSREFS
KEYWORD
nonn
AUTHOR
Christian G. Bower, Jun 09 2000
STATUS
approved