OFFSET
1,2
COMMENTS
Equals A127569 * [1, 2, 3, ...]. - Gary W. Adamson, Jan 19 2007
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..10000
Wenchang Chu, Jordan's totient function and trigonometric sums, Studia Scientiarum Mathematicarum Hungarica 57 (1), 40-53 (2020).
Nikolai Osipov, Problem 12003, Amer. Math. Monthly, 124(8) (2017), p. 754.
FORMULA
a(n) = Sum_{d|n} d^2*phi(n/d). - Vladeta Jovovic, Jul 31 2002
a(n) = Sum_{k=1..n} gcd(n, k)^2. - Vladeta Jovovic, Aug 27 2003
Dirichlet g.f.: zeta(s-2)*zeta(s-1)/zeta(s). - R. J. Mathar, Feb 03 2011
a(n) = n*Sum_{d|n} J_2(d)/d, where J_2 is A007434. - Enrique Pérez Herrero, Feb 25 2012.
G.f.: Sum_{n >= 1} phi(n)*(x^n + x^(2*n))/(1 - x^n)^3 = x + 5*x^2 + 11*x^3 + 22*x^4 + .... - Peter Bala, Dec 30 2013
Multiplicative with a(p^e) = p^(e-1)*(p^e*(p+1)-1). - R. J. Mathar, Jun 23 2018
Sum_{k=1..n} a(k) ~ Pi^2 * n^3 / (18*zeta(3)). - Vaclav Kotesovec, Sep 18 2020
a(n) = Sum_{k=1..n} (n/gcd(n,k))^2*phi(gcd(n,k))/phi(n/gcd(n,k)). - Richard L. Ollerton, May 07 2021
From Peter Bala, Dec 26 2023: (Start)
For n odd, a(n) = Sum_{k = 1..n} gcd(k,n)/cos(k*Pi/n)^2 (see Osipov and also Chu, p. 51).
It appears that for n odd, Sum_{k = 1..n} (-1)^(k+1)*gcd(k,n)/cos(k*Pi/n)^2 = n. (End)
a(n) = Sum_{1 <= i, j <= n} gcd(i, j, n). Cf. A360428. - Peter Bala, Jan 16 2024
Sum_{k=1..n} a(k)/k ~ Pi^2 * n^2 / (12*zeta(3)). - Vaclav Kotesovec, May 11 2024
MATHEMATICA
A069097[n_]:=n^2*Plus @@((EulerPhi[#]/#^2)&/@ Divisors[n]); Array[A069097, 100] (* Enrique Pérez Herrero, Feb 25 2012 *)
f[p_, e_] := p^(e-1)*(p^e*(p+1)-1); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 18 2020 *)
PROG
(PARI) for(n=1, 100, print1((sumdiv(n, k, k*sigma(k)*moebius(n/k))), ", "))
CROSSREFS
KEYWORD
easy,nonn,mult
AUTHOR
Benoit Cloitre, Apr 05 2002
STATUS
approved