OFFSET
1,2
COMMENTS
Comment from N. J. A. Sloane, Mar 22 2020: (Start)
Theorem: Sum_{ 1<=i<=n, 1<=j<=n, gcd(i,j)=1 } i*j = a(n).
Proof: From the Apostol reference we know that:
Sum_{ 1<=i<=n, gcd(i,n)=1 } i = n*phi(n)/2 (see A023896).
We use induction on n. The result is true for n=1.
Then a(n) - a(n-1) = 2*Sum_{ i=1..n-1, gcd(i,n)=1 } n*i = n^2*phi(n). QED (End)
REFERENCES
T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 48, problem 16, the function phi_1(n).
LINKS
Seiichi Manyama, Table of n, a(n) for n = 1..10000
FORMULA
a(n) ~ 3*n^4 / (2*Pi^2).
MATHEMATICA
Accumulate[Table[k^2*EulerPhi[k], {k, 1, 50}]]
PROG
(PARI) a(n) = sum(k=1, n, k^2*eulerphi(k)); \\ Michel Marcus, Sep 12 2018
CROSSREFS
KEYWORD
nonn
AUTHOR
Vaclav Kotesovec, Sep 10 2018
STATUS
approved