Moment of a function

Define the $m$th moment of a function $f(n)$ as

$F_{(m)}(n)=\displaystyle\sum\limits_{k=1}^n k^mf(k)$

Now the same procedure also relates the moments of the three functions, since

$f(n)*g(n)=h(n)\implies n^mf(n)*n^mg(n)=n^mh(n)$

For example, since we know that $\varphi(n)*1=n$, we can use this to calculate the Summatory Totient moments.

Choosing $f(n)=\varphi(n)$ and $g(n)=1$, we have $n^k\varphi(n)*n^k=n^{k+1}$. The corresponding summation functions are

$F_k(n)=\displaystyle\sum\limits_{m=1}^n m^k\varphi(m)$, $G(n)=S_k(n)$ and $H(n)=S_{k+1}(n)$. The corresponding results are

$F_k(n)=S_{k+1}(n)-\displaystyle\sum\limits_{m=2}^n m^k F_k\left(\left\lfloor\frac{n}{m}\right\rfloor\right)$

$F_k(n)=S_{k+1}(n)-\displaystyle\sum\limits_{m=1}^{n/(u+1)}m^k\varphi(m)S_k\left(\left\lfloor\frac{n}{m}\right\rfloor\right)-\sum_{m=2}^u m^kF_k\left(\left\lfloor\frac{n}{m}\right\rfloor\right)+S_k(u)F_k\left(\left\lfloor\frac{n}{u+1}\right\rfloor\right)$


Yours Aye
Me