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Is the formula for nth term of sequence http://oeis.org/A057660 is multiplicative when the numbers are coprime ? If yes how to prove it? And what is the answer when they are not coprime.

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Your function is multiplicative.

First look at the function $g(n)=n\phi(n)$. This function is multiplicative, using the fact that $\phi$ is multiplicative: For relatively prime $m,n$ we have $g(mn)=mn\phi(mn)=mn\phi(m)\phi(n)=g(m)g(n)$.

But then using the result if a function $f(n)$ is multiplicative, then so is $\sum_{d\mid n} f(d)$ (see for instance https://proofwiki.org/wiki/Sum_Over_Divisors_of_Multiplicative_Function), we get that your function, which is equal to $\sum_{d\mid n} g(d)$, is also multiplicative.

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  • $\begingroup$ Thanks but what will be the relation for non coprimes ? $\endgroup$ – Vishvajeet Patil Apr 14 '15 at 11:24
  • $\begingroup$ Well, it won't be multiplicative for non-coprimes. One can determine the values at prime powers, which then determines the values of the function. $\endgroup$ – paw88789 Apr 14 '15 at 11:43

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