More fun in 2015, Part 39

Note that if $dg = n$ then the expression is equivalent to $\sum_{g |n}{\mu(g) f(n/g)}$ which by Mobius Inversion, means that $f(n) = \sum_{d|n}{d}$ or the sum of the divisors of $n$ . The answer follows by summing the divisors of 2015.

Here is a solution for those who know the dirichlet convolution , $(f*g)(n)=\sum_{d|n}f(n/d)g(d)$ . This product turns out to be associative, with an identity $e$ given by $e(1)=1,e(n)=0$ for $n>1$ . It is easy to see that $I*\mu=e$ where $I(n)=1$ for all $n$ ; this simply means that $\sum_{d|n}\mu(d)=0$ for $n>1$ .

We are given the equation $id=\mu*f$ , the identity function. To solve for $f$ , let's multiply both sides with $I$ to find $I*id=I*\mu*f=e*f=f$ , meaning that $f(n)=\sum_{d|n}d$ , the sum of all divisors of $n$ , a multiplicative function.

Since $f(p)=p+1$ for primes, we have $f(2015)=f(5*13*31)=6*14*32=\boxed{2688}$ .

For prime powers we have $f(p^m)=1+p+p^2+...+p^m=\frac{p^{m+1}-1}{p-1}$ .

First, it is easy to see that $f(1) = 1$

Then we have $f(p) = p+1$ for every prime p because $\mu(p)f(1)+\mu(1)f(p) = p$ , that means $f(p)-f(1)=p$

Next we will prove that if $n=p_1*p_2*...*p_k$ with $p_i$ are prime, then $f(n)=f(p_1)*f(p_2)*...*f(p_k)$

From the condition we have
$\mu(1)f(n)+\sum_{p_i|n}\mu(p_i)f(\frac{n}{p_i})+\sum_{p_i,p_j|n}\mu(p_i*p_j)f(\frac{n}{p_i*p_j}) ... = n$

or

$f(n)-\sum_{p_i|n}f(\frac{n}{p_i})+\sum_{p_i,p_j|n}f(\frac{n}{p_i*p_j}) - ...+... = p_1*p_2*...*p_k = [f(p_1)-1][f(p_2)-1]...[f(p_k)-1]$

When expand the RHS, all the sum in LHS will be cancel out and that leaves $f(n)=f(p_1)*f(p_2)*...*f(p_k)$

Apply that to 2015 we have $f(2015)=f(5)*f(13)*...*f(31) = 6*14*32 = \boxed{2688}$

Note: In general, if $n=p_1^{\alpha_1}p_2^{\alpha_2}...p_k^{\alpha_k}$ :

$f(n) = \prod_{i=1}^{k}\frac{p_i^{\alpha_i+1}-1}{p_i-1}$ (sum of all divisors of n)

i am quite certain this function is multiplicative, but i dont have proof(i was trying but it became too lengthy). lets use easy method:

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put n=1 to get f(1)=1. now put n=5) to find $f(5)-f(1)=5$ , from this we find $f(5)=6$ . similarly we have $f(13)=14$ , $f(31)=32$ . now put $n=65=13*5$ , we get: $f(65)-f(5)-f(13)+f(1)=65$ . from this we get $f(65)=84$ , it seems so multiplicative! similarly we have $f(155)=192$ , $f(403)=448$ . put n=2015 to get that [f(2015)-f(65)-f(155)-f(403)+f(5)+f(13)+f(31)-f(1)=2015). plug in values to find $f(2015)=\boxed{2688}$ . as for wether a function like this exists or not, try $f(n)=n\prod_{p|n} \left(1+\dfrac{1}{p}\right)$ . this is just a function satisfying, but any other function satisfying the equation will also have f(2015)=2688.