Factor Sum Is Odd!

The function $S$ is multiplicative and so, if we have the prime factorisation $n = p_1^{a(1)}p_2^{a(2)} \cdots p_m^{a(m)}$ , where $p_1,p_2,...,p_m$ are distinct primes, then $S(n) \; = \; \prod_{k=1}^m S\big(p_k^{a(k)}\big) \; = \; \prod_{k=1}^m \big(1 + p_k + p_k^2 + \cdots + p_k^{a(k)}\big)$ Note that $S(2^n) = 2^{n+1}-1$ is odd for all $n \ge 1$ , while $S(p^n)$ is odd precisely when $n$ is even for any odd prime $p$ .

If $S(n)$ is to be odd, we must have $S\big(p_k^{a(k)}\big)$ odd for all $k$ , and hence the index of any odd prime factor of $n$ must be even. Thus the set of numbers $n$ for which $S(n)$ is odd is the set of perfect squares, together with the set of twice the perfect squares. Thus we need to evaluate $\sum_{n=1}^\infty \frac{1}{n^2}+ \sum_{n=1}^\infty \frac{1}{2n^2} \; = \; \tfrac32\zeta(2) \; = \; \boxed{\tfrac14\pi^2}$

For $S(n)$ to be odd, $n$ must be either a square or twice a square.

Sketch of Proof: If $n$ is a square with 2 as a factor, it has an odd number of odd factors -- all those with no factor of 2. If $n$ is a square without 2 as a factor it has an odd number of factors and all of its factors are odds. If you multiply this $n$ by $2$ you don't change the number of odd factors. If $n$ is not of these forms it has an even number of odd factors and the sum will be even.

Euler proved what is known as The Basel Problem: the sum of the reciprocals of the squares equals $\frac{\pi^{2}}{6}$ .

The sum of the reciprocals of twice squares is then half of this or $\frac{\pi^{2}}{12}$ .

Add these together and you get $\frac{\pi^{2}}{4}\approx\boxed{2.4674011}$ .

We can see which numbers $n$ make $S(n)$ odd by splitting $n$ into a power of 2 and an odd factor, then considering each group of factors. For example, if $n=120$ we can split 120 into $8 \cdot 15$ and consider all of the factors of 8 (other than 1) and all the factors of 15 giving the two groups $(2,4,8) (1,3,5,15)$ . All the factors of $n$ will be given by all the numbers in the first group, all the numbers in the second group, and all the numbers obtained by multiplying together a number from each group. All the numbers in the first group are even and all the numbers obtained by crossing the two groups are also even. We can pair all of the factors in the second group with each other where the product for each pair is the odd factor of $n$ (in our example we pair 1 and 15 and pair 3 and 5). This means that we will have an even number of factors in the second group, (and all of those factors are odd) so $S(n)$ is even. The only time this argument breaks down is when we have a perfect square for the odd factor of $n$ because one number could only pair with itself and we would have an odd number of factors. Thus the only $n$ for which $S(n)$ is odd, are of the form $a^2$ (when $n$ has an even number of factors of 2) and $2a^2$ (when $n$ has an odd number of factors of 2). Thus the value we want is equal to $(1+\frac 12)(1+\frac 14+\frac 19+\frac 1{16}+\cdots)$ Here is an elementary proof that the infinite sum is equal to $\frac{\pi^2}{6}$ , so the answer is $\boxed{\frac {\pi^2}{4}}$

The function $S(n)$ is multiplicative only if the factors in the parantheses are coprime

$S(a \cdot b ) = S(a) \cdot S(b) \Longleftrightarrow mcd(a,b) =1$

For a prime $p$ , the value of the function $S(p^{k})$ is equal to the sum of all the powers of $p$ with exponent from 1 to $k$ .

$S(p^{k}) = \left( 1 + p +p^{2} + \dots + p^{k} \right)$

now it is possible to distinguish two cases:

• $p \neq 2$ . In this case $p$ is always. To obtain $S(p^{k})$ odd, it is necessary that the sum of the formula above have a number of odd terms. Then $k$ has to be even. Since $S$ ) is multiplicative, $S$ is odd if $n$ is a perfect square odd number

• $p = 2$ . In this case for every k, S(2^{k}) is a odd number. In fact $S(2^{k}) = 1+2+2^{2}+ \dots + 2^{k} = 1+2 \left( 1+2+\dots+2^{k-1} \right) = 1+ 2q$

Then combine the two cases it is possible to affirm that ${ S(n) }$ is odd if and only if n has the form $n = 2^{i} \cdot (2j+1)^{2}$ for every i and j integer non negative

Then the solution is possible to write in this way

$\sum_{\substack{n=0 \\ m=0} }^\infty \frac{1}{2^m}\frac{1}{(2n+1)^{2}} = \sum_{n=0}^\infty \frac{1}{2^n} \cdot \sum_{n=0 }^\infty \frac{1}{(2n+1)^2} = 2 \cdot \frac{3}{4} \frac{\pi^2}{6} = \frac{\pi^2}{4} \approx \boxed{2.467}$

If the prime factorization of an integer is $p_1^{e_1}p_2^{e_2} \dots p_n^{e_m}$ , then the number of positive factors is $(e_1 + 1) (e_2 + 1) \dots (e_m + 1)$ . Similarly, the number of positive factors formed only from the primes $p_i, p_{i + 1}, \dots p_m$ is $(e_i + 1) (e_{i + 1} + 1) \dots (e_m + 1)$ (including the factor 1). An integer is odd iff it can be written as a sum of integers with an odd number of odd integers in the sum. Thus, in order for $S(n)$ to be odd, there must be an odd number of odd factors of $n$ . The odd factors are precisely those which do not contain the prime number 2, so if the exponents of all primes in $n$ other than 2 are $e_1, e_2, \dots, e_m$ , then it must be the case that $(e_1 + 1) (e_2 + 1) \dots (e_m + 1)$ is odd, which means each individual exponent from $e_1$ to $e_m$ must be even. If all the exponents on the odd primes are even, then their product is the square of an odd number. So, for $S(n)$ to be odd, it must be equal to a power of 2 times the square of an odd number. The possible values of $n$ are thus of the form $(2m + 1)^2, 2(2m + 1)^2, 2^2(2m + 1)^2, \dots$ for all nonnegative integers $m$ . The sum of the reciprocals of all these values is $\sum_{i = 0}^{\infty} \sum_{m = 0}^{\infty} \frac{1}{2^i (2m + 1)^2} = \left(\sum_{m = 0}^{\infty} \frac{1}{(2m + 1)^2}\right) \sum_{i = 0}^{\infty} \frac{1}{2^i} = \left(\sum_{m = 0}^{\infty} \frac{1}{(2m + 1)^2}\right) 2$ The sum of the reciprocals of the squares of all odd positive integers can be found using the identity $\sum_{m = 1}^{\infty} \frac{1}{m^2} = \frac{\pi^2}{6}$ This sum is equal to the sum of the reciprocals of the even squares plus the sum of the reciprocals of the odd squares, or $\sum_{m = 1}^{\infty} \frac{1}{(2m)^2} + \sum_{m = 0}^{\infty} \frac{1}{(2m + 1)^2} = \frac{\pi^2}{6}$ which is equivalent to $\frac{1}{4} \sum_{m = 1}^{\infty} \frac{1}{m^2} + \sum_{m = 0}^{\infty} \frac{1}{(2m + 1)^2} = \frac{\pi^2}{6}$ which simplifies to $\frac{1}{4} \cdot \frac{\pi^2}{6} + \sum_{m = 0}^{\infty} \frac{1}{(2m + 1)^2} = \frac{\pi^2}{6}$ so $\sum_{m = 0}^{\infty} \frac{1}{(2m + 1)^2} = \frac{3}{4} \cdot \frac{\pi^2}{6}$ The final answer is thus $(\frac{3}{4} \cdot \frac{\pi^2}{6})2 = \frac{3}{2} \cdot \frac{\pi^2}{6} = \frac{\pi^2}{4} \approx 2.47$ .

It can be shown by prime factorization, S(n) is odd for n=(2^r)((2s+1)^2). Summing infinite series for (1/n), from r,s=0 to infinity gives:

Answer = (π^2)/4 ~ 2.4674