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Let $[n]:=\{1,\cdots,n\}$. It is known that $\{\log(p) \mid p \text{ is prime }\}$ is linearly independent over $\mathbb{Q}$. For a subset $A \subset [n]$ we can consider the matrix $L(A):=(\log(x) \mid x \in A)$ with $\operatorname{rank}_{\mathbb{Q}}(L(A)) = \omega(\prod_{x \in A} x )$, where $\omega(n)$ counts the distinct prime divisors of $n$. Then we have:

$$\sum_{ A \subset [n]} (-1)^{|A|} \omega\left(\prod_{x \in A} x\right) = 0$$

Which denoting $(n;k) := \#\{A \subset [n]\mid \omega(\prod_{x \in A} x) = k\}$ for $k=0,\cdots,n$ might be rewritten as:

$$\sum_{k=0}^n (-1)^k (n;k)=0$$

My question is, if this last quantity can be interpreted as one Euler characteristic [https://en.wikipedia.org/wiki/Euler_characteristic#Topological_definition]?

Here are some numbers $(n;k)$ computed:

1 [2]
2 [2, 2]
3 [2, 4, 2]
4 [2, 8, 6, 0]
5 [2, 10, 14, 6, 0]
6 [2, 10, 30, 22, 0, 0]
7 [2, 12, 40, 52, 22, 0, 0]
8 [2, 20, 80, 108, 46, 0, 0, 0]
9 [2, 24, 148, 232, 106, 0, 0, 0, 0]
10 [2, 24, 180, 488, 330, 0, 0, 0, 0, 0]
11 [2, 26, 204, 668, 818, 330, 0, 0, 0, 0, 0]
12 [2, 26, 332, 1308, 1714, 714, 0, 0, 0, 0, 0, 0]
13 [2, 28, 358, 1640, 3022, 2428, 714, 0, 0, 0, 0, 0, 0]
14 [2, 28, 390, 2280, 5646, 5884, 2154, 0, 0, 0, 0, 0, 0, 0]

Thanks for your help.

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  • 1
    $\begingroup$ +1, I had never seen this identity pretty cool it if I understand correctly it gives 0 because it is an inclusion-exclusion formula, I read this question and came back to see if someone answered but it was deleted so is nice too see it back $\endgroup$
    – Dabed
    Commented Mar 8, 2020 at 20:04
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    $\begingroup$ @DanielD. thanks for your comment $\endgroup$
    – user6671
    Commented Mar 8, 2020 at 20:08
  • $\begingroup$ we can write $\omega( \prod_{x\in A} x ) $ as the sum over $p<n$ of $1$ if $p$ divides $x$ for some $x$ in $A$ and $0$ otherwise. Exchanging the order of summation proves your identity. This makes me think, if it's an Euler characteristic, this will just be a space that is a disjoint union over the primes. $\endgroup$
    – Will Sawin
    Commented Mar 9, 2020 at 1:58
  • $\begingroup$ Euler characteristic is based on the equation between the alternative sums of ranks of two kinds of groups (chains vs homology). Do you have two seemingly unrelated kinds of objects, etc. ? $\endgroup$
    – Wlod AA
    Commented Mar 9, 2020 at 10:50
  • $\begingroup$ $\sum_{k=0}^n(-1)^k\cdot\binom nk = 0\ $ because $(n-1)$-simplex is acyclic. Also, $\ \sum_{k=0}^n (-1)^k\cdot 2^{n-k}\cdot\binom nk = 1\ $ because $n$-cube is acyclic, etc. $\endgroup$
    – Wlod AA
    Commented Mar 9, 2020 at 11:24

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