This answer is dedicated to prove the following theorem (complementary to the accepted answer).

**Theorem 1**: At rank $\le 5$, $\varphi(f)>0$.

*Notation*: Let $d_{i_1i_2\dots i_r}:=f(\{ i_1,i_2,\dots, i_r \})$. Let $s_i:=\frac{f(\hat{1})}{f(\{i \}^{\complement})} \in \mathbb{N}_{\ge 2}$.

*Proposition 1*: $\varphi(f)>0$ at rank $1$.

*Proof*: $\varphi(f) = d_1- 1 \ge 2-1 = 1$. $\square$

*Proposition 2*: $\varphi(f)>0$ at rank $2$.

*Proof*: $\varphi(f) = d_{12}- d_1-d_2+1 \ge d_{12}- d_{12}/2-d_{12}/2+1 = 1$. $\square$

*Lemma 1*: For any $i$, and for any subset $S \subset \{1,2,\dots,n\}$ with $i\not \in S$, we have:

$$ \frac{f(S \cup \{i\})}{f(S)} \le s_i $$
*Proof*: By assumption $f(S \cup \{i\})f(\{i \}^{\complement}) \le f(\{1,2,\dots,n\}) f(S)$. $\square$

*Corollary 1*: $f(\hat{1}) \le \prod_i s_i$.

*Proof*: $f(\hat{1}) = \prod_i \frac{f(\{1,2,\dots,i\})}{f(\{1,2,\dots,i-1\})}$. The result follows by Lemma 1. $\square$

*Lemma 2*: If there is $i$ such that $s_i = 2$, then $\varphi(f) = \varphi(f_{| \ [\hat{0},\{i \}^{\complement}]})$.

*Proof*: By Lemma 1, for any $S \not \ni i$, $\frac{f(S \cup \{i\})}{f(S)} = 2$, so $\varphi(f) = 2\varphi(f_{| \ [\hat{0},\{i \}^{\complement}]})-\varphi(f_{| \ [\hat{0},\{i \}^{\complement}]})$. $\square$

*Proposition 3*: $\varphi(f)>0$ at rank $3$.

*Proof*: If there is $i$ such that $s_i=2$, then $\varphi(f)>0$ by Lemma 2 and Proposition 2.

Else, for all $i$, $s_i \ge 3$. Then $\forall i<j$, $d_{ij}\le d_{123}/3$, so $$\varphi(f) := d_{123}-d_{12}-d_{13}-d_{23}+d_1+d_2+d_3-1 \ge $$ $$ d_{123}-d_{123}/3-d_{123}/3-d_{123}/3 +d_1+d_2+d_3-1\ge 8$$ The result follows. $\square$

*Proposition 4*: $\varphi(f)>0$ at rank $4$.

*Proof*: First we can assume $s_i \ge 3$, for all $i$, by Lemma 2 and Proposition 3. Next, if $\sum_i s_i^{-1} \le 1$, then we can use the argument of the proof of Proposition 3. So we can assume that $\sum_i s_i^{-1} > 1$. We are reduced to the following cases: $(s_1,s_2,s_3,s_4) = $

- $(3,3,3,r)$, for $r \in \mathbb{N}_{\ge 3}$.
- $(3,3,4,r)$, for $r=4,5, \dots, 11$. Then $f(\hat{1}) \le 3 \cdot 3 \cdot 4 \cdot 11 = 396$, by Corollary 1.
- $(3,3,5,r)$, for $r=5,6,7$; and so $f(\hat{1}) \le 3 \cdot 3 \cdot 5 \cdot 7 = 315$.
- $(3,4,4,r)$, for $r=4,5$; and so $f(\hat{1}) \le 3 \cdot 4 \cdot 4 \cdot 5 = 240$.

We quickly check by this code that if $\forall i$, $s_i \ge 3$, $\sum_i s_i^{-1} > 1$ and $f(\hat{1}) \le 400$, then $\varphi(f) \ge 5$:

```
gap> CombinatEuler4(400);
[ [ [ 1 ], [ 2, 2, 2, 2 ], [ 4, 4, 4, 4, 4, 4 ], [ 12, 12, 12, 12 ], [ 36 ] ], 5 ]
```

So we are reduced to the first case, for $r \ge 400/27 > 14$. Then $d_{123} = d_{1234}/r < d_{1234}/14$, but $d_{14} \ge d_{1234}/9$, so $d_{14} > d_{123}$ ; it follows easily that $\varphi(f) \ge 13$. $\square$

*Remark 1*: At rank $\le 8$, if $\sum_i s_i^{-1} \le 1$, then $\varphi(f)>0$, because ${n \choose 3}/{n \choose 2} \le 2$ iff $n \le 8$.

*Proposition 5*: $\varphi(f)>0$ at rank $5$.

*Proof*: We can take $s_1 \le s_2 \le \cdots \le s_5$, and by Lemma 2 and Proposition 4, we can assume that $3 \le s_1$. Moreover, by Remark 1, we can assume $\sum_{i=1}^5 s_i^{-1} > 1$. Then, we check by this code that if $f(\hat{1}) \le 10000$, we have $\varphi(f) \ge 1$:

```
gap> CombinatEuler5(10000);
[ [ [ 1 ], [ 2, 2, 2, 2, 2 ], [ 4, 4, 4, 4, 4, 4, 4, 4, 4, 4 ], [ 8, 8, 8, 8, 8, 8, 8, 8, 8, 8 ], [ 24, 24, 24, 24, 24 ], [ 72 ] ], 1 ]
```

If $\sum_{i=1}^4 s_i^{-1} \le 1$ then $d_{12345} \ge d_{2345} + d_{1345} + d_{1245} + d_{1235}$. If moreover $s_5 \ge s_1s_2$, then $d_{345} \ge d_{1234}$, and so $\varphi(f) > 0$; so we can assume $s_5 < s_1s_2$, but then we can check by the following code that $\prod_{i=1}^5 s_i \le 5808$, realized by $(s_1,s_2,s_3,s_4,s_5) = (3,4,4,11,11)$.

```
r:=5800;
for s1 in [3..4] do for s2 in [s1..6] do for s3 in [s2..9] do for s4 in [s3..s1*s2-1] do for s5 in [s4..s1*s2-1] do
if (1/s1 + 1/s2 + 1/s3 + 1/s4 <= 1) and (1/s1 + 1/s2 + 1/s3 + 1/s4 + 1/s5 > 1) then r0:=s1*s2*s3*s4*s5;
if r0>r then r:=r0; Print([[s1,s2,s3,s4,s5],r]);
fi; fi; od; od; od; od; od;
```

But $f(\hat{1}) \le \prod_{i=1}^5 s_i \le 5808 < 10000$, so by the checking above, $\varphi(f) \ge 1$.

So, we can assume $\sum_{i=1}^4 s_i^{-1} > 1$. If $s_4 \ge s_1s_2$ and $s_5 \ge s_1s_3$, then $d_{245} \ge d_{1234}$ and $d_{345} \ge d_{1235}$, so that $\varphi(f) > 0$; so we can assume that $s_4 < s_1s_2$ or $s_5 < s_1s_3$. If $s_4 < s_1s_2$ and $s_5 < s_1s_3$, then we check as before that $\prod_{i=1}^5 s_i \le 4410$, realized by $(3, 3, 5, 7, 14)$. If $s_4 \ge s_1s_2$ and $s_5 < s_1s_3$, then we check as before that $\prod_{i=1}^5 s_i \le 4356$, realized by $(3, 3, 4, 11, 11)$. So we can assume $s_4 < s_1s_2$ and $s_5 \ge s_1s_3$. Now let $\varphi_1$ be $d^{-1}_5 \varphi(f_{| \ [\{5 \},\hat{1}]})$ and let $\varphi_2$ be $\varphi(f_{| \ [\hat{0},\{5 \}^{\complement}]})$. Then $\varphi(f) = d_5 \varphi_1 - \varphi_2$. But, $$s_5 d_{1234} = d_{12345} \le (\prod_{i=1}^4 s_i)d_5. $$ Now $ s_i \ge 3$ and $\sum_{i=1}^4 s_i^{-1} > 1$, so by the proof of Proposition 4, we have $\varphi_1 \ge 5$, moreover we observe that $$\varphi_2 \le d_{1234} – d_{234} \le d_{1234} (1-1/s_1).$$ It follows that $$\varphi(f) = d_5 \varphi_1 - \varphi_2 \ge d_{1234}(\frac{5s_5}{\prod_{i=1}^4 s_i} - \frac{s_1-1}{s_1})$$

So, if $s_5 > (s_1-1)s_2s_3s_4/5 $ then $\varphi(f) > 0$; so we can assume $s_5 \le (s_1-1)s_2s_3s_4/5.$

Now $d_{1234} \le d_1 s_2 s_3 s_4$, and if $d_1 = s_1$, then (as for the proof of Lemma 2) $$\varphi(f) = (s_1 – 1) \varphi(f_{| \ [\hat{0},\{1 \}^{\complement}]}) >0$$ by Proposition 4; so we can assume that $d_1 \le s_1-1$, so that $d_{1234} \le (s_1-1) s_2s_3s_4.$ Then, $$d_{12345}= s_5 d_{1234} \le [(s_1-1)s_2s_3s_4]^2/5,$$ but for $s_4 < s_1s_2$ and $\sum_{i=1}^4 s_i^{-1} > 1$, we check as before that $(s_1-1)s_2s_3s_4 \le 210$, realized by $(s_1,s_2,s_3,s_4) = (3,3,5,10)$, so $$f(\hat{1}) = d_{12345} \le 210^2/5 = 8820 < 10000,$$ and the result follows by the checking above. $\square$

**About rank 6**

After some investigations with this code, it seems that if $\varphi(f) \le 0 $, then there is an atom $a$ such that all the maximal chains of $[\hat{0}, a^{\complement}]$ are labeled by $(1,2,4,8,16,48)$ expect $k ( = 0, 1 )$ of them with $(1,2,4,8,24,48)$, and those of $[a, \hat{1}]$ by $(2r,4r,8r,16r,48r,144r)$ with $r \ge 1$. Then, $\varphi(f)=2r +8k-17$ with $1 \le r \le 8$ for $k=0$ and $1 \le r \le 4$ for $k=1$, and $f(\hat{1})\le 1152$.

*Question 1*: Is it the complete classification at rank $6$ with $\varphi(f) \le 0 $ ?

It is checked for $f(\hat{1})\le 500$. It would follow that, at rank $6$, $\varphi(f) \ge -15 $ and $\varphi(f) \neq 0$.

Note that these series generalizes as follows :

There is an atom $a$ such that all the maximal chains of $[\hat{0}, a^{\complement}]$ are labeled by $$(1,m,m^2, \dots , m^{n-2},m^{n-2}x)$$ expect $k$ of them with $$(1,m,m^2, \dots , m^{n-3},m^{n-3}x, m^{n-2}x),$$ and those of $[a, \hat{1}]$ by $$(mr,m^2r, \dots , m^{n-2}r, m^{n-2}rx, m^{n-2}rx^2),$$ with $2 \le m \le x$, $k \ge 0$ and $r \ge 1$. Then $$\varphi(f) = (m-1)^{n-1}(rm-1) + m^{n-3}(x-m)(m[r(m+x+1-n)-1)]+ k).$$ Note that $\varphi(f) \equiv (-1)^n \mod m$, which proves that $\varphi(f) \neq 0$. A necessary condition for having $\varphi(f) < 0 $ is $m < x$ and $n \ge m+x+1$, so the smallest possible $n$ is $6$ with $m=2$ and $x=3$ (and $k=0,1$), which is exactly the series above.