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The answer is yes, it was pointed out to me by Vitaly Bergelson that it follows from Theorem 6.1 (d) of a paper by Bergelson, Hindman and Kra (Trans. Amer. Math. Soc. 348 (1996), no. 3, 893–912). The notation $g_{\alpha,\gamma}[A]$ is defined in the previous section and means $$g_{\alpha,\gamma}[A]:=\{\lfloor \alpha n+\gamma\rfloor:n\in A\}$$ where $\lfloor\... 4 The basic idea is to separate out the action of each prime, to the maximal extent possible. In the example given in the above comments, the idea was to ignore (the remainder modulo)$2$as long as possible, ignore (the remainder modulo)$3$as long as possible given that$2$was being ignored, etc. I was inspired by the examples you posted, since almost all ... 2 It's false in general, there is a counter-example of rank$6$. Consider the labeling$f$on a boolean lattice of rank$n$such that for every maximal chain $$\hat{0} = a_0 < a_1< \cdots <a_n = \hat{1}$$ we have $$(f(a_0), f(a_1), \dots , f(a_n)) =(1,k,k^2, \dots, k^{n-2}, k^{n-2}x, k^{n-2}x^2)$$ with$k,x \in \mathbb{N}_{\ge 2}$and$k \le x $. ... 2 This answer is dedicated to prove the following theorem (complementary to the accepted answer). Theorem 1: At rank$\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) =...