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  • $\begingroup$ For $T=\emptyset$, is there a guess what are $s$ divisors of the form $1+q^i$? $\endgroup$ Commented Jun 7 at 5:43
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    $\begingroup$ @FedorPetrov, if $n$ has binary expansion $2^{a_1} + \cdots + 2^{a_s}$ with $a_k = k-1$ and either $k=s$ or $a_{k+1} > k$ then $$(q^{2^{a_1}} + 1)\cdots (q^{2^{a_k}} + 1)(q^{2^{a_{k+1}-1}}+1)\cdots(q^{2^{a_s}-1}+1)$$ $\endgroup$ Commented Jun 7 at 10:17
  • $\begingroup$ If we assume that the replacement is $q^{f(k_S, k_T)} \binom{n}{k_S}_q$ and that the factors are as mentioned in my previous comment then we have $f(0,0) \not\equiv f(1,0) \pmod 2$ and $f(0,1) \equiv f(1,1) \pmod 2$, so $f$ can't be independent of $k_T$. However, calculating possible sequences for $f(k_S, 1) \pmod 8$ there are over 1000 candidates for the first eight values in the sequence, so it's not a simple guessing game. $\endgroup$ Commented Jun 10 at 9:59