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Take a standard finite Heisenberg group with two standard generators and let's consider its growth polynomial - the polynomial which coefficients are equal to the sphere sizes. For example for $H_3(Z/2)$ , polynomial is given by $p(t) = 1+2t+2t^2+2t^3+t^4$ : one element on distance zero, two elements on distances 1,2,3 and one on distance 4.

Question: Do we know the general form of these polynomials ? Are the generalizations  for the higher-dimensional Heisenberg groups known?

We have some pictures with roots of these polynomials for the different groups ( https://www.kaggle.com/code/mixnota/growth-polynomial-analysis ):

Can patterns seen on this data be helpful to guess properties of polynomials ? enter image description here

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PS

Googling does not help to find the answer, similar question on MSE remains unanswered: https://math.stackexchange.com/q/4868762/21498

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  • $\begingroup$ It should be that $P(t) = (t^{N}P_1(t) + P_2(t) )/Q(t) $ , where $P_2(t)/Q(t)$ is growth for INFINITE $H_3(Z)$ and N is large for large "n" ($Z/n$) . (See g.f. for oeis.org/A063810 for $P_2(t)/Q(t)$ ) . But it is NOT clear how much $P_1(t)$ depends on "n" ($Z/n$). At least it quite might depend on n mod 2 or 4 , because n mod 2 it is already present for the more simple case of the commutative group $Z/n$. $\endgroup$
    – Alexander Chervov
    Commented Feb 29 at 6:15
  • $\begingroup$ In the other words we hope we know the limit of $P_n(t)$ in t-adic topology - it is naturally the series for $H_3(Z)$ (i.e. $n=\infty$) , but what we miss - is the control of the error term. If error term reduces to some finite number of options (polynomial(s)) - then we can find it by just by brute force computing finite number of cases $H_3(Z/n)$ and solving linear equations. $\endgroup$
    – Alexander Chervov
    Commented Feb 29 at 8:43
  • $\begingroup$ Surprising: "The diameter of the Cayley graph for the finite Heisenberg group over Z/n is not explicitly provided in the given sources" scienceos.ai/… If diameter is not known, then polynomial also. But may be that AI-search is not just powerful enough. Searh request was "what is the diameter of the cayley graph for finite heisenberg group over Z/n" $\endgroup$
    – Alexander Chervov
    Commented Mar 6 at 8:07
  • $\begingroup$ With Bing AI chat: "Unfortunately, I don’t have the exact diameter for this group over (\mathbb{Z}/n\mathbb{Z}) in my current knowledge base. " bing.com/chat Query: "what is diameter of the finite Heisenberg group over Z/n" . So should be good question to benchmark AI searches :) :) $\endgroup$
    – Alexander Chervov
    Commented Mar 6 at 12:50
  • $\begingroup$ We have counted that $N=\lceil2\sqrt{n}\rceil$ , this rule is valid for all numbers n greater than 14 for which we have calculated the growth polynomial. $\endgroup$
    – Mikhail Evseev
    Commented Mar 17 at 15:48

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