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Binomial products of groups (1 + g q)

I'm a recreational mathematician (as opposed to a real one) who's got a program related to finding products of binomials of groups, as in related to zeta functions. I'm light on the terminology, but the code is solid.
This multiplies binomials of all the members of a group, g, times a constant, q, as in ( 1 + g q ). And it takes this product repeatedly to get the terms to cancel into a real polynomial.
To get it working now, copy all the code and run every function in the order it's written. The list of numbers in the following algorithm can be changed to any integers to make nice polynomials. The larger number at the end is the precision, and 500 should be mostly okay.
Simplify[ hiBinomialProduct[ hiWorkingGroupPrim[{1, 2, 3}], 500] ] 

I had the functions split up this way for debugging, but it works well for explaining what's going on, too. Notes are included with each function. The code blocks are formatted with the function, then the test case, followed by the result you should expect.

hiNewEigenrot yields what I think is called a simple group. Or close enough to it. A clock-arithmetic group.
'fram' is best-used as two consecutive '1's, or two wrapping around as below. If there are an even number of members of 'fram', then one of the '1's is negative in the wrap-around. The 'cycl' term I think is called the generator.
hiNewEigenrot[fram_, cycl_] := Table[fram/cycl*cind, {cind, 0, cycl - 1}]; hiNewEigenrot[{-1, 0, 0, 1}, 4] (* {{0, 0, 0, 0}, {-(1/4), 0, 0, 1/4}, {-(1/2), 0, 0, 1/2}, {-(3/4), 0, 0, 3/4}} *) 
This does the same thing, except it only includes primitive roots, and results down-the-way in smaller polynomials.
hiNewEigenrotPrim[fram_, cycl_] := Table[If[CoprimeQ[cind, cycl], fram/cycl*cind, Nothing], {cind, 0, cycl - 1}]; hiNewEigenrotPrim[{-1, 0, 0, 1}, 4] (* {{-(1/4), 0, 0, 1/4}, {-(3/4), 0, 0, 3/4}} *) 
You can choose to create a group for each consecutive pair. So above, you'd have a group with a 'fram' of each of:
{1, 1, 0, 0} {0, 1, 1, 0} {0, 0, 1, 1} {-1, 0, 0, 1} 

hiMultiplyEigenrots multiplies two of the above groups together. Order is irrelevant, and it nests:
hiMultiplyEigenrots[eigenSet_, eigenLayer_] := Flatten[Table[esnd + eind, {esnd, eigenSet}, {eind, eigenLayer}], 1]; hiMultiplyEigenrots[{{0, 0}, {1/2, 1/2}}, {{0, 0}, {1/2, 3/2}}] (* {{0, 0}, {1/2, 3/2}, {1/2, 1/2}, {1, 2}} *) 
Think of these as the product of Cis functions, where each imaginary variable is unique. So the second term above is ( Cos( 1/2 π ) + i Sin( 1/2 π ) ) ( Cos( 3/2 π ) + j Sin( 3/2 π ) ).

hiTuplate takes the above group and negates some of the angles. This is done so that things cancel into the reals, but a quick explanation escapes me on this. Nested complex-conjugates?
hiTuplate[angles_] := DeleteDuplicates[ Times[angles, #] & /@ Tuples[{1, -1}, Length[angles]], SameQ[#1, -#2] &]; hiTuplate[{a, b, c}] (* {{a, b, c}, {a, b, -c}, {a, -b, c}, {a, -b, -c}} *) 

hiEigenList takes a length and makes a list of 'fram's for the hiWorkingGroup functions.
hiEigenList[len_] := Table[ Table[If[inr == eig, 1, If[inr == eig + 1, 1, If[inr == 1 && eig == len, If[EvenQ[eig], -1, 1], 0], 0], 0], {inr, len}], {eig, len}]; hiEigenList[4] (* {{1, 1, 0, 0}, {0, 1, 1, 0}, {0, 0, 1, 1}, {-1, 0, 0, 1}} *) 

hiWorkingGroup will automatically create a group with a list of generators [numbers].
hiWorkingGroup[genList_] := Flatten[hiTuplate[#] & /@ Fold[hiMultiplyEigenrots, Table[ hiNewEigenrot[hiEigenList[Length[genList]][[rot]], genList[[rot]]], {rot, Length[genList]}]], 1]; hiWorkingGroup[{1, 2, 1}] (* {{0, 0, 0}, {0, 1/2, 1/2}, {0, 1/2, -(1/2)}, {0, 1/2, 1/2}, {0, 1/2, -(1/2)}} *) 
hiWorkingGroupPrim is the same thing, but only includes primitive roots.
hiWorkingGroupPrim[genList_] := Flatten[hiTuplate[#] & /@ Fold[hiMultiplyEigenrots, Table[ hiNewEigenrotPrim[hiEigenList[Length[genList]][[rot]], genList[[rot]]], {rot, Length[genList]}]], 1]; hiWorkingGroupPrim[{2, 3, 5}] (* {{7/10, 5/6, 8/15}, {7/10, 5/6, -(8/15)}, {7/10, -(5/6), 8/15}, {7/10, -(5/6), -(8/15)}, {9/10, 5/6, 11/15}, {9/10, 5/6, -(11/15)}, {9/10, -(5/6), 11/15}, {9/10, -(5/6), -(11/15)}, {11/10, 5/6, 14/15}, {11/10, 5/6, -(14/15)}, {11/10, -(5/6), 14/15}, {11/10, -(5/6), -(14/15)}, {13/10, 5/6, 17/15}, {13/10, 5/6, -(17/15)}, {13/10, -(5/6), 17/15}, {13/10, -(5/6), -(17/15)}, {7/10, 7/6, 13/15}, {7/10, 7/6, -(13/15)}, {7/10, -(7/6), 13/15}, {7/10, -(7/6), -(13/15)}, {9/10, 7/6, 16/15}, {9/10, 7/6, -(16/15)}, {9/10, -(7/6), 16/15}, {9/10, -(7/6), -(16/15)}, {11/10, 7/6, 19/15}, {11/10, 7/6, -(19/15)}, {11/10, -(7/6), 19/15}, {11/10, -(7/6), -(19/15)}, {13/10, 7/6, 22/15}, {13/10, 7/6, -(22/15)}, {13/10, -(7/6), 22/15}, {13/10, -(7/6), -(22/15)}} *) 

hiAnglestoVars converts terms in the group into complex-ish numbers. If there's no name for them, I like "hyperimaginary" because it's easy. The imaginary number is represented as 'iN' followed by an index. The "NHoldAll" attribute is to keep Mathematica from evaluating the index.
hiAnglestoVars[th_] := Product[Cos[Pi*th[[thind]]] + iN[thind]*Sin[Pi*th[[thind]]], {thind, Length[th]}]; SetAttributes[iN, NHoldAll]; hiAnglestoVars[{1/2, 1/2}] (* iN[1] iN[2] *) 

hiSquarestoNeg is used in the algorithm below to repeatedly reduce squares of imaginary numbers to (-1).
hiSquarestoNeg[expr_] := ReplaceAll[{iN[_]^2 -> (-1)}][Expand[expr]]; hiSquarestoNeg[iN[5]^2 + iN[7]] (* -1 + iN[7] *) 

CleanPolynomial is another quality-of-life function. It might need some adjustment for very large inputs.
CleanPolynomial[polyn_] := Rationalize[ Collect[Chop[hiSquarestoNeg[Expand[Rationalize[polyn]]]], q]]; 

qNomial just gives a polynomial in q quickly from a list of coefficients. It's for messing with the results to see if you can get pieces to multiply through.
qNomial[colist_] := Sum[colist[[t + 1]]*q^t, {t, 0, Length[colist] - 1}]; qNomial[{-1, 5, 6, 1}] (* -1 + 5 q + 6 q^2 + q^3 *) 

hiBinomialProduct uses the group in 'testSet' to be the coefficients binomials in q, then multiplies the binomials together. The result returned is cleaned, but the result is stored dirty in hiBinoResult. 'prec' loosely controls the amount of memory you're using. 'prec'=~500 should be good for initial tests.
hiBinomialProduct[testSet_, prec_] := { hiBinoResult = N[(1 + q*hiSquarestoNeg[hiAnglestoVars[testSet[[1]]]]), prec]; For[indx = 2, indx <= Length[testSet], indx++, hiBinoResult = N[hiSquarestoNeg[ hiBinoResult*(1 + q*N[hiSquarestoNeg[hiAnglestoVars[testSet[[indx]]]], prec])], prec]; ]; CleanPolynomial[hiBinoResult]}; 

And to throw it all together and get the abelian group you want into the hiBinomialProduct function, here's the easy-button with a bunch of test data.
Simplify[ hiBinomialProduct[ hiWorkingGroupPrim[{2, 2}], 500] ] (* { (-1 + q)^2 } *) 

Simplify[ hiBinomialProduct[ hiWorkingGroupPrim[{2, 3}], 500] ] (* { (-1 + q)^2 (1 - q + q^2) } *) 

Simplify[ hiBinomialProduct[ hiWorkingGroup[{2, 1, 4}], 500] ] (* { (-1 + q)^8 (1 + q)^13 (1 + q^2)^4 } *) 

Simplify[ hiBinomialProduct[ hiWorkingGroup[{2, 4, 6}], 500] ] (* {(-1 + q)^48 (1 + q)^45 (1 + q^2)^16 (1 - q^2 + q^4)^4 (1 + q^2 + q^4)^12} *) 

Simplify[ hiBinomialProduct[ hiWorkingGroup[{3, 5, 7}], 500] ] (* {(1 + q)^69 (1 - q + q^2)^36 (1 - q + q^2 - q^3 + q^4)^22 (1 - q + q^2 - q^3 + q^4 - q^5 + q^6)^16 (1 + 2 q + q^2 - q^3 - 2 q^4 - 2 q^5 - q^6 + q^10 + 2 q^11 + 2 q^12 + q^13 - q^15 - 2 q^16 - 2 q^17 - q^18 + q^20 + q^21 + q^24 + q^25 - q^27 - q^28 + q^30 + q^31 - q^33 - q^34 - q^35 - q^36 + q^38 + q^39 + q^40 + q^41 + q^42 + q^43 - 2 q^45 - 3 q^46 - 2 q^47 + q^49 + q^50 + q^51 + q^52 + q^53 + q^54 - q^56 - q^57 - q^58 - q^59 + q^61 + q^62 - q^64 - q^65 + q^67 + q^68 + q^71 + q^72 - q^74 - 2 q^75 - 2 q^76 - q^77 + q^79 + 2 q^80 + 2 q^81 + q^82 - q^86 - 2 q^87 - 2 q^88 - q^89 + q^90 + 2 q^91 + q^92)} *) 
To omit all the subgroups, use the hiWorkingGroupPrim function instead:
Simplify[ hiBinomialProduct[ hiWorkingGroupPrim[{3, 5, 7}], 500] ] (* {(1 - q + q^2)^24 (1 - q + q^2 - q^3 + q^4)^12 (1 - q + q^2 - q^3 + q^4 - q^5 + q^6)^8 (1 - q + q^2 + q^5 - q^6 + 2 q^7 - q^8 + q^9 + q^12 - q^13 + q^14 - q^15 + q^16 - q^17 - q^20 - q^22 - q^24 - q^26 - q^28 - q^31 + q^32 - q^33 + q^34 - q^35 + q^36 + q^39 - q^40 + 2 q^41 - q^42 + q^43 + q^46 - q^47 + q^48)} *) 
I promised more binomials in the result, and I want to demonstrate the use of the qNomial function a bit.
Expand[Times[(1 - q + q^2 + q^5 - q^6 + 2 q^7 - q^8 + q^9 + q^12 - q^13 + q^14 - q^15 + q^16 - q^17 - q^20 - q^22 - q^24 - q^26 - q^28 - q^31 + q^32 - q^33 + q^34 - q^35 + q^36 + q^39 - q^40 + 2 q^41 - q^42 + q^43 + q^46 - q^47 + q^48), qNomial[{1, 1}], qNomial[{1, 0, 0, -1, 0, 0, 1}], qNomial[{1, 0, 0, 0, 0, -1, 0, 0, 0, 0, 1}], qNomial[{1, 0, 0, 0, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0, 1}]]] (* 1 + q^9 - q^35 - q^44 + q^70 + q^79 *) 

I'm working on some other things and I've maybe put another restriction on zeros in the Riemann zeta function, like how the duplication formula for Gamma works. And when you're doing that and you're not finding anything in literature about it, it's time to say something. This is the piece that I think is responsible. I'm having difficulty explaining it in writing, but the compiler doesn't get lost after five solid pages of math.
So I'm guessing this is one of two things: somebody's going to say "this is Derpy's theorem, we know about this already, and it's way more complicated because....", or "this is Derpy's theorem, this is unproven, where'd you get this???". I've scratched the surface of group theory, barely, and it seems overcomplicated for what it's doing. So this is either a specific case or a generalization, and I'm not sure which, and I don't think I'll figure it out anytime this year.
Otherwise, just have fun with it. Take, use, modify, enjoy.
Thanks!
( all tested in v 12.0.0.0 64-bit )

Addition:
In regards to the Riemann zeta function, there's an obscure function called the Burgess zeta function. Outside of me, I think there's about 3 references to it anywhere on the Internet, so the notation is going to just be close-enough:

Burgess zeta function, modified notation
I'm using the above to find (more) multiplication formulae. There's code included, but I think the error term gets out of control quickly.
As before, you can copy and run the code. It's not linked to the earlier section, and I might not do that ever.

bzSumModZeta converts a Hurwitz zeta function from (Σ (p/q)^-s + (1+p/q)^-s + (2+p/2)^-s + ...) into (Σ (p)^-s + (q+p)^-s + (2q+p)^-s + ...)
bzSumModZeta[s_, r_, d_] = Zeta[s, d]*d^-s; 

bzTestZetaSum takes the sum of (n^-s) of a set of n
bzTestZetaSum[s_, set_] := Sum[n^-s, {n, set}]; 

bzFiniteSetBurgZeta returns the members of the summation of a Burgess zeta function, p%q, up to the max. If this code were a machine, it would be held together with band-aids and bubblegum, but it works. This, in conjunction with bzTestZetaSum, is for testing the accuracy of the relationships that follow east of Re(s)=1.
bzFiniteSetBurgZeta[p_, q_, max_] := If[p != 1, Prepend[#, 1], #] [email protected] (Last[#] & /@ (Select[ MapThread[{#1, #2} &, { Thread[Mod[#, q]] & /@ (First[#] & /@ FactorInteger[#] & /@ Range[max]), Range[max]}], DeleteDuplicates[First[#]] == {p} &])); bzFiniteSetBurgZeta[2, 3, 100] (* {1, 2, 4, 5, 8, 10, 11, 16, 17, 20, 22, 23, 25, 29, 32, 34, 40, 41, 44, 46, 47, 50, 53, 55, 58, 59, 64, 68, 71, 80, 82, 83, 85, 88, 89, 92, 94, 100} *) 

bzBurgZetaFinite is for testing, and only converges east of Re(s)=1.
bzBurgZetaFinite[s_, p_, q_, setPrec_] := bzTestZetaSum[s, bzFiniteSetBurgZeta[p, q, setPrec]]; N[bzBurgZetaFinite[2 + 3 I, 1, 3, 150]] (* 1.0175405101 + 0.0031633343783 I *) 

The binomial product thing above is designed to do the heavy-lifting for coming up with multiplication formulae. It'll take some pages to explain, and I'm way out of my depth, so here's some easier-to-interpret results instead:


Here's a short list of duplication formulae, to get the idea across:

Burgess zeta duplication formulae
And code to verify with. bzHyperModZeta2of3 is the expression on the left-hand side for %3 above. bzHyperModZeta2of3Finite is the right-hand side, which is only convergent east of Re(s)=1. The final expression is an accuracy test.
bzHyperModZeta2of3[s_] = bzSumModZeta[s, 1, 3] - bzSumModZeta[s, 2, 3]; bzHyperModZeta2of3Finite[s_, setPrec_] := bzBurgZetaFinite[s, 1, 3, setPrec]* bzBurgZetaFinite[2*s, 2, 3, setPrec]/ bzBurgZetaFinite[s, 2, 3, setPrec]; N[Abs[bzHyperModZeta2of3[#] - bzHyperModZeta2of3Finite[#, 500]]] [email protected](2 + 3 I) (* 0.00015547545069 *) 
Same format for %4:
bzHyperModZeta3of4[s_] = bzSumModZeta[s, 1, 4] - bzSumModZeta[s, 3, 4]; bzHyperModZeta3of4Finite[s_, setPrec_] := bzBurgZetaFinite[s, 1, 4, setPrec]* bzBurgZetaFinite[2*s, 3, 4, setPrec]/ bzBurgZetaFinite[s, 3, 4, setPrec]; N[Abs[bzHyperModZeta3of4[#] - bzHyperModZeta3of4Finite[#, 500]]] [email protected](2 + 3 I) (* 0.000058898498961 *) 

And %5:
bzHyperModZeta4of5[s_] = bzSumModZeta[s, 1, 5] - bzSumModZeta[s, 2, 5] - bzSumModZeta[s, 3, 5] + bzSumModZeta[s, 4, 5]; bzHyperModZeta4of5Finite[s_, setPrec_] := bzBurgZetaFinite[s, 1, 5, setPrec]* bzBurgZetaFinite[s, 4, 5, setPrec]* bzBurgZetaFinite[2*s, 2, 5, setPrec]* bzBurgZetaFinite[2*s, 3, 5, setPrec]/ (bzBurgZetaFinite[s, 2, 5, setPrec]* bzBurgZetaFinite[s, 3, 5, setPrec]); N[Abs[bzHyperModZeta4of5[#] - bzHyperModZeta4of5Finite[#, 500]]] [email protected](2 + 3 I) (* 0.00012260036466 *) 

And a couple triplication formulae:

Burgess zeta triplication formulae
Followed by code for %7 for verification:
bzHyperModZeta2of7[s_] = bzSumModZeta[s, 1, 7]^2 + bzSumModZeta[s, 2, 7]^2 + bzSumModZeta[s, 3, 7]^2 + bzSumModZeta[s, 4, 7]^2 + bzSumModZeta[s, 5, 7]^2 + bzSumModZeta[s, 6, 7]^2 - bzSumModZeta[s, 1, 7]*bzSumModZeta[s, 2, 7] - bzSumModZeta[s, 1, 7]*bzSumModZeta[s, 3, 7] - bzSumModZeta[s, 1, 7]*bzSumModZeta[s, 4, 7] - bzSumModZeta[s, 1, 7]*bzSumModZeta[s, 5, 7] + 2*bzSumModZeta[s, 1, 7]*bzSumModZeta[s, 6, 7] - bzSumModZeta[s, 2, 7]*bzSumModZeta[s, 3, 7] - bzSumModZeta[s, 2, 7]*bzSumModZeta[s, 4, 7] + 2*bzSumModZeta[s, 2, 7]*bzSumModZeta[s, 5, 7] - bzSumModZeta[s, 2, 7]*bzSumModZeta[s, 6, 7] + 2*bzSumModZeta[s, 3, 7]*bzSumModZeta[s, 4, 7] - bzSumModZeta[s, 3, 7]*bzSumModZeta[s, 5, 7] - bzSumModZeta[s, 3, 7]*bzSumModZeta[s, 6, 7] - bzSumModZeta[s, 4, 7]*bzSumModZeta[s, 5, 7] - bzSumModZeta[s, 4, 7]*bzSumModZeta[s, 6, 7] - bzSumModZeta[s, 5, 7]*bzSumModZeta[s, 6, 7]; bzHyperModZeta2of7Finite[s_, setPrec_] := (bzBurgZetaFinite[s, 1, 7, setPrec]* bzBurgZetaFinite[s, 6, 7, setPrec])^2* bzBurgZetaFinite[3*s, 2, 7, setPrec]* bzBurgZetaFinite[3*s, 3, 7, setPrec]* bzBurgZetaFinite[3*s, 4, 7, setPrec]* bzBurgZetaFinite[3*s, 5, 7, setPrec]/(bzBurgZetaFinite[s, 2, 7, setPrec]* bzBurgZetaFinite[s, 3, 7, setPrec]* bzBurgZetaFinite[s, 4, 7, setPrec]* bzBurgZetaFinite[s, 5, 7, setPrec]); N[Abs[bzHyperModZeta2of7[#] - bzHyperModZeta2of7Finite[#, 500]]] [email protected](2 + 3 I) (* 0.000072972989042 *) 

[ Quick edits correcting info from here. More complete ones to come when I have some more time. Basically the following equalities were wrong. This is what I get for doubting Wolfram. ]
And another for what I believe is the Klein four-group:

Burgess zeta multiplication formulae
I mentioned something about an error term and recovering pure Burgess zeta functions. This is edited out. Here's some code though for proving the above equalities work:

bzHyperModZetaKleinof8[s_] = (bzSumModZeta[s, 1, 8]^2 - bzSumModZeta[s, 3, 8]^2 - bzSumModZeta[s, 5, 8]^2 + bzSumModZeta[s, 7, 8]^2 - 2 bzSumModZeta[s, 1, 8]*bzSumModZeta[s, 7, 8] + 2 bzSumModZeta[s, 3, 8]*bzSumModZeta[s, 5, 8])* (bzSumModZeta[s, 1, 8] - bzSumModZeta[s, 3, 8] - bzSumModZeta[s, 5, 8] + bzSumModZeta[s, 7, 8]); bzHyperModZetaKleinof8Finite[s_, setPrec_] := bzBurgZetaFinite[s, 1, 8, setPrec]^3* bzBurgZetaFinite[2 s, 7, 8, setPrec]^2* bzBurgZetaFinite[2 s, 3, 8, setPrec]^2* bzBurgZetaFinite[2 s, 5, 8, setPrec]^2/ (bzBurgZetaFinite[s, 3, 8, setPrec]* bzBurgZetaFinite[s, 5, 8, setPrec]* bzBurgZetaFinite[s, 7, 8, setPrec]); N[Abs[bzHyperModZeta2by2for7of8[#] - bzHyperModZeta2by2for7of8Finite[#, 500]]] [email protected](2 + 113 I) (* 0.000022287674137 *) 

Nifty meaningless graph of a bunch of zeta functions
[ End of the editing ]

Also, there's implications that the concept of this post has on locations of the Riemann zeta zeros that I'm not finding in literature. And I think I'm finding papers saying this function can only be estimated up to the imaginary axis.
I am way out of my depth, and was really just expecting to get to the maw of the abyss with this problem and turn back, but never quite got there. I have no idea if these relationships are known, but I'm highly inclined to believe that they aren't, and the work will never see the light of day if not posted.
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