Ice cream and orbifold RR

Anita Buckley, Miles Reid and Shengtian Zhou

Addenda

This website accompanies the paper.

This note consists of some homework exercises and their solutions, some
computer algebra routines, as well as some remarks and speciment
calculations that are alternatives that we played with before discarding
in favour of the current draft, or just too long to fit in the main paper.
We checked all our statements repeatedly in Magma. All the
routines we give here run on the Magma online calculator
  http://magma.maths.usyd.edu.au/calc


Function to calculate the Dedekind sum polynomial by the formula of
Theorem 2.8.

R<t> := PolynomialRing(Rationals());
function DedekindPolynomial(r,L); // r an integer, L a nonempty list of integers
A := prod (1-t^ai);
h := hcf(1-t^r, A);
F = (1-t^r) div h;
return h*t*InverseMod(h*t*A,F);
end function;
DedekindPolynomial(14,[1,2,5,7]);

/* returns the answer as stated in (2.22)
-1/7*t^14 - 1/7*t^13 - 1/14*t^12 + 1/28*t^11 - 1/28*t^9 + 1/14*t^8 + 1/7*t^7 +
    1/7*t^6 + 1/14*t^5 - 1/28*t^4 + 1/28*t^2 - 1/14*t
*/


Algorithm
To calculate Porb(1/r(a1,.. an), k).

Includes sanity check: verify k + sum ai == 0 mod r.
Set A = prod (1-t^ai); set h = hcf(1-t^r, A) and F = (1-t^r)/h;

R<t> := PolynomialRing(Rationals());
function Porb(r,LL,k)
 L := [Integers() | i : i in LL]; // this allows empty list
 if (k + &+L) mod r ne 0
    then error "Error: Canonical weight not compatible";
 end if;
n := #L;
S := [GCD(a,r) : a in L];
D := (1-t^r) * &*[1-t^s : s in S]; // Denom
A := &*[(1-t^(L[i])) div (1-t^(S[i])) : i in [1..n]];
F := (1-t^r) div GCD(1-t^r, &*[1-t^s: s in S]);
dF := Degree(F);
shift0 := Ceiling((k + 1 + &+[s : s in S] + r - dF)/2);
de := Maximum(0,Ceiling(-shift0/r));
shift := shift0+de*r;
 G, al, be := XGCD(t^shift*A, F);
 return t^shift*al/(D*t^(de*r));
end function;

The tricky bit is to get the shift right. When r and ai
are not coprime, the denominator is the product of 1-t^si
where si = hcf(ai,r). It has degree r + sum si. The
denominator must be symmetric of degree k + r + sum si.
It must be a smallest residue modulo (1-t^r) modulo its
common factors (1-t^si) with (1-t^ai).


Exercise 2.1

The numerator of Porb(1/r(a), k) is either

 (I) the sum of the b terms t^{ia} for i in [0,.. b-1]
written in the interval (1.11); or

 (II) the sum of r-b terms -t^{ia+1} for i in [0,.. r-b-1]
written in the interval (1.11).

The two set of exponents are complementary, so the two
polynomials are congruent modulo (1-t^r)/(1-t).

Examples: r = 7, a = 2 and k = -2. Then
  (I) is 1 + t^2 + t^4 + t^6. The interval (1.11)
  (centred at (k+r+1)/2 and of length <= r-2) is
  [t .. t^5] so no translate of (I) fits, and the
  answer is (II) -t - t^3 - t^5.

r = 7, a = 2 and k = 5. The solution here is to shift 1 and t^2
up by 7 to give
  t^4 + t^6 + t^7 + t^9. (The interval is centred
  at 13/2 of length r-2, so [t^4.. t^9]).

r = 8, a = 3 and k = -3. Then
  (I) is 1 + t^3 + t^6, which works on the nose.

r = 8, a = 3 and k = 5. Then
  (II) as given is -t-t^4-t^7-t^10-t^13,
  and has a symmetric folding to
  -t^4-t^5-t^7-t^9-t^10, which is the answer
  (-t is promoted to -t^9 and -t^{13} demoted to -t^5).

r = 8, a = 5 and k = -5. Then the interval is centred at
(r+k+1)/2 = 2 and of length r-2, so [t^-1 .. t^5]. The
expression (I) is 1 + t^5 + t^10 + t^15 + t^20, which
is == t^-1 + 1 + t^2 + t^4 + t^5 mod 1-t^r.

r = 8, a = 5 and k = 3. The interval is centred at 6
and of length <= 6, so [t^3,..,t^9]. The terms t^10
and t^15 in (I) are not in this interval, so (I) does
not work, and we use (II) folded to -t^3-t^6-t^9.

For lots more examples, plug the following lines into the online Magma calculator:

for r in [8..12], a in [-2*r+1..0] do
  if IsCoprime(r,a) then
    k := -a; b := InverseMod(a,r);
    if ((((a mod r)*(b-1)-(k+r+1)) mod (2*r)) div r) eq 0 then ep := "(I)"; else ep := "(II)"; end if;
    MyRange := [Ceiling((k+1)/2)+1 .. Floor((k+1)/2)+r-1];
    MyNum := R!(Qorb(r,[a mod r],-a)*(1-t^r)*(1-t));
    MyExs := [Degree(t) : t in Terms(MyNum)];
    printf "Case 1/%o(%o) with k = %o. b = %o, %o. Range = %o, Exponents = %o, %o, Numerator = %o\n", r,a mod r, k, b, ep, MyRange, Min(MyExs), Max(MyExs), MyNum;
  end if;
end for;


Whether the numerator is a sum of positive or a sum of
negative terms depends on the following binary decision
tree:

Case r is even
 use (I)  <=> \overline{b(k+1)/2} \ge b
 use (II) <=> \overline{b(k+1)/2} < b.

Case r is odd and k is even
 use (I)  <=> b < r/2
 use (II) <=> b > r/2

Case r is odd and k is odd
 use (I)  <=> b is even
 use (II) <=> b is odd

(In this case, the midpoint and endpoints of the interval are half-integers,
so the expression has evenly many terms -- either b positive terms as in (I)
or r-b negative terms as in (II).)



\begin{exc} \label{exc!ice} \rm One of the following two constructions
gives the numerator of $P\orb(\frac1r(a),k)$ (here $ab\equiv 1\mod r$
and $\overline{\phantom{m}}$ denotes residue modulo $r$):

\begin{enumerate}
\renewcommand{\labelenumi}{(\Roman{enumi})}
\item $\frac{1-t^{ab}}{1-t^a} = \sum_{i=0}^{b-1} t^{ia}$. Modulo $1-t^r$
this is
\begin{equation}
\sum t^j \quad\hbox{with $j\in\bigl\{\overline{ia}\bigm| \hbox{for $0 \le i \le b-1$}\bigr\}$}.
\end{equation}
It is a sum of $b$ terms, and occurs with the natural symmetric degree
$a(b-1)\equiv k+r+1\mod r$, because $ab\equiv 1$ and $k\equiv -a$.

\item $-t\frac{1-t^{(r-a)b}}{1-t^b}=-\sum_{i=0}^{r-a-1} t^{ib+1}$.
Modulo $1-t^r$ this is
\begin{equation}
-\sum t^j \quad\hbox{with $j\in\bigl\{\overline{ib+1}\bigm| \hbox{for $0
\le i \le r-a-1$}\bigr\}$}.
\end{equation}
It is minus a sum of $r-a$ terms, and occurs with symmetric degree
$(r-a-1)b+2\equiv k+r+1\mod r$.
\end{enumerate}
The two sets are complimentary because $ia+1\equiv (i+b)a\mod r$.
Depending on the parity choices specified below, we can fold exactly one
of (I) or (II) modulo $r$ into an interval of length $r-2$ while
preserving the symmetry.

When $r$ is even
\begin{equation}
 \hbox{(I) works} \iff \overline{b(k+1)/2}\ge b, \enspace
 \hbox{(II) works} \iff \overline{b(k+1)/2}<b.
\end{equation}

When $r$ is odd and $k$ is odd
\begin{equation}
 \hbox{(I) works} \iff \hbox{$b$ is even}, \quad
 \hbox{(II) works} \iff \hbox{$b$ is odd}.
\end{equation}

When $r$ is odd and $k$ is even
\begin{equation}
 \hbox{(I) works} \iff b<r/2, \quad
 \hbox{(II) works} \iff b>r/2.
\end{equation}
\end{exc}



Check that summing the terms  1/(2*r)*m*mkbar*(r-mkbar)
in [Buckley and Szendroi] as a power series gives III_C in
[BRZ], Theorem 4.1

MAGMA

1/23(4,19)  [Buckley and Szendroi] involves 6 = InvMod(4,23).

r:=23;
X := &+[ - 1/(2*r)*m*mkbar*(r-mkbar)*t^m where mkbar is (6*m) mod r : m in [1..10*r]];
Y := &+[ u : u in Terms(X*(1-t^23)^2) | Degree(u) le 3*r];
// replace the truncated power series X by the rational function Y/(1-t^23)^2;
// Y is -51/23*t^45 - 132/23*t^44  - .. - 135/23*t^3 - 132/23*t^2 - 51/23*t

// On the other hand, calculate the Dedekind sums:

K23<ep> := CyclotomicField(23);
Mu23 := [ep^j : j in [0..22]];
Si := [1/23* Rationals()!  &+ [ e^i/((1-e^4)*(1-e^19)) : e in Mu23 | e ne 1] : i in [0..22]];

// Our sum is (noting that Si[1] is si_0 because of the way Magma does arrays).

&+ [ j*(Si[j+1] - Si[1]) * t^j  : j in [1..22]]/(1-t^23) + 23*&+ [ (Si[j+1] - Si[1]) * t^(j+23)  : j in [1..22]]/(1-t^23)^2;
$1 eq Y/(1-t^23)^2;  // gives true

&+ [ (Si[j+1]-Si[1])*(j*t^j + (23-j)*t^(23+j)) : j in [1..22]]/(1-t^23)^2;
$1 eq Y/(1-t^23)^2;  // gives true

\sum (si_j - si_0) * (jt^j + (r-j)t^(r+j)) /(1-t^23)^2




The calculation for Ex 4.3. These assume Magma and My_magmarc
(but it is easy to rewrite so that they will run in online
Magma calculator).

\begin{verbatim}
P := (1-t^40)/Denom([2,5,8,10,15]);
I := 1 + 13/75*t/(1-t)^2 + 1/1800*(t+4*t^2+t^3)/(1-t)^4;
A := Denom([2,5,8]);
h := GCD(A,1-t^15);
F := (1-t^15) div (1-t^5);
De := t*h*InverseMod(t*h*A,F);
II := De/(1-t^15);
III2 := -1/8*t*(1+t^2)/(1-t^2)^2;
III5 := 4/3*(1/5*t^3 + 1/5*t^2 + 3/5)/(1-t^5)^2;
IV5 := (4/25*t^4-32/75*t^3+4/25*t^2-4/25*t-4/5)/(1-t^5);
P eq I + II + III2 + III5 + IV5;
\end{verbatim}

\begin{verbatim}
  P := (1-t^40)/Denom([2,5,8,10,15]);
  PI := (1-4*t+7*t^2-4*t^3+t^4)/(1-t)^4;
  // or P := 1 + t^2/(1-t)^4;
  P eq PI + Qorb(15,[2,5,8],0)
    + Qorb(5,[2,3],5)/(1-t^5) + Qorb(2,[1,1],2)/(1-t^2)
    + (-3*t^3 + 2*t^4 - 3*t^5)/Denom([1,1,1,5]);
\end{verbatim}


Other examples now omitted.

\begin{exa} \rm
Consider the Calabi--Yau 3-fold $X_{66}\subset\PP(5,6,11,11,33)$. It has
$\frac13(2,2,2)$ and $\frac15(1,1,3)$ points and a $\frac1{11}(5,6)$
curve $\Ga$ of degree $\frac2{11}$ that is pure (no dissident points). We
parse its Hilbert series as
\begin{equation*}\frac{1-t^{66}}{\prod_{a\in[5,6,11,11,33]} (1-t^a)}=
P_I+P\orb(\tfrac13)+P\orb(\tfrac15)+P\per(\Ga)+P\grow(\Ga),
\end{equation*}
where our standard terms are
\begin{equation*}
P_I=1, \quad
P\orb(\tfrac13)=-\frac{t^3}{(1-t)^3(1-t^3)}, \quad
P\orb(\tfrac15)=\frac{t^3(1+t^2)}{(1-t)^3(1-t^5)},
\end{equation*}
and the contributions from $\Ga$ are
\begin{align*}
P\per(\Ga) &= -\frac{t^6+t^8}{(1-t)^3(1-t^{11})} \\
P\grow(\Ga) &= \frac{2t^8(1+t+t^2+t^3-t^4+t^5+t^6+t^7+t^8)}{(1-t)^2(1-t^{11})^2} \\
 &=2\times \frac{P\orb(\frac{1}{11}(5,6),11)}{1-t^{11}}.
\end{align*}
Moreover,
\begin{equation*}
\begin{aligned}
-t^6-t^8 &=\tfrac2{11} \cdot \Num_{D_C}- \tfrac1{11}\cdot \Num_{N_C} \\
&=\tfrac{2t}{11}(2 t^2+2 t^3+2 t^4-4 t^5-t^6-4 t^7+2 t^8+2 t^9+2 t^{10}) \\
&\qquad-\tfrac1{11}(4 t^3+4 t^4+4 t^5+3 t^6-2 t^7+3 t^8+4 t^9+4 t^{10}+4 t^{11}).
\end{aligned}
\end{equation*}
\end{exa}

\begin{exa} \rm
A harder case $X_{80}\subset\PP(3,4,15,20,38)$ (a random choice from the
vast lists of Kreuzer and Skarke). It has orbifold loci:
\begin{itemize}
\item a $\frac12$ curve $C_{80}\subset\PP(4,20,38)$
\item a dissident point $\frac1{38}(3,15,20)$ on $C_{80}$
\item $4\times\hbox{dissident points}$ $\frac14(3,3,2)$ on $C_{80}$
\item a $\frac13(1,2)$ curve $\PP(3,5)$ with normal monomials $ux^{14}$, $tz^4$
\item a dissident point $\frac1{15}(3,4,8)$ on $\frac13$ curve
\item an isolated point 1/5(3,4,3)
\end{itemize}

\end{exa}

\begin{verbatim}
PI := 1; // because the first two plurigenera are zero.
P := (1-t^80)/Denom([3,4,15,20,38]); P;
P1 := P - PI; P1;
P2 := P1 - Qorb(38,[3,15,20],0) - Qorb(15,[3,4,8],0)
  - 4*Qorb(4,[3,3,2],0) - Qorb(5,[3,4,3],0); P2;
 // The denominator drops on subtracting each term.
PartialFractionDecomposition(P2*(1-t)^4);
 // This hints as to what remaining degree of the curve is.
 // I experiment by subtracting off different degrees
P3:=P2-Qorb(2,[1,1],2)/(1-t^2); P3;
P3:=P2-Qorb(2,[1,1],2)/(1-t^2)-Qorb(3,[1,2],3)/(1-t^3); P3;
P3:=P2-Qorb(2,[1,1],2)/(1-t^2)+Qorb(3,[1,2],3)/(1-t^3); P3;
PartialFractionDecomposition(P3*(1-t)^4);
P4 := P3 - t^3/Denom([1,1,1,3]); P4;

P eq PI + Qorb(38,[3,15,20],0) + Qorb(15,[3,4,8],0)
  + 4*Qorb(4,[3,3,2],0) + Qorb(5,[3,4,3],0)
  + Qorb(2,[1,1],2)/(1-t^2) - Qorb(3,[1,2],3)/(1-t^3)
  + t^3/Denom([1,1,1,3]);

Qorb(38,[3,15,20],0)*Denom([1,1,2,38]);
t^38 - 2*t^37 + 2*t^36 - 2*t^34 + 2*t^33 + t^32 - 3*t^31
 + 3*t^30 - t^29 - t^28 + 2*t^27 - t^26 - t^25 + 3*t^24
 - 2*t^23 + t^21 - 2*t^19 + 3*t^18 - t^17 - t^16 + 2*t^15
 - t^14 - t^13 + 3*t^12 - 3*t^11 + t^10 + 2*t^9 - 2*t^8
 + 2*t^6 - 2*t^5 + t^4
The numerator is Gor symm of same degree 42 and support
[t^4 .. t^38], which has length < 38 - 2

Qorb(15,[3,4,8],0)*Denom([1,1,3,15])
t^15 - t^13 + 2*t^12 - t^10 + 2*t^8 - t^7 + t^5
The numerator in [t^5 .. t^15] of length 10 < 15 - 3

\end{verbatim}

From Theorem 4.1. The formula comes from [BuSz]. We could
spend 2 or 3 pages explaining the derivation. There are many
intermediate forms, incl. the following:

\begin{equation}
\III_C= \left( \frac{\sum_{j=1}^{s-1} j(\si_j-\si_0)t^j}{1-t^s}
+\frac{s t^s \sum_{j=1}^{s-1} (\si_j-\si_0)t^j}{(1-t^s)^2}\right)DC
\end{equation}

\begin{equation}
\III_C=\frac{\sum_{j=1}^{s-1}\, \bigl(\si_j
- \si_0\bigr)\bigl(jt^j + (s-j)t^{s+j}\bigr)}{(1-t^s)^2}DC
\end{equation}

The current version of 4.4--4.5 could probably be further
simplified: the sums are really Dedekind polynomials for
1/s(a,-a) and their derivatives.

From Theorem 4.2. This is an intermediate form of
the contribution $B_C$ (not completely correct in
this form)
\begin{equation*}
P\per = \frac1{(1-t)^3(1-t^s)}\cdot \left( \deg D|_C\cdot
\Num_{D_C}-\frac{N_C}{2 s}\cdot \Num_{N_C} \right),
\end{equation*}
and the numerators $\Num_{D_C}$, $\Num_{N_C}$ are palindromic
polynomials supported in $\left[3,\dots,s\right]$. Moreover, for
$j=3,\dots,\ru{\frac s2}+1$ the $t^j$ term in $\Num_{D_C}$ equals
\begin{equation}
\sum_{i=0}^3 (-1)^i \textstyle\binom 3i \bigl( (j-i) \si_{-j+i}-
(2-i) \si_{-2+i} \bigr),
\end{equation}
and $\Num_{N_C}$ is computed as an ice cream function from
\begin{equation}
\Num_{N_C}\times \frac{(1-t^a)(1-t^{s-a})^2}{(1-t)^3}\equiv
1+t^{s-a}\mod \frac{1-t^s}{1-t}. \qed
\end{equation}

The main thing still to untangle is the NC of the Buckley--Szendroi. I
believe that it is  a*c1(L_a) - a*c1(L_{-a}), which would have an
obvious naive generalisation to the general case of 1-dim orbifold locus
C.



========

I messed around with the formulas (4.4) and (4.5)
without checking that they are true. (1) They may
be slightly wrong. (2) It may be possible to improve
them further. (3) I think the quantity B is closely
related to t \De'. (4) It would be useful to understand
why taking the derivative of De = InvMod(A,F) gives
a solution to A^2 x B == something simple mod F.
(5) It would be useful to write down a complete
derivation of Theorem 4.1 from [BuSz] and of
Theorem 4.2 from 4.1. It would also be useful to
know in detail how this follows from [Zh], Section 5.1.

========

integrate 1 in deg 0 only -> sums to 1
chi in every degree >= 1 sums to t/(1-t) chi. Gor symm implies chi = 0.
m^2 x (-1/4 DKX^2) sums to (t+t^2)/(1-t)^3 x (-1/4 DKX^2) Gor symm implies = 0.
m (Dc_2/12 + DK^2/12) sums to D(c2/12 + DK^2) x t/(1-t)^2.

So the RR part I is

I = 1  +  D*(c1^2+c2)/12* t/(1-t)^2  +  D^3/6 *(t + 4t^2 + t^3)/(1-t)^4



Ex 2.1, long solution

The question is: when we calculate
Porb(r,[a],k) with k = r-a or 2*r-a, it is
IceCream(b/r) where a = InvMod(b,r). When do
we get (I) a positive sum of terms or
(II) negative?

r := 13;
[Numerator(Qorb(r,[a],r-a)) : a in [0..25]];
[
 0:   0,
 1:   t^13,
 2:   -t^18 - t^16 - t^14 - t^11 - t^9 - t^7,
 3:   -t^17 - t^14 - t^10 - t^7,
 4:   t^17 + t^16 + t^15 + t^13 + t^12 + t^11 + t^10 + t^8 + t^7 + t^6,
 5:   -t^16 - t^14 - t^11 - t^8 - t^6,
 6:   -t^14 - t^7,
 7:   t^13 + t^7,
 8:   -t^15 - t^14 - t^12 - t^10 - t^9 - t^7 - t^5 - t^4,
 9:   t^13 + t^9 + t^5,
10:   t^13 + t^10 + t^7 + t^4,
11:   t^13 + t^11 + t^9 + t^7 + t^5 + t^3,
12:   t^13 + t^12 + t^11 + t^10 + t^9 + t^8 + t^7 + t^6 + t^5 + t^4 + t^3 + t^2,
13:   0,
14:   -t^12 - t^11 - t^10 - t^9 - t^8 - t^7 - t^6 - t^5 - t^4 - t^3 - t^2 - t,
15:   -t^11 - t^9 - t^7 - t^5 - t^3 - t,
16:   -t^10 - t^7 - t^4 - t,
17:   -t^9 - t^5 - t,
18:   t^11 + t^10 + t^8 + t^6 + t^5 + t^3 + t + 1,
19:   -t^7 - t,
20:   t^7 + 1,
21:   t^10 + t^8 + t^5 + t^2 + 1,
22:   -t^11 - t^10 - t^9 - t^7 - t^6 - t^5 - t^4 - t^2 - t - 1,
23:   t^10 + t^7 + t^3 + 1,
24:   t^11 + t^9 + t^7 + t^4 + t^2 + 1,
25:   -t
]

Continued fraction

function CF(x);
tmp := [];
while x ne Floor(x) do
Append(~tmp,Floor(x));
x := 1/(x-Floor(x));
end while;
Append(~tmp,Floor(x));
return tmp;
end function;

\begin{exc} \label{exc!ice} \rm One of the following two constructions
gives the numerator of $P\orb(\frac1r(a),k)$ (here $ab\equiv 1\mod r$
and $\overline{\phantom{m}}$ denotes residue modulo $r$):

\begin{enumerate}
\renewcommand{\labelenumi}{(\Roman{enumi})}
\item $\frac{1-t^{ab}}{1-t^a} = \sum_{i=0}^{b-1} t^{ia}$. Modulo $1-t^r$
this is
\begin{equation}
\sum t^j \quad\hbox{with $j\in\bigl\{\overline{ia}\bigm| \hbox{for $0 \le i \le b-1$}\bigr\}$}.
\end{equation}
It is a sum of $b$ terms, and occurs with the natural symmetric degree
$a(b-1)\equiv k+r+1\mod r$, because $ab\equiv 1$ and $k\equiv -a$.

\item $-t\frac{1-t^{(r-a)b}}{1-t^b}=-\sum_{i=0}^{r-a-1} t^{ib+1}$.
Modulo $1-t^r$ this is
\begin{equation}
-\sum t^j \quad\hbox{with $j\in\bigl\{\overline{ib+1}\bigm| \hbox{for $0
\le i \le r-a-1$}\bigr\}$}.
\end{equation}
It is minus a sum of $r-a$ terms, and occurs with symmetric degree
$(r-a-1)b+2\equiv k+r+1\mod r$.
\end{enumerate}
The two sets are complimentary because $ia+1\equiv (i+b)a\mod r$.
Depending on the parity choices specified below, we can fold exactly one
of (I) or (II) modulo $r$ into an interval of length $r-2$ while
preserving the symmetry.

When $r$ is even
\begin{equation}
 \hbox{(I) works} \iff \overline{b(k+1)/2}\ge b, \enspace
 \hbox{(II) works} \iff \overline{b(k+1)/2}<b.
\end{equation}

When $r$ is odd and $k$ is odd
\begin{equation}
 \hbox{(I) works} \iff \hbox{$b$ is even}, \quad
 \hbox{(II) works} \iff \hbox{$b$ is odd}.
\end{equation}

When $r$ is odd and $k$ is even
\begin{equation}
 \hbox{(I) works} \iff b<r/2, \quad
 \hbox{(II) works} \iff b>r/2.
\end{equation}
\end{exc}


==========

// Shengtian's thesis, Program 3.2.3
// calculates Dedekind sum si_i(1/r(b1,.. bn))

function Contribution(r,LL)
QQ := Rationals();
Poly<t> := PolynomialRing(QQ);
L := [Integers()|i : i in LL]; n := #LL;
pi := &*[1-t^i : i in L]; A := (1-t^r) div (1-t);
G := GCD(pi,A); dG := Degree(G);
B := A div G; dB := Degree(B);
a,be,c := XGCD(pi,B); dbe := Degree(be);
R<[v]> := PolynomialRing(QQ,dG+2);
va := Name(R,dG+2);
bnew := &+[Coefficient(be,i)*va^i : i in [0..dbe]];
RR := &+[v[i]*va^(i-1) : i in [1..dG+1]];
Bnew := &+[Coefficient(B,i)*va^i : i in [0..dB]];
AA := bnew-RR*Bnew;
S := [Coefficient(AA,va,0)] cat [Coefficient(AA, va, r-i) : i in [1..r-1]];
empty := [];
for a in L do
 dd := GCD(a,r); tt := r/dd;
 relations := empty cat [&+[S[dd*l+i] : l in [0..tt-1]] : i in [1..dd]];
 empty := relations;
end for;
Mat := Matrix(QQ,[[Coefficient(empty[i],v[j],1) : j in [1..dG+1]]
 : i in [1..#empty]]);
zero := [0 : i in [1..dG+2]];
V := -Vector(QQ,[Evaluate(empty[i],zero) : i in [1..#empty]]);
MF := Transpose(Mat); x,y,z := IsConsistent(MF,V);
yy := &+[y[i+1]*va^(i) : i in [0..dG]];
sigma := bnew-yy*Bnew;
Sigma := [QQ!Coefficient(sigma, va, 0)] cat
  [QQ!Coefficient(sigma, va, i) : i in [1..r-1]];
return Sigma;
end function;

==========



Section 4, other examples now omitted.

\begin{exa} \rm
Consider the Calabi--Yau 3-fold $X_{66}\subset\PP(5,6,11,11,33)$. It has
$\frac13(2,2,2)$ and $\frac15(1,1,3)$ points and a $\frac1{11}(5,6)$
curve $\Ga$ of degree $\frac2{11}$ that is pure (no dissident points). We
parse its Hilbert series as
\begin{equation*}\frac{1-t^{66}}{\prod_{a\in[5,6,11,11,33]} (1-t^a)}=
P_I+P\orb(\tfrac13)+P\orb(\tfrac15)+P\per(\Ga)+P\grow(\Ga),
\end{equation*}
where our standard terms are
\begin{equation*}
P_I=1, \quad
P\orb(\tfrac13)=-\frac{t^3}{(1-t)^3(1-t^3)}, \quad
P\orb(\tfrac15)=\frac{t^3(1+t^2)}{(1-t)^3(1-t^5)},
\end{equation*}
and the contributions from $\Ga$ are
\begin{align*}
P\per(\Ga) &= -\frac{t^6+t^8}{(1-t)^3(1-t^{11})} \\
P\grow(\Ga) &= \frac{2t^8(1+t+t^2+t^3-t^4+t^5+t^6+t^7+t^8)}{(1-t)^2(1-t^{11})^2} \\
 &=2\times \frac{P\orb(\frac{1}{11}(5,6),11)}{1-t^{11}}.
\end{align*}
Moreover,
\begin{equation*}
\begin{aligned}
-t^6-t^8 &=\tfrac2{11} \cdot \Num_{D_C}- \tfrac1{11}\cdot \Num_{N_C} \\
&=\tfrac{2t}{11}(2 t^2+2 t^3+2 t^4-4 t^5-t^6-4 t^7+2 t^8+2 t^9+2 t^{10}) \\
&\qquad-\tfrac1{11}(4 t^3+4 t^4+4 t^5+3 t^6-2 t^7+3 t^8+4 t^9+4 t^{10}+4 t^{11}).
\end{aligned}
\end{equation*}
\end{exa}

For Magma online calculator:
\begin{verbatim}
// This can be plugged directly into the online Magma
// calculator http://magma.maths.usyd.edu.au/calc
RR<t> := PolynomialRing(Rationals());
function Denom(L) return &*[1-t^a : a in L]; end function;
function DedekindPolynomial(r,L)
return h*t*InverseMod(h*t*A,(1-t^r) div h) where h is GCD(A, 1-t^r) where A := Denom(L);
end function;

P := (1-t^66)/Denom([5,6,11,11,33]);
I := 1 +7/30*t/(1-t)^2 + 1/(1815*6)*(t+4*t^2+t^3)/(1-t)^4;
De3 := DedekindPolynomial(3,[2,2,2]);
II3 := De3/(1-t^3);
De5 := DedekindPolynomial(5,[1,1,3]);
II5 := De5/(1-t^5);
De11 := DedekindPolynomial(11,[5,6]); si0 := Coefficient(De11,11);
III11 := 2/11*(11*t^11*De11/(1-t^11)^2 + t*Derivative(De11)/(1-t^11) - si0*t/(1-t)^2);

BB := (-21/121*t^9 - 35/121*t^8 - 30/121*t^7 - 10/121*t^6 + 5/121*t^5 - 
       10/121*t^4 - 30/121*t^3 - 35/121*t^2 - 21/121*t)*11*(1-t);
IV11 := 1/11*BB/(1-t^11);
P - I - II3 - II5 - III11 - IV11;
\end{verbatim}


s := 19;
a := 3;
B1 := InverseMod(Denom([a,a,s-a]), (1-t^s) div (1-t));
B2 := InverseMod(Denom([a,s-a,s-a]), (1-t^s) div (1-t));
(B1*(1-t^a)^2*(1-t^(s-a))^2) mod ((1-t^s) div (1-t);
((B1-B2)*(1-t^a)^2*(1-t^(s-a))^2) mod (1-t^s);


P := (1-t^24)/Denom([4, 4, 5, 5, 6]);
PartialFractionDecomposition(P);

C2 curve of 1/2 of degree 12/2.2.3, but with 6 x 1/4(1,1,2)
dissident points.

PI := 1;
B5 := (-2*t^3-2*t^4-2*t^5)/Denom([1,1,1,5]);  // trial and error
P-PI-6*Qorb(4,[1,1,2],0)+2*Qorb(2,[1,1],2)/(1-t^2)-Qorb(5,[1,4],5)/(1-t^5)-B5;


\begin{verbatim}

// Check the assertion of Theorem 4.1 in Ex 4.3, X(40) in PP(2,5,8,10,15)

P := (1-t^40)/Denom([2,5,8,10,15]);
I := 1 + 1/3*t/(1-t)^2 + 1/1800*(t+4*t^2+t^3)/(1-t)^4;
De := DedekindPolynomial(15,[2,5,8]); IIQ := De/(1-t^15);
De2 := DedekindPolynomial(2,[1,1]); si0 := Coefficient(De2,2);
III2 := 1/2*(2*t^2*De2/(1-t^2)^2 + t*Derivative(De2)/(1-t^2) - si0*t/(1-t)^2);
De5 := DedekindPolynomial(5,[2,3]); si0 := Coefficient(De5,5);
III5 := 4/15*(5*t^5*De5/(1-t^5)^2 + t*Derivative(De5)/(1-t^5) - si0*t/(1-t)^2);
IV5 := 2/75*(-6*t+12*t^2-12*t^3+6*t^4)/(1-t^5);
P-I-IIQ-III2-III5-IV5;

\end{verbatim}

\begin{verbatim}
  P := (1-t^40)/Denom([2,5,8,10,15]);
  PI := (1-4*t+7*t^2-4*t^3+t^4)/(1-t)^4;
  // or P := 1 + t^2/(1-t)^4;
  P eq PI + Qorb(15,[2,5,8],0)
    + Qorb(5,[2,3],5)/(1-t^5) + Qorb(2,[1,1],2)/(1-t^2)
    + (-3*t^3 + 2*t^4 - 3*t^5)/Denom([1,1,1,5]);
\end{verbatim}


====

Much bigger case:

RR<t> := PolynomialRing(Rationals());

function Denom(L) return &*[1-t^a : a in L]; end function;

function DedekindPolynomial(r,L)
return h*t*InverseMod(h*t*A,(1-t^r) div h) where h is GCD(A, 1-t^r) where A := Denom(L);
end function;

P := (1-t^73)/Denom([3,7,11,21,31]);

I := 1 + (34301/902286)*t/(1-t)^2+ 1/6*73/(3*7*11*21*31)*(t+4*t^2+t^3)/(1-t)^4;

De31 := DedekindPolynomial(31,[3,7,21]);
II31 := De31/(1-t^31);

De11 := DedekindPolynomial(11,[3,21,31]);
II11 := De11/(1-t^11);

De21 := DedekindPolynomial(21,[3,7,11]);
II21 := De21/(1-t^21);

De7 := DedekindPolynomial(7,[3,4]); si0 := Coefficient(De7,7);
III7 := 1/21*(7*t^7*De7/(1-t^7)^2 + t*Derivative(De7)/(1-t^7) + si0*t/(1-t)^2);

De3 := DedekindPolynomial(3,[1,2]); si0 := Coefficient(De3,3);
III3 := 1/21*(3*t^3*De3/(1-t^3)^2 + t*Derivative(De3)/(1-t^3) + si0*t/(1-t)^2);

B1 := 5*t-2*t^2-5*t^3+5*t^4+2*t^5-5*t^6;
IV7 := -25/147*B1/(1-t^7);

B3 := t-t^2;
IV3 := 23/189*B3/(1-t^3);

P-I-II31-II11-II21 - III7 - III3 - IV7 - IV3;

================


\begin{exa} \rm
A harder case $X_{80}\subset\PP(3,4,15,20,38)$ (a random choice from the
vast lists of Kreuzer and Skarke). It has orbifold loci:
\begin{itemize}
\item a $\frac12$ curve $C_{80}\subset\PP(4,20,38)$
\item a dissident point $\frac1{38}(3,15,20)$ on $C_{80}$
\item $4\times\hbox{dissident points}$ $\frac14(3,3,2)$ on $C_{80}$
\item a $\frac13(1,2)$ curve $\PP(3,5)$ with normal monomials $ux^{14}$, $tz^4$
\item a dissident point $\frac1{15}(3,4,8)$ on $\frac13$ curve
\item an isolated point 1/5(3,4,3)
\end{itemize}

\end{exa}

\begin{verbatim}
PI := 1; // because the first two plurigenera are zero.
P := (1-t^80)/Denom([3,4,15,20,38]); P;
P1 := P - PI; P1;
P2 := P1 - Qorb(38,[3,15,20],0) - Qorb(15,[3,4,8],0)
  - 4*Qorb(4,[3,3,2],0) - Qorb(5,[3,4,3],0); P2;
 // The denominator drops on subtracting each term.
PartialFractionDecomposition(P2*(1-t)^4);
 // This hints as to what remaining degree of the curve is.
 // I experiment by subtracting off different degrees
P3:=P2-Qorb(2,[1,1],2)/(1-t^2); P3;
P3:=P2-Qorb(2,[1,1],2)/(1-t^2)-Qorb(3,[1,2],3)/(1-t^3); P3;
P3:=P2-Qorb(2,[1,1],2)/(1-t^2)+Qorb(3,[1,2],3)/(1-t^3); P3;
PartialFractionDecomposition(P3*(1-t)^4);
P4 := P3 - t^3/Denom([1,1,1,3]); P4;

P eq PI + Qorb(38,[3,15,20],0) + Qorb(15,[3,4,8],0)
  + 4*Qorb(4,[3,3,2],0) + Qorb(5,[3,4,3],0)
  + Qorb(2,[1,1],2)/(1-t^2) - Qorb(3,[1,2],3)/(1-t^3)
  + t^3/Denom([1,1,1,3]);

Qorb(38,[3,15,20],0)*Denom([1,1,2,38]);
t^38 - 2*t^37 + 2*t^36 - 2*t^34 + 2*t^33 + t^32 - 3*t^31
 + 3*t^30 - t^29 - t^28 + 2*t^27 - t^26 - t^25 + 3*t^24
 - 2*t^23 + t^21 - 2*t^19 + 3*t^18 - t^17 - t^16 + 2*t^15
 - t^14 - t^13 + 3*t^12 - 3*t^11 + t^10 + 2*t^9 - 2*t^8
 + 2*t^6 - 2*t^5 + t^4
The numerator is Gor symm of same degree 42 and support
[t^4 .. t^38], which has length < 38 - 2

Qorb(15,[3,4,8],0)*Denom([1,1,3,15])
t^15 - t^13 + 2*t^12 - t^10 + 2*t^8 - t^7 + t^5
The numerator in [t^5 .. t^15] of length 10 < 15 - 3

\end{verbatim}


===

From [BuSz] to Theorem 4.1.

Check that summing the terms  1/(2*r)*m*mkbar*(r-mkbar)
in [Buckley and Szendroi] as a power series gives III_C in
[BRZ], Theorem 4.1

MAGMA

1/23(4,19)  [Buckley and Szendroi] involves 6 = InvMod(4,23).

r:=23;
X := &+[ - 1/(2*r)*m*mkbar*(r-mkbar)*t^m where mkbar is (6*m) mod r : m in [1..10*r]];
Y := &+[ u : u in Terms(X*(1-t^23)^2) | Degree(u) le 3*r];
// replace the truncated power series X by the rational function Y/(1-t^23)^2;
// Y is -51/23*t^45 - 132/23*t^44  - .. - 135/23*t^3 - 132/23*t^2 - 51/23*t

// On the other hand, calculate the Dedekind sums:

K_r<ep> := CyclotomicField(r);
Mu_r := [ep^j : j in [0..r-1]];
Si := [1/r* Rationals()!  &+ [ e^i/((1-e^4)*(1-e^19)) : e in Mu_r | e ne 1] : i in [0..r-1]];

// Our sum is (noting that Si[1] is si_0 and Si[i+1] is si_i
// because of the way Magma does arrays).

&+ [ j*(Si[j+1] - Si[1]) * t^j  : j in [1..22]]/(1-t^23) + 23*&+ [ (Si[j+1] - Si[1]) * t^(j+23)  : j in [1..22]]/(1-t^23)^2;
$1 eq Y/(1-t^23)^2;  // gives true

&+ [ (Si[j+1]-Si[1])*(j*t^j + (23-j)*t^(23+j)) : j in [1..22]]/(1-t^23)^2;
$1 eq Y/(1-t^23)^2;  // gives true

De := h*t*InverseMod(h*t*(1-t^4)*(1-t^19), (1-t^r) div h) where h is (1-t);

Y/(1-t^r)^2 - r*t^r*De/(1-t^r)^2 - r*Si[1]*t^r*(1+t)/(1-t)/(1-t^r);

X := &+[ - 1/(2*r)*m*mkbar*(r-mkbar)*t^m where mkbar is (6*m) mod r : m in [1..10*r]];
Y := &+[ u : u in Terms(X*(1-t^23)^2) | Degree(u) le 3*r];
Y/(1-t^r)^2 - r*t^r*De/(1-t^r)^2 - t*Derivative(De)/(1-t^r) + Si[1]*t/(1-t)^2;
Y/(1-t^r)^2 eq r*t^r*De/(1-t^r)^2 + t*Derivative(De)/(1-t^r) - Si[1]*t/(1-t)^2;


P1 := &+ [ Si[i+1]*i*t^i : i in [1..r-1]];
P2 := &+ [ -Si[1]*i*t^i : i in [1..r-1]];
P3 := &+ [ r*t^r*Si[i+1]*t^i : i in [1..r-1]];
P4 := &+ [ -t^r*Si[i+1]*i*t^i : i in [1..r-1]];
P5 := &+ [ -r*t^r*Si[1]*t^i : i in [1..r-1]];
P6 := &+ [ t^r*Si[1]*i*t^i : i in [1..r-1]];
P1 eq t*Derivative(De) - r*t^r*Si[1];
P2 eq - Si[1]*((t - r*t^r + (r-1)*t^(r+1)) div (1-t)^2);                        
P3 eq r*De*t^r - r*Si[1]*t^(2*r);
P4 eq -t^r*P1;
P5 eq -r*Si[1]*t^r*(t-t^r) div (1-t);
P6 eq -t^r*P2;
P1+P4 eq (1-t^r)*(t*Derivative(De) - Si[1]*r*t^r);
P2+P6 eq -Si[1]*((t*(1-t^r)^2 - r*(1-t^r)*t^r*(1-t)) div (1-t)^2);
P3+P5 eq r*t^r*(De -(Si[1]) * t*((1-t^r) div (1-t)));
P2+P6+P3+P5 eq r*t^r*De - Si[1]*t*((1-t^r)^2 div (1-t)^2) + Si[1]*r*t^r*(1-t^r);
P1+P2+P3+P4+P5+P6 eq t*(1-t^r)*Derivative(De) + r*t^r*De - Si[1]*t*((1-t^r)^2 div (1-t)^2);




From Theorem 4.1. The formula comes from [BuSz]. We could
spend 2 or 3 pages explaining the derivation. There are many
intermediate forms, incl. the following:

\begin{equation}
\III_C= \left( \frac{\sum_{j=1}^{s-1} j(\si_j-\si_0)t^j}{1-t^s}
+\frac{s t^s \sum_{j=1}^{s-1} (\si_j-\si_0)t^j}{(1-t^s)^2}\right)DC
\end{equation}

\begin{equation}
\III_C=\frac{\sum_{j=1}^{s-1}\, \bigl(\si_j
- \si_0\bigr)\bigl(jt^j + (s-j)t^{s+j}\bigr)}{(1-t^s)^2}DC
\end{equation}

The current version of 4.4--4.5 could probably be further
simplified: the sums are really Dedekind polynomials for
1/s(a,-a) and their derivatives.

From Theorem 4.2. This is an intermediate form of
the contribution $B_C$ (not completely correct in
this form)
\begin{equation*}
P\per = \frac1{(1-t)^3(1-t^s)}\cdot \left( \deg D|_C\cdot
\Num_{D_C}-\frac{N_C}{2 s}\cdot \Num_{N_C} \right),
\end{equation*}
and the numerators $\Num_{D_C}$, $\Num_{N_C}$ are palindromic
polynomials supported in $\left[3,\dots,s\right]$. Moreover, for
$j=3,\dots,\ru{\frac s2}+1$ the $t^j$ term in $\Num_{D_C}$ equals
\begin{equation}
\sum_{i=0}^3 (-1)^i \textstyle\binom 3i \bigl( (j-i) \si_{-j+i}-
(2-i) \si_{-2+i} \bigr),
\end{equation}
and $\Num_{N_C}$ is computed as an ice cream function from
\begin{equation}
\Num_{N_C}\times \frac{(1-t^a)(1-t^{s-a})^2}{(1-t)^3}\equiv
1+t^{s-a}\mod \frac{1-t^s}{1-t}. \qed
\end{equation}

The main thing still to untangle is the NC of the Buckley--Szendroi. I
believe that it is  a*c1(L_a) - a*c1(L_{-a}), which would have an
obvious naive generalisation to the general case of 1-dim orbifold locus
C.



========

I messed around with the formulas (4.4) and (4.5)
without checking that they are true.

(1) They may be slightly wrong.

(2) It may be possible to improve them further.

(3) I think the quantity B is closely related to t \De'.

(4) It would be useful to understand why taking the derivative
of De = InvMod(A,F) gives a solution to
  A^2 x B == something simple mod F.

(5) It would be useful to write down a complete
derivation of Theorem 4.1 from [BuSz] and of
Theorem 4.2 from 4.1. It would also be useful to
know in detail how this follows from [Zh],
Section 5.1.

========

integrate 1 in deg 0 only -> sums to 1
chi in every degree >= 1 sums to t/(1-t) chi. Gor symm implies chi = 0.
m^2 x (-1/4 DKX^2) sums to (t+t^2)/(1-t)^3 x (-1/4 DKX^2) Gor symm implies = 0.
m (Dc_2/12 + DK^2/12) sums to D(c2/12 + DK^2) x t/(1-t)^2.

So the RR part I is

(1) Implication that chi = 0. Implication that DKX^2 = 0.

(2) The result that Part I equals

I = 1  +  D*(c1^2+c2)/12* t/(1-t)^2  +  D^3/6 *(t + 4t^2 + t^3)/(1-t)^4

as in paper, (4.4).

The term sum cQ(m)t^m looks exactly the same as the Dedekind polynomial
De; the sum has ep^-ai etc, corresponding to the sidestep in De of
putting \si_{r-j}*t^j. A sanity check consists of asking De/(1-t^r) has
same r periodicity as the Qorb

for i in [1..50] do
r := Random([30..70]);
a := Random([1..r-1]); b := Random(Exclude([a..r-1],r-a)); c := (-a-b) mod r; L:= [a,b,c]; L;
X := PartialFractionDecomposition(Qorb(r,L,0));
h := GCD(Denom(L),1-t^r); Y := PartialFractionDecomposition(h*t*InverseMod(h*t*Denom(L),(1-t^r) div h)/(1-t^r))[1]; Y in X;
end for;

Correction II is equal to De/(1-t^r), not minus the same.

I want to sum as a Hilbert series the 3 remaining terms:

−m*mkbar(r-mkbar)/2r * DC

mkbar(r − mkbar)/4r * KC

mkbar(r − mkbar)(r − 2 · mkbar) * NC/(12r2tau_C)

r := 23; k := 5; 

I observe the following curious point:
In [BS], the last term in the formula for sC(mD) is
  mkbar * (r- mkbar) * (r-2*mkbar)  times NC/(12*r^2*tau_C)
I sum this as a Hilbert series to get
  sum_m   mkbar * (r- mkbar) * (r-2*mkbar)  t^m  =:  IV_C / (1-t^r)
Then the numerator IV_C is a palindromic polynomial with support
in [1 .. r-1]  determined by
  Num*(1-t^b)^2*(1-t^(r-b))^2  mod (1-t^r)  eq 6*r*(t^b - t^(r-b));
where b = InverseMod(k,r);


r := 23;
for k in [1..11] do
b := InverseMod(k,r);
Q2 := &+[ mkbar*(r-mkbar)*(r-2*mkbar)*t^m where mkbar is (m*k) mod r :
m in  [1..4*r]];
Num := &+[ m : m in Terms(Q2*(1-t^r)) | Degree(m) le 2*r ];
Num*(1-t^b)^2*(1-t^(r-b))^2 mod (1-t^r) eq 6*r*(t^b - t^(r-b));
end for;

Maybe there is some expression that is simpler still.

The quantity sum -1/2r mkbar*(r-mkbar)*t appearing in
the second term of s_C(mD) multiplying - KC/2r
is Num/(1-t^r)
where Num is InverseMod((1-t^b)(1-t^(r-b), (1-t^r)/(1-t), 2)

r:=23; k := 5;
for k in [1..11] do
X := &+[ -1/(2*r)*mkbar*(r-mkbar)*t^m where mkbar is (m*k mod r) : m in [0..5*r]];
Y := &+[ u : u in Terms(X*(1-t^r)) | Degree(u) lt 3*r ];
  // i.e. this term is InvMod((1-t^b)*(1-t^{r-b}) mod ((1-t^r) div (1-t))
b := InverseMod( k, r); 
Y eq t*InverseMod(t*(1-t^b)*(1-t^(r-b)), ((1-t^r) div (1-t)));
end for;

Y*(1-t^9)*(1-t^14) mod ((1-t^r) div (1-t));
Numerator(Qorb(r,[5,18],r));

//  verification of III
r:=37;
for a in [1..18] do
b:=InverseMod(a,r);
De := h*t*InverseMod(h*t*(1-t^a)*(1-t^(r-a)), (1-t^r) div h) where h is (1-t);
Si := [Coefficient(De,r-i) : i in [0..r-1]];
X := &+[ - 1/(2*r)*m*mkbar*(r-mkbar)*t^m where mkbar is (b*m) mod r : m in [1..10*r]];
Y := &+[ u : u in Terms(X*(1-t^r)^2) | Degree(u) le 3*r];
Y/(1-t^r)^2 - r*t^r*De/(1-t^r)^2 - t*Derivative(De)/(1-t^r) + Si[1]*t/(1-t)^2;
Y/(1-t^r)^2 eq r*t^r*De/(1-t^r)^2 + t*Derivative(De)/(1-t^r) - Si[1]*t/(1-t)^2;
end for;







s1:=5; s2:=7; s3:=11;
a:=1; b:=1; c:= 1;
A := [a*s2*s3, b*s1*s3, c*s1*s2];
P := 1/Denom(A);
function DedekindPolynomial(r,L)
return h*t*InverseMod(h*t*A,(1-t^r) div h) where h is GCD(A, 1-t^r) where A := Denom(L);
end function;
P;
Degree(Denominator($1));
P-DedekindPolynomial(A[1],[A[2],A[3]])/(1-t^A[1]);
Degree(Denominator($1));
P - DedekindPolynomial(A[1],[A[2],A[3]])/(1-t^A[1])
 - DedekindPolynomial(A[2],[A[1],A[3]])/(1-t^A[2]);
 - DedekindPolynomial(A[3],[A[1],A[2]])/(1-t^A[3]);
Degree(Denominator($1));

X := P - DedekindPolynomial(A[1],[A[2],A[3]])/(1-t^A[1])
 - DedekindPolynomial(A[2],[A[1],A[3]])/(1-t^A[2])
 - DedekindPolynomial(A[3],[A[1],A[2]])/(1-t^A[3]);

X - 5/385*DedekindPolynomial(5,[77])/(1-t^5)^2
 - 7/385*DedekindPolynomial(7,[55])/(1-t^7)^2
 - 1/35*DedekindPolynomial(11,[35])/(1-t^11)^2;

> PartialFractionDecomposition($1);
[
    <1, 1, 211/385>,
    <t - 1, 1, -468/21175>,
    <t - 1, 2, 82/148225>,
    <t - 1, 3, -1/148225>,
    <t^4 + t^3 + t^2 + t + 1, 1, 78/1925*t^3 + 3/35*t^2 + 124/1925*t + 243/1925>,
    <t^6 + t^5 + t^4 + t^3 + t^2 + t + 1, 1, -23/385*t^5 - 156/2695*t^4 - 6/385*t^3 + 124/2695*t^2 + 57/539*t + 
        384/2695>,
    <t^10 + t^9 + t^8 + t^7 + t^6 + t^5 + t^4 + t^3 + t^2 + t + 1, 1, 5/121*t^9 + 64/847*t^8 + 397/4235*t^7 + 
        122/847*t^6 + 558/4235*t^5 + 808/4235*t^4 + 596/4235*t^3 + 852/4235*t^2 + 449/4235*t + 136/847>
]

These bits need to fit together as terms like
III_C and IV_C in Theorem 4.1.




If  ar  is odd then (I) and (II) are palindromic polynomials
of symmetric degree  ab-a  and  ar-ab-a+2  respectively. There
numbers are congruent to  1-a  modulo r, but not modulo 2r.
According to which of  ab-a  and  ar-ab-a+2  belongs to
  1+r-a  and  1+2r-a  modulo 2r,
(I) and (II) fold into the numerators of Porb(1/r(a),k) for
  k = 2ir-a, k = (2i+1)r-a.

for al in [0..10] do a := (2*al+1); b := InverseMod(a,r); ((a*b-1) mod (2*r)) div r; Qorb(r,[a],-a); end for;
for al in [12..22] do a := (2*al+1); b := InverseMod(a,r); ((a*b-1) mod (2*r)) div r; Qorb(r,[a],-a); end for;