The polynomial forms and their common factors
We construct 𝑈𝑛,𝑉𝑛 as integral
polynomials before evaluating them. Cancelling their common factors at
this stage lowers the degree and hence the power of 𝑏 required at a rational base 𝑎/𝑏.
For each 𝑛 ≥1, the coefficient
formulas below are rational functions of an indeterminate 𝑋. We evaluate them at real 𝑥 >1 and put 𝑞 =𝑥−1 when using the convergent
series. Set
𝑎0=14𝑛+1,𝑎1=12𝑛+1,𝑎2=14𝑛+1,𝛽𝑛=27𝑛+2,𝑁=15𝑛,𝑀𝑛=266𝑛2+34𝑛+1.
The symbol 𝛽𝑛 is the parameter 𝑏 of [8], the exponent in the lower parameter
𝑞𝑏 of the Heine series (13) on
p. 157; the letter 𝑏 continues to
denote the denominator of the base. In this construction alone, 𝑁 =15𝑛 is the cyclotomic cutoff. The
exponent 𝑀𝑛 is the integer of that
paper’s (16) on p. 157. Write (𝑧;𝑞)𝑚 =∏𝑚−1𝑗=0(1 −𝑧𝑞𝑗) for
the 𝑞-Pochhammer symbol (the
shifted 𝑞-factorial). Zudilin’s
positive remainder is
𝐻𝑛(𝑥)=∑𝑡≥0(𝑞𝑡+1;𝑞)𝑎1−1(𝑞;𝑞)𝑎1−1(𝑞;𝑞)𝛽𝑛−𝑎2−1(𝑞𝑎2+𝑡;𝑞)𝛽𝑛−𝑎2𝑞𝑎0𝑡.(2)
The denominator length 𝛽𝑛 −𝑎2 agrees with the gamma
expression (7) and residues (8) of [8]. The unnumbered display of 𝑅(𝑇) on that page instead has length
𝛽𝑛 −𝑎2 −1; we use the
normalisation fixed by the numbered identities. Then [8] gives 𝐻𝑛 =𝐴𝑛𝐹 −𝐵𝑛 with 𝐴𝑛,𝐵𝑛 ∈ℚ(𝑋). We need their explicit
formulas to compute degrees and bound coefficients. Let [𝑚𝑟]𝑋 =∏𝑟𝑗=1(1 −𝑋𝑚−𝑟+𝑗)/(1 −𝑋𝑗)
for 0 ≤𝑟 ≤𝑚. This Gaussian
binomial, or 𝑞-binomial, polynomial
has degree 𝑟(𝑚 −𝑟) and nonnegative
coefficients summing to (𝑚𝑟). For 𝑎2 ≤𝑘 ≤𝛽𝑛 −1 put
𝑐𝑛,𝑘(𝑋)=(−1)𝑎1+𝑎2+𝑘+1𝑋𝑒𝑛,𝑘[𝑘−1𝑎1−1]𝑋[𝛽𝑛−𝑎2−1𝛽𝑛−𝑘−1]𝑋,(3)
𝑒𝑛,𝑘=𝑎1(𝑎1−1)2−(𝛽𝑛−𝑎2)(𝛽𝑛−𝑎2−1)2+(𝛽𝑛−𝑘)(𝛽𝑛−𝑘−1)2.
Then
𝐴𝑛(𝑋)=𝛽𝑛−1∑𝑘=𝑎2𝑐𝑛,𝑘(𝑋)𝑋𝑎0𝑘,𝐵𝑛(𝑋)=𝛽𝑛−1∑𝑘=𝑎2𝑐𝑛,𝑘(𝑋)𝑋𝑎0𝑘(𝑘−𝑎1∑𝑙=11𝑋𝑙−1+𝑎0−1∑𝑗=1𝑋−𝑗(𝑘−𝑎1)𝑋𝑗−1).(4)(5)
Put
𝐷𝑁(𝑋)=𝑁∏𝑙=1Φ𝑙(𝑋),𝜈𝑛,𝑙=𝜔(𝑛/𝑙),Ω𝑛(𝑋)=𝑁∏𝑙=2Φ𝑙(𝑋)𝜈𝑛,𝑙,(6)
with Φ𝑙
the 𝑙th cyclotomic polynomial; by
Lemma 2.1 every
𝜈𝑛,𝑙 is 0 or 1, and 𝜈𝑛,𝑙 =0 for 𝑙 >15𝑛 because the intervals all begin
at or above 1/14.
Zudilin’s Lemma 7 [8] applies at the parameters above: the
tuple is 𝑐 =𝑛 ⋅(13,14,12,14,15,13) with maximum
𝑚(𝑐) =15𝑛 and difference 𝑠(𝑐) =𝑛 >0, and the source’s (14) on
p. 157 holds because 12𝑛 +1 ≤14𝑛 +1
and 26𝑛 +2 ≤27𝑛 +2 ≤28𝑛 +2. Its
conclusion is the inclusion
𝑈𝑛:=𝑋−𝑀𝑛𝐷𝑁Ω𝑛𝐴𝑛∈ℤ[𝑋],𝑉𝑛:=𝑋−𝑀𝑛𝐷𝑁Ω𝑛𝐵𝑛∈ℤ[𝑋].(7)
Two points about that citation matter. First,
the conclusion used is the one in ℤ[𝑋], not integrality at integer
arguments: an integer-valued polynomial such as 𝑋(𝑋 −1)/2 shows that the two statements
differ. The source’s proof supplies the polynomial form. Its Lemma 4
(pp. 157–159) clears the two rational coefficients separately by
Gaussian-binomial identities in ℤ[𝑋]. Lemma 5 (p. 160) makes the
normalised series invariant under the order-twelve group generated by
Heine’s parameter transformation and the exchange of 𝑎1 and 𝑎2. For the divisibility argument,
however, only its order-six subgroup preserving 𝑠 >0 is used: one application of
Heine’s transformation reverses the sign of 𝑠. Under that subgroup the normalising
factorial product takes the three forms displayed in the source.
Comparing their cyclotomic valuations gives the maximum of three terms
in 𝜔. Lemma 7 (p. 161) removes
the resulting common cyclotomic factors by polynomial divisibility; the
divisor Ω𝑛 is monic. These
steps take place before the integer specialisation in (24) on
p. 162.
Second, the module inclusion gives a polynomial coefficient pair, and
uniqueness identifies it with the displayed cancelled source pair. If
two pairs differed, subtracting their identities would express 𝐹 as a rational function of 𝑋. This is impossible: as ℎ ↓0, splitting the original
Lambert sum 𝐹(𝑒ℎ) =∑𝑚≥1(𝑒ℎ𝑚 −1)−1 at
𝐿 =⌊1/ℎ⌋ gives an
initial sum ℎ−1∑𝑚≤𝐿𝑚−1 +𝑂(𝐿), since 𝑦−1 −1 ≤(𝑒𝑦 −1)−1 ≤𝑦−1 for
0 <𝑦 ≤1. The remaining sum is at
most 𝑒−ℎ(𝐿+1)/((1 −𝑒−1)(1 −𝑒−ℎ)) =𝑂(ℎ−1).
Hence 𝐹(𝑒ℎ) =ℎ−1log(1/ℎ) +𝑂(ℎ−1), so
(𝑋 −1)𝐹(𝑋) →∞ and (𝑋 −1)2𝐹(𝑋) →0 as 𝑋 ↓1, which no rational function
does. Thus the source’s module inclusion yields the specific pair
in (7),
not merely some unspecified polynomial representation of the same
function. This functional nonrationality says nothing about an
individual value. For 𝑥 >1, write
the cancelled remainder as
Λ𝑛(𝑥)=𝑈𝑛(𝑥)𝐹(𝑥)−𝑉𝑛(𝑥)=𝑥−𝑀𝑛𝐷𝑁(𝑥)Ω𝑛(𝑥)𝐻𝑛(𝑥).