We compare the two normalisations before considering the sums over
pure prime powers. The subseries assertions in this subsection have
ordinary proofs and remain unformalised. For 𝑃 ={2,3,5} the two tail recurrences have
the same bases, 𝑏𝜏𝑎 =𝑃𝑎+1/𝑃𝑎 =𝑏𝑎, but their
coefficients behave differently:
Prime-power subseries.
The comparison of two tails applies to each prime-power subseries.
For 𝑝 ∈𝑃 put
𝐸𝑝=∑𝛼≥01L(𝑝𝛼)=∑𝛼≥01H𝑃(𝑝𝛼),
the sub-sum of R𝑃 over
the terms at which the smooth integer is a power of 𝑝. For 𝑃 ={𝑝,𝑞} with 𝑝 <𝑞 these are the numbers 𝐴 =𝐸𝑝 and 1 +𝐵∗ =𝐸𝑞 of the proof of
Theorem 1.2.
Proof. (i) Each term 1/𝑄𝑘 of D𝑃 with 𝑘 ≥1 is a term of 𝐸𝑞𝑘 and of no other 𝐸𝑞, and each 𝐸𝑞 also contains the term 1. (ii) and (iii) The exponent of 𝑝 in H𝑃(∏𝑞∈𝑃𝑞𝑒𝑞)
is 𝑒𝑝 +⌊∑𝑞≠𝑝𝑒𝑞log𝑝𝑞⌋, and its exponent in ∏𝑞∈𝑃H𝑃(𝑞𝑒𝑞)
is 𝑒𝑝 +∑𝑞≠𝑝⌊𝑒𝑞log𝑝𝑞⌋. For 𝑚
nonnegative reals the floor of the sum exceeds the sum of the floors by
an integer in {0,…,𝑚 −1}.
With two primes 𝑚 =1, so the running
LCMs factor, and summing H𝑃(𝑝𝑖)−1H𝑃(𝑞𝑗)−1
over 𝑖,𝑗 ≥0 gives 𝐸𝑝𝐸𝑞. ◻
With three primes R𝑃
is therefore the product of the three single-prime sums twisted by the
carries 𝜅, and the plain
product 𝐸𝑝𝐸𝑞𝐸𝑟 differs from
it.
For 𝑃 ={𝑝,𝑞}, 𝐸𝑝 is 𝐴 or 1 +𝐵∗ in the proof of Theorem 1.2, an
affine function of the Hecke–Mahler value 𝐹𝜃(1/𝑝,1/𝑞) with nonzero slope; for
𝑝 <𝑞, 𝐸𝑝 =𝑝/(𝑝 −1) −(𝑞 −1)𝐹𝜃(1/𝑝,1/𝑞)
by (6).
It is transcendental by the theorem of Bugeaud and Laurent [15], in the case
due to Loxton and van der Poorten [16]. We cite these sources for this case
and have not re-verified its earlier attribution history.
Proof. We modify the word argument so that it counts only
occurrences of 𝑝. For an infinite
word 𝑢 and a finite word 𝜎 =𝜎1⋯𝜎𝑛, put
𝑉𝑝(𝑢)=∑𝑖:𝑢𝑖=𝑝1𝑢1⋯𝑢𝑖,𝑓𝑝(𝜎)=∑𝑗:𝜎𝑗=𝑝𝜎𝑗+1⋯𝜎𝑛.
Then
𝑉𝑝(𝜎𝑢) =(𝑓𝑝(𝜎) +𝑉𝑝(𝑢))/Π(𝜎). Writing 𝑦𝑘 =𝑉𝑝(𝑞𝑘+1𝑞𝑘+2⋯), we have
𝐸𝑝=1+∑𝑘:𝑞𝑘=𝑝1𝑄𝑘,𝑄𝑘𝐸𝑝=𝑄𝑘+∑𝑗≤𝑘, 𝑞𝑗=𝑝𝑄𝑘𝑄𝑗+𝑦𝑘.
In an
actual suffix, the prefix product at the 𝑗th occurrence of 𝑝 is at least 𝑝𝑗, with strict inequality after another
prime has occurred. Hence 0 <𝑉𝑝(𝑢) <1/(𝑝 −1) ≤1, and
rationality 𝐸𝑝 =𝑁/𝐾 would imply
𝐾𝑦𝑘 ∈ℤ. The proof of
Remark 5.3
therefore applies to 𝑉𝑝,𝑓𝑝 as
well. On a chain of distinct primes containing 𝑝, only one term remains in 𝑓𝑝: the product of the primes following
𝑝. Unique factorisation detects a
change in that set, without recovering its order. This weaker
requirement avoids the collisions of the full block numerator 𝑓.
For each 𝑞 ≠𝑝, the form 𝜃𝑝 −𝜃𝑞 is nonzero on the
rational subspace 𝐿 used in
Section 5.1. We may
therefore choose 𝑣 ∈𝐿 with |𝑣𝑞| ≤1 for every 𝑞 and 𝑣𝑞 ≠𝑣𝑝 for 𝑞 ≠𝑝, where
𝑣𝑞 =𝜃𝑞(𝑣), and put 𝜇 =min𝑞≠𝑝|𝑣𝑝 −𝑣𝑞| >0. Suppose
𝐸𝑝 =𝑁/𝐾. Choose a jump 𝑡𝑘 labelled 𝑝, and choose 𝑇,𝜈,𝑔,𝛽 as in the general
irrationality proof, with 2𝜈 ≥𝐾. Set 𝜂 =1/10 and first
fix
0<𝜀<min{𝛽,𝑔/2,log2/(2|𝑃|),𝜂/4,𝜂𝜇/4}.
Dirichlet’s theorem then supplies arbitrarily large 𝑛 and integers 𝑚𝑞 ≥1 such that 𝛿𝑝 =0 and 0 <|𝛿𝑞| =|𝑛log𝑝 −𝑚𝑞log𝑞| <𝜀 for 𝑞 ≠𝑝. We compare 𝑡𝑘 with
𝑡𝑚 =𝑡𝑘𝑝𝑛 and cut crossings at
gaps greater than 2𝜀. As
before, each chain consists of distinct primes and is merely rearranged
by the phase shift. The common prefix of 𝜈 letters and the integral-tail
argument give 𝑦𝑘 =𝑦𝑚, and hence
𝑓𝑝(𝐶) =𝑓𝑝(𝐶′) for every chain
and its rearrangement.
We shall contradict this equality near the origin of 𝑇0. Put
𝑏0=min𝑞≠𝑝|𝛿𝑞||𝑣𝑝−𝑣𝑞|,𝐵={𝑞≠𝑝:𝛿𝑞<0},𝐴+={𝑞≠𝑝:𝑣𝑞<𝑣𝑝},𝐴−={𝑞≠𝑝:𝑣𝑞>𝑣𝑝}.
Since 0 <𝑏0 ≤𝜀/𝜇 <𝜂/4 and
𝐴+,𝐴− partition the nonempty set
𝑃 ∖{𝑝}, one of these
two sets, say 𝐴𝑠 with 𝑠 ∈{ +1, −1}, differs from 𝐵. Take 𝑏 =𝑠𝑏0/2 and consider 𝑧 =𝑏𝑣 +Θ( −𝜂/2). For each 𝑞, its crossing nearest 𝜂/2 occurs at
𝜏𝑞=𝜂/2−𝑏𝑣𝑞∈(3𝜂/8,5𝜂/8),𝜏𝑞−𝜏𝑝=𝑏(𝑣𝑝−𝑣𝑞).
Every other 𝑞-crossing lies outside ( −𝜂/4,5𝜂/4), and for 𝑞 ≠𝑝 the displayed difference is
nonzero with absolute value at most |𝛿𝑞|/2. We choose a neighbourhood
𝑈 of 𝑧 in 𝑇0 on which there are no crossings in
[ −𝜂/4,𝜂/4] ∪[3𝜂/4,5𝜂/4],
each prime has a crossing in (𝜂/4,3𝜂/4), and every difference
𝜏𝑞 −𝜏𝑝 retains its sign and
has absolute value less than |𝛿𝑞|. These conditions are open and
hold strictly at 𝑧.
By density, 𝜃𝑘 +Θ(𝑡∗) ∈𝑈 for some 𝑡∗ >𝑇 +2. The
resulting crossings near 𝑡∗ form
one chain: each is within 𝜀 of the 𝑝-crossing and every other crossing is
more than 𝜂/2 >2𝜀
away. In this chain the followers of 𝑝 form 𝐴𝑠. After the shift, 𝑞 follows 𝑝 precisely when 𝜏𝑞 −𝜏𝑝 >𝛿𝑞, which, by the
bound |𝜏𝑞 −𝜏𝑝| <|𝛿𝑞|, is
equivalent to 𝛿𝑞 <0. Its
followers now form 𝐵 ≠𝐴𝑠,
contradicting 𝑓𝑝(𝐶) =𝑓𝑝(𝐶′). ◻
This argument uses a change of followers and so needs no two-wall
lemma. Either of 𝐴+,𝐴− may be
empty, but they are distinct when |𝑃| ≥2. For 𝑃 ={𝑝} the sum is 𝑝/(𝑝 −1), and this is precisely where the
choice of two distinct follower sets fails. The strict conditions
defining 𝑈 are needed to carry the
construction to an actual orbit.
For 𝑃 ={2,3,5}, 𝑝 =2, 𝑡𝑘 =220 and 𝑛 =65, the first rearranged chain is (2,3) against (3,2) after 131 letters, with 𝑓2 equal to 3 and 1, and the difference of the two tails,
computed exactly to 300 letters,
has log10|𝑦𝑘 −𝑦𝑚| = −58.098.
For the repeated sum R𝑃, changing the scale from 𝑡𝑘 to 𝑡𝑘𝑝𝑛 brings in early terms from the
rays 𝑝𝛼𝑤, with 𝑤 supported on 𝑃 ∖{𝑝}. The comparison
above accounts only for the ray 𝑤 =1, so it gives no corresponding
estimate for the difference of the full repeated tails. In the case
𝑃 ={2,3,5} those normalised tails
are unbounded, as the lower bounds below show.