Keywords

1 Introduction

The Thue–Morse sequence

$$\mathbf {t}=(t(n))_{n\ge 0}=0,1,1,0,1,0,0,1,1,0,0,1,0,1,1,0,\ldots $$

can be defined via

$$\begin{aligned} t(n)=s_2(n)\!\!\!\!\!\!\,\mod \!2, \end{aligned}$$
(1)

where \(s_2(n)\) denotes the number of one bits in the binary expansion of n, or equivalently, the sum of digits of n in base 2. This sequence can be found in various fields of mathematics and computer science, such as combinatorics on words, number theory, harmonic analysis and differential geometry. We refer the reader to the survey articles of Allouche and Shallit [2], and of Mauduit [14] for a concise introduction to this sequence. As is well-known, Thue–Morse is 2-automatic and can be generated by the morphism \(0\mapsto 01\), \(1\mapsto 10\). It is also the prime example of an overlapfree sequence.

The overall distribution of the symbols 0 and 1 in Thue–Morse is trivial since the sequence consists exclusively of consecutive blocks of the forms 01 and 10, thus there are “as many 0’s as 1’s” in the sequence. The investigation of Thue–Morse along subsequences can be said to have started with an influential paper by Gelfond in 1967/68 [9]. He proved, via exponential sums techniques, that \(\mathbf {t}\) is uniformly distributed along arithmetic progressions, namely,

$$\#\{n<N: \quad t(an+b)=\varepsilon \} \sim \frac{N}{2},\qquad (\varepsilon =0 \text{ or } 1)$$

with an explicit error term. Gelfond’s result shows that there are “as many 0’s as 1’s” in the sequence also regarding arithmetic progressions. His result, however, gives no information on how long one actually has to wait to “see” the first, say, “1” along a specific arithmetic progression. Newman [19] showed that there is a weak preponderance of the 0’s over the 1’s in the sequence of the multiples of three. More precisely, he showed that

$$\#\{n<N: \quad t(3n)=0\}=\frac{N}{2}+C(N),$$

with \(c'_1 N^{\log _4 3} < C(N)< c'_2 N^{\log _4 3}\) for all \(N\ge 1\) and certain positive constants \(c'_1, c'_2\). For the multiples of three one has to wait for 7 terms to “see the first 1”, i.e.,

$$\min \{n: \quad t(3n)=1\}=7.$$

Morgenbesser, Shallit and Stoll [18] proved that for \(p\ge 1\),

$$\min \{n:\quad t(pn)=1\}\le p+4,$$

and this becomes sharp for \(p=2^{2r}-1\) for \(r\ge 1\) (Note that \(3=2^{2\cdot 1}-1\) is exactly of that form). A huge literature is nowadays available for classes of arithmetic progressions where such Newman-type phenomena exist and many generalizations have been considered so far (see [3, 4, 8, 10, 12, 22] and the references given therein). Still, a full classification is not yet at our disposal.

Most of the results that hold true for Thue–Morse in the number-theoretic setting of (1) have been proven for the sum of digits function in base q, where q is an integer greater than or equal to 2, and where the reduction in (1) is done modulo an arbitrary integer \(m\ge 2\). We refrain here from the general statements and refer the interested readers to the original research papers.

Following the historical line, Gelfond [9] posed two challenging questions concerning the distribution of the sum of digits function along primes and along polynomial values instead of looking at linear subsequences. A third question was concerned with the simultaneous distribution when the sum of digits is taken to different bases; this question has been settled by Kim [13]. In recent years, this area of research gained much momentum due to an article by Mauduit and Rivat [15] who answered Gelfond’s question for primes with an explicit error term. In a second paper [16], they also answered Gelfond’s question for the sequence of squares. Their result implies that

$$\#\{n<N: \quad t(n^2)=\varepsilon \} \sim \frac{N}{2}.$$

In a very recent paper, Drmota, Mauduit and Rivat [6] showed that \(\mathbf {t}\) along squares gives indeed a normal sequence in base 2 meaning that each binary block appears with the expected frequency. This quantifies a result of Moshe [17] who answered a question posed by Allouche and Shallit [1, Problem10.12.7] about the complexity of Thue–Morse along polynomial extractions.

We are still very far from understanding

$$\#\{n<N: \quad t(P(n))=\varepsilon \},$$

where \(P(x)\in \mathbb {Z}[x]\) is a polynomial of degree \(\ge 3\). Drmota, Mauduit and Rivat [7] obtained an asymptotic formula for \(\#\{n<N: \; s_q(P(n))=\varepsilon \pmod m\}\) whenever q is sufficiently large in terms of the degree of P. The case of Thue–Morse is yet out of reach of current technology. The currently best result is due to the author [23], who showed that there exists a constant \(c=c(P)\) depending only on the polynomial P such that

$$\begin{aligned} \#\{n<N: \quad t(P(n))=\varepsilon \}\ge c N^{4/(3 \deg P+1)}. \end{aligned}$$
(2)

This improves on a result of Dartyge and Tenenbaum [5] who had \(N^{2/(\deg P)!}\) for the lower bound. The method of proof for (2) is constructive and gives an explicit bound on the minimal non-trivial n such that \(t(n^h)=\varepsilon \) for fixed \(h\ge 1\). Since \(t(n^h)=1\) for all \(n=2^r\), and \(t(0^h)=0\), we restrict our attention to

$$\mathcal {A}=\{n:\quad n\ne 2^r,\; r\ge 0\}=\{3,5,6,7,9,10,11,12,13,14,15,17,\ldots \}.$$

From the proof of (2) follows that

$$\min (n\in \mathcal {A}:\quad t(n^h)=\varepsilon )\le 64^{h+1}\left( 8 h\cdot 12^h\right) ^{3h+1}.$$

Hence, there exists an absolute constant \(c_1>0\) such that

$$\begin{aligned} \min (n\in \mathcal {A}:\quad t(n^h)=\varepsilon )\le \exp (c_1h^2). \end{aligned}$$
(3)

With some extra work, a similar result can be obtained for a general polynomial P(n) instead of \(n^h\), where there corresponding constant will depend on the coefficients of P.

The joint distribution of the binary digits of integer multiples has been studied by J. Schmid [20] and in the more general setting of polynomials by Steiner [21]. The asymptotic formulas do not imply effective bounds on the first n that realizes such a system and it is the aim of this paper to prove effective bounds in the case of two equations for integer multiples and for monomials.

Our first result is as follows.

Theorem 1

  Let \(p>q\ge 1\) be odd integers. Then there exists an absolute constant \(c_2>0\) such that

$$\min (n:\; t(pn)=\varepsilon _1, \; t(qn)=\varepsilon _2)\le \exp (c_2 \log p) \qquad (\varepsilon _1,\varepsilon _2\in \{0,1\}).$$

Remark 1

Note that for \(p=2^r+1\) with \(r\ge 1\) we have \(t(pn)=0\) for all \(n<p-1\), so that there is no absolute bound for the minimal n.

There are examples that show that sometimes one has to “wait” quite some time to see all of the four possibilities for \((\varepsilon _1,\varepsilon _2)\) when the extraction is done along two monomial sequences. For instance, we have

$$\min \{n\in \mathcal {A}: \quad t(n^{130})=\varepsilon _1)\}\le 7\quad \text {and}\quad \min \{n\in \mathcal {A}: \quad t(n^{53})=\varepsilon _2)\}\le 5$$

for \(\varepsilon _1, \varepsilon _2=0,1\) but

$$\min \left\{ n:\quad (t(n^{130}),t(n^{53}))=(0,0)\right\} =113.$$

The construction that we will use to prove Theorem 1 will not be useful to study the minimal n along polynomial subsequences since in this case we would need to keep track of the binary digits sum of various binomial coefficients. Instead, we will use ideas from work of Hare, Laishram and the author [11] to show the following result.

Theorem 2

  Let \(h_1>h_2\ge 1\) be integers. Then there exists an absolute constant \(c_3>0\) such that

$$\min (n\in \mathcal {A}:\; t(n^{h_1})=\varepsilon _1, \; t(n^{h_2})=\varepsilon _2)\le \exp (c_3 h_1^3) \qquad (\varepsilon _1,\varepsilon _2\in \{0,1\}).$$

Remark 2

The method also allows to treat general monic polynomials \(P_1(x)\), \(P_2(x)\in \mathbb {Z}[x]\) of different degree \(h_1, h_2\) in place of \(x^{h_1}, x^{h_2}\). Even more generally, we can deal with non-monic polynomials \(P_1(x), P_2(x)\in \mathbb {Z}[x]\) provided \(h_1\) is odd. As we will see in the proof (compare with the remark after (14)), the latter condition relies on the fact that for odd h the congruence \(x^h\equiv a \pmod {16}\) admits a solution mod 16 for each odd a, while this is not true in general if h is even.

We write \(\log _2 \) for the logarithm to base 2. Moreover, for \(n=\sum _{j=0}^{\ell } n_j 2^j\) with \(n_j\in \{0,1\}\) and \(n_\ell \ne 0\) we write \((n_\ell ,n_{\ell -1},\cdots ,n_1,n_0)_2\) for its digital representation in base 2 and set \(\ell =\ell (n)\) for its length. To simplify our notation, we allow to fill up by a finite number of 0’s to the left, i.e., \((n_\ell n_{\ell -1}\cdots n_1 n_0)_2=(0n_\ell n_{\ell -1} \cdots n_1 n_0)_2=(0\cdots 0 n_\ell n_{\ell -1} \cdots n_1 n_0)_2\).

The paper is structured as follows. In Sect. 2 we prove Theorem 1 and in Sect. 3 we show Theorem 2.

2 Thue–Morse at Distinct Multiples

The proof of Theorem 1 is based on the following lemma.

Lemma 1

Let p, q be odd positive integers with \(p>q\ge 1\) and let \((\varepsilon _1,\varepsilon _2)\) be one of (0, 0), (0, 1), (1, 0), (1, 1). Then we have

$$\min \left\{ n:\; (t(pn),t(qn))=(\varepsilon _1,\varepsilon _2)\right\} \le C_{\varepsilon _1,\varepsilon _2}(p,q),$$

where

$$ C_{0,0}(p,q)=C_{1,1}(p,q)=4 p,\quad C_{0,1}(p,q)=\frac{2^{23}p^{11}}{(p-q)^6}\quad C_{1,0}(p,q)=\frac{2^6p^3}{(p-q)^2}.$$

Proof

Recall that for \(1\le b<2^k\) and \(a,k\ge 1\), we have

$$\begin{aligned} s_2(a2^k+b)=s_2(a)+s_2(b),\qquad s_2(a2^k-b)=s_2(a-1)+k-s_2(b-1). \end{aligned}$$
(4)

In the sequel we will make frequent use of these splitting formulas. We first deal with the two cases when \((\varepsilon _1,\varepsilon _2)\) is one of (0, 0), (1, 1). If \(2^k>p>q\) then \(s_2(p(2^k-1))=s_2(q(2^k-1))=k\). Moreover, since \(k\equiv 0\) or 1 mod 2 and

$$2^k-1\le 2^{\log p/\log 2 +2}-1=4p-1<4p,$$

we get that \(C_{0,0}(p,q),C_{1,1}(p,q)\le 4p\). Finding explicit bounds for \(C_{0,1}(p,q)\) and \(C_{1,0}(p,q)\) is more involved. To begin with, we first claim that there exists \(n_1\ge 1\) with the following two properties:

  1. (a)

    \(\ell (pn_1)>\ell (qn_1)\),

  2. (b)

    \(pn_1\equiv 1 \mathrm mod4\).

As for (a), we need to find two integers \(a, n_1\) such that \(2^a\le pn_1\) and \(2^a> qn_1\). This is equivalent to

$$\begin{aligned} \frac{2^a}{p}\le n_1<\frac{2^a}{q}. \end{aligned}$$
(5)

For odd k, n either \(kn\equiv 1\) (mod 4) or \(k(n+2)\equiv 1\) (mod 4), so provided \(2^a\left( \frac{1}{q}-\frac{1}{p}\right) \ge 4\), we can find an odd \(n_1\) that satisfies both (a) and (b). By taking a to be the unique integer with

$$\begin{aligned} \frac{4pq}{p-q}\le 2^a<2\cdot \frac{4pq}{p-q}, \end{aligned}$$
(6)

we get an \(n_1\) with

$$n_1<\frac{8p}{p-q}.$$

Now, define \(n_2=2^{\ell (pn_1)}+1\). Since both p and \(n_1\) are odd we have \(n_2\le pn_1\). Then

$$\begin{aligned} s_2(pn_1n_2)=s_2(pn_1 2^{\ell (pn_1)}+pn_1)=2s_2(pn_1)-1\equiv 1\pmod 2, \end{aligned}$$
(7)

since there is exactly one carry propagation from the most significant digit of \(pn_1\) to the digit at digit place \(\ell (pn_1)\) of \(pn_1 2^{\ell (pn_1)}\) which stops immediately after one step because of property (b). On the other hand, (a) implies that

$$\begin{aligned} s_2(qn_1n_2)=s_2(qn_1 2^{\ell (pn_1)}+qn_1)=2s_2(qn_1)\equiv 0\pmod 2, \end{aligned}$$
(8)

because the terms \(qn_1\) and \(qn_1 2^{\ell (pn_1)}\) do not interfere and there is therefore no carry propagation while adding these two terms. We therefore can set

$$n=n_1n_2\le \left( \frac{8p}{p-q}\right) ^2 p$$

to get that \(C_{1,0}(p,q)\le 2^6\cdot \frac{p^3}{(p-q)^2}\). For \((\varepsilon _1,\varepsilon _2)=(0,1)\), take m to be the unique odd integer with

$$pn_1n_2<2^m\le 4pn_1n_2$$

and put

$$n_3=2^{2m}+2^m-1\le 2(4pn_1n_2)^2\le 2^5 (pn_1)^4.$$

Then by (7),

$$\begin{aligned} s_2(pn_1n_2n_3)&=s_2\left( (p n_1n_2 2^m+pn_1n_2)2^m-pn_1n_2\right) \\&=s_2\left( pn_1n_2 2^m+pn_1n_2-1\right) +m-s_2\left( pn_1n_2-1\right) \\&=s_2\left( pn_1n_2-1\right) +s\left( pn_1n_2\right) +m-s_2\left( pn_1n_2-1\right) \\&\equiv m+1\equiv 0 \pmod 2. \end{aligned}$$

A similar calculation shows by (8) that

$$s\left( qn_1n_2n_3\right) \equiv m\equiv 1 \pmod 2.$$

We set \(n=n_1n_2n_3\) and get

$$n\le \left( \frac{8p}{p-q}\right) ^2 p \cdot 2^5p^4 \cdot \frac{(8p)^4}{(p-q)^4}=2^{23}\cdot \frac{p^{11}}{(p-q)^6}.$$

This completes the proof.    \(\square \)

Proof

(Theorem 1 ). This follows directly from Lemma 1 and

$$2^{23}\cdot \frac{p^{11}}{(p-q)^6}\le 2^{17} p^{11}.$$

   \(\square \)

3 Thue–Morse at Two Polynomials

This section is devoted to the proof of a technical result which implies Theorem 2. Considering the extractions of Thue–Morse along \(n^{h_1}\) and \(n^{h_2}\), it is simple to get two and not too difficult to get three out of the four possibilities for \((\varepsilon _1,\varepsilon _2)\). However, to ensure that we see all of the four possibilities, we need a rather subtle construction. The difficulty is similar to that one to get \(C_{0,0}(p,q)\) in the proof of Theorem 1. The idea is to shift two specific blocks against each other while all other terms in the expansions are non-interfering. Via this procedure, we will be able to keep track of the number of carry propagations. In the proof of Theorem 1 we have used the blocks \(pn_1\) and \(qn_1\). In the following, we will make use of \(u_1=9=(1001)_2\) and \(u_2=1=(0001)_2\). Then

$$\begin{aligned} s_2(u_1+u_2)&=2,\\ s_2(u_1+2u_2)&=3,\\ s_2(u_1+2^2u_2)&=3,\\ s_2(u_1+2^3 u_2)&=2, \end{aligned}$$

and mod 2 we get the sequence (0, 1, 1, 0). These particular expansions and additions will be of great importance in our argument (compare with (16)–(19)).

Lemma 2

Let \(h_1\), \(h_2\) be positive integers with \(h_1>h_2\ge 1\) and let \((\varepsilon _1,\varepsilon _2)\) be one of (0, 0), (0, 1), (1, 0), (1, 1). Then we have

$$\min \left\{ n\in \mathcal {A}:\; (t(n^{h_1}),t(n^{h_2}))=(\varepsilon _1,\varepsilon _2)\right\} \le C,$$

where

$$ C=32^{3h_1+5}\left( \frac{1}{15} \left( \frac{150}{h_1}\,8^{h_1}\right) ^2\right) ^{5h^2_1+h_1-5}.$$

Proof

Let \(h\ge 1\) and put

$$\begin{aligned} l&= \left\lfloor \log _2 \left( \frac{1}{15} \left( \frac{150}{h} \,8^h\right) ^2\right) \right\rfloor +1,\\ a&= (2^{lh}-1) 2^4+1,\nonumber \\ b&= \left\lfloor \frac{150}{h}\, 8^h a^{1-\frac{1}{h}}\right\rfloor ,\nonumber \\ M&= \left\lfloor 15 a^{1-\frac{1}{h}}\right\rfloor +1,\nonumber \\ k&= lh^2+3h+4-l.\nonumber \end{aligned}$$
(9)

It is straightforward to check that for all \(h\ge 1\), we have

$$\begin{aligned} l\ge 12,\quad a\ge 65521,\quad b\ge 1200,\quad M\ge 16,\quad k\ge 7. \end{aligned}$$
(10)

Obviously, we have \(b\ge M\). Moreover,

$$a=16\cdot 2^{lh}-15 \ge 15\cdot 2^{lh}\ge \left( \frac{150}{h}\,8^h\right) ^h$$

and therefore \(a\ge b\). Furthermore, for \(h\ge 2\), we have \(aM>b^2\) since by

$$a> \left( \frac{1}{15}\left( \frac{150}{h}\, 8^h\right) ^2\right) ^h$$

we have

$$aM-b^2>a\cdot 15 a^{1-\frac{1}{h}}-\left( \frac{150}{h}\, 8^h a^{1-\frac{1}{h}}\right) ^2>0.$$

Let

$$T(x)=ax^5+bx^4+Mx^3+Mx^2-x+M$$

and write \(T(x)^h=\sum _{i=0}^{5h} \alpha _i x^i\). Obviously, \(\alpha _0>0\) and \(\alpha _1<0\). We claim that for \(h\ge 1\) we have \(\alpha _i>0\) for \(2\le i\le 5h\). To see this we write

$$\begin{aligned} T(x)^h=\left( ax^5+bx^4+Mx^3+Mx^2+M\right) ^h+r(x), \end{aligned}$$
(11)

with

$$r(x)=\sum _{j=1}^h \left( {\begin{array}{c}h\\ j\end{array}}\right) (-x)^j \left( ax^5+bx^4+Mx^3+Mx^2+M\right) ^{h-j}=\sum _{j=1}^{5(h-1)}d_j x^j.$$

Since \(a\ge b\ge M\) the coefficient of \(x^i\) in the first term in (11) is \(\ge M^h\). On the other hand,

$$|d_j|<h2^h(5a)^{h-1},$$

and

$$M^h\ge \left( 15 a^{1-1/h}\right) ^h> \left( 3\cdot 5 a^{1-1/h}\right) ^h> \left( 2 h^{1/h} (5a)^{1-1/h}\right) ^h=|d_j|$$

which proves the claim. Next, we need a bound on the size of \(\alpha _i\). The coefficients \(\alpha _i\), \(0\le i\le 5h-2\), are bounded by the corresponding coefficients in the expansion of \((ax^5+bx^4+M(x^3+x^2+x+1))^h\). Since \(aM>b^2\) and \(M\le 16 a^{1-1/h}\), each of these coefficients is bounded by

$$\left( a^{h-1}M\right) \cdot 6^h<a^{h-1}M 8^h\le a^{h-\frac{1}{h}}\cdot 16 \cdot 8^h,$$

and therefore

$$\begin{aligned} |\alpha _i| \le a^{h-\frac{1}{h}}\cdot 16 \cdot 8^h,\qquad i=0,1,2,\ldots , 5h-2. \end{aligned}$$
(12)

Moreover, we have

$$\begin{aligned} \alpha _{5h-1}=hba^{h-1} = \left\lfloor \frac{150}{h}\, 8^h a^{1-\frac{1}{h}}\right\rfloor h a^{h-1} \end{aligned}$$
(13)

and

$$149 \cdot 8^h a^{h-\frac{1}{h}}\le \alpha _{5h-1}\le 150 \cdot 8^h a^{h-\frac{1}{h}},$$

which is true for all \(a\ge 1\) and \(h\ge 1\). Note that both the bound in (12) and the coefficient \(\alpha _{5h-1}\) are increasing functions in h. From now on, suppose that \(h\ge 2\). We further claim that

$$\begin{aligned} a^{h-\frac{1}{h}}\cdot 16 \cdot 8^h < 2^k,\qquad 9\cdot 2^k\le \alpha _{5h-1}<10\cdot 2^k, \end{aligned}$$
(14)

which will give us the wanted overlap for the digital blocks of \(\alpha _{5h-1}\) and \(\alpha _{5h}\). By (14) the binary expansion of \(\alpha _{5h-1}\) is \((1001\cdots )_2\) and interferes with the digital block coming from \(\alpha _{5h}= a^h\) which is \((\cdots 0001)_2\) since \(a\equiv 1 \pmod {16}\). To prove (14), we show a stronger inequality that in turn implies (14), namely,

$$\begin{aligned} 144\cdot 8^h a^{h-\frac{1}{h}}<9\cdot 2^k \le 149 \cdot 8^h a^{h-\frac{1}{h}}. \end{aligned}$$
(15)

Passing to logarithms, this is equivalent to

$$(k-3h-4)-\delta \le \left( h-\frac{1}{h}\right) \log _2 a < k-3h-4$$

with

$$\delta =\log _2 149-\log _29 -4= 0.04924\cdots >\frac{1}{25}.$$

We rewrite

$$\left( h-\frac{1}{h}\right) \log _2 a=\left( h-\frac{1}{h}\right) \left( lh+\log _2\left( 1-\left( \frac{1}{2^{lh-4}}-\frac{1}{2^{lh}}\right) \right) \right) ,$$

which on the one hand shows that

$$\left( h-\frac{1}{h}\right) \log _2 a< \left( h-\frac{1}{h} \right) lh=lh^2-l=k-3h-4,$$

and on the other hand by \(-x/(1-x)<\log (1-x)<0\) for \(0<x<1\) that

$$\left( h-\frac{1}{h}\right) \log _2 a> lh^2-l-\frac{1}{\log 2}\cdot \frac{(2^4-1)\cdot \left( h-\frac{1}{h}\right) }{2^{lh}-2^4+1}.$$

Finally, we easily check that for all \(h\ge 2\) and \(l\ge 5\) we have

$$\frac{1}{\log 2}\cdot \frac{15\left( h-\frac{1}{h}\right) }{2^{lh}-15}<\frac{1}{25},$$

which finishes the proof of (15) and thus of (14).

After this technical preliminaries we proceed to the evaluation of the sum of digits. First note that for all \(h\ge 1\) by construction none of \(n=T(2^k)\), \(T(2^{k+1})\), \(T(2^{k+2})\), \(T(2^{k+3})\) is a power of two and therefore \(n\in \mathcal {A}\) in these four cases. Let \(h=h_1\ge 2\) and define a, b, M, l, k according to (9). To begin with, by (12)–(14) and the splitting formulas (4), we calculate

$$\begin{aligned} s_2(T(2^k)^{h_1})=s_2\left( \sum _{i=0}^{5h_1} \alpha _i 2^{ik}\right) =A_1-1+A_2+k+A_3, \end{aligned}$$
(16)

where

$$\begin{aligned} A_1&=s_2(\alpha _{5h_1})+s_2(\alpha _{5h_1-1}),\\ A_2&=s_2\left( \sum _{i=3}^{5h_1-2} \alpha _i \right) +s_2(\alpha _0),\\ A_3&=s_2(\alpha _{2}-1)-s_2(\alpha _1-1). \end{aligned}$$

Note that the summand k in (16) comes from formula (4) due to the negative coefficient \(\alpha _1\), and the \(-1\) comes from the addition of \((\cdots 0001)_2\) and \((1001)_2\) (the ending and starting blocks corresponding to \(\alpha _{5h_1}\) and \(\alpha _{5h_1-1}\)) which gives rise to exactly one carry. A similar calculation shows that

$$\begin{aligned} s_2(T(2^{k+1})^{h_1})&=A_1+A_2+(k+1)+A_3,\end{aligned}$$
(17)
$$\begin{aligned} s_2(T(2^{k+2})^{h_1})&=A_1+A_2+(k+2)+A_3,\end{aligned}$$
(18)
$$\begin{aligned} s_2(T(2^{k+3})^{h_1})&=A_1+A_2-1+(k+3)+A_3. \end{aligned}$$
(19)

In (17) we add \((1001)_2\) to \((\cdots 00010)_2\) which gives no carry. The same happens for the addition of \((1001)_2\) to \((\cdots 000100)_2\) in (18). Finally, in (19) we again have exactly one carry in the addition of \((1001)_2\) to \((\cdots 0001000)_2\). If we look mod 2 this shows that

$$\left( t(T(2^{k})^{h_1}),t(T(2^{k+1})^{h_1}),t(T(2^{k+2})^{h_1}),t(T(2^{k+3})^{h_1})\right) $$

is either (0, 0, 1, 1) or (1, 1, 0, 0). For \(h_2<h_1\) with \(h_2\ge 1\) all coefficients are non-interfering. To see this, consider the coefficients of \(T(x)^{h_2}=\sum _{i=0}^{5h_2} \alpha '_i x^i\). By (12), they are clearly bounded in modulus by \(a^{h_1-\frac{1}{h_1}}\cdot 16 \cdot 8^{h_1}<2^k\) for \(i=0,1,2,\ldots ,5h_2-2\). Also, by (10) and \(h_1\ge 2\),

$$ \alpha '_{5h_2-1}\le 150\cdot 8^{h_2} a^{h_2-\frac{1}{h_2}}<\frac{150}{8}\cdot 8^{h_1} a^{h_1-1}<a^{h_1-\frac{1}{h_1}}\cdot 16\cdot 8^{h_1}<2^k,$$

and thus we don’t have carry propagations in the addition of terms in the expansion of \(T(2^k)^{h_2}\). Similarly, we show that

$$\left( t(T(2^{k})^{h_2}),t(T(2^{k+1})^{h_2}),t(T(2^{k+2})^{h_2}),t(T(2^{k+3})^{h_2})\right) $$

is either (0, 1, 0, 1) or (1, 0, 1, 0). This yields in any case that

$$\{(t(T(2^{k+i})^{h_1}),t(T(2^{k+i})^{h_2})):\quad i=0,1,2,3\}=\{(0,0),(0,1),(1,0),(1,1)\},$$

which are the four desired values. Finally,

$$\begin{aligned} T(2^{k+3})&\le 2^{5(k+3)}\cdot 2^{lh_1+4}\\&\le 2^{l(5h_1^2+h_1-5)}\cdot 2^{5(3h_1+4)+4}\\&\le 32^{3h_1+5}\left( \frac{1}{15} \left( \frac{150}{h_1}\,8^{h_1}\right) ^2\right) ^{5h^2_1+h_1-5}, \end{aligned}$$

which completes the proof.    \(\square \)

Proof

(Theorem 2 ). This follows from Lemma 2 and

$$32^{3h_1+5}\left( \frac{1}{15} \left( \frac{150}{h_1}\,8^{h_1}\right) ^2\right) ^{5h^2_1+h_1-5}\le \exp \left( c_3 h_1^3\right) $$

for some suitable positive constant \(c_3\).    \(\square \)