UDC 519.1

MATHEMATICS

A. A. EVDOKIMOV

ON STRONGLY ASYMMETRIC SEQUENCES GENERATED BY A FINITE NUMBER OF SYMBOLS

(Presented by Academician A. I. Mal’tsev on 31 V 1967)

We consider sequences composed of a finite number of symbols. \(i\) consecutive symbols of such a sequence form a segment of length \(i\) \((i = 1, 2, 3, \ldots)\). Two consecutive segments are called equal if they consist of identical symbols and each symbol occurs in both segments the same number of times. The result of the present note is the construction of an infinite sequence, composed of a finite number of symbols, in which no two consecutive segments are equal. The question of the possibility of constructing such a sequence using a minimal number of symbols was posed as a problem by P. Erdős in paper \((^1)\) and arose, apparently, as a natural generalization of some previously known problems \((^{2-4})\).

1. Consider the sequence of 5 symbols

\[ abcdeacbcebdedaedcba. \tag{*} \]

We agree that the \(i\)-section of the sequence \((*)\) leaves on the left \(i\) terms of the sequence \((i = 1, 2, \ldots, 19)\). Introduce the column

\[ L_i = \begin{pmatrix} a_a\\ a_b\\ a_c\\ a_d\\ a_e \end{pmatrix}, \]

where \(a_x\) is the number of occurrences of the symbol \(x\) in the left part of the \(i\)-section of the sequence \((*)\). Similarly, the column \(P_i\) of the occurrences of symbols in the right part of the \(i\)-section of the sequence \((*)\) is defined.

Lemma 1. The sequence \((*)\) has the following properties:

a) it contains no consecutive equal segments;

b) \(L_i\) and \(P_i\) for \(i \in [2, 18]\) (i.e., \(i\) takes natural values from the indicated interval) contain at least two odd numbers, and for \(i = 1, 19\) one odd number \(a_a\);

c) the column \((P_i - L_i)\) for \(i \in [2, 18]\) contains at least two numbers equal in absolute value to 2, and for \(i = 1, 19\) one number \(|a_a| = 2\) (this property follows easily from b);

d) the column \((P_i + L_j)\) contains only even numbers if and only if either \(i = j\), or \(i + j = 20\) and \(i \notin [7, 13]\);

e) the column \([L_{(10+i)\bmod 20} - L_i]\) for \(i \ne 5, 15\) contains at least two odd numbers, and for \(i = 5, 15\) all numbers are equal in absolute value to 2.

1. Let \(\{x_0, x_1, \ldots, x_{n-1}\}\) be a set of symbols. Form \(n\) subsets

\[ M_i = \{x_i, x_{(i+1)\bmod n}, x_{(i+3)\bmod n}, x_{(i+7)\bmod n}, x_{(i+12)\bmod n}\} \qquad (i = 0, 1, \ldots, n-1). \]

Lemma 2. If \(n \ge 25\), then \(M_i \cap M_j\) for \(i \ne j\) contains no more than one symbol.

Indeed, if \(M_i\) and \(M_j\) contain at least two common symbols, then

\[ (i+p)\bmod n=(j+q)\bmod n,\qquad (i+u)\bmod n=(j+v)\bmod n, \tag{1} \]

where \(p,q,u,v\) take values from the set \(\{0,1,3,7,12\}\) and \(p\ne u\), \(q\ne v\), \(p\ne q\), \(u\ne v\).

From (1) it follows that

\[ q+u\equiv p+v \pmod n . \tag{2} \]

A direct check shows that, under the restrictions indicated above, identity (2) is impossible.

1. Let the symbols \(\{x_0,x_1,\ldots,x_{n-1}\}\) \((n\ge 25)\) be cyclically ordered

\[ x_0\to x_1\to \cdots \to x_{n-1}\to x_0. \tag{3} \]

From the symbols \(x_0,x_1,x_3,x_7,x_{12}\), as from \(a,b,c,d,e\), respectively, form the sequence \((*)\). We shall call this sequence the zero block. The \(i\)-block \((i=1,2,3,\ldots,n-1)\) is formed from the \((i-1)\)-blocks by a cyclic replacement of symbols according to (3). Since each of the \(n\) blocks is obtained by some replacement of the symbols in \((*)\), each block has properties a)—e) of Lemma 1. It is easy to see that the \(i\)-block is composed of the symbols of the set \(M_i\), and therefore distinct blocks have no more than one common symbol.

Denote by \(S_i\) the column of occurrences of symbols in the \(i\)-block,

\[ S_i= \begin{Vmatrix} \alpha_{i0}\\ \alpha_{i1}\\ \vdots\\ \alpha_{i(n-1)} \end{Vmatrix}, \]

where \(\alpha_{ij}\) is the number of occurrences of the \(j\)-th symbol in the \(i\)-block. By the construction of the \(i\)-block we have
\[ \alpha_{ii}=\alpha_{i,i+1}=\alpha_{i,i+7}=\alpha_{i,i+12}=4, \]
and all the other numbers in the column \(S_i\) are equal to zero. Addition and multiplication of the columns \(S_i\) by numbers are defined in the usual way.

Lemma 3. The collection of columns \(\{S_i\}\) \((i=0,1,\ldots,n-1)\) forms a basis of the \(n\)-dimensional vector space over the rational field.

It is enough to show that the determinant \(D\) of the matrix \(\|\alpha_{ij}\|\) of order \(n\), whose columns are the columns \(\{S_i\}\), is nonzero. By the construction of the blocks the matrix \(\|\alpha_{ij}\|\) is circulant (5), and therefore the value of its determinant is equal to

\[ D=f(\varepsilon_0)f(\varepsilon_1)\cdots f(\varepsilon_{n-1}), \]

where

\[ f(x)=4+4x+4x^3+4x^7+4x^{12}, \]

and \(\varepsilon_k\) are all the values of \(\sqrt[n]{1}\) \((n\ge 25)\). But it is not hard to show that the polynomials \(f(x)\) and \(x^{25}-1\) are relatively prime, i.e. \(f(\varepsilon_k)\ne 0\) for \(k=0,\ldots,24\).

1. Let \(N_0\) be an arbitrary sequence without consecutive equal segments, composed of the symbols \(\{x_0,x_1,\ldots,x_{n-1}\}\). Replace in \(N_0\) each symbol \(x_i\) by the \(i\)-block. We shall call the resulting sequence the 1-iteration and denote it by \(N_1\). Obviously, \(N_1\) consists of the symbols \(x_0,x_1,\ldots,x_{n-1}\). In general, the \(m\)-iteration \(N_m\) \((m=2,3,\ldots)\) is formed from the \((m-1)\)-iteration by replacing in it the symbols \(x_i\) by the \(i\)-blocks. The iterative process described makes it possible to construct arbitrarily long sequences composed of a finite number of symbols. If one takes the symbol \(x_0\) as \(N_0\), then \(N_1\) is the zero block, and by induction it is easy to prove that every sequence \(N_m\) is an initial segment of \(N_{m+1}\), i.e. in this case the iterative process defines the construction of an infinite sequence.

Theorem. The sequence \(N_m\) \((m=1,2,3,\ldots)\) contains no consecutive equal segments.

It is proved that the presence in \(N_m\) of consecutive equal segments either entails their presence in \(N_{m-1}\), or contradicts Lemmas 1, 2, 3. The assumed equality in \(N_m\) of consecutive segments is represented schematically (see Fig. 1). The rectangles represent the blocks composing

sequence \(N_m\). The vertical lines determine the boundaries of consecutive equal segments. The letters \(\alpha\) and \(i\) at a line mean that this line makes an \(i\)-section of an \(\alpha\)-block (i.e., to the left of the line there are \(i\) symbols of the \(\alpha\)-block). As in 1, for every \(\alpha\)-block the columns \(L_i^\alpha\) and \(P_i^\alpha\) of occurrences of symbols in the left and right parts of the \(i\)-section are introduced. \(L_i^\alpha+P_i^\alpha=S_\alpha\). By construction of the blocks, \(L_i^\alpha\) and \(P_i^\alpha\) differ from \(L_i\) and \(P_i\) for \((*)\) only by zeros indicating the absence in the \(\alpha\)-block of the corresponding symbols from \(\{x_0,x_1,\ldots,x_{n-1}\}\).

Fig. 1

The column \(F=(P_i^\alpha+L_r^\beta-P_r^\beta-L_j^\gamma)\) characterizes the asymmetry of the symbols created by the sections \(i,r,j\) of the blocks \(\alpha,\beta,\gamma\). We consider all possible cases of equality of segments in \(N_m\)

\[ \begin{array}{ll} \text{I.}\quad (i=j=0) & \begin{cases} r=0; & (11)\\ r=10. & (12) \end{cases} \end{array} \]

\[ \text{II.}\quad (i=0,\ j\ne0)\quad (i\ne0,\ j=0\ \text{is proved analogously}). \]

\[ \begin{array}{lll} \text{III.}\quad (i\ne0,\ j\ne0) & \begin{cases} (\alpha\ne\gamma) \begin{cases} i\in[2,18]\ (j\in[2,18]); & (\mathrm{III}\,1)\\ i=1,19;\quad j=1,19; & (\mathrm{III}\,2) \end{cases}\\[6pt] (\alpha=\gamma) \begin{cases} i=j; & (\mathrm{III}\,3)\\ i+j=20;\quad i\notin[7,14]; & (\mathrm{III}\,4)\\ i,j\ \text{do not satisfy }(\mathrm{III}\,3)\ \text{and }(\mathrm{III}\,4). & (\mathrm{III}\,5) \end{cases} \end{cases} \end{array} \]

(11) \(F\) is the zero column. Remove identical blocks from both consecutive segments. If, after removal of the blocks, nothing remains in the segments, this means that each block entered both segments the same number of times, but then the corresponding consecutive segments in \(N_{m-1}\) are equal, which is impossible. Suppose that after deletion, in the left segment there remain \(i_1\)-, \(i_2\)-, \(\ldots\), \(i_p\)-blocks with multiplicities of occurrence \(a_1,a_2,\ldots,a_p\), respectively, and in the right segment \(j_1\)-, \(j_2\)-, \(\ldots\), \(j_q\)-blocks with multiplicities of occurrence \(b_1,b_2,\ldots,b_q\). The condition of equality of the segments is then written in the form

\[ \sum_{l=1}^{q} b_l S_{j_l}-\sum_{l=1}^{p} a_l S_{i_l}=F, \tag{**} \]

where \(a_l\) and \(b_l\) are natural numbers.

But in our case \(F\) is the zero column, and therefore \((**)\) means a linear dependence of the columns \(\{S_i\}\), which is impossible by Lemma 3.

(12) \(L_{10}^{\beta}-P_{10}^{\beta}\), by property c) of Lemma 1, contains numbers whose absolute values are equal to 2. Under this condition the equality \((**)\) is impossible, since the columns \(\{S_i\}\) consist only of zeros and fours, and consequently any linear combination of them with integer coefficients is a column of numbers divisible by four.

(II) Let us note that from the condition of equality of consecutive segments it necessarily follows that the column \((P_i^\alpha+L_j^\alpha)\) consists only of even numbers. In our case, by property b) of Lemma 1, this is not satisfied.

(III 1) \(P_i^\alpha\), by property b) of Lemma 1, contains at least two odd numbers. But, by Lemma 2, the \(\alpha\)- and \(\gamma\)-blocks \((\alpha\ne\gamma)\) have no more than one common symbol.

one common symbol, and therefore, when adding \(\Pi_i^\alpha\) to \(\Pi_j^\gamma\), at least one of the odd numbers of the column \(\Pi_i^\alpha\) will be added to a zero of the column \(\Pi_j^\alpha\), i.e. \((\Pi_i^\alpha+\Pi_j^\gamma)\) contains an odd number.

(III 2) As in (III 1), \(\Pi_i^\alpha+\Pi_j^\gamma\), in view of \(\alpha\ne\gamma\) and \(i=1,19,\ j=1,19\), contains two odd numbers.

(III 3) Four subcases are possible:

\[ \begin{aligned} (r=i)=&\begin{cases} \longrightarrow \alpha=\beta;\\ \longrightarrow \alpha\ne\beta; \end{cases} &&\begin{aligned} &\text{(III 3a)}\\ &\text{(III 3b)} \end{aligned}\\[6pt] r=(10+i)\bmod 20=&\begin{cases} \longrightarrow \alpha=\beta;\\ \longrightarrow \alpha\ne\beta. \end{cases} &&\begin{aligned} &\text{(III 3c)}\\ &\text{(III 3d)} \end{aligned} \end{aligned} \]

We shall prove only one of them.

(III 3c) The case \(i=10\) is proved by (I 2). Let \(i\ne10\).

\[ F=\Pi_i^\alpha+\Pi_{(10+i)\bmod 20}^\alpha-\Pi_{(10+i)\bmod 20}^\alpha-\Pi_i^\alpha =2(\Pi_{(10+i)\bmod 20}^\alpha-\Pi_i^\alpha). \]

If \(i\ne5,15\), then, by property e) of Lemma 1, \(F\) contains either the numbers \(\pm2\), or \(\pm6\), i.e. equality \((**)\) is impossible. If, however, \(i=5,15\), then, by property e) of Lemma 1, \(F=\pm S_\alpha\). In view of the uniqueness of the expansion of \(F\) with respect to the basis \(\{S_i\}\), in condition \((**)\) the coefficient of \(S_\alpha\) is equal to \(\pm1\), and all the others are equal to zero, i.e. after deleting identical nonintersecting blocks from both segments, only one \(\alpha\)-block remains in the right or left segment. It is not hard to see that then in \(N_m\) there exist two consecutive equal segments for which \(i=j=r=0\), which is impossible by (I 1).

(III 4) Can be reduced to case (I 1) or (I 2).

(III 5) By property d) of Lemma 1, the column \(\Pi_i^\alpha+\Pi_j^\alpha\) contains odd numbers.

The cases enumerated exhaust all possible dispositions of consecutive segments in \(N_m\).

1. Remark. The number of symbols necessary for the indicated construction of the sequence may be reduced. It is possible that \(n=5\). It is easy to show that \(n\ne3\).

The author expresses gratitude to V. V. Glagolev, who made a number of valuable remarks.

REFERENCES

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3. S. E. Arshon, Matem. sborn., 2, 769 (1937).
4. M. Morse; A. Hedlund, Duke Math. J., 11, 1 (1944).
5. I. V. Proskuryakov, Collection of Problems in Linear Algebra, Moscow, 1957.