Additive irreducibles in α-expansions

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1 Publ. Math. Debrecen Manuscript Additive irreducibles in α-expansions By Peter J. Grabner and Helmut Prodinger * Abstract. The Bergman number system uses the base α = + 5 2, the digits 0 and, and the condition that adjacent ones are forbidden. We are interested in those positive integers such that replacing one or more of the ones never results again in a positive integer; they are called additively irreducible. These numbers are characterised in terms of the positions of their ones. Further, the number of irreducible positive integers below a given bound is considered and evaluated asymptotically, as the bound goes to infinity. The periodic function that appears is analysed in detail.. Introduction Let α = be the golden ratio. It is well known that each natural number n has a unique α-expansion n = k Zε k α k with the digits ε k {0,}, such that no nonzero digits are adjacent to each other, and only a finite number of digits is different from 0 cf. [, 4]. It is sometimes called Bergman s number system. If one considers a standard q-ary expansion ε k q k 2 Received by the editors August 3, 20. Mathematics Subject Classification: A63, K6, M4. Key words and phrases: Bergman expansion, non-integer base. This author is supported by the project S9605-N2 of the Austrian Science Fund. * This author is supported by an incentive grant from the South African National Research Foundation. k 0

2 2 P. J. Grabner and H. Prodinger of a natural number, with digits ε k in the set {0,...,q } then one still gets a natural number if one replaces some of the digits by zero. We are interested in positive integers such that all the numbers L I = L k= K ε k α k, k= K ε k α k for K < K and L < L are no integers if they are not zero. The integers I with this property can be seen as the additive building blocks of the integers in the Bergman-representation. Positive integers that cannot be decomposed any further in the above sense will be called additive irreducibles, or α-irreducibles for short. It is easy to see that the decomposition into α-irreducibles is unique. Here is a short list of the first few natural numbers, decomposed into additive irreducibles:,2,3,4= +3,5,6,7,8= +7,9 = 2+7,0 = 3+7, = 4+7,2, 3,4,5= +4,6,7,8,9= +8,20 = 2+8,2 = 3+8, 22 = 4+8,23 = 5+8,24 = 6+8,25 = 7+8,26 = +7+8,..., = ,.... The original motivation for studying these integers was that every positive integer n can be decomposed uniquely into a sum of Is; n = I + +I k. The sum-of-digits function in Bergman-representation of integers s B L l= K ε l α l = is additive with respect to this decomposition L l= K s B I + +I k = s B I + +s B I k. The sum-of-digits function s B has been studied in [2]. In this paper, we will prove the following theorems, a characterisation of the α-irreducibles m in terms of the α-expansion, and an irreducible number counting theorem, giving an asymptotic formula for the number of α-irreducibles below a given number n. ε l

3 Additive irreducibles in α-expansions 3 Theorem. An integer m > with α-expansion m = α e + + α e l + α f + +α f h with e > > e l, and 0 > f > > f h is an α-irreducible, iff e,...,e l are odd. Equivalently, m is an α-irreducible, iff f,...,f h are odd and f h is even. In words, the positive exponents must all be odd, with the possible exception of the smallest one. Since m = is α-irreducible by definition, α 0 cannot occur in the Bergman-expansion of an α-irreducible. Theorem 2. ThenumberAnofα-irreduciblesamongthenumbers,...,n satisfies for n an asymptotic formula An = Φ 2 log αn n ρ +Ologn, with ρ = log2 2logα = and Φ a continuous periodic function of period. Primes.eps Fourier50.eps Figure. Plot of the function Φ compared to the first 50 terms of its Fourier series. 2. Proofs Lemma. Let ε K,...,ε L {0,} be a sequence of digits satisfying the condition that no two consecutive digits are both and assume that n = Then K = 2 L 2 holds. L l= K ε l α l N with ε K = ε L =.

4 4 P. J. Grabner and H. Prodinger Proof. Since n is an integer, it is not changed by applying the conjugation α α in the ring Z[α]. This gives n = L l= K ε l α l = K l ε l α l. l= L The highest occurring power in this sum has to be even, since otherwise the number represented by the sum would be negative. For even K we have α K α K 3 α K 5 = α K < n < α K +α K 2 + = α K+. On the other hand α L < n < α L+, which gives the inequalities L < K + and L+ > K. Thus we have L K L+, which together with the fact that K is even implies K = 2 L 2. The following Lemma explains how the positive powers in the expansion of an integer determine the negative powers. Lemma 2. Let ε 0,...,ε L {0,} be a sequence of digits satisfying the condition that no two consecutive digits are both. Let 0 < 2l < 2l 2 <... < 2l m be the strictly positive even indices of non-zero digits set m = 0, if there are none and set l 0 = 0 and l m+ =. Furthermore, let 2l j < 2k j + < 2l j+ 0 j m be the largest odd index of a non-zero digit between the two consecutive even indices 2l j and 2l j+. Set k j = l j, if all digits ε 2lj+,...,ε 2lj+ are zero. Then L ε l α l l=0 L m = ε l α l + l=0 j=0 { for ε 2l+ α 2l +α 2kj 2 k0 =, 0 otherwise, k j 3 where the sum in the first parenthesis is 0 if k 0 =. Proof. We first notice that the sum over the negative powers of α is an admissible expansion no two consecutive digits are and therefore less than.

5 Additive irreducibles in α-expansions 5 Let L k = α k + α k denote the k-th Lucas-number. Then we have ε 2lj α 2lj + k j ε 2l+ α 2l+ + = ε 2lj α 2lj + = ε 2lj L 2lj + k j k j ε 2l+ k j ε 2l+ α 2l +α 2kj 2 α 2l+ α 2l + ε 2l+ L 2l+ N k j α 2l +α 2kj 2 for j > 0; a slight modification is necessary for j = 0. Thus the number in 3 is an integer which is at most larger than L l=0 ε lα l, which proves the lemma. Proof of Theorem. The decomposition of an integer given in Lemma 2 shows that the only possibility for an integer to be additively irreducible is that there is only one summand in the decomposition 3. Notice that by Lemma every expansion of an integer has to end with the digit in an even position. Thus the summands in 3 cannot be decomposed any further. The only possibilities to have only one summand in the decomposition 3 is to either have m = 0 and therefore no non-zero digit in an even position, or m = and k 0 =. The first possibility is the case when e,...,e l are all odd, the second possibility is the case when e,...,e l are odd and e l is even. Proof of Theorem 2. Let Then define the set A by { L } A = ε 2l+ α 2l+ l : ε 2l+ {0,}. l=0 A = α 2k α 2 A + {}, 4 k= where the union is disjoint. Furthermore, we define A = A A \ {0}. The set of α-irreducibles is then formed by completing A with appropriate digits after the comma by Theorem and Lemma 2. We now count the number of elements of A less than X = 2L+ l=0 ε l α l, 5

6 6 P. J. Grabner and H. Prodinger where we assume that either ε 2L+ = or ε 2L =. We denote X k = 2L+ l=k ε l α l and define M = max{l ε 2l = } set M = 0 if the set is empty. Then we have #{a A a < X} = 2L+ k=2m The condition X k+ a < X k implies that and a = k δ l α l + l=0 #{a A X k+ a < X k }. 2L+ l=k+ k δ l α l < ε k α k. l=0 ε l α l Thus a A can only hold if the digits ε 2L+,...,ε k+ have the property that all non-zero digits have odd index, except possibly for the index k+. This explains why the sum starts with k = 2M. For k 2M all non-zero digits in ε 2L+,...,ε k+ have odd index; then the digits δ k,...,δ 0 can be chosen so that all non-zero digits have odd index, except possibly for the index of the last non-zero digit. For k the number of possible choices for δ k,...,δ 0 is then k 2 ε k l= 2 k 2 l +2 k 2 = ε k 3 2 k 2, where each summand in the sum corresponds to those choices of digits, which end with a non-zero digit at the even position 2l, whereas the second term counts the choices which only have non-zero digits in odd positions. For k = 0 the number of possible choices is ε 0 ; this can only occur, if M = 0. For x = ε k α k [α 2,] k=

7 Additive irreducibles in α-expansions 7 define px = min{k ε k = and k even} and px φx = ε k 2 k+ 2. k= Then the number of α-irreducibles less than X = α 2L+2 y with y [α 2,] equals min{py,2l} k= ε 2L+ k 3 2 L k 2 +ε 2L+ = 2 L 3φy+OL. Defining Φt = 2 t 3φα 2t 2 for 0 t and writing #{a A a < X} = 2 L 3φXα 2L 2 +OL = X ρ Xα 2L 2 ρ 3φXα 2L 2 +OL = X Φ ρ 2 log αx L +OlogX gives the number of truncated irreducibles. The function φ is continuous on the interval [α 2,]: we only have to check that two different representations of a number x give the same value for φx. For that let L L x = ε l α l = ε l α l + α L+2k, l= where ε L =. Then we have l= px < L: φx = px l= ε l2 l+ 2, since the digits with index < L are the same for both representations px = L: in this case we have φx = L l= k= L ε l 2 l+ 2 = ε l 2 l+ 2 + l= k= 2 L+2k 2, since the last sum equals 2 L 2 = 2 L+ 2, because L is even.

8 8 P. J. Grabner and H. Prodinger px > L: in this case we have φx = L l= because L+ is even. L ε l 2 l+ 2 = ε l 2 l L+ 2, l= Let x,y be two real numbers with α k x y < α k. Then either the digital expansions of x and y to base α agree up to the k-th digit, which yields φx φy < 2 k 2; or the two numbers have less than k digits in common. In this case there exists a number ξ between x and y which has two different representations to base α, which have k digits in common with x and y respectively. Then φx φy φx φξ + φξ φy 2 k 2 + and we have We now have for N = X and X as in 5 φx φy 2 x y ρ. 6 AN = #{n < N n irreducible} = #{a A a < X}+O = X Φ ρ 2 log αx +OlogX = N +X N ρ Φ 2 log α N +O N ρ Φ 2 log αn Φ 2 log αx +OlogN = N ρ Φ 2 log α N +OlogN, 7 where we have used 6 to bound Φ 2 logαn Φ 2 logαx by ON ρ. This proves the theorem. 3. Dirichlet series and Fourier coefficients We will explain now how to compute the Fourier coefficients of Φ numerically to high precision. For this purpose we study the Dirichlet series ζ A s = a s, ζ A s = a s, and ζ A s = a s. 8 a A a A a A a 0

9 Additive irreducibles in α-expansions 9 Clearly, we have ζ A s = ζ A s+ζ A s and ζ A s = = α 2ks k= + a A a 0 +α 2 a s α 2s +α 2s ζ A s+α 2s l= s α 2l ζ a s+l l 9 by 4. Furthermore, by the decomposition we have A \{0} = α 2 A \{0} α+α 2 A \{0} {α} ζ A s = α 2s ζ A s+ a A a 0 α+α 2 a s +α s = 2α 2s ζ A s+α s +α 2s This gives the functional equation for ζ A s ζ A s = α 2s α s + 2 l= l= s α l ζ A s+l. l s α l ζ A s+l. 0 l This functional equation provides the analytic continuation of ζ A s to the left of Rs = ρ. It is similar in its form to the trivial functional equation of the Riemann zeta function and also in the way it is derived. This functional equation can be used to compute values of ζ A s numerically tohighprecisionbytheobservationthatforlargel, ζ A s+l α s l = Oα 3l with an explicit O-constant. Thus the series in 0 is geometrically convergent and the error can be controlled. Starting with ζ A s+k with k large enough to meet the prescribed error bound, we can use the functional equation to compute ζ A s+k,...,ζ A s by rapidly convergent series, where we can control the error bounds. For a detailed description in a similar instance we refer to [5]. From 0 we read off that ζ A s has simple poles at the points s = ρ+ kπi logα with residues Res ζ A s = k 2+ s=ρ+ kπi 4logα log α ρ kπi l= logα l α l ζ A ρ+l+ kπi, logα

10 0 P. J. Grabner and H. Prodinger which can be computed numerically by the procedure described above. By 9 ζ A s has the same poles with residues Res ζ A s = Res ζ A s. s=ρ+ kπi 2 log α s=ρ+ kπi log α We now introduce the Dirichlet generating function of all irreducible elements: ηs = n s. n N n irreducible Since the irreducible integers differ at most by from the elements of A by Lemma 2, this function has the same poles and residues as ζ A s. The counting function of irreducible elements can now be expressed by the Mellin-Perron formula from classical analytic number theory cf. [6]: BN = n<n n irreducible n = N 2πi 2+i 2 i ηsn s ds ss+. Shifting the line of integration to Rs = ρ ε, where the integral is still convergent, we obtain by residue calculus BN = N ρ ρ+ kπi kπi k Z logα ρ++ logα Res = N Ψ ρ 2 log αn +ON ρ ε s=ρ+ kπi log α ζ A se kπilog α N +ON ρ ε with a continuous periodic function Ψ. For more details on the application of this technique to digital functions we refer to [3, 5]. The counting functions AN defined in 7 and BN are related by partial Abel summation: BN = An. N The periodic function Φ in Theorem 2 is Hölder continuous with exponent ρ > 2 by6. Therefore its Fourier series is absolutely convergent by Bernstein s theorem cf. [7]. Thus the Fourier coefficients of Φ can be related to those of Ψ by Φk = n N ρ++ kπi Ψk = logα ρ+ kπi logα Res ζ A s s=ρ+ kπi log α

11 Additive irreducibles in α-expansions by [3] or [5, Proposition 5]. Wehaveusedthenumericalmethodforthecomputationofζ A sasdescribed above to compute the first Fourier coefficients Φ0 = , Φ = i, Φ2 = i. The plot of the first 50 terms of the Fourier series compared to the function Φ is shown in Figure. Acknowledgement. The authors are grateful to two anonymous referees for their valuable comments and suggestions. References [] G. M. Bergman, A number system with an irrational base, Math. Mag. 3957/58, [2] C. Cooper and R. E. Kennedy, The first moment of the number of s function in the β- expansion of the positive integers, J. Inst. Math. Comput. Sci. Math. Ser , [3] P. Flajolet, P. J. Grabner, P. Kirschenhofer, H. Prodinger, and R. F. Tichy, Mellin transforms and asymptotics: digital sums, Theor. Comput. Sci , [4] C. Frougny, How to write integers in a non-integer base, LATIN 92, São Paulo, Lecture Notes in Computer Science, vol. 583, Springer, Berlin 992. [5] P. J. Grabner and H.-K. Hwang, Digital sums and divide-and-conquer recurrences: Fourier expansions and absolute convergence, Constructive Approximation , [6] G. Tenenbaum, Introduction to analytic and probabilistic number theory, Cambridge Studies in Advanced Mathematics, no. 46, Cambridge University Press, 995. [7] A. Zygmund, Trigonometric Series. Vol. I, II, Cambridge University Press, Cambridge, 988. P. Grabner INSTITUT FÜR ANALYSIS UND COMPUTATIONAL NUMBER THEORY, TECHNISCHE UNIVERSITÄT GRAZ, STEYRERGASSE 30, 800 GRAZ, AUSTRIA Peter.Grabner@tugraz.at H. Prodinger DEPARTMENT OF MATHEMATICS UNIVERSITY OF STELLENBOSCH 7602 STELLENBOSCH SOUTH AFRICA hproding@sun.ac.za