Infinite Products Associated with Counting Blocks in Binary Strings

Transcription

1 Ifiite Products Associated with Coutig Blocks i Biary Strigs J.-P. Allouche U. A. 226, Mathématiques et Iformatique J. O. Shallit Departmet of Computer Sciece Uiversité de Bordeaux Uiversity of Chicago 351, cours de la Libératio 1100 E. 58th St TALENCE Cedex Chicago, IL Frace USA AMS (1985) Subject Classificatio Codes: 05A15, 11A63, 68C05. Abstract. Let w be a strig of 0 s ad 1 s, ad let a w () be the fuctio which couts the umber of (possibly overlappig) occurreces of w i the biary expasio of. We show that there exists a effectively computable ratioal fuctio b w () such that 0 log 2 (b w ())X a w() = 1 1 X. By settig X = 1 ad expoetiatig, we recover previous results ad also obtai some ew oes; for example, ( 2 )( 1) a 0() 2 = Our work is a geeralizatio of previous results of D. Woods, D. Robbis, H. Cohe, M. Medès Frace, ad the authors. 1

2 I. Itroductio. Let s q () deote the sum of the digits of the oegative iteger whe writte i base q. WOODS ad ROBBINS [9] [7] showed that 0 ( ) ( 1) s 2 () = 2 2 (1) This formula was geeralized by the secod author to bases other tha 2 usig methods of real aalysis [8]. Later, the first author ad H. COHEN foud more geeral results usig Dirichlet series [1]. I a joit paper with COHEN ad M. MENDÈS FRANCE [3] the authors showed that ( ) X sq() + 1 log q q q + q = 1 (2) 1 X 0 for X i a suitable regio of covergece, from which the results of WOODS ad ROBBINS easily follow upo lettig X = 1 ad q = 2. I [3] it was also show that 0 ( ) X u() (2 + 1) 2 log 2 ( + 1)(4 + 1) = 1 1 X, where u() is the Rudi-Shapiro fuctio, which couts the umber of occurreces of 11 i the biary expasio of. This formula aturally suggests the existece of similar formulas correspodig to the coutig of ay biary strig. The existece of such formulas is the questio we address i this paper. For ay fiite oempty block w of 0 s ad 1 s, we defie a w () as the umber of occurreces of w i the biary expasio of. With this quatity we associate a ifiite series of the form log 2 (b w ())X a w() whose sum is also 1 1 X. This allows us to evaluate some ovel ifiite products, as well as recover previous results. II. Notatio. Let w be a strig or block of 0 s ad 1 s (i. e. w (0 + 1) ). Let v : (0 + 1) N be the map that assigs to w its value whe iterpreted i base 2; e. g. v(0101) = 5. Let w deote the legth of w, i. e. the umber of symbols i the strig w. Let w be oempty ad let a w () cout the umber of (possibly overlappig) occurreces of the block w i the biary expasio of. For example, a 11 (15) = 3. Some clarificatio is eeded i the case where w starts with a 0; if w 0 j, the i evaluatig 2

3 a w () we assume that the biary expasio of starts with a arbitrarily log prefix of 0 s. Thus a 010 (5) = 1, sice we cosider the biary expasio of 5 to be This defiitio of a w () is appropriate for all cases except w = 0 j ; i this case we use the biary expasio of which starts with a 1. Thus a 00 (4) = 1. If w ad z are strigs, let a w(z) cout the umber of (possibly overlappig) occurreces of w i z. Note that here the strig z is ot exteded with leadig zeroes. Fially, we defie III. Some ifiite series. ( ) x L(x) = log 2. x + 1 Our goal is to prove the result metioed at the ed of the itroductio. however, we will prove the followig First, Lemma 1. Let w be a oempty strig of 0 s ad 1 s ad let a w () be as defied above. Let g ad h be itegers. The for all k 0, the series coverges. 1 a w (g+h)=k 1 Proof. It suffices to prove that 1 a w (g+h)=k 1 g + h coverges. But this is majorized by 1 a w ()=k 1, so it suffices to prove that this last series coverges. Let b = 2 w, let N be a positive iteger, ad let l be defied by b l 1 N b l 1, so that l 1 + log N log b. We may suppose l k. Let us write s w() for the fuctio which couts the umber of occurreces of the digit w whe is writte i base b. The S(N) = 0 N a w ()=k 1 0 b l 1 s w () k 3 1 = 0 j k ( ) l (b 1) l j. j

4 Now boud (b 1) l j by (b 1) l ad ( ) l j by l j j! lk. Thus S(N) = 0 N a w ()=k 1 (k + 1)l k (b 1) l c k (log N) k N log(b 1) log b. Now put We see that a() = S(N) = { 1 if aw () = k, 0 otherwise. 0 N a w ()=k 1 = 0 N a(), from which we get 1 N a w ()=k 1 = 1 N a() = 1 N = S(0) + S(N) N + S() S( 1) 1 N 1 S() ( + 1) ad, usig the boud determied above for S(), this quatity teds to a fiite limit as N. Theorem 2. Let w be a oempty strig of 0 s ad 1 s ad v(w) g = 2 w 1, h =. 2 The, for all k 0, a w (g+h)=k L(2g + v(w)) = 1, (3) where the sum is over 1 if w = 0 j ad 0 otherwise. Proof. Note that we claim that the sum is 1, idepedet of k. Sice L(2g + v(w)) = ( ) 1 1 2(log 2)g + O 2, 4

5 Lemma 1 esures covergece of these series. The proof is divided ito three cases: (I) w eds i a 1; (II) w eds i a 0 but w 0 j ; ad (III) w = 0 j. Case I: w eds i a 1. Let d w (k) be defied by d w (k) = 0 a w ()=k L(2 + 1). By writig = gr + m, with r 0 ad 0 m g 1, we see that d w (k) = g 1 m=0 r 0 a w (gr+m)=k L(2gr + 2m + 1). Similarly, if we let the e w (k) = e w (k) = g 1 m=0 0 a w (2+1)=k r 0 a w (2gr+2m+1)=k L(2 + 1), L(2gr + 2m + 1). Now we claim that a w (2g + 2m + 1) a w (g + m) = { 1 if m = v(w)/2, 0 otherwise. This is easy to see, sice the biary expasio of 2g + 2m + 1 is the same as that for g + m, except that there is a extra 1-bit o the ed. Hece if a w (2g + 2m + 1) > a w (g + m) the the last w bits of 2g +2m+1 must coicide with the strig w, ad so m = v(w)/2, sice w eds i 1. Thus we see that all but oe of the terms i the sums for d w (k) ad e w (k) are idetical, ad hece, o recallig h = v(w)/2, d w (k) e w (k) = r 0 a w (gr+h)=k L(2gr + v(w)) r 0 a w (gr+h)=k 1 L(2gr + v(w)). (4) Now if we could show that d w (k) = e w (k) for k > 0, the it would follow from equatio (4) that the value of the sum 5

6 r 0 a w (gr+h)=k L(2gr + v(w)) is idepedet of k ad hece equal to d w (0) e w (0). I fact, we ow show that d w (k) = e w (k) + E k where Hece For we have 1 a w ()=k e w (k) = L() = E k = 0 a w (2+1)=k { 1 if k = 0, 0 if k 1. L(2) + 1 a w (2)=k L(2 + 1) = 0 a w (2+1)=k 1 a w ()=k L(2 + 1). (L() L(2)) = ( E k ) + 0 a w ()=k This completes the proof of case I. L(2 + 1) = d w (k) E k. ad Case II: w eds i a 0, but w 0 j. First, we defie fuctios d w ad e w similar to those defied above: d w(k) = 1 a w ()=k L(2) e w(k) = L(2). 1 a w (2)=k Followig the method used for Case I, we easily fid d w(k) e w(k) = r 0 a w (g+h)=k L(2gr + v(w)) 6 r 0 a w (g+h)=k 1 L(2gr + v(w)).

7 As before, we shall obtai the relatio betwee d w(k) ad e w(k). We have 1 a w ()=k ( L() = 1 a w (2)=k ) ( L(2) + 1 a w (2+1)=k ) L(2 + 1) + E k. Hece e w(k) = L(2) = ( E k ) + 1 a w (2)=k = ( E k ) + 1 a w ()=k 1 a w ()=k L(2) = d w(k) E k. (L() L(2 + 1)) This completes the proof of Case II. Case III: w = 0 j. Left to the reader. This completes the proof of Theorem 2. IV. Some uusual power series. We ow modify Theorem 2 to obtai some uusual power series: Theorem 3. Let w be a strig of 0 s ad 1 s, ad v(w) g = 2 w 1, h =, 2 ad let X be a complex umber with X < 1. The X a w(g+h) L(2g + v(w)) = 1 1 X, where the sum is over 1 for w = 0 j ad 0 otherwise. Proof. For X < 1, the series 0 X is absolutely coverget, ad the quatities L(2g + v(w)) 7

8 are all egative; thus we have 1 1 X = X k ( 1) k 0 = X k k 0 a w (g+h)=k L(2g + v(w)) = X a w(g+h) L(2g + v(w)), where we have used Theorem 2. This result, although appealig, is usatisfactory i two ways. First, the expoet of X is a w (g + h) istead of a w (). Secod, i order to obtai the ifiite products metioed i the itroductio, we must show that Theorem 3 i fact holds for all X 1, X 1. The first of these problems is corrected i this sectio, while the questio of covergece o the uit circle is addressed i the ext sectio. We ow show how to modify Theorem 2 so that the summatio is over all with a w () = k. To do this, we eed the otio of a suffix of a strig. Let x ad y be two strigs. The we say x is a suffix of y if there exists a third strig z such that y = zx. Now we prove the followig Lemma 4. Let t be a iteger with biary expasio t = b 1 b 2 b r b r+1 b s. (A) If b 1 b 2 b r is ot a suffix of w, the L(2 s + t) = a w (2 r +v(b 1 b r ))=k a w (2 r 1 +v(b 1 b r 1 ))=k (B) If, however, b 1 b 2 b r is a suffix of w, the L(2 s + t) = a w (2 r +v(b 1 b r ))=k L(2 s + t). L(2 s 1 + t 1 ) a w (2 r 1 +v(b 2 b r ))=k L(2 s + t 2 ) a w (2 r 1 +v(b 1 b 2 b r 1 ))=k where t 1 = v(b 2 b 3 b r b r+1 b s ), t 2 = v(b 1 b 2 b r b r+1 b s ), ad by b we mea the complemet of the bit b; i. e. 0 = 1; 1 = 0. 8

9 Proof. Part (A) of the lemma ca be left to the reader. To prove part (B), we start with the summatio L(2 s 1 + t 1 ) a w (2 r 1 +v(b 2 b r ))=k ad break the sum ito two parts, correspodig to = 2m + b 1 ad = 2m + b 1, where b 1 is a sigle bit. We get a w (2 r 1 +v(b 2 b r ))=k L(2 s 1 + t 1 ) = L(2 s m + t) + m a w (2 r m+v(b 1 b r ))=k L(2 s m + t 2 ). m a w (2 r m+v(b 1 b 2 b r ))=k Now b 1 b 2 b r is a suffix of w, so b 1 b 2 b r caot be a suffix of w. Thus we may chage the idex of summatio i the rightmost sum to a w (2 r 1 m + v(b 1 b 2 b r 1 )), ad the result follows. Lemma 5. There is a ratioal fuctio b w () such that for all k 0 we have a w ()=k log 2 (b w ()) = 1. (5) (The summatio is over 1 for w = 0 j ad 0 otherwise.) This fuctio b w () is effectively computable, ad the degree d w of the umerator ad deomiator of b w () satisfies d w 2 w 1. Proof. To prove the first statemet, we start with Theorem 2 ad successively apply Lemma 4. At each step, we covert a sum over a w (2 r + x 1 ) to either oe or two sums over a w (2 r 1 +x 2 ). Thus, after at most w 1 stages (ad a total of at most 2 w 1 ivocatios of Lemma 4), we will reduce the origial sum (3) to a sum of sums of the form (5), which ca be combied i a sigle sum usig the properties of the logarithm. The degree d w of the umerator ad deomiator correspod to the umber of ivocatios of Lemma 4 (B), ad hece d w 2 w 1. 9

10 Let us give a example. For w = 1010, Theorem 2 shows that 0 a 1010 (8+5)=k L( ) = 1. By successive applicatios of Lemma 4, we fid 0 a 1010 (8+5)=k L( ) = 0 a 1010 ()=k ( ) L(4 + 2) L(8 + 6) L(8 + 2)+L( ). Puttig this all together, we coclude that, for all k 0, ( ) (4 + 2)(8 + 7)(8 + 3)( ) log 2 = 1. (4 + 3)(8 + 6)(8 + 2)( ) 0 a 1010 ()=k Commet. It ca be show that max w x d w = O(x 10.1 ); from this it easily follows that the algorithm to calculate b w () is actually a polyomial-time algorithm. For the details, see [4]. V. Behaviour O the Uit Circle. I this sectio, we examie the covergece of X aw() L(2g + v(w)) (6) o the uit circle. We will show that (6) coverges uiformly o each radius of the uit disc, with the exceptio of the radius lyig o the positive reals. It suffices to show uiform covergece for the series 1 X a w() For this, we follow the techique used previously i [3]. Write T w (N) = T w (N, X) =. 0 <N X a w() so that 1 N X a w() = T w(n + 1) N T w (1) + 1 N 1 T w ( + 1) ( + 1). 10

11 Thus it suffices to show that, for each ray from the origi to a poit ( 1) o the uit circle, there exists α < 1 such that uiformly i X o the ray. T w (N, X) = O(N α ) (7) We do this for w 0 j, leavig the case w = 0 j to the reader. For w = 1, this easily follows from the fact that T 1 (2 m, X) = (X + 1) m. Thus let us assume that w 2. We write A = 2 w 1 ad defie A differet sums, P 0 (m) through P A 1 (m), by P i (m) = X aw(a+i), 0 i < A. 0 <A m Clearly at least oe of the sums P i (m) coicides with T w (A m ); if w = w 1 w 2 w k, we may take i = v(w k w k w }{{ k ). } k Thus to boud T w (A m ), it suffices to give a upper boud for P (m) = P (m), where the colum vector P (m) is defied by P 0 (m). P A 1 (m), ad the orm v is defied by v = 1 i k v i. First, we observe that the vector P (m) ca be writte as a liear trasformatio of the vector P (m 1). Recall that a w(z) couts the umber of occurreces of the strig w i the strig z. The we have the followig Lemma 6. Let M w (X) be the A A matrix with M w (X) = [M ij ] = [ X a w (y jy i ) ], where y j ad y i are strigs such that v(y j ) = j, v(y i ) = i, y j = y i = w 1. The P (m) = M w (X)P (m 1). 11

12 Proof. We have P i (m) = 0 <A m X aw(a+i) = 0 j<a = 0 j<a = 0 j<a 0 k<a m 1 X aw(a(ak+j)+i) X aw(ak+j)+a w (y jy i ) 0 k<a m 1 X a w (y jy i ) P j (m 1). Thus to boud P (m), it suffices to determie a good boud for the matrix orm M w (X), where M = max M ij. 1 i 1 j Lemma 7. (A) The matrix M w (X) cotais at least oe row of all 1 s. (B) At least oe row of M w (X) is ot idetically 1. I every such row, there is at least oe elemet 1 ad at least oe elemet X. Proof. (A) Write w = w 1 w 2 w k. Let i = v(w k w k w }{{ k ). } k 1 The it is easily verified that w is ot a substrig of ay strig of the form y j y i, ad the result follows from the descriptio of Lemma 6. (B) To see that at least oe row of M w (X) is ot idetically 1, let i = v(w 2 w 3 w k ). The i row i, ay colum j whose least sigificat bit equals w 1 will cotai a o-1 etry. A argumet similar to (A), but for colums, shows that every row cotais at least oe elemet 1. Fially, let i be the idex of a row of M = M w (X) that is ot idetically 1. The w must be a substrig of some strig of the form y j y i. Choose j so that this substrig 12

13 appears as far to the right as possible; the y j y i = xwz for some strigs x ad z. The either x = 0, i which case M ij = X, or x = p > 0. I this latter case, let x = w 1 w 1 w }{{} 1. p The w matches the strig x wz i exactly oe place, ad this strig correspods to y j y i for some value of j ; i. e. a elemet X i the i th row. Now let us recall the basic facts about matrix orms [5]. Let M be a matrix. Let σ(m) be the set of all eigevalues of M. The spectral radius of M, r σ (M), is defied as Defie the matrix orm r σ (M) = where M deotes the cojugate traspose. max λ. λ σ(m) M 2 = r σ (MM ), Lemma 8. Let X 1, X 1. The M w (X) 2 < A. Proof. Let X 1, X 1. Write M = M w (X) ad M = MM. Note that each elemet M ij of M satisfies M ij 1. Fix a row i of the matrix M. If it is idetically 1, the by Lemma 7, there exists a colum j of M which cotais a elemet 1 ad a elemet X, the cojugate of X. Hece M ij A 2 + X + 1 < A. If row i is ot idetically 1, the by Lemma 7, it must cotai a elemet 1 ad a elemet X. Let j be a colum of M which is idetically 1. The agai M ij < A. Thus we see that each row of M cotais a elemet M ij with M ij < A. Thus for each i, 0 i < A we have M ij < A 2. 0 j<a Thus M < A 2. By a well-kow theorem [5, Theorem 7.8], ad the result follows. r σ (M ) M, 13

14 Lemma 9. There exists α < 1 such that T w (N, X) = O(N α ) uiformly i X o ay ray from the origi to a poit ( 1) o the uit circle. Proof. Write c(x) = M w (X) 2. Lemma 8 shows that c(x) < A. Fix a agle θ, 0 < θ < 2π, ad cosider c(x) as a fuctio of X = re iθ, 0 r 1. Sice c(x) is a cotiuous fuctio of r, ad the iterval 0 r 1 is compact, c(x) must attai its maximum o this iterval, which is ecessarily a positive costat c < A. Thus there are costats c 1, c 2 such that uiformly i X o the ray. T w (A m, X) P (m) c 1 P (m) 2 Now writig α = log c log A = c 1 M w (X) m P (0) 2 = c 1 M w (X) 2 m P (0) 2 c 2 c m, < 1, we have proved T w (N, X) = O(N α ) uiformly i X o the ray, for N = A m. We ow sketch how to exted this boud to all itegers N, as was doe previously i [2]. To do this, we observe that T w (N, X) ca be writte as a sum over at most A log A N terms of the form Q d,i (m) = X aw(a+i). d A m <(d+1) A m It is easily verified that the vector Q d (m) = Q d,0 (m). Q d,a 1 (m) ca be writte as a liear trasformatio of the vector Q d (m 1), usig the same matrix that appeared i Lemma 6. Repeatig the argumet i Lemmas 6-8, we see that Q d,i (m) is bouded i the same way as P i (m). Thus where α < 1. T w (N, X) = O((log N)N α ) = O(N α ) 14

15 We have show Theorem 10. Theorem 3 also holds for all X o the uit circle except X = 1. There is a ratioal fuctio b w () (which is effectively computable usig Lemmas 4 ad 5) such that, for all X 1 with X 1, we have log 2 (b w ())X a w() = 1 1 X. (Here the summatio is over 1 for w = 0 j ad 0 otherwise.) Commet. After the origial versio of this paper was completed, we leared of the results of Boyd, Cook, ad Morto, i which they study the partial sums 0 N ( 1) a w() for arbitrary w (0 + 1). This correspods to the case X = 1 i equatio (7). See [6]. VI. Some cosequeces. Let us put w = 0, X = 1 i Theorem 10 ad expoetiate. We get a ice compaio formula to (1); amely, ( 2 )( 1) a 0() 2 = Similarly, usig the example from sectio IV, we fid ( (4 + 2)(8 + 7)(8 + 3)( ) (4 + 3)(8 + 6)(8 + 2)( ) 0 ) ( 1) a 1010 () = 2 2. VII. Ackowledgemets. Much of this work was doe while the secod author was visitig the Mathematics Departmet at the Uiversity of Bordeaux. Both authors wish to thak Michel Medès Frace for his helpful advice. The secod author ackowledges with thaks coversatios with Eric Bach ad Howard Karloff. Todd Dupot helped with the proof of Lemma 8. 15

16 Refereces [1] J.-P. Allouche ad H. Cohe, Dirichlet series ad curious ifiite products, Bull. Lod. Math. Soc. 17 (1985), [2] J.-P. Allouche ad M. Medès Frace, O a extremal property of the Rudi-Shapiro sequece, Mathematika 32 (1985), [3] J.-P. Allouche, H. Cohe, M. Medès Frace ad J. O. Shallit, De ouveaux curieux produits ifiis, Acta Arithmetica 49 (1987), [4] J.-P. Allouche, P. Hajal ad J. O. Shallit, Aalysis of a ifiite product algorithm, Uiversity of Chicago Departmet of Computer Sciece Techical Report , (November 1987); submitted for publicatio. [5] K. E. Atkiso, A itroductio to umerical aalysis, Joh Wiley & Sos, New York, [6] D. W. Boyd, J. Cook, ad P. Morto, O sequeces of ±1 s defied by biary patters, Dissertatioes Mathematicae, to appear. [7] D. Robbis, Solutio to Problem E 2692, Am. Math. Mothly 86 (1979), [8] J. O. Shallit, O ifiite products associated with sums of digits, J. Number Theory 21 (1985), [9] D. R. Woods, Elemetary Problem Proposal E 2692, Am. Math. Mothly 85 (1978),