INTRINSIC PALINDROMES

Transcription

1 Antonio J. Di Scala Politecnico di Torino, Dipartamento di Matematica, Corso Duca degli Aruzzi, Torino, Italy Martín Somra Université de Lyon 1, Laoratoire de Mathématiques, Appliquées de Lyon, 21 Avenue Claude Bernard, Villeuranne Cedex, France (Sumitted July 2001 Final Revision March 2002) To our friend Cristian Jansenson Natural numers occur everywhere in our daily life: a us ticket, a car plate, an id numer, a timetale, etc. These numers are mostly expressed in the decimal system. That is, for a natural numers n N we write n = (a k 1... a 0 ) 10 for some 0 a i 9 and a k 1 0, which means that n = a k 1 10 k 1 + a k 2 10 k a 0. Of particular interest are the so-called palindromic numers. These are the numers whose decimal expansion is the same when read from left to right, and from right to left, that is (a k 1... a 0 ) 10 = (a 0... a k 1 ) 10. This kind of numer appears already in the Ganitasârasamgraha, a sanskrit manuscript dated around 850 AD. Therein the Indian mathematician Mahâvîrâchârya descried as the quantity which eginning with one until it reaches six, then decreases in reverse order [1, p. 399]. This is a curious palindromic numer, and in particular it is the square of another palindrome: = There are many ways of generating palindromic numers (see for instance [2]) including an interesting conjectural one [3]. Running across a palindrome y chance is a rare occurrence: for instance, the proaility of picking at random a numer 10 4 n < 10 5 and otaining a palindrome is 1/10 2, a fact well-known y the collectors of palindromic us tickets. Hence we tend to feel pretty lucky when one of these rare numers crosses our path. And the more digits it has, the luckier we feel, and the luckier the numer itself seems to e. But, is this feeling really justified? The truth has to e said: this property is not intrinsic to the numer, ut also depends on the ase used to express it. So the numer n := ( ) 10 is lucky (or palindromic) in ase 10, ut is not in ase 13 as n = ( ) 13. Hence each time we encounter we have to at least in principle thank heaven for this occurrence together with the fact that the human race has five fingers in each hand. However, we can easily get independent of the ase y defining a numer to e intrinsically palindromic if it is palindromic in some ase. A little reflection shows that this definition is meaningless as it stands: every numer is palindromic in any ase m > n, as n = (n) m. Indeed it is much more natural to take into account the numer of digits. So we define a numer n N to e k-palindromic if there is a ase such that the -expansion of n is palindromic of length k. The previous oservation shows that every numer if 1-palindromic. Also note that n = (11) n 1 provided that n 3, that is every n 3 is 2-palindromic. What aout k-palindromic numers for k 3? Our thesis is that very few numers are k-palindromic, at least for k 4. The proaility of a numer in an appropriate range eing, 76

2 say, intrinsically 9-palindromic is small, and indeed quite close to the proaility of eing 9-palindromic in ase 10. This justifies our first impression that is lucky, regardless of the ase chosen to represent it. To write down our results we first need to introduce the following counting functions. Take k, N, N and set Φ k (N, ) := #{n N; n is k-palindromic in ase }. Also set Φ k (N) := #{n N; n is k-palindromic}. Then Φ k (N, )/N and Φ k (N)/N represent the density (or proaility) of numers elow N which are k-palindromic in ase and intrinsically k-palindromic, respectively. Theorem 1: Let k 4, and write k = 2i + r with i N and r = 0, 1. Then Φ k (N) 4(N + 1) i+r+1 k. Proof: A ase contriutes to Φ k (N) if and only if Φ k (N, ) > 0, namely if and only if there exists a numer n = a k 1 k a 0 N palindromic in ase of length k, that is such that 0 a j 1 and a k j = a j for j = 0,..., i + r, and a k 1 0. Then k n N, and hence contriutes to Φ k (N) if and only if (N 1) 1 k 1. We consider separately the cases (N + 1) 1 k and (N + 1) 1 k < (N 1) 1 k 1. In the first case, the largest k-palindrome in ase is ( 1)( k ) = k 1 N, and so Φ k (N, ) = Φ k (, ) = ( 1) i+r 1. Now for the second case we let θ() N e the largest integer such that θ()( k 1 +1) N, that is θ() = [N/( k 1 + 1)] N/ k 1. Then (θ() + 1)( k 1 + 1) > N and so every k- palindromic numer in ase egins with a k 1 θ(). Hence Φ k (N, ) θ() i+r 1 N/ i. Set L := [(N + 1) 1 1 k ] and M := [(N 1) k 1 ]. We split the sum Φ k (N) = ϕ + χ + ψ with ϕ := L 1 =2 Φ k(n, ), ψ := M =L+2 Φ k(n, ), and χ := Φ k (N, L) + Φ k (N, L + 1). From the previous considerations we conclude that On the other hand ψ ϕ L+2 L 1 =2 i+r L 2 t i+r dt Li+r+1 i + r + 1 i+r+1 (N + 1) k. i + r + 1 N M i+r+1 i N dt L+1 t i N (N + 1) k. (i 1)(L + 1) i 1 i 1 Finally χ L i+r + N/(L + 1) i 2(N + 1) i+r k ( Φ k (N) 2 (N + 1) 1 k + 1 i + r i 1 and thus ) (N + 1) i+r+1 k 4(N + 1) i+r+1 k. 77

3 Let k = 2i + r 4, and let 2 e a ase. Set N := k 1, so that N is larger than every numer whose representation in ase has length k. Then Φ k (N, ) = Φ k (, ) = ( 1) i+r 1, and so the density of numers elow N which are k-palindromic in ase is ( 1) i+r 1 /N 1/ i. On the other hand, the previous result shows that the density of intrinsic k-palindromes elow N is ounded y 4( k ) i+r+1 k /( k 1) 4/ i 1. For instance, the proaility of a numer n < 10 9 eing 9-palindromic in ase 10 is , while the proaility of it eing 9-palindromic in any ase is elow From the point of view of proaility, the situation is then in most cases quite clear. For k 2 every numer is k-palindromic, while for k 4 almost every numer is not. The critical case is k := 3. Consider the following tale: Φ 3 ( ) Φ 3 (100) = 61 Φ 3 ( ) Φ 3 (10 3 ) = 70 Φ 3 ( ) Φ 3 (10 4 ) = 83 Φ 3 ( ) Φ 3 (10 5 ) = 86 Φ 3 ( ) Φ 3 (10 6 ) = 89 Φ 3 ( ) Φ 3 (10 7 ) = 94 This suggests that almost every numer is 3-palindromic, ut not every sufficiently large numer. To tackle this prolem we consider the following reformulation. We recall that {ξ} := ξ [ξ] [0, 1) denotes the fractional part of a real numer ξ R. Lemma 2: Let n, N such that the -expansion of n has length 3. Then n is 3-palindromic in ase if and only if { } (n + 1) 2 < Proof: First note that the hypothesis that the -expansion of n has length 3 is equivalent to the fact that n 3 1. Now n is 3-palindromic in ase if and only if there exists 0 < e < and 0 f < such that n = e( 2 + 1) + f. Solving the associated Diophantine linear equation n = x( 2 + 1) + y with respect to x, y we see that the aove representation is equivalent to the existence of l Z satisfying 0 < n l < and 0 l( 2 + 1) n <. The second pair of inequalities is equivalent to n/( 2 + 1) l < (n + 1)/( 2 + 1), and so this implies that {(n + 1)/( 2 + 1)} < /( 2 + 1). Then this condition is necessary for n to e 3-palindromic in ase. Let s check that it is also sufficient. The integer l := [(n+1)/( 2 +1)] satisfies the second pair of inequalities. Then it only remains to prove that it also satisfies the first pair, which is equivalent to l < n/ < l + 1. This follows from the inequalities (n + 1) < n < n , which are in turn a consequence of the hyposthesis n

4 After the first version of this paper was written, N.J.A. Sloane included the sequences of k-palindromic numers for k = 3,..., 9 in his on-line Encyclopedia of Integer Sequences ( njas/sequences). We refer to it for this additional numerical data. Now we egin to look at palindromicity as an intrinsic property not attached to any particular ase nothing stops us from considering the fact that a given numer can e palindromic in several different ases. For instance 3074 = (44244) 5 = (22122) 6. Common sense dictates that multiple palindromicity should e a much more rare occurrence than a simple one, which is also rare as we have already shown. In fact it even seems unclear whether there are numers which are k-palindromic in as many ases as desired. To formalize this, let µ k (n) := #{; n is k-palindromic in ase }. We first propose the prolem of determining whether µ k is unounded or not. Again the cases k = 1, 2 are easy. In the first case n = (n) m for any ase m > n, and so µ 1 (n) = for every n. In the second case, set n := 2 2u+1 for some u N. Then n = 2 v (2 w 1) + 2 v = (2 v, 2 v ) 2 w 1 for v < w such that v + w = 2u + 1. Then µ 2 (n) u. The following solves the case k = 3. Theorem 3: There exists an infinite sequence n 1 < n 2 < n 3 <... such that µ 3 (n j ) 1 7 log(n j + 1). Proof: Take N 0 and assume that µ 3 (n) < (1/7) log(n + 1) for all n N, so that N Φ 3 (N, ) = µ 3 (n) < 1 N log(n + 1). (1) 7 n=1 We will soon see that this is contradictory: Set L := [(N + 1) 1 3 ] and M := [(N 1) 1 2 ]. For L M we let ζ() N e the largest integer such that ζ()( 2 + 1) + ( 1) N. Then 1 ζ() 1, and also every 3-palindromic numer n := e 2 + f + e with e ζ() is less or equal that N. Hence Φ 3 (N, ) ζ() which implies that ( ) N Φ 3 (N, ) Φ 3 (N, ) ζ() , =L as ζ() + 2 N/( 2 + 1). We have that =L =L ( ) N M t N L t dt 2M 2 N 2 (log((m + 1)2 + 1) log(l 2 + 1)) 2N. We have that (M + 1) N and L N 2/3 and thus we conclude p Φ 3(N, p) > N 6 log N 2N log 2, which contradicts Inequality (1) for N large enough. It follows that for each (sufficiently large) N N there exists an n N such that =L µ 3 (n) 1 7 log(n + 1) 1 log(n + 1). 7 79

5 The fact that µ 3 (n) < implies that the set of such n s is infinite. Here is some sample data for the cases k := 4, 5. µ 4 (624) = µ 4 (910) = 2, µ 4 (19040) = 3, µ 5 (2293) = 2. For k, l, N N, we let Φ k,l (N) e the numer of n N which are k-palindromes in l different ases, that is Φ k,l (N) := #{n N; µ k (n) l}. In particular Φ k,1 = Φ k. The following tale gives some more informative data. k l N Φ k,l (N) This suggests that for k 4, l 2 and k + l 8, there are no k-palindromic numers with multiplicity l whatsoever. Finally we can also consider µ k (n) := #{; n is j-palindromic in ase for some j k} = j k µ j (n), that is the numer of different ases in which n is a palindrome of length at least k. It is easy to see that this function is unounded; we have that n L := 2 2L 1 = 2 L l {}}{ (2 2l 1,..., 2 2l 1) 2 2 l and so n L is 2 L l -palindromic in ase 2 2l for l = 0,..., L. Hence µ k (n L ) L log 2 k. A further prolem is to determine the density of k-palindromic numers in l different ases. From this point of view, Theorem 1 is an important advance towards the solution of the cases k 4 and l = 1. The cases when l 2 seem to e much more elusive, ut also interesting. A solution of them would allow us, for instance, to know how lucky we are when the numer of the taxi-ca we are riding is the = (8888) 13 = (5995) 15 = (2, 14, 14, 2) 19. ACKNOWLEDGMENT A. Di Scala supported y a EPSRC Grant (England) and a CONICET Fellowship (Argentina). Also partially supported y Secyt-UNC and CONICOR (Argentina). M. Somra supported y a post-doctoral Marie Curie Fellowship of the European Community program Improving the Human Research Potential and the Socio-Economic Knowledge Base. Also partially supported y CONICET and CyT-UNLP (Argentina). 80

6 REFERENCES [1] G. Ifrah. The Universal History of Numers: From Prehistory to the Invention of the Computer. John Wiley & Sons, New York, [2] W.J. Reichmann. The Fascination of Numers. Methuen, London, [3] C.W. Trigg. Palindromes y Addition. Math. Mag. 40 (1967): AMS Classification Numers: 11A63 81