ON TRANSCENDENTAL NUMBERS GENERATED BY CERTAIN INTEGER SEQUENCES

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1 iauliai Math. Semin., 8 (16), 2013, 6369 ON TRANSCENDENTAL NUMBERS GENERATED BY CERTAIN INTEGER SEQUENCES Soichi IKEDA, Kaneaki MATSUOKA Graduate School of Mathematics, Nagoya University, Furocho, Chikusaku, Nagoya , Japan; s: m10004u@math.nagoya-u.ac.jp, m10041v@math.nagoya-u.ac.jp Abstract. By generalizing the technique of Dresden [2], we prove a theorem which gives a sucient condition for the transcendence of the numbers generated by certain integer sequences. In the last section, we consider the numbers generated by the last non-zero digits of n n, n nn, n nnn, etc. and the number of trailling zeros of n j, j N, and 10 j, as examples. Key words and phrases: decimal expansion, last non-zero digits, number of trailing zeros, Roth's theorem, transcendental number Mathematics Subject Classication: 11J Introduction There are some transcendental numbers whose transcendence is proved by using their decimal expansion. For example, Champernowne's number [5] and Liouville's number [4] 1 2 n! =

2 64 S. Ikeda, K. Matsuoka are well-known. On the other hand, Dresden proved the transcendence of lnzd(n n )10 n = , (1) where lnzd(x) means that the last non-zero digit of x for x N [2]. For example, lnzd(123) = 3, lnzd(100) = 1 and lnzd(9002) = 2. Incidentally, (1) can be found in [8]. In the present paper, generalizing the technique in Dresden's proof of the transcendence of (1), we prove a theorem which gives us a sucient condition for the transcendence of numbers generated by certain integer sequences. As applications of our theorem, we prove the transcendence of (1), lnzd(n nn )10 n, lnzd(n nnn )10 n, etc., by our theorem. In addition, we also prove the transcendence of (ntz(n j )( mod 10)10 n, j N, 10 j, (2) where (x( mod 10)) {0, 1,..., 9} for x N, and the denition of the function ntz follows: Definition 1. Let n N. We dene ntz(n) = max{l Z 0 : 10 l n}. This denition is the same as in [7]. Incidentally, we can nd the sequences related with ntz in [9] and [10]. Since the function ntz was not considered in Dresden's works [1] and [2], (2) is a new example. This indicates that our theorem can be used for many numbers which were not covered by previous work. 2. Proof of main theorem Definition 2. Let {d n } be a sequence of integers. Let r N with r 2. We dene { d D(n, k, r) = r k 1 n if r n, 0 if r n for n, k N.

3 On transcendental numbers generated by certain integer sequences 65 In order to prove the main theorem, we use Roth's theorem [6]. Theorem A (Roth's theorem). Let α be an algebraic number with deg α 2. For any ϵ > 0, there exist only nitely many rational numbers p q such that α p q < 1 q 2+ε. Now we state the main theorem. Theorem 1. Let g N with g 2. Let {d n } be a sequence of the numbers belonging to the set {0, 1,..., g 1}. If (1) α = i=1 d ig i is an irrational number, (2) there exist r N 2 and a N such that D(n + ar, k, r) = D(n, k, r) holds for any k, n N, (3) there exist innitely many n N such that d r n i = d r n (i+a) holds for any i {1,..., a}, then α is a transcendental number. Proof. For simplicity, we write D(n, k) = D(n, k, r). If we dene s k = D(i, k)g rk 1 i i=1 for k N, then α = k=1 s k and s k Q. From condition (2), we can take g ark 1 as a denominator of s k. Since (g ark 1) (g ark+1 1), we can take g arn 1 as a denominator of In addition, if we dene u n = a i=1 j=0 t n = n s k. k=1 d r n ig (arn j+r ni) = a i=1 j=0 d r n ig rn (aj+i) for n N, then we can also take g arn 1 as a denominator of u n. If we write t n + u n = d ig i, i=1

4 66 S. Ikeda, K. Matsuoka then d i = d i, 1 i (2a + 1)r n 1), holds for any n N which satises condition (3). Moreover, we set q n = g arn 1 and p n = q n (t n + u n ) for n N which satises condition (3). Then α p n = q n (d i d i)g i < (g 1)g i i=(2a+1)r n i=(2a+1)r n = g < g g (2a+1)rn qn 2+1/a 1 q 2+1/2a n holds for suciently large n. Moreover, there exist innitely many p n and q n from condition (3). Therefore, α must be a transcendental number by Roth's theorem. 3. Some examples In this section, we prove the transcendence of some concrete numbers by Theorem 1. (i) The numbers constructed from lnzd(n n ), lnzd(n nn ), etc. First, we dene a function g l (n). Definition 3. Let f(x, y) = x y. For n N, we dene { f(n, 1) = n if l = 1, g l (n) = f(n, g l 1 (n)) if l 2. Note that g 2 (n) = n n, g 3 (n) = n nn and g 4 (n) = n nnn. Our rst purpose is to prove the following proposition. Proposition 1. Let l N with l 2. If we dene α l = then α l is a transcendental number. lnzd(g l (n)) 10 n, Note that Dresden's result is the case l = 2. We set g = r = 10, a = 2, α = α l and d n = d n,l = lnzd(g l (n)) for l N, and prove that α and {d n } satisfy the conditions in Theorem 1 by the following lemmas. Then Theorem 1 implies Proposition 1. We can easily see that the following lemma holds.

5 On transcendental numbers generated by certain integer sequences 67 Lemma 1. For any n, x, y N, (i) if x y( mod 4) and 2 n, then n x n y ( mod 4); (ii) if x y( mod 4), then n x n y ( mod 10); (iii) if 10 x and x y( mod 10) then lnzd(x) = lnzd(y). Lemma 2. For any l, m, n N, holds. g l (n + 4m) g l (n)( mod 4) Proof. In the case l = 1, the assertion of the lemma is trivial. In addition, if l 2 and 2 n, then g l (n + 4m) g l (n) 0( mod 4) holds. Therefore, we may assume that 2 n and l 2. From Lemma 1 (i), in the case l = 2, we have g 2 (n + 4m) = (n + 4m) n+4m n n+4m n n ( mod 4). If g l 1 (n + 4m) g l 1 (n)( mod 4) holds for l 3, then g l (n + 4m) f(n, g l 1 (n + 4m)) f(n, g l 1 (n)) = g l (n)( mod 4) holds by Lemma 1 (i). Lemma 3. For any k, n N, D(n + 20, k) = D(n, k). Proof. In the case 10 n, the assertion of the lemma is trivial. Otherwise, from Lemmas 1 and 2, we obtain and D(n + 20, k) = d 10 k 1 (n+20) = lnzd(f(10 k 1 (n + 20), g l 1 (10 k 1 (n + 20)))) = lnzd(f(n, g l 1 (10 k 1 n k ))) = lnzd(f(n, g l 1 (10 k 1 n))), D(n, k) = lnzd(f(10 k 1 n, g l 1 (10 k 1 n))) = lnzd(f(n, g l 1 (10 k 1 n))). Therefore, D(n + 20, k) = D(n, k). From Lemma 1 (ii) and (iii), we can easily obtain the next lemma.

6 68 S. Ikeda, K. Matsuoka Lemma 4. There exist innitely many n N such that d 10 n i = d 10 n (i+2) for i = 1, 2. Lemma 5. For any l 2, is an irrational number. α l = d n,l 10 n Proof. By contradiction, assume that α l is a rational number. This implies that there exist T, N N such that d n+mt = d n lnzd(g l (n + mt )) = lnzd(g l (n)) holds for any n N and m N. We take j N such that 10 j > max{1000t, N}. Then we have If lnzd(t ) = 5, then we obtain lnzd(g l (10 j + mt )) = lnzd(g l (10 j )) = 1. (3) lnzd(g l (10 j + mt )) = 5 by taking m = 1. This contradicts (3). Otherwise, we take m = 20, and obtain lnzd(g l (10 j + mt )) = lnzd(f(2t, g l 1 (10 j + 20T ))) = 6. This also contradicts (3). (ii) The numbers constructed from ntz(n j ). Our second purpose is to prove the following proposition. Proposition 2. Let j N with 10 j. If we dene α j = (ntz(n j )( mod 10)) 10 n, then α j is a transcendental number, where x( mod 10) {0, 1,..., 9}. Note that if 10 j, then α j = 0. We set g = r = 10, a = 1, α = α j and d n = d n,j = (ntz(n j )( mod 10)), and prove that α and {d n } satisfy the conditions in Theorem 1. Since conditions (2) and (3) in Theorem 1, clearly, hold by the denition of ntz, the remaining task is to prove condition (1). We can easily see that the following lemma holds.

7 On transcendental numbers generated by certain integer sequences 69 Lemma 6. Let a N with 10 a. For any k N, there exist innitely many n N such that ak an( mod 10). Lemma 7. If 10 j, then α j is an irrational number. Proof. By contradiction, assume that α j is rational. This implies that there exist N, T N such that d n,j = d n+t,j for any n N. By Lemma 6, we can choose an l N such that 10 l N, 10 l 1 > T, and jntz(t ) jl( mod 10) hold. For this l, we obtain d 10 l,j = ntz(10 lj )( mod 10) = jl( mod 10), d 10 l +T,j = ntz((10 l + T ) j )( mod 10) = jntz(t )( mod 10). Hence jl( mod 10) = jntz(t )( mod 10). jntz(t )( mod 10). References However, this contradicts jl [1] G. Dresden, Two irrational numbers from last non-zero digits of n! and n n, Math. Mag., 74, (2001). [2] G. Dresden, Three transcendental numbers from the last non-zero digits of n n, F n, and n!, Math. Mag., (2007). [3] R. Euler, J. Sadek, A number that gives the unit degit of n n, J. Recreat. Math., 29(3), (1998). [4] A.O. Gelfond, Transcendental and Algebraic Numbers, Dover, New York, [5] K. Mahler, Arithmetische Eigenschaften einer Klasse von Dezimalbrüchen, Proc. Konin. Neder. Akad. Wet. Ser. A., 40, (1937). [6] K.F. Roth, Rational approximations to algebraic numbers, Mathematika, 2, 120 (1955). [7] H.S. Warren Jr., Hacker's Delight, Addison, Wesley Professional, [8] On-Line Encyclopedia of Integer Sequences, [9] [10] Received 29 November 2012