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On the Constant Congruence Speed of Tetration

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On the Constant Congruence Speed of Tetration Notes on Number Theory and Discrete Mathematics Print ISSN 1310–5132, Online ISSN 2367–8275 Vol. 26, 2020, No. 3, 245–260 DOI: 10.7546/nntdm.2020.26.3.245-260 On the constant congruence speed of tetration Marco Ripà sPIqr Society, World Intelligence Network Rome, Italy e-mail: [email protected] Received: 11 September 2019 Revised: 11 August 2020 Accepted: 13 August 2020 Abstract: Integer tetration, the iterated exponentiation for , is characterized by fascinating periodicity properties involving its rightmost figures, −in 0,any 1 numeral system. Taking into account a radix-10 number system, in the book “La strana coda della serie n ^ n ^ ... ^ n” (2011), the author analyzed how many new stable digits are generated by every unitary increment of the hyperexponent , and he indicated this value as or “congruence speed” of . A few conjectures about arose. If is sufficiently() large, the congruence speed 0(moddoes not 10) depend on , taking on a (strictly() positive) unique value. We derive the formula that describes for every ending in . Moreover, we claim that for any () 5 and () = 1 otherwise. Finally,(mod we25 )show 2, the 3, 4, size 6, 8, of 9, 11,the 12,fundam 13, 14,ental 16, period 17, 19, 21, for 22, any 23 of the remaining() ≥ 2 values of the congruence speed: if , then . ( )!" Keywords: Number theory, Power tower, Tetration,() ≥ 2 Hyperoperati(()on,) = Charmichael 10 function, Euler’s totient function, Primitive root, Exponentiation, Integer sequence, Congruence speed, Modular arithmetic, Stable digit, Rightmost digit, Cycle, Periodicity. 2010 Mathematics Subject Classification: 11A07, 11F33. 1 Introduction In the present paper, we study recurrence properties involving the rightmost digits of the tetration #$ ( -times) [6], observing that, when the hyperexponent is sufficiently large and = , the number of new stable digits generated by any (modunitary 25)increment 0, 1,of 5, 7,is 10, unitary 15, 18, as 20,well: 24 it depends only on the congruence modulo 25 of the base [9]. In Sections 4 and 5, we extend the aforementioned relation to the remaining values of . These new results would contribute to improve big numbers rightmost digits calculations, opening new scenarios in cryptography/cryptanalysis, too [13]. 245 2 Congruence speed of It is well known that, for any arbitrarily%(&'( large )*), originates a string of stable figures [10], thus we can say that is well-defined modulo+ , for any [9,+ 11, 12]. , - It is possible to observe the same peculiarity cons10 idering many ≥ different(+, ) numeral systems [4], but we only take into account the decimal one (radix-10). Let us now introduce the definition of “congruence speed” as it was originally presented by Ripà in his book about the righmost digits of [6]. Definition 1. Let be an arbitrary base which is not a multiple of and let be such that − 1 , where10 . ." 0 ." 0!" We define − 0, 1 to be the non-negative / (mod integer 10 such) 1 that (mod 10 ) 2 (, ) . !" 0!( ) !" 0!( )!" For simplicity, from / here 3mod on out, 10 we refer4 1 to as 3mod the congruence 10 4speed of the base of the tetration . () 5 0(mod 10) 3 Conjectures about the congruence speed In this section we present the conjectures and a few remarks to point out their main implications, specifying that indicates Euler’s totient function (which counts the number of positive integers up to a6(+) given that are relatively prime to ), while repesents the Charmichael lambda function+ (the − 0reduced totient function given by+ the smallest7(+) positive divisor of that satisfies the conclusion of the well known Euler’s totient theorem). 6(+)Before starting to discuss the main results, let us introduce a lemma that we will use later. Lemma 1. Referring to Definition 1, , . If , then . Proof . If , the lemma is trivially8 verified, ≥ 2 (3, ) = 1 = 1 (3, ) = 0 . : In order =to 1show that , we need = 1 to 9 prove 3 3 that(mod 10 ) 9 3 7(mod 10 ) , ≥ 2 9 (3) = 1 ; !" ." !" and we start with the8 first ≥ 2 congruence3 / 3 relation.(mod 10 ) 1 3 3(mod 10 ) We prove by induction on that , . !" !" 3 / 3(mod 2 ) 8 ≥ 2 . ;< Let and assume = 2 9 27 / 3 (mod .8) Since 9 3 / and 3(mod are 8) coprime, we can invoke !" !" Euler’s totient ≥ 2 theorem. It is3 well/ known 3(mod that, 2 ) , 3 10 (proof follows from !; !" ( ) Euclid’s lemma). Thus, > >@A 8 6 2 , and= 2 we can rewrite it as = :? = :? !; , which3 concludes/ 3 the=mod proof 6( 2of B the 2 inductive)? step. !" !; !; Similarly,3 / 3 (wemod can 2 show) that , . ." !" Base case: 5 C 3 − 3 8 ≥ 2 . ;< Induction step: = let 2 9 27 / and 3 assume(mod 5) 9 2 / 2(mod 5) . !" ." ≥ 2 3 /246 3(mod 5 ) As shown before, we have by Euler’s !" !" !; theorem. Since 3 / 3(mod 6(5 )) 9 3 / (see 3(mod [6, p. 5 66]),) it follows !" ." !" !; ." that 3 / 3(mod 5 ) 9 3 / 3.(mod This confirms4 B 5 ) that the inductive step is !" ." !" !; also true.3 / 3(mod 5 ) 9 3 / 3(mod 5 ) Therefore, we have proved that . !" DEF(!", .") !" ." We complete the3 / proof 3(mod of Lemma 10 1 showing )that9 3 / 3(mod 10 , ) . !" In order to prove that such that 3 , 3(modit is sufficient 10 ) 8 to show − 0, that 1 !" , .G We prove that 3 / 3(mod, 10 ) by induction on (as !" !" usual).3 3(mod 5 ) 8 5 H 3 − 3 8 − 0, 1 Let . is true, since . ;< We need = 2to3 prove 27(modthe induction 25) step. 12 2(mod 25) Let and assume . This is true since is the largest power of !" ." that divides ≥ 2 . Otherwise,3 3(mod 5 would) divide 5 , which implies that5 !" !" “ 3 is a 3 multiple− 34 of ”,5 but the statement 3is false 3 − since 34 is a primitive root !" !" modulo3 3 − 3for4 any , and6( 5we )deduce this from [5, Theorem 1].3 Assuming , [5, Theorem5 1] states that ≥ if 1 is an odd prime (and it is) and if is a primitive root of I = 0, (J;) then is also a primitive root5 of . 3 5 !" Thus,3 we need to check that 5 is a primitive root of , and it is so (since has a total ; ; of primitive roots: 3 ). 5 = 25 5 8Therefore, for 2, 3,, the 8, 12, conguence 13, 17, 22, speed 23 of is constant ( ). In particular, we =have 3 shown that (for any ) using8 ≥ the 2 primitive root analysis, and this concludes the proof. (3) = 1 K 1 £ The primitive root argument is a key point at the bottom of many results that we will introduce in the next sections and represents a central topic discussed in [6] as well. Property 1. , such that, , - - - 8 − 1, (see 2 5 A317905 0(mod of the10 ) OEISL ruling − 0out 5 the Mfirst term of the sequence8 ≥ [9]).(, ) = () − 0 − Remark 1. Referring to the aforementioned property and considering all the bases , we point out that this is a peculiarity of tetration (as for the exponentiation 5 0( modif 10 ) ): from pentation (hyper-5) and beyond, , the number of stable5 digits /0 will(mod increase 10 ) for any unitary increment of the hyperexponent.8 − Perhaps0, 1 we should coin the term “congruence acceleration” (rather than congruence speed), speculating that hyper-5 may be characterized by a constant congruence acceleration, for any given . The −special1 5 feature 0(mod of 10tetration) captured by Property 1 has been widely analyzed in [6] and confirmed for specific values of [4, 10], including (for any ) [1, p. 148]. (2) = 1 5 ≥ 3 247 Lemma 1 proves that such that, , is a strictly positive integer, and we are absolutely convincedL that the8 constancy ≥ 2 (,) of t =he () congruence speed (when is sufficiently large) is a general property [3], involving every base which is not a multiple of . Therefore, in the rest of this paper, we will assume Property 1 as a general axiom [6, 11].10 Conjecture 1. Let be such that . Now, 0 !" 0!" !" assume that 2 N. 10 C 3 − , 4we 1 10 have H 3 − 4 (e.g., 5 ≥ 3 8 − 1 5 0(mod) 10 and,) ( − 2, )weB (have) O 2that O ( P 1) B () =292=(−2)B(()=1)). 8Q − 0 RBS = 2 92=(P1)B( Remark 2. It is trivial to point out that Conjecture 1 implies that , . !" .; Thus, 8for any − 1, 5we know 0(mod that 10 the) 1rightmost 8 5 ≥ 3figures / of the(mod integer 10 tetration) are stable digits: they form ≥ 3 the same final string of T , where − 2 . Hence, we get a conservative general upper bound forU the hyperexponent 5 U K , applied to any V base (not a multiple of ), which assures us that all the rightmost figures of T are stable digits, for any [6]:10 2 U ≥ V (1) 0 V = W( )X P 2 (where the ceiling denotes the function that takes the rational number as input and gives as output the least integerYZ[ greater than or equal to ). Z Therefore, , (e.g., if Z and , then 0 \;] 8 ≥ 3 () O .; = 143 ≥ 5 0 P 6 P 6 P 5 P ^_`] 4 17 P ( − 4) B 4 4 = () O = is true, while and − 2 − 2 are trivially verified). \;] "< \;] Furthermore, = if 4 Conjecture 9 (143 1 )Oholds,; it follows = 3 9that (143 the maximum) O 12 number of missed stable digits, for any given value of and for a sufficiently large , is . 3 B () We can consistently define V(1) = 0, extending the domain of (by Definition 1 and Property 1), if we observe that, for any a , (, ) = () 8 such ≥ that P 1 2 G .
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