Additive Number Theory and Automata Jeffrey Shallit School of Computer Science, University of Waterloo Waterloo, ON N2L 3G1 Canada Joint work with Carlo Sanna Daniel Kane P. Madhusudan Dirk Nowotka Aayush Rajasekaran Tim Smith Additive number theory Let S; T be subsets of the natural numbers N = 0; 1; 2;::: . f g The principal problem of additive number theory is to determine whether every element of T (or every sufficiently large element of T ) can be written as the sum of some constant number of elements of S, not necessarily distinct. Often (but not always) T = N and S is a relatively sparse subset of N. Lagrange's theorem Probably the most famous example is Lagrange's theorem (1770): (a) every natural number is the sum of four squares; and (b) three squares do not suffice for numbers of the form 4a(8k + 7). (Conjectured by Bachet in 1621.) Goldbach's conjecture Let P = 2; 3; 5;:::; be the prime numbers. f g Goldbach's conjecture (1742): every even number 4 is the sum of two primes. ≥ So here T = 2N. Zwillinger's conjecture (1979): every even number > 4208 is the sum of 2 numbers, each of which is part of a twin-prime pair. Additive bases Let S N. ⊆ We say that a subset S is an basis of order h if every natural number can be written as the sum of h elements of S, not necessarily distinct. We say that a subset S is an asymptotic basis of order h if every sufficiently large natural number can be written as the sum of h elements of S, not necessarily distinct. Usual convention: 0 S. 2 Gauss's theorem for triangular numbers A triangular number is a number of the form n(n + 1)=2. Gauss wrote the following in his diary on July 10 1796: i.e., The triangular numbers form an additive basis of order 3 Waring's problem for powers Edward Waring (1770) asserted, without proof, that every natural number is { the sum of 4 squares { the sum of 9 cubes { the sum of 19 fourth powers { \and so forth". Waring's problem Let g(k) be the least natural number m such that every natural number is the sum of m k'th powers. Let G(k) be the least natural number m such that every sufficiently large natural number is the sum of m k'th powers. Proving that g(k) and G(k) exist, and determining their values, is Waring's problem. By Lagrange we know g(2) = G(2) = 4. Hilbert proved in 1909 that g(k) and G(k) exist for all k. (Later, proof simplified by Vinogradov.) By Wieferich and Kempner we know g(3) = 9. We know that 4 G(3) 7, but the true value is still unknown. ≤ ≤ Other additive bases? What other sets can be additive bases? Not the powers of 2 { too sparse. Need a set whose natural density is at least N1=k for some k. But this is not enough: consider the set S = 22n + i : n 1 and 0 i < 2n : f ≥ ≤ g Its density is Ω(N1=2). But S does not form an additive basis of any finite order, because adding k elements of S in decreasing order can only result in at most 2k + 1 \one" bits in the highest-order positions. Palindromes I How about numbers with palindromic base-b expansions? I A palindrome is any string that is equal to its reversal I Examples are radar (English), ressasser (French), and 10001. I We call a natural number a base-b palindrome if its base-b representation (without leading zeroes) is a palindrome I Examples are 16 = [121]3 and 297 = [100101001]2. I Binary palindromes (b = 2) form sequence A006995 in the On-Line Encyclopedia of Integer Sequences (OEIS): 0; 1; 3; 5; 7; 9; 15; 17; 21; 27; 31; 33; 45; 51; 63;::: 1=2 I They have density Θ(N ). The problem Do the base-b palindromes form an additive basis, and if so, of what order? William Banks (2015) showed that every natural number is the sum of at most 49 base-10 palindromes. (INTEGERS 16 (2016), #A3) Javier Cilleruelo, Florian Luca, and Lewis Baxter (2017) showed that for all bases b 5, every natural number is the≥ sum of three base-b palindromes. (Math. Comp. (2017), to appear) What we proved However, the case of bases b = 2; 3; 4 was left unsolved. We proved Theorem (Rajasekaran, JOS, Smith) Every natural number N is the sum of 4 binary palindromes. The number 4 is optimal. For example, 10011938 = 5127737 + 4851753 + 32447 + 1 = [10011100011111000111001]2 + [10010100000100000101001]2 + [111111010111111]2 + [1]2: 4 is optimal: 10011938 is not the sum of 2 binary palindromes. EVERY NATURAL NUMBER IS THE SUM OF FORTY-NINE PALINDROMES 5 In the case that 10 m 43, we write m 10a b with digits a, b , ď ď “ ` P D a 0. Using (2.2) we have ‰ L 1 L 2 n q a, b n 10 ´ a 10 ´ b a b ´ L,0p q“ ´p ` ` ` q L 2 n 10 ´ m a b “ ´p ` ` q L 1 ´ j L 2 10 δj 10 ´ 10δL 1 δL 2 6 a b ´ ´ “ j 0 ´ p ` ´ q´ ´ ÿ“ L 3 L 2 ´ j 6 10 ´ 10 δj a b, “ ¨ ` j 0 ´ ´ ÿ“ and the latter number lies in NL 1,0 5`; c , where c δ0 a b mod 10. Since Previous proofs were complicated´ p q (1)”p ´ ´ q qL,0 a, b is the sum of two palindromes, we are done in this case as well. p q Excerpt from Banks (2015): 2.4. Inductive passage from Nℓ,k 5`; c1 to Nℓ 1,k 1 5`; c2 . p q ´ ` p q Lemma 2.4. Let ℓ,k N, ℓ k 6, and cℓ be given. Given n Nℓ,k 5`; c1 , one P ě ` P D P p q can find digits a ,...,a , b ,...,b 0 and c such that the number 1 18 1 18 P Dzt u 2 P D 18 n qℓ 1,k aj, bj ´ ´ j 1 p q ÿ“ lies in the set Nℓ 1,k 1 5`; c2 . ´ ` p q ℓ 1 . Proof. Fix n Nℓ,k 5`; c1 , and let δj j´0 be defined as in (1.1) (with L . ℓ). P p q t u “ “ Let m be the three-digit integer formed by the first three digits of n; that is, . m . 100δℓ 1 10δℓ 2 δℓ 3. “ ´ ` ´ ` ´ Clearly, m is an integer in the range 500 m 999, and we have ď ď ℓ 1 ℓ 4 ´ j ℓ 3 ´ j n 10 δj 10 ´ m 10 δj. (2.4) “ j k “ ` j k ÿ“ ÿ“ Let us denote . 19, 29, 39, 49, 59 . S “ t u In view of the fact that 9 . 171, 181, 191,..., 531 , S “ S `¨¨¨` S “ t u nine copies it is possible to find an element h 9 for which m 80 2h m 60. With P S ´ ă ď ´ h fixed, let s ,...,s be elements of such that 1 9 S s s h. 1 `¨¨¨` 9 “ Finally, let ε ,...,ε be natural numbers, each equal to zero or two: ε 0, 2 1 9 j P t u for j 1,..., 9. A specific choice of these numbers is given below. “ Previous proofs wereEVERY POSITIVE complicated INTEGER IS A SUM OF THREE (2) PALINDROMES 15 II.1 cm = 1. We do nothing and the temporary configuration becomes the final one. Excerpt from Cilleruelo et al. (2017) II.2 cm = 0. We distinguish the following cases: II.2.i) y = 0. m 6 δm δm 1 δm δm 1 − − 0 0 1 1 y −→ y 1 ∗ m ∗ m − ∗ ∗ ∗ ∗ II.2.ii) ym = 0. II.2.ii.a) ym 1 = 0. − 6 δm δm 1 δm 2 δm δm 1 δm 2 − − − − 0 0 1 1 ∗ ∗ ym 1 0 ym 1 −→ ym 1 1 g 2 ym 1 1 − − − − − − − zm 1 zm 1 zm 1 + 1 zm 1 + 1 ∗ − − ∗ − − The above step is justified for zm 1 = g 1. But if zm 1 = g 1, then − 6 − − − cm 1 (ym 1 + zm 1)/g 1, so cm = (zm 1 + cm 1)/g = (g 1+ 1)/g = 1, − ≥ − − ≥ − − − a contradiction. II.2.ii.b) ym 1 = 0, zm 1 = 0. − − 6 δm δm 1 δm 2 δm δm 1 δm 2 − − − − 0 0 0 0 ∗ ∗ 0 0 0 −→ 1 1 1 zm 1 zm 1 zm 1 1 zm 1 1 ∗ − − ∗ − − − − II.2.ii.c) ym 1 = 0, zm 1 = 0. − − If also cm 1 = 0, then δm 1 = 0, which is not allowed. Thus, cm 1 = 1. − − − This means that xm 1 g 1, g 2 . Since xi 0, 1, 2 for i 3, it − ∈ { − − } ∈ { } ≥ follows that m = 3 and we are in one of the cases A.5) or A.6). Further, δ2 = 1. In this case we change the above configuration to: δm+1 δm δm 1 δm 2 − − xm 1 1 1 1 xm 1 1 − − − − g 1 g 4 g 1 ∗ − − − 0 2 2 ∗ II.3 cm = 2. In this case it is clear that zm 1 = ym = g 1 (otherwise − − cm = 2). Note also that if ym 1 = 0, then cm 1 = 2 and then cm = 2. Thus, 6 − − 6 6 Previous proofs were complicated (3) I Proofs of Banks and Cilleruelo et al. were long and case-based I Difficult to establish I Difficult to understand I Difficult to check, too: the original Cilleruelo et al.