The sum of digits of polynomial values in arithmetic progressions
aa r X i v : . [ m a t h . N T ] O c t THE SUM OF DIGITS OF POLYNOMIAL VALUES INARITHMETIC PROGRESSIONS
THOMAS STOLL
Abstract.
Let q, m ≥ m, q −
1) = 1. Denote by s q ( n )the sum of digits of n in the q -ary digital expansion. Further let p ( x ) ∈ Z [ x ]be a polynomial of degree h ≥ p ( N ) ⊂ N . We show that there exist C = C ( q, m, p ) > N = N ( q, m, p ) ≥
1, such that for all g ∈ Z and all N ≥ N , { ≤ n < N : s q ( p ( n )) ≡ g mod m } ≥ CN / (3 h +1) . This is an improvement over the general lower bound given by Dartyge andTenenbaum (2006), which is CN /h ! . Introduction
Let q, m ≥ s q ( n ) the sum of digits of n in the q -arydigital expansion of integers. In 1967/68, Gelfond [1] proved that for nonnegativeintegers a , a with a = 0, the sequence ( s q ( a n + a )) n ∈ N is well distributed inarithmetic progressions mod m , provided ( m, q −
1) = 1. At the end of his paper,he posed the problem of finding the distribution of s q in arithmetic progressionswhere the argument is restricted to values of polynomials of degree ≥
2. Recently,Mauduit and Rivat [8] answered Gelfond’s question in the case of squares.
Theorem 1.1 (Mauduit & Rivat (2009)) . For any q, m ≥ there exists σ q,m > such that for any g ∈ Z , as N → ∞ , { ≤ n < N : s q ( n ) ≡ g mod m } = Nm Q ( g, d ) + O q,m ( N − σ q,m ) , where d = ( m, q − and Q ( g, d ) = { ≤ n < d : n ≡ g mod d } . The proof can be adapted to values of general quadratic polynomial insteadof squares. We refer the reader to [7] and [8] for detailed references and furtherhistorical remarks. The case of polynomials of higher degree remains elusive so far.The Fourier-analytic approach, as put forward in [7] and [8], seems not to yieldresults of the above strength. In a recent paper, Drmota, Mauduit and Rivat [4]applied the Fourier-analytic method to show that well distribution in arithmeticprogressions is obtained whenever q is sufficiently large.In the sequel, and unless otherwise stated, we write p ( x ) = a h x h + · · · + a for an arbitrary, but fixed polynomial p ( x ) ∈ Z [ x ] of degree h ≥ p ( N ) ⊂ N . Date : December 1, 2018.2010
Mathematics Subject Classification.
Primary 11A63; Secondary 11N37, 11N69.
Key words and phrases.
Sum of digits, polynomials, Gelfond’s problem.
Theorem 1.2 (Drmota, Mauduit & Rivat (2011)) . Let q ≥ exp(67 h (log h ) ) be a sufficiently large prime number and suppose ( a h , q ) = 1 . Then there exists σ q,m > such that for any g ∈ Z , as N → ∞ , { ≤ n < N : s q ( p ( n )) ≡ g mod m } = Nm Q ⋆ ( g, d ) + O q,m,p ( N − σ q,m ) , where d = ( m, q − and Q ⋆ ( g, d ) = { ≤ n < d : p ( n ) ≡ g mod d } . It seems impossible to even find a single “nice” polynomial of degree 3, say, thatallows to conclude for well distribution in arithmetic progressions for small bases,let alone that the binary case q = 2 is an emblematic case. Another line of attackto Gelfond’s problem is to find lower bounds that are valid for all q ≥
2. Dartygeand Tenenbaum [3] provided such a general lower bound by a method of descenton the degree of the polynomial and the estimations obtained in [2].
Theorem 1.3 (Dartyge & Tenenbaum (2006)) . Let q, m ≥ with ( m, q −
1) = 1 .Then there exist C = C ( q, m, p ) > and N = N ( q, m, p ) ≥ , such that for all g ∈ Z and all N ≥ N , { ≤ n < N : s q ( p ( n )) ≡ g mod m } ≥ CN /h ! . The aim of the present work is to improve this lower bound for all h ≥
3. Moreimportantly, we get a substantial improvement of the bound as a function of h .The main result is as follows. Theorem 1.4.
Let q, m ≥ with ( m, q −
1) = 1 . Then there exist C = C ( q, m, p ) > and N = N ( q, m, p ) ≥ , such that for all g ∈ Z and all N ≥ N , { ≤ n < N : s q ( p ( n )) ≡ g mod m } ≥ CN / (3 h +1) . Moreover, for monomials p ( x ) = x h , h ≥ , we can take N = q h + m ) (cid:16) hq (6 q ) h (cid:17) h +1 ,C = (cid:16) hq (6 q ) h · q (24 h +12 m ) / (3 h +1) (cid:17) − . The proof is inspired from the constructions used in [5] and [6] that were helpfulin the proof of a conjecture of Stolarsky [9] concerning the pointwise distributionof s q ( p ( n )) versus s q ( n ). As a drawback of the method of proof, however, it seemsimpossible to completely eliminate the dependency on h in the lower bound.2. Proof of Theorem 1.4
Consider the polynomial(2.1) t ( x ) = m x + m x − m x + m , Gelfond’s work and Theorem 1.1 give precise answers for linear and quadratic polynomials, sowe do not include the cases h = 1 , HE SUM OF DIGITS OF POLYNOMIAL VALUES IN ARITHMETIC PROGRESSIONS 3 where the parameters m , m , m , m are positive real numbers that will be chosenlater on in a suitable way. For all integers l ≥ T l ( x ) = t ( x ) l = l X i =0 c i x i to denote its l -th power. (For the sake of simplicity we omit to mark the dependencyon l of the coefficients c i .) The following technical result is the key in the proof ofTheorem 1.4. It shows that, within a certain degree of uniformity in the parameters m i , all coefficients but one of T l ( x ) are positive. Lemma 2.1.
For all integers q ≥ , l ≥ and m , m , m , m ∈ R + with ≤ m , m , m < q, < m < l − (6 q ) − l we have that c i > for i = 0 , , , . . . , l and c i < for i = 1 . Moreover, for all i , (2.3) | c i | ≤ (4 q ) l . Proof.
The coefficients of T l ( x ) in (2.2) are clearly bounded above in absolute valueby the corresponding coefficients of the polynomial ( qx + qx + qx + q ) l . Since thesum of all coefficients of this polynomial is (4 q ) l and all coefficients are positive,each individual coefficient is bounded by (4 q ) l . This proves (2.3). We now showthe first part. To begin with, observe that c = m l > c = − lm m l − whichis negative for all m >
0. Suppose now that 2 ≤ i ≤ l and consider the coefficientof x i in(2.4) T l ( x ) = ( m x + m x + m ) l + r ( x ) , where r ( x ) = l X j =1 (cid:18) lj (cid:19) ( − m x ) j (cid:0) m x + m x + m (cid:1) l − j = l − X j =1 d j x j . First, consider the first summand in (2.4). Since m , m , m ≥ x i in the expansion of ( m x + m x + m ) l is ≥
1. Note also that all the powers x , x , . . . , x l appear in the expansion of this term due to the fact that every i ≥ i = 3 i + 2 i with non-negative integers i , i .We now want to show that for sufficiently small m > x i inthe first summand in (2.4) is dominant. To this end, we assume m < m > m j for 2 ≤ j ≤ l . Using (cid:0) lj (cid:1) < l and a similar reasoning as above we getthat | d j | < l l m (3 q ) l = l (6 q ) l m , ≤ j ≤ l − . This means that if m < l − (6 q ) − l then the powers x , . . . , x l in the polynomial T l ( x ) indeed have positive coefficients. This finishes the proof. (cid:3) To proceed we recall the following splitting formulas for s q which are simpleconsequences of the q -additivity of the function s q (see [5] for the proofs). THOMAS STOLL
Proposition 2.2.
For ≤ b < q k and a, k ≥ , we have s q ( aq k + b ) = s q ( a ) + s q ( b ) ,s q ( aq k − b ) = s q ( a −
1) + k ( q − − s q ( b − . We now turn to the proof of Theorem 1.4. To clarify the construction we considerfirst the simpler case of monomials, p ( x ) = x h , h ≥ . (We here include the cases h = 1 and h = 2 because we will need them to deal withgeneral polynomials with linear and quadratic terms.) Let u ≥ t ( x )in (2.1) by q u − . Lemma 2.1 then shows that for all integers m , m , m , m with(2.5) q u − ≤ m , m , m < q u , ≤ m < q u / ( hq (6 q ) h ) , the polynomial T h ( x ) = ( t ( x )) h = p ( t ( x )) has all positive ( integral ) coefficients withthe only exception of the coefficient of x which is negative. Let u be an integersuch that(2.6) q u ≥ hq (6 q ) h and let k ∈ Z be such that(2.7) k > hu + 2 h. For all u with (2.6) the interval for m in (2.5) is non-empty. Furthermore, rela-tion (2.7) implies by (2.3) that q k > q hu · q h ≥ (4 q u ) h > | c i | , for all i = 0 , , . . . , h, where c i here denotes the coefficient of x i in T h ( x ). Roughly speaking, the use ofa large power of q (i.e. q k with k that satisfies (2.7)) is motivated by the simplewish to split the digital structure of the h -power according to Proposition 2.2. Bydoing so, we avoid to have to deal with carries when adding terms in the expansionin base q since the appearing terms will not interfere. We also remark that this isthe point where we get the dependency of h in the lower bound of Theorem 1.4.Now, by c , | c | ≥ s q ( t ( q k ) h ) = s q h X i =3 c i q ik + c q k − | c | q k + c ! = s q h X i =3 c i q ( i − k + c q k − | c | ! + s q ( c )= s q h X i =3 c i q ( i − k ! + s q ( c −
1) + k ( q − − s q ( | c | −
1) + s q ( c )= h X i =3 s q ( c i ) + s q ( c −
1) + k ( q − − s q ( | c | −
1) + s q ( c )= k ( q −
1) + M, (2.8)where we write M = h X i =3 s q ( c i ) + s q ( c − − s q ( | c | −
1) + s q ( c ) . HE SUM OF DIGITS OF POLYNOMIAL VALUES IN ARITHMETIC PROGRESSIONS 5
Note that M is an integer that depends (in some rather obscure way) on the quanti-ties m , m , m , m . Once we fix a quadruple ( m , m , m , m ) in the ranges (2.5),the quantity M does not depend on k and is constant whenever k satisfies (2.7).We now exploit the appearance of the single summand k ( q −
1) in (2.8). Since byassumption ( m, q −
1) = 1, we find that(2.9) s q ( t ( q k ) h ) , for k = hu + 2 h + 1 , hu + 2 h + 2 , . . . , hu + 2 h + m, runs through a complete set of residues mod m . Hence, in any case, we hit a fixedarithmetic progression mod m (which might be altered by M ) for some k with hu + 2 h + 1 ≤ k ≤ hu + 2 h + m .Summing up, for u with (2.6) and by (2.5) we find at least(2.10) ( q u − q u − ) ( q u / ( hq (6 q ) h ) − ≥ (1 − /q ) hq (6 q ) h q u integers n that in turn by (2.1), (2.5), (2.7) and (2.9) are all smaller than q u · q hu +2 h + m ) = q h + m ) · q u (3 h +1) and satisfy s q ( n h ) ≡ g mod m for fixed g and m . By our construction and bychoosing k > hu + 2 h > u all these integers are distinct. We denote N = N ( q, m, p ) = q h + m ) · q u (3 h +1) , where u = (cid:6) log q (cid:0) hq (6 q ) h (cid:1)(cid:7) ≤ log q (cid:0) hq (6 q ) h (cid:1) . Then for all N ≥ N we find u ≥ u with(2.11) q h + m ) · q u (3 h +1) ≤ N < q h + m ) · q ( u +1)(3 h +1) . By (2.10) and (2.11), and using (1 − /q ) ≥ / q ≥
2, we find at least(1 − /q ) hq (6 q ) h q u ≥ (cid:16) hq (6 q ) h · q (24 h +12 m ) / (3 h +1) (cid:17) − N / (3 h +1) integers n with 0 ≤ n < N and s q ( n h ) ≡ g mod m . We therefore get the statementof Theorem 1.4 for the case of monomials p ( x ) = x h with h ≥
3. The estimates arealso valid for h = 1 and h = 2.The general case of a polynomial p ( x ) = a h x h + · · · + a of degree h ≥ h ≥
1) follows easily from what we have already proven.Without loss of generality we may assume that all coefficients a i , 0 ≤ i ≤ h ,are positive, since otherwise there exists e = e ( p ) depending only on p such that p ( x + e ) has all positive coefficients. Note that a finite translation can be dealt withchoosing C and N appropriately in the statement. Since Lemma 2.1 holds for all l ≥ x , we have thatthe polynomial p ( t ( x )) has again all positive coefficients but one where the negativecoefficient again corresponds to the power x . It is then sufficient to suppose that k > hu + 2 h + log q max ≤ i ≤ h a i in order to split the digital structure of p ( t ( q k )). In fact, this implies that q k > (cid:18) max ≤ i ≤ h a i (cid:19) · (4 q u ) h , THOMAS STOLL and exactly the same reasoning as before yields ≫ q,p q u distinct positive integersthat are ≪ q,m,p q u (3 h +1) and satisfy s q ( p ( n )) ≡ g mod m . This completes the proofof Theorem 1.4. Acknowledgements.
This research was supported by the Agence Nationale de laRecherche, grant ANR-10-BLAN 0103 MUNUM.
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