Primitive polynomial subsequences

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PRIMITIVE POLYNOMIAL SUBSEQUENCES I. ANDERSON, S. D. COHEN AND W. W. STOTHERS 1. Introduction. A sequence {«„} of integers is said to be primitive if a^aj whenever / # j . For example, if n is any positive integer, the sequence {m: n < m 0, there exists an infinite primitive sequence {a,) such that hm sup n-»oo

> i — e, II

where A(n) denotes the number of a, < n. In this paper we take a polynomial/with integer coefficients and consider primitive subsequences of {/(I), ...,/(«)} and infinite primitive subsequences of {/(l),/(2), ...}. In §2 we shall provide an extension of Besicovitch's result (for which/(x) = x) which applies to all linear polynomials. The results for general polynomials of degree ^ 2 are of a different nature; these are given in §§3-5. The following is a brief description of our notation. I(n) will denote the set {1, 2, ..., n}. If / is a subset of N; the set of natural numbers, we shall u s e / ( / ) to denote the sequence {/(;): i el}. For example, f(N) = {/(l),/(2), ...}. In order to describe the relative density of an infinite subsequence V of f(N) in the sequence / (N), we introduce the idea of the upper sequence density of £f, 3 / ( y ) , defined by

Further, when the limit

exists, it is called the sequence density of £f and is denoted by df{£f). In particular, if/(x) = x, then df(Sf) = d(£P), the ordinary density of £f. Finally, we shall use B(I) to denote the sequence of all multiples of the integers in / . 2. Arithmetic progressions. We first consider/(x) = ax + b where, without loss of generality, a > 0 and (a, b) — 1. The following simple lemma is basic. LEMMA

1. If f(x) = ax + b, then f(x) \f(y) o y = x(modf(x)).

Proof. f(y) = O(mod/0)) of(y) o y = x (modf(x)).

s f(x)(modf(x)) o ay s ax (mod ax+b)

Our first theorem shows that, for large n, the longest primitive subsequence of /(/(«)) has length essentially (a/(a + 1))«. To avoid the complications which arise [MATHEMATIKA 21 (1974), 239-247]

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I. ANDERSON, S. D. COHEN AND W. W. STOTHERS

when the sequence begins with negative terms, we state it under the assumption that a + b > 0 so t h a t / i s positive on N. We also assume that n > b since it is easy to see, using Lemma 1, that /(/(«)) is itself primitive ifn 0, a + b > 0 and (a, b) = 1. Then for any n ^ b, a maximal primitive subsequence off(l(rij) has length L where L is the least integer 5= (an + b)/(a + 1).

Proof. Note that, in integral part notation, = n - M, say, where M(> 0) is the greatest integer such that M +/(M) < n. Now the sequence/(J), where J = {M + 1, ...,«} has length L and is primitive since, by Lemma \,f(x) \f{y) => y ^ x +f(x) => y > n if x > M. To show that there is no longer primitive subsequence of/(/(«)), observe first that, for every x, f(x)\f{x +f(x)). Next put g(x) = x +f(x) and partition the distinct elements of the set / = I(M) u g(l(M)) ( s /(«)) into a family of (say) r chains of the form {i, g(i), g(g(i)), •••}. Since/is positive on N, each of these chains is a strictly increasing sequence in which precisely the last member exceeds M. It follows that the number of distinct elements in / is M + r. However, our initial remark indicates that the value under/ of at most one member of each chain can be included in a primitive subsequence of/(/(«)), and so at least M members off (I) must be excluded. This completes the proof. The next theorem concerns the maximal upper sequence density of a primitive subsequence of f(N), and is the extension of Besicovitch's result. THEOREM 2. Let f(x) = ax + b where (a, b) = 1. Given e > 0, there exists an infinite primitive subsequence £f off(N) such that

Proof. For any real numbers u, v, let (w,;;] here denote the set of integers in the real interval (u, v]. Besicovitch's result follows from the theorem of Erdos [2, p. 256] which asserts that, as T -> oo, the density of the multiples of integers in (T, 2T] -> 0, i.e. d{B((T,2T^j -> 0. Clearly we have (

as T-»oo, for anyfixedc1,c2. In precisely the same way, Theorem 2 follows from the next lemma in which df denotes the sequence density with respect to/(x) = ax + b. LEMMA

2. Letf(x)

= ax + b and JT = (T, (a + l)T]. Then df(B(JT)) -> 0 as

T -> oo. Proof. By Lemma 1, df(B(JT)) is the density of the union of congruence classes x(mod ax + b), x e Jr. One form of the Chinese remainder theorem [3, p. 34] asserts that two of these congruences have a common solution, if, and only if, (ax +b,ay

+b)\(x-

y)

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for all x, y. For chosen x,y, let h = (ax + b,ay + b). Then h\a{x—y). But (h, a) = 1 since (a, b) = 1. Thus h\(x — y). The congruences therefore have a cpmmon solution, and the density of the union of the congruence classes is therefore the density of the congruence classes 0 (mod ax + b), x e JT, i.e. the density of the set of multiples of some of the integers between aT + b and a(a + l)T + b. By Erdos's result, this tends to 0 as T -» oo. 3. Primitive subsequences of/(/(«)) ; / non-linear. Let / be a given polynomial of degree d > 2 with leading coefficient af. From the nature of the problem it is clear that we may assume that the coefficients o f / a r e coprime, that af > 0 and that / i s not simply a power of a linear polynomial. Define pf(n) to be the minimal cardinality of a subset/ of /(«) such that/(/(«) — /) is primitive. Thus a maximal primitive subsequence of/(/(«)) has length n — pr{n). Our task is to estimate Pf(n). We first obtain a lower bound. For this purpose define F to be the integral polynomial which is the product of the distinct irreducible factors o f / a n d such that > 0. Let the degree of F be D ( < d). By our assumptions we know that D > 2. THEOREM 3. With the above notation and assumptions let f be such that D ^ 2. Then, for large n and some constant E depending onf

p,(n) > («/ is replaced in (1) by ag where g(x) = x + F(x).) Proof. Let T be the least non-negative integer such that F(x + T) is positive and strictly increasing on M and let h(x) = f(x + T). Then Pf(n)

> ph(n - T) 3* ph(n) - T.

It is therefore sufficient to assume that F is positive and strictly increasing on N. In this case let g(x) = x + F(x) and M = Mf(n) be the greatest integer for which g(M) < n ; thus for large n Mf(n) = (n/aFy>D + 0(1). We show that pf(n) > Mf(n). Note that by Taylor's theorem

/(*(*))= £ (F(*))'/«(*)/i!, i= 0

where/ ( i ) denotes the i-th derivative of/. Now by considering the precise power of each irreducible factor o f / we see t h a t / ( x ) is a factor of (F(x))if 2 and that f is not the d-th power of a linear polynomial. Then for sufficiently large n, = O(H«H-1>') o gidax + b) \ giday + b) then /({x 1 ( ..., xu}) is a primitive subsequence of/(/(«)), i.e. u ^ n — P/(«). Hence pfin) < pgidan + b) and the theorem will be true for /provided it is true for g. From now on assume that n is large enough to ensure t h a t / ( x ) is positive and strictly increasing for x > H*. We require two simple lemmas. 3. Let f have the form (4) and suppose that k, x, y e N are such that = kfix) where ndn'1+1) < x < y < n. Then, for sufficiently large n,

LEMMA

fiy)

i < yd/kxd < 2. Proof.

Let A = max \at\.

Then clearly

| / - kx 1 and so, by Lemma 3, for sufficiently large n T < x < 2y\0

< 2n/k1/d,

and

243

= kf(x)

for Some v < « and

2 < k < 2yd/xd < 2nd/Td.

Hence | / | does not exceed the sum over all k with 2 < k < 2nd/Td of the number of solutions (x, y) of f(y)

T < x < 2n/k1/d,

= kf(x),

x < y < n.

(5)

Accordingly we shall from now on consider only solutions (x, y) of (5) for k in the range [2, 2nd/Td). Put A = max | a, | . Then, by Lemma 1, for large n we have

\yd-kxd\

< Admax(yr,kxr) < 2Adkxr.

Since trivially it follows that — - k1" x > nd'(d+i)

To prove (2), choose T = nid+i)l(d+2) r < d — 2, (6) then implies that y 1.

... k /

X

X

2AdkUd

< 2

so that k < 2ndl(d+2). 1/d 2Adk :


0). Then s = 0 or for any / = 0, ..., s—1 l

y> >

HI.

X; ^ x i + 1 Xj

4n 2

and hence

A_A > xs

x0

An2

However, (7) implies that, if 5 ^ 1, then

A_i± < x

s

X

It follows that s < \6dAn2l(d+2)k~lld. solutions of (5) for this & is < d(s+1). have

0

Now, by Lemma 4, the total number of Hence, putting sk for s and m for 2nd/(d+2\ we

Cm]

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I. ANDERSON, S. D. COHEN AND W. W. STOTHERS

This proves (2). For (3), assume that r ^ d - 3 and put T = nmd+1} This time (6) implies that a solution of (5) satisfies 1/d

so that k < 2ndKd+1\

1/d

2Adk 2Adk l_ _ kUd ,9/4 n x Hence, if (xlt yx) and (x2, y2) are solutions of (5) with yi/xt < y2/x2, then

—

_J_ 0, there exists a primitive subsequence £f off(N) such that ciAS?) > 1 — e.

Let 9f(m) denote the number of solutions of the congruence/(x) = 0 (mod m). In order to construct £f we need a bound for 0f(m) for an arbitrary modulus m. For squarefree/, an appropriate bound is provided by the next lemma. LEMMA

5. Let f be a square-free polynomial. Then 6f(m) < C(f)d"im\

where C(J) is a positive constant independent ofm and oo(m) is the number of distinct primes dividing m. Proof. It is easy to see that the coefficients of/ may be assumed to be relatively prime. Since / and its derivative / ' are coprime as polynomials, there exist integer polynomials u(x), v(x) and a positive integer c such that u(x)f(x) + v(x)f'(x) = c.

(8)

2

We shall prove the lemma with C(f) = dc , where d is the degree of/. For any prime p, let ps(s > 0) be the exact power of p dividing c. Since 0f is multiplicative, it suffices to prove that, for any r, 6f(pr) < dp2s. Now trivially df(p2s+1) < p2sof(p) < dP2s, because / i s a non-zero polynomial (modp). To complete the proof we show that 6f(pr) is decreasing for r > 2s + 1.

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Now note that, by (8) and the definition of s, r > s and/(x) = 0 (modp r ) implies From this it follows easily by Hensel's lemma that Ofitf) = Ofif'1) ps+i Jff'(x). for all r ^ 2s + 2. So we have 6f(pr) < dp2s for all r, and the lemma is proved. We deduce from Lemma 5 the following inelegant but adequate bound for 6f(m) in general. LEMMA

6. Let f be as in Theorem A. Then for sufficiently large m 0f(m)
T, then 9f(f(x)) = o(xP2 Inserting this in (9) and putting

we obtain B = O[Tk

E

x-"\

=O\Tk\

x-

for sufficiently large T. Summarising the above, we see that, for a large enough T — T(e) and every keN, \