Sequences Generated by Polynomials

E. F. Cornelius Jr. and Phill Schultz

1. INTRODUCTION. For which integer sequences of length n is there a poly-nomial (of arbitrary degree) having integer coefficients whose sequence of values( f (1), . . . , f (n)) equals the given sequence?

The restriction to integer coefficients significantly limits the sequences of integersthat can be generated in this way by polynomials. For example, there is no polynomialwith integral coefficients which will generate the sequence (1, 0, 0) because if f (x) issuch a polynomial, then (x − 2)(x − 3) is a factor of f (x) and hence (1 − 2)(1 − 3) =2 divides f (1), a contradiction. On the other hand, the sequence (1, 0, 1) is generatedby the polynomial (x − 2)2.

It is easy to see that each sequence of length 1 is the image f (1) of a constant poly-nomial and each sequence (a, b) of length two is ( f (1), f (2)) for the linear polyno-mial f (x) = (b − a)x + (2a − b). In fact we shall see that “most” finite sequencescannot be generated by polynomials, but “many” can.

Denote by Zn the set of all integer sequences of length n, by Z[x] the set of all

polynomials with integer coefficients and by Z[x]n the set of all polynomials in Z[x]of degree < n. Let Pn denote the set of all a = (a1, . . . , an) ∈ Z

n for which thereexists some f (x) ∈ Z[x] such that f (1) = a1, f (2) = a2, . . . , f (n) = an . We callsequences in Pn polynomial sequences, and the object of this paper is to characterizepolynomial sequences.

2. POLYNOMIAL SEQUENCES. Let a = (a1, . . . , an) ∈ Zn . Then the Lagrange

Interpolation Theorem produces a polynomial

�a(x) =n∑

j=1

a j

∏i �= j

x − i

j − i

of degree ≤ n − 1 for which (�a(1), . . . , �a(n)) = (a1, . . . , an), the given sequence.To see why this is so, note that �a(x) is a sum of n terms, each of which is a poly-

nomial of degree n − 1. In evaluating �a(k), where k is an integer between 1 and n,each term except the kth has a factor x − k, so evaluates to 0. The kth term evaluatesto ak

∏i �=k(k − i)/(k − i) = ak , as required. The polynomial �a(x) may have degree

< n − 1 because the terms of degree n − 1 in different summands may cancel.The objection, of course, is that the coefficients of �a(x) are rationals which need

not be integers. We first show the somewhat surprising result that a sequence a =(a1, . . . , an) is in Pn if and only if the polynomial �a(x) determined by the LagrangeInterpolation Theorem has integral coefficients, which implies that a has a unique gen-erating polynomial of degree ≤ n − 1.

Theorem 2.1. Let a = (a1, a2, . . . , an) ∈ Zn. Then a ∈ Pn if and only if �a(x) ∈

Z[x]n. Furthermore, �a(x) is the unique polynomial of degree < n with real coeffi-cients that generates a.

Proof. Suppose that a ∈ Pn and let f (x) ∈ Z[x] satisfy f ( j) = a j for j = 1, . . . , n.Let pn(x) = (x − 1)(x − 2) · · · (x − n), a polynomial of degree n satisfying pn(i) = 0

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for i = 1, 2, . . . , n. By the division algorithm, f (x) = q(x)pn(x) + r(x) for uniqueq(x) ∈ Z[x] and r(x) ∈ Z[x]n . Evaluating both sides of this equation at 1, 2, . . . , nshows that r(x) generates a. This means that r(x) and �a(x) are both polynomials withrational coefficients of degree < n which agree at n different points. Hence they areidentical.

The converse is trivial. Moreover, for any real polynomial g(x) of degree < n, ifg(i) = ai for all i = 1, 2, . . . , n, then g(x) agrees with �a(x) at n points, so g(x) =�a(x).

We can now identify those sequences a which satisfy the condition of Theorem 2.1.In order to do this, it is convenient to express �a(x) as a linear combination of the poly-nomials p j (x) introduced in the proof of Theorem 2.1; that is, p0(x) = 1 and p j (x) =(x − 1)(x − 2) · · · (x − j) for all j ≥ 1. We shall also need the following lower trian-gular n × n rational matrices:

1. An has (i, j)-entry(i

j

)for all i, j = 0, 1, . . . , n − 1, where

(ij

) = 0 if i < j . An

is well known as Pascal’s matrix [1]. As shown in [1], and as is easily checked,An is invertible with lower triangular inverse (ai j ) where ai j = (−1)i+ j

(ij

).

2. Cn has (i, j)-entry (i) j for all i, j = 0, 1, . . . , n − 1. Here, (i) j is the fallingfactorial defined by

(i) j =

⎧⎪⎨⎪⎩

i(i − 1) · · · (i − j + 1) if i ≥ j ≥ 11 if j = 00 if i < j .

Note that p j (i + 1) = (i) j .3. Dn is the diagonal matrix whose j th diagonal entry is j ! for j = 0, 1, . . . , n − 1.

Since (i) j = (ij

)j !, Cn = An Dn .

4. Bn is C−1n = D−1

n A−1n = (bi j ) where

bi j = (−1)i+ j

i !(

i

j

)= (−1)i+ j

j !(i − j)!for all i ≥ j and bi j = 0 for i < j .

For example, in the case n = 5 we have

B5 =

⎡⎢⎢⎢⎢⎣

1 0 0 0 0−1 1 0 0 0

12 −1 1

2 0 0− 1

612 − 1

216 0

124 − 1

614 − 1

61

24

⎤⎥⎥⎥⎥⎦ , C5 =

⎡⎢⎢⎢⎣

1 0 0 0 01 1 0 0 01 2 2 0 01 3 6 6 01 4 12 24 24

⎤⎥⎥⎥⎦ .

Lemma 2.2. Let a = (a1, a2, . . . , an) be a sequence of n integers. Let La(x) =b0 p0(x) + b1 p1(x) + · · · + bn−1 pn−1(x) where

bk =k∑

j=0

(−1)k+ j

j ! (k − j)! a j+1

for k = 0, . . . , n − 1. Then La(x) = �a(x).

February 2008] NOTES 155

Proof. By Theorem 2.1, �a(x) is the unique polynomial with real coefficients and de-gree < n generating a. Thus it suffices to show that La( j) = a j for 1 ≤ j ≤ n. Let bbe the sequence (b0, b1, . . . , bn−1) written as a column vector. Express a as a columnvector so that by definition, Bna = b.

Now let pn(x) be the row vector (p0(x), p1(x), . . . , pn−1(x)). A routine multipli-cation shows that La(x) = pn(x)b. Since the i th row of Cn is pn(i + 1), it followsthat

(La(1), La(2), . . . , La(n))T = Cnb = Cn Bna = a.

Thus for all i = 1, 2, . . . , n, La(i) = ai as required.

Notice that in Lemma 2.2, if the coefficients of �a(x) are integers, then in particularbn−1, which is the coefficient of xn−1, is an integer. But then bn−2, which is the co-efficient of xn−2 in the polynomial La(x) − bn−1 pn−1(x), is also an integer. Similarreasoning shows that bi is an integer for all i = 0, . . . , n − 1.

Corollary 2.3. Let f (x) ∈ Z[x]n . Then there are unique integers b0, b1, . . . , bn−1

such that f (x) = b0 p0(x) + b1 p1(x) + · · · + bn−1 pn−1(x).

Proof. Let a be the integer sequence ( f (1), f (2), . . . , f (n)). Then f (x) and La(x)

are both polynomials of degree < n which agree at n points. Hence they are identical.

In order to find necessary and sufficient conditions for La(x) and hence �a(x) to bein Z[x], we use the characterization of Lemma 2.2.

Corollary 2.4. Let a ∈ Zn. Then a is a polynomial sequence if and only if for all

k = 0, 1, . . . , n − 1,

bk =k∑

j=0

(−1)k+ j

j ! (k − j)! a j+1

is an integer.

Proof. The result follows immediately from Lemma 2.2, since �a(x) has integer coef-ficients if and only if La(x) does.

For example, for any sequence a ∈ Zn , Lemma 2.2 states that b0 = a1 and b1 =

a2 − a1, so the condition of Corollary 2.4 that b0 and b1 be integers is vacuous. On theother hand, b2 = a1/2 − a2 + a3/2, so the Corollary explains once again why (1, 0, 1)

is a polynomial sequence but (1, 0, 0) is not.Another application of Lemma 2.2 shows that if a ∈ Z

n, then some integral multipleof a is generated by a polynomial in Z[x]n .

Theorem 2.5. Let a = (a1, . . . , an) ∈ Zn. Then (n − 1)! a = ((n − 1)! a1, . . . ,

(n − 1)! an) ∈ Pn. Moreover, (n − 1)! is the least positive integer for which this istrue for every sequence of length n.

Proof. Denote the sequence (n − 1)!a by (a′1, a′

2, . . . , a′n) and let b′

0, b′1, . . . , b′

n−1 bethe corresponding sequence of coefficients defined in Lemma 2.2.

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Note that for all j and k satisfying 0 ≤ j ≤ k ≤ n − 1, k! = j ! (k − j)! (kj

)so that

k! and hence (n − 1)! is a multiple of j ! (k − j)!. Hence by Corollary 2.4 each b′k is an

integer, so (n − 1)! a ∈ Pn .To see that this result is the best possible, consider the sequence a = (1, 0, . . . , 0).

Then for all k = 0, . . . , n − 1, bk = (−1)k/k! and in particular,

bn−1 = (−1)n−1

(n − 1)! .

Hence the least positive integer m such that ma is generated by an integral polynomialis m = (n − 1)!.

3. FINITE ABELIAN GROUPS. Recall that every non-zero finite abelian groupG is isomorphic to Z/n1Z ⊕ Z/n2Z ⊕ · · · ⊕ Z/nmZ for unique positive integers mand ni > 1 such that each ni except nm is a multiple of ni+1 [3, Theorem 9.3]. Thisexpression is called the Smith Normal Form of G.

For example, Z/6Z ⊕ Z/3Z is a Smith Normal Form decomposition of Z/3Z ⊕Z/3Z ⊕ Z/2Z, but the latter is not in Smith Normal Form. The sequence (n1, n2,

. . . , nm) is called the Smith Invariant of the group G. Notice that the order of G is|G| = n1n2 · · · nm . Moreover, n1 is the exponent of G; that is, the least positive integerm for which ma = 0 for all a ∈ G.

Let us see how this is related to the polynomial sequence problem. Let v : Z[x] →Z

n denote the valuation map defined by

v( f (x)) = ( f (1), f (2), . . . , f (n))

for each f (x) ∈ Z[x]. The function v is an additive homomorphism with image Pn

and kernel the set of polynomials in Z[x] for which f (1) = f (2) = · · · = f (n) = 0.The First Fundamental Homomorphism Theorem [3, Theorem 13.3] states that Pn

is isomorphic to the factor group Z[x]/ ker v. We show now that Pn is also isomorphicto Z[x]n .

Theorem 3.1. With the notation above, the homomorphism v restricts to an isomor-phism of Z[x]n onto Pn.

Proof. Since every f (x) ∈ ker v vanishes at each of x = 1, 2, . . . , n, by the DivisorTheorem [3, Theorem 31.1], f (x) = pn(x)q(x) for some q(x) ∈ Z[x]. Hence ker v isthe ideal of Z[x] generated by pn(x). By the division algorithm, every g(x) ∈ Z[x] hasa unique decomposition as g(x) = pn(x)q(x) + r(x) for some r(x) ∈ Z[x]n . HenceZ[x]n is a complementary summand in Z[x] to ker v. Thus v restricted to Z[x]n is anisomorphism onto Pn .

It follows that Pn is a subgroup of the free abelian group Zn of the same rank n,

and we wish to characterise the factor group Zn/Pn . First recall that we have shown in

Theorem 2.5 that for every a ∈ Zn, (n − 1)!a ∈ Pn , and (n − 1)! is the least positive

integer for which this is true for every a. This means that G = Zn/Pn is a finitely gen-

erated torsion abelian group of exponent (n − 1)!. Hence G is a finite abelian group,so it remains only to describe the Smith Normal Form of G.

We first deal with the cases n = 1 or 2. We have seen that in fact Z1 = P1 and

Z2 = P2, so that Z

1/P1 and Z2/P2 are trivial. This is in accordance with the fact that

0! = 1! = 1. Thus we shall assume from now on that n ≥ 3.

February 2008] NOTES 157

By Corollary 2.3, the n polynomials {p0(x), p1(x), . . . , pn−1(x)} form a free basisfor the free abelian group Z[x]n . Since the valuation map v : Z[x]n → Pn is an iso-morphism, it follows that the images {v(p j (x)) : j = 0, . . . , n − 1} form a free basisfor Pn . Recall from the remarks preceding Lemma 2.2 that the elements of this basisare the columns of the matrix Cn .

For j = 0, . . . , n − 1, let e( j) be the j th column of An . We have seen that An isinvertible with integral inverse, so {e( j) : j = 0, . . . , n − 1} forms a free basis for Z

n .Note that since Cn = An Dn , we have v(p j (x)) = j !e( j) for j = 0, . . . , n − 1.

We proceed to establish the following relationship:

Theorem 3.2. Zn/Pn

∼= Z/2!Z ⊕ Z/3!Z ⊕ · · · ⊕ Z/(n − 1)!Z.

Proof. For each j = 0, 1, . . . , n − 1, the j th basis element v(p j (x)) of Pn is a multiplej !e( j) of the j th basis element of Z

n . Such bases are called stacked bases [2], and it isnot hard to see that

Zn

Pn= 〈e(0)〉

0!〈e(0)〉 ⊕ 〈e(1)〉1!〈e(1)〉 ⊕ · · · ⊕ 〈e(n − 1)〉

(n − 1)!〈e(n − 1)〉 .

The first two summands, isomorphic to Z/0! Z and Z/1! Z respectively, are trivial,leaving the required n − 2 proper summands.

Corollary 3.3. Zn/Pn is a finite abelian group with Smith Normal Form

Z/(n − 1)! Z ⊕ · · · ⊕ Z/3! Z ⊕ Z/2! Zand Smith Invariant ((n − 1)!, . . . , 3!, 2!). Thus |Zn/Pn| = ∏n−1

i=2 i !.One way of viewing this last result heuristically is that an integer sequence of length

n chosen at random has probability only 1/k of being a polynomial sequence, wherek = ∏n−1

i=2 i !. More precisely, for each N > 0, let Prob(N ) be the probability that aninteger sequence drawn uniformly at random from among all integer sequences in[−N , N ] is generated by a polynomial. Then as N tends to infinity, Prob(N ) tends to1/

∏n−1i=2 i !.

ACKNOWLEDGMENTS. The authors wish to thank Professor Rodney Forcade of Brigham Young Univer-sity for his kind assistance, and the MONTHLY referees, who have improved the exposition.

REFERENCES

1. G. S. Call and D. J. Velleman, Pascal’s matrices, this MONTHLY 100 (1993) 372–376.2. J. M. Cohen and H. Gluck, Stacked bases for modules over principal ideal domains, Journal of Algebra

14 (1970) 493–505.3. J. B. Fraleigh, A First Course in Abstract Algebra, Addison Wesley, Reading, MA, 1967.

College of Engineering and Science, University of Detroit Mercy, Detroit, MI 48221-3038efcornelius@comcast.net

School of Mathematics and Statistics, The University of Western Australia, Nedlands, Australia 6009schultz@maths.uwa.edu.au

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