proceedings of the
american mathematical
Volume 72, Number
1, October
„_„
society
1978
REPRESENTATIONFUNCTIONSOF SEQUENCES
IN ADDITIVENUMBER THEORY
MELVYN B. NATHANSON1
Abstract. Let â be a set of nonnegative integers, and let rf(n) denote the
number of representations of n in the form n = a, + a¡ with a¡, a¡ 6 &■The
set & is periodic if a e & implies a + m £ â for some m > 1 and all
a > N. It is proved that if & is not periodic, then for every set ® ¥= & there
exist infinitely many n such that rf(n) =£ rf(n). Moreover, all pairs of
periodic sets & and 9> are constructed that satisfy rf(n) = rf(n) for all but
finitely many n.
Let & be a set of nonnegative integers. Let rf(ri) denote the number of
representations of « as a sum of h elements of &. If fj[z) = 2aeSz" is the
generating function for &, then f^zf = '2^0rh\n)zn.
Let r%n) denote the
number of representations of « as a sum of an arbitrary number of elements
of éE. If 0 g &, then r&(n) = 2£L,rf(/i) is finite for all n. Representation
functions have been studied by various authors [l]-[7].
In this note I consider the question: To what extent do the sequences rf(n)
and r&(n) determine the set 6Ü?I shall prove that if & and ® are sets of
nonnegative integers such that rf(ri) = r®(n) for some h > 1 and all n > 0, or
if r&(n) = r\n) for all n > 0, then éE = $. However, there do exist sets â
and <S such that rf(n) = rf(n) for all sufficiently large n, but 6B^ %. All
such pairs of sets & and ÇBwill be constructed explicitly. An infinite set 6Eof
integers is called periodic if there exist integers m > 1 and N such that a E &
implies a + m E & for all a > N. It will be shown that if the set 6B is not
periodic, then for every set % ^ & we must have rf(n) ^= rf(ri) for infinitely
many n.
Theorem 1. Let (E and % be sets of nonnegative integers, and let rf(n) and
rf(n) denote the number of representations of n as a sum of h elements of & and
%, respectively. If r%(n) = r®(ri) for all n > 0, then & = %.
Proof. If rf(«) = rf(n) for all n > 0, then
CO
0= 2rf(«)z"n=0
00
^rf(n)z"=U(z)h-U(z)h
n=0
=(/«w - h {*))(/«
w*-' +/*w~%«+•••+/• <*)*•)•
Received by the editors February 14, 1977.
/IMS (A/OS) sufyecfclassifications(1970).Primary 10L15,10L05,10J99.
/fey words and phrases. Representation functions, addition of sequences, sum sets.
'This paper was written in July, 1976, at the Institute for Advanced Study, Princeton, New
Jersey. The author wishes to thank the Institute for its hospitality.
O American Mathematical Society 1978
16
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17
REPRESENTATION functions
But f,A[z)h ' + • • ■ + fq,(z)h ' is a nonzero power series with nonnegative
coefficients, and so/^z) - /9(z) = 0, that is, & = <$>. D
Theorem 2. Le/ 6Ea/i*/ íB fe se/s of positive integers, and let rs(n) and r*(/i)
denote the number of representations of n as a sum of an arbitrary number of
elements of & and <$, respectively. If rs(n) = r%(n) for all n > 1, then
& = %.
Proof. Let r&(0)= zffü) = 1 and let rg(n) = 0 for all n > 1. Clearly,
r&(n)= f 4(n)
for all « > 0.
/i = 0
Therefore,
S r&(n)z
= 22
4in)z" =22
71= 0 /l = 0
71= 0
A= 0
Consequently, if rfy) = r\n)
/9(z))~',andsoffi = S. D
rf(n)z»
71=071 = 0
' ~ h\z)
for all « > 0, then (1 -/^z))"1
= (1 -
Theorem 3. Let & and % be sets of positive integers, and let pa(n) andp%(n)
denote the number of representations of n as the sum of an arbitrary number of
elements of & and ®, respectively, where representations differing only in the
arrangement of their summands are not counted separately. Ifp&(n) = p (ri)for
all n > 1, then & = %.
Proof. Let p®(0) = p\0)
form
= 1. The generating function for p<\n) has the
00
CO
!
2 ps(n)z"= n s ** - n r^-p.
71=0
\\p\n)
= p\ri)
for all n>\,
aeS
k=0
aeS
'
Z
then Wa&¿\ - za) = \\*4\
- zb), and so
â = %. n
If 6B and ÍB are sets of nonnegative integers such that rf(«) = rf(n) for
some h > 1 and all sufficiently large n, then it does not follow that â = %.
For example, rf(n) = 1 if n E & and rf(n) = 0 if n 6 &, and so rff» =
rf(n) for all sufficiently large n if and only if & and 1 eventually coincide. If
(£ = {/,> 1} and ® = (0) U [n > 2), then rf(n) = rf(n) for all n > 3, but
& ¥= iß. This construction can be generalized in the following way.
Let A, B, and T be finite sets of integers. If each residue class modulo m
contains exactly the same number of elements of A as elements of B, then we
write A = B (mod m). If the number of solutions of the congruence a + t =
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18
M. B. NATHANSON
n (mod m) with a E A, t E T, equals the number of solutions of the congruence b + t = n (mod m) with b E B, t E T, for each residue class n
modulo m, then we write A + T = B + T (mod m).
Let A, B, and T be finite sets of integers with A \j B c [0, 1,2, . . ., N)
and T c {0, 1, . . . , m - 1} such that A + T = B + T (mod m) for some
m > 1. Define the periodic set 6 by
C = {c > N\c = t (mod m) for some /ET}.
Let & = ^ u C and <S = 5 u C. If >4^ 5, then 6? ¥= ®. We shall prove
that /f(n) = rf (ai)for all n > 2N.
Let sA(ri) (resp. íb(/i)) denote the number of representations of n in the
form n = a + c with a E A, c E Q (resp. n = b + c with b E B, c E G). If
ax,a2E A, then ax < N and a2 < N, hence a, + a2 < 27V.Consequently, if
n > 2N then the only representations of « as a sum of two elements of
6? = A u G are of the form a + c, c + a, and c + c', where a E /I and
c, c' E 6. Since ^ n C = 0, this implies that rf(n) = r^n) + 2sA(n). Simi-
larly, rf(n) = rffn) + 2sB(n).
Therefore, rf(n) = rf(ri) if and only if sA (ri) = sB(n).
But if « > 2/V, then sA(n) is exactly the number of solutions of the
congruence n = a + t (mod m) with aG/l,
t E T. For if n s a + t
(mod m), then n - a = t (mod m) and n — a > 27V— a > TV,hence n — a
= c E 6, and « = a + (n — a) = a + c. Conversely, if n — a + c with a E
A, c E 6, then c = n — a > 2N - a > N, and so c = t (mod m) for some
/ E T, and n = a + t (mod m).
Similarly, JB(«) is the number of solutions of the congruence n = b + t
(mod m) with b E B, t E T. Since A + T = B + T (mod m), this implies
that sA(n) = sB(ri), and so rf(/i) = rf(ri) for all n > 27V. The following
theorem shows that this construction produces all pairs of sets <£ and % such
that rf(n) = rf(ri) for all sufficiently large n.
Theorem 4. Let & and % be sets of nonnegative integers such that r^ri) =
rf(n) for all sufficiently large n. Then there exist finite sets A, B, and T with
A u B c {0, 1, . . . , TV}and T c {0, 1, . . . , m - 1} such that A + T = B
+ T (mod m), and & = A u 6 and % = B u 6, wAtre <2= {c > 7V|c= í
(mod m) for some t E T).
Proof. To every representation 2« = a, + a¡ with a„ Oj E 61 and a¡ ^ a,
corresponds a second representation 2/i = a, + a,. Therefore, rf(2«) is odd if
and only if n E &. Similarly, n E % if and only if rf(n) is odd. If rf(n) =
rf (n) for all n > 7V0,then for all n > 7V0we have n E &■if and only if
nE®.
Let/^z)
« 2aegz"
and/a(z)
= 2ftei6z*. Then/^z)
- /9(z) = Q(z)
is a polynomial of degree at most 7V0,and/^z) + /^(z) is a power series such
that, for n > 7V0,the coefficient of z" is either 0 (if n Ç £ U $) or 2 (if
« E fi n $). Then
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19
REPRESENTATIONFUNCTIONS
ß (*)(/« (*) + h i*)) = (/« (z) - /a (z))(/a (z) + f9 (z))
-M*f-U(*f-
2 (rfin) - rf(n))z"
71=0
= Í(r?in)-r?(n))z"
= P(z),
71= 0
where P (z) is a polynomial of degree at most N0. Therefore, f^z) + f^(z) =
P(z)/Q(z)
is a rational function, and so the sequence of coefficients of the
power series /^z) + f^(z) eventually satisfies a linear recurrence relation.
These coefficients are eventually either 0 or 2. But a sequence defined by a
linear recurrence on a finite set must be eventually periodic. This implies that
for some modulus m > 1 there exists a set T c {0, 1, 2, . . ., m - 1} and
an integer N > N0 such that, for all n > N, we have n E & n ÍB if and only
if n = f (mod m) for some / G T.
Let
t4 = {a < N\a E &}, B = {b < N\b E %},
and
Q = {c > N\c E & n $} = {c > N\c = t (mod m) for some / Ê 71}.
Then 6E= A u 6 and ® = fi u ß.
If n>2JV, then zf(/z) - 4(n) + 2sA(n) and rf(n) = rf(ri) + 2sB(n),
where sA(n) (resp. jä(«)) is the number of representations of n in the form
n = a + c with a E A, c G 6 (resp. /z = zb+ c with 6 E B, c E Q). Since
4(n) = /f(n) for zz > IN, it follows that sA(ri) = jä(/j). But ^(/z) (resp.
sB(n)) is also the number of solutions of the congruence n = a + t (mod m)
with a E A, t E T (resp. n = b+t (mod m) with b E B, t E T). Therefore,
A + T = B + T (mod m). □
Corollary,
/f the set & of nonnegative integers is not eventually periodic,
then for every set 9> =>=3, we have rf(n) =*=rf(ri)for infinitely many n.
Remarks. The converse of this corollary is false. If & is the sequence of all
nonnegative integers, then r2(n) = n + 1 for all n > 0. If ÍB =?=&, then
/•^(zz)< « for all sufficiently large n. But the sequence 6B is periodic with
period 1.
It is an open problem to determine those sets & and $ such that
rf(ri) = rf(n) for some h > 3 and all sufficiently large n.
The following problem arises from Theorem 4. Let A, B, and T be finite
nonempty sets of nonnegative integers with T c {0, 1, 2,.. ., m - 1} for
some m > 1 such that A + T = B + T (mod m). Assume the modulus m is
"reduced" in the sense that there does not exist a divisor m, of m and a set
Tx c {0, 1, . . . , m, - 1) such that T = {/ E (0, 1, . . . , m - l}|í 5 /,
(mod mx) for some /, E Tx). Does this imply that A = B (mod m)? This is
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20
M. B. NATHANSON
obviously true if m = 1. The following result proves this is also true if the
modulus m is prime.
Theorem 5. Let p be aprime number, and let A, B,T be finite, nonempty sets
of nonnegative integers with T c {0, 1, ...,/>1}. // T = {0, 1, . . . ,p -
1}, then A + T = B + T (modp) if and only if \A\ - \B\. If Tc
{0, I, ... ,p - I), then A + T = B + T (mod p) if and only if A =*B
(modp).
Proof. If T = (0, 1,... ,p - 1}, then the number of solutions of the
congruence a + t = n (modp) with a E A, t E T (resp. b + t = n (modp)
with b E B, t E T) is exactly \A\ (resp. \B\) for all n. Therefore, A + T = B
+ 7(modp) if and only if \A\ = \B\.
Suppose that T c {0, 1, ...,/>1}. Write fA(z) = 2aeAza, fB(z) -
2beBzb, and fT(z) = 2ieTz'. If A + T = B + T(modp), then
Ia (*)/r W = Îb (*)/t (*) (mod z» - \),
that is, zp - 1 divides (fA(z) - fB(z))fT(z). The cyclotomic polynomial z" 1 is the product of two irreducible factors
zp -\=(z
- 1)(1 + z + z2 + ■ ■ ■ + z"-1).
The degree offT(z) is at most/? - 1, and so, if 1 + z + • • • + zp~l divides
fT(z), then fT(z) = 1 + z + • • • + zp~x and T = {0, 1, ... ,p - 1), which
is false. Therefore, 1 + z + • • • + zp~x divides fA(z) - fB(z). If z - 1 divides fT(z), then fT(z) = (z — l)g(z) for some polynomial g(z). But this
implies that/r(l)
= 0. However, since fT(z) = 2,e7-z', it follows that/r(l)
=
\T\ ¥= 0 since T¥=0. Therefore, z - 1 also divides fA(z) —fB(z). Consequently, zp — 1 divides fA(z) - fB(z). But this means precisely that A = B
(modp).
□
Corollary.
Let p be a prime, and let A, B, and T be nonempty, proper
subsets of {0, 1, . . . ,p - 1}. If A + T = B + T (modp), then A = B.
References
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number theory, Proc. Amer. Math. Soc. 14 (1963),278-284.
3. P. Erdös and W. H. J. Fuchs, On a problem of additive number theory, J. London Math. Soc.
31 (1956),67-73.
4. P. Erdös, B. Gordon, L. A. Rubel and E. G. Straus, Tauberian theorems for sum sets, Acta
Arith.9(1964), 177-189.
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Department of Mathematics, Southern Illinois University, Carbondale, Illinois 62901
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