Table Of ContentSTOLARSKY’S CONJECTURE AND THE SUM OF
DIGITS OF POLYNOMIAL VALUES
0
KEVIN G. HARE, SHANTA LAISHRAM, AND THOMAS STOLL
1
0
2
n
a
J Abstract. Let sq(n) denote the sum of the digits in the q-ary
3 expansion of an integer n. In 1978, Stolarsky showed that
2 s2(n2)
liminf =0.
] n→∞ s2(n)
T He conjectured that, as for n2, this limit infimum should be 0
N
for higher powers of n. We prove and generalize this conjecture
h. showing that for any polynomial p(x)=ahxh+ah−1xh−1+···+
at a0 ∈Z[x] with h≥2 and ah >0 and any base q,
m s (p(n))
q
liminf =0.
[ n→∞ sq(n)
For any ε > 0 we give a bound on the minimal n such that the
1
v ratio sq(p(n))/sq(n) < ε. Further, we give lower bounds for the
9 number of n<N such that s (p(n))/s (n)<ε.
q q
6
1
4
.
1 1. Introduction
0
0
Let q ≥ 2 and denote by s (n) the sum of digits in the q-ary repre-
1 q
: sentation of an integer n. In recent years, much effort has been made
v
to get a better understanding of the distribution properties of s re-
i q
X
garding certain subsequences of the positive integers. We mention the
r
ground-breaking work by C. Mauduit and J. Rivat on the distribution
a
of s of primes [9] and of squares [10]. In the case of general polynomi-
q
als p(n) of degree h ≥ 2 very little is known. For the current state of
knowledge, we refer to the work of C. Dartyge and G. Tenenbaum [3],
who provided some density estimates for the evaluation of s (p(n)) in
q
arithmetic progressions. The authors [7] recently examined the special
case when s (p(n)) ≈ s (n).
q q
K.G. Hare was partially supported by NSERC.
Computational support provided by CFI/OIT grant.
Th. StollwaspartiallysupportedbyanAPART grantofthe AustrianAcademy of
Sciences.
1
2 KEVIN G. HARE,SHANTALAISHRAM,AND THOMAS STOLL
A problem of a more elementary (though, non-trivial) nature is to
study extremal properties of s (p(n)). Here we will always assume that
q
(1) p(x) = a xh +a xh−1 +···+a ∈ Z[x]
h h−1 0
is a polynomial of degree h ≥ 2 with leading coefficient a > 0.
h
In the binary case when q = 2, B. Lindstr¨om [8] showed that
s (p(n))
2
(2) limsup = h.
log n
n→∞ 2
In the proof of (2), Lindstr¨om uses a sequence of integers n with many
1’s in their binary expansions such that p(n) also has many 1’s. The
special case p(n) = n2 of (2) has been reproved by M. Drmota and
J. Rivat [5] with constructions due to J. Cassaigne and G. Baron.
On the other hand, it is an intriguing question whether it is possi-
ble to generate infinitely many integers n such that p(n) has few 1’s
compared to n. If this is possible, then this is indeed a rare event. It
is well-known [4, 12] that the average order of magnitude of s (n) and
q
s (nh) is
q
1 q −1
(3) s (n) ∼ s (nh) ∼ N logN.
q q
h 2logq
n<N n<N
X X
In particular, the average value of s (nh) is h times larger than the
q
average value of s (n).
q
In 1978, K. Stolarsky [14] proved several results on the extremal
values of s (p(n))/s (n) for the special case when q = 2 and p(n) = nh.
q q
He showed that the maximal order of magnitude is
c(h)(log n)1−1/h,
2
wherec(h)onlydependsonh. Thisresultisbestpossible, whichfollows
from the Bose-Chowla theorem [2, 6]. His proof can be generalized to
base q and to general polynomials p(n). Although this generalization is
straightforward, we include it here for completeness. Recall that p(n)
may have negative coefficients as well.
Theorem 1.1. Let p(x) ∈ Z[x] have degree at least 2 and positive
leading coefficient.
(1) If p(n) has only nonnegative coefficients then there exists c ,
1
dependent only on p(x) and q, such that for all n ≥ 2,
s (p(n))
q ≤ c (log n)1−1/h.
s (n) 1 q
q
STOLARKSY’S CONJECTURE 3
This is best possible in that there is a constant c′, dependent
1
only on p(x), such that
s (p(n))
q > c′(log n)1−1/h
s (n) 1 q
q
infinitely often.
(2) If p(n) has at least one negative coefficient then there exists c
2
and n , dependent only on p(x) and q, such that for all n ≥ n ,
0 0
s (p(n))
q
≤ c log n.
s (n) 2 q
q
This is best possible in that for all ε > 0 we have
s (p(n))
q
> (q −1−ε)log n
s (n) q
q
infinitely often.
The proof of this result along with some useful preliminary results
are given in Section 2.
Fortheminimalorderofs (p(n))/s (n), Stolarskytreatedthespecial
q q
case q = 2 and p(n) = n2. He proved that there are infinitely many
integers n such that
s (n2) 4(loglogn)2
2
(4) ≤ .
s (n) logn
2
He conjectured that an analogous result is true for every fixed h ≥ 2
but he did “not see how to prove this”.
Conjecture 1.2 (Stolarsky [14], 1978). For fixed h ≥ 2,
s (nh)
2
liminf = 0.
n→∞ s2(n)
By naive methods, it can be quite hard to find even a single value
n such that s (nh) < s (n) for some h, let alone observe that the
2 2
limit infimum goes to 0. For example, an extremely brute force cal-
culation shows that the minimal n such that s (n3) < s (n) is n =
2 2
407182835067 ≈ 239.
In Section 3 we prove and generalize Conjecture 1.2.
Theorem 1.3. We have
s (p(n))
q
liminf = 0.
n→∞ sq(n)
In view of our generalization, it is natural to ask how quickly we can
expect this ratio to go to zero. Recall that h = degp.
4 KEVIN G. HARE,SHANTALAISHRAM,AND THOMAS STOLL
Theorem 1.4. There exist explicitly computable constants B and C,
dependent only on p(x) and q, such that for all ε with 0 < ε < h(4h+1)
there exists an n < B ·C1/ε with
s (p(n))
q
< ε.
s (n)
q
The proof of this result along with an explicit construction for B and
C is given in Section 4. As a nice Corollary to this result we have
Corollary 1.5. There exists a constant C , dependent only on p(x)
0
and q, such that there exists infinitely many n with
s (p(n)) C
q 0
≤ .
s (n) logn
q
This is an improvement and generalization upon (4).
Proof. By solving for ε in n < B · C1/ε, one easily sees that ε <
logC . Without loss of generality we may assume that B > 1,
logn−logB
hence we can take C = logC. (cid:3)
0
One might expect that the ratio s (p(n))/s (n) is small only rarely,
q q
with most of its time being spent near h = degp. It turns out that this
ratio is small somewhat more often than expected.
Theorem 1.6. For any ε > 0 there exists an explicitly computable
α > 0, dependent only on ε, p(x) and q, such that
s (p(n))
# n < N : q < ε ≫ Nα
s (n)
(cid:26) q (cid:27)
where the implied constant also only depends on ε, p(x) and q.
The proof of this result is given in Section 5.
In Section 6 we collect together questions raised in this paper and
pose some further lines of inquiry for this research.
2. Preliminaries and Proof of Theorem 1.1
First we prove some preliminary results about s which we need in
q
the proofs. Recall (cf. [8]) that terms are said to be noninterfering if
we can use the following splitting formulæ:
Proposition 2.1. For 1 ≤ b < qk and a,k ≥ 1,
(5) s (aqk +b) = s (a)+s (b),
q q q
(6) s (aqk −b) = s (a−1)+(q −1)k −s (b−1).
q q q
STOLARKSY’S CONJECTURE 5
Proof. Relation (5) is a consequence of the (strong) q-additivity of s .
q
For (6) we write b−1 = k−1b qi with 0 ≤ b ≤ q −1. Then
i=0 i i
s (aqk −b) = s ((a−P1)qk +qk −b) = s (a−1)+s (qk −b)
q q q q
k−1
= s (a−1)+s (q −1−b )qi
q q i
!
i=0
X
k−1
= s (a−1)+ (q −1−b )
q i
i=0
X
(cid:3)
implying (6).
Proposition 2.2. The function s is subadditive and submultiplicative,
q
i.e., for all a,b ∈ N we have
(7) s (a+b) ≤ s (a)+s (b),
q q q
(8) s (ab) ≤ s (a)s (b).
q q q
Proof. The proof follows on the lines of [13, Section 2]. As for (7), an
even stronger result is true, namely that s (a+b) = s (a)+s (b)−(q−
q q q
1)·r where r is the number of “carry” operations needed when adding
a and b. Writing b = k−1b qi we also have
i=0 i
P k−1 k−1
s (ab) = s a b qi ≤ s (ab )
q q i q i
!
i=0 i=0
X X
k−1 k−1
= s (a+···+a) ≤ s(a) b ,
q i
Xi=0 bi times Xi=0
where we used twice the subad|ditiv{izty o}f s and we get (8). (cid:3)
q
Proof of Theorem 1.1. This is an almost direct generalization of Sto-
larsky’s proof (see [14, Section 2]) and Propositions 2.1 and 2.2. First,
supposethat p(n)hasonlynonnegative coefficients. Then using Propo-
sition 2.2 we see that s (p(n)) ≤ p(s (n)). Therefore
q q
s (p(n)) min{(q −1) log p(n)+1 ,p(s (n))}
q q q
≤
s (n) s (n)
q (cid:0) q (cid:1)
min{log n,s (n)h}
q q
(9) ≤ c ·
1
s (n)
q
6 KEVIN G. HARE,SHANTALAISHRAM,AND THOMAS STOLL
where c only depends on p(x) and q. If log n ≤ s (n)h then we have
1 q q
(log n)1/h ≤ s (n). From this and (9), we get that
q q
s (p(n)) log n
q ≤ c · q = c (log n)1−1/h.
s (n) 1 (log n)1/h 1 q
q q
Alternately, if log n > s (n)h then we have (log n)1/h > s (n) and
q q q q
s (p(n))
q ≤ c ·s (n)h−1 ≤ c (log n)1−1/h.
s (n) 1 q 1 q
q
For the lower bound, set
(10) k = ⌊log (λ(h+1)!)⌋+1,
q
where λ = max{a : 0 ≤ i ≤ h}. By Stolarsky’s use of the Bose-Chowla
i
Theorem, there are infinitely many integers M ≥ 3(k + 1) such that
there are integers y ,y ,...,y with N := ⌊(M +1)/(k+1)⌋−1, with
1 2 N
the following three properties:
(i) 1 ≤ y < y < ··· < y ≤ Mh,
1 2 N
(ii) y ≡ 0 mod (k +1),
i
(iii) all sums y +···+y are distinct (distinct sum property); here
j1 jh
j ,j ,...,j ∈ {1,2,...,N} with possible repetition.
1 2 h
Note that (iii) implies the distinct sum property for all y +···+y
j1 ji
with 1 ≤ i ≤ h. Now set
N
n = qyi,
i=1
X
such that
h h
(11) p(n) = a ni = ′a α(i;h ,...,h )qy1h1+···+yNhN
i i 1 N
i=0 i=0
X XX
′
where thesummation isover allvectors (h ,...,h ) satisfying h +
1 N 1
···+h = i, and α(i;h ,...,h ) denote the multinomial coefficients
N 1 N
P
i!/(h !...h !) bounded by i!. Consider (11) as a polynomial in q. By
1 N
the distinct sum property (iii) we have for all 0 ≤ i ≤ h that
N +i−1
#{y h +···+y h : h +···+h = i} = .
1 1 N N 1 N
N −1
(cid:18) (cid:19)
Thus the coefficients of qy1h1+···+yNhN = qR with h + ··· + h = h
1 N
in (11) are nonzero and bounded by
(12) a h!+a (h−1)!+···+a ≤ λ(h+1)h! < qk.
h h−1 0
STOLARKSY’S CONJECTURE 7
By (12) and (ii), the sums y h + ··· + y h ≡ 0 mod (k + 1) and
1 1 N N
hence the powers qR are noninterfering and we get
s (p(n)) N +h−1 1 Nh−1
q
≥ · ≥ .
s (n) N −1 N h!
q (cid:18) (cid:19)
By construction,
log n ≤ y +1 ≤ 2h+1Nh(k +1)h.
q N
The claim now follows by observing that k is largest for q = 2.
Secondly suppose that p(n) has at least one negative coefficient.
Thenthefirst claimfollowsby observing that s (p(n)) ≤ ⌊log p(n)⌋+1
q q
for sufficiently large n. For the lower bound, denote by a the negative
j
coefficient with smallest index j, i.e., a < 0 and a ≥ 0 for 1 ≤ l ≤ j.
j j−l
Then for all sufficiently large k we have
s (p(qk)) = s (a qhk +···+a q(j+1)k +a qjk +a q(j−1)k +···+a )
q q h j+1 j j−1 0
j−1
= s (a q(h−j)k +···+a qk +a )+ s (a )
q h j+1 j q l
l=0
X
≥ k(q −1)−s(−a −1)
j
> k(q −1−ε).
Here we have used Proposition 2.1. As s (qk) = 1 and log (qk) = k,
q q
(cid:3)
the result follows. This completes the proof of Theorem 1.1.
3. Proof of Theorem 1.3
The proof of Theorem 1.3 will use a construction of a sequence with
noninterfering terms. First assume that p(x) = xh, h ≥ 2 and define
the polynomial
t (x) = mx4 +mx3 −x2 +mx+m
m
where m ∈ Z with m ≥ 3. By consecutively employing (5) and (6) we
see that for all k with qk > m,
(13) s (t (qk)) = (q −1)k +s (m−1)+3s (m).
q m q q
The appearance of k in (13) is crucial. The next lemma lies at the
heart of the proofs. We will use it to see that s (t (qk)h), h ≥ 2, is
q m
independent of k whenever k is sufficiently large. Furthermore, we will
exploit the fact that the coefficients of [xi] in t (x)h are polynomials
m
in m with alternating signs.
8 KEVIN G. HARE,SHANTALAISHRAM,AND THOMAS STOLL
Lemma 3.1. For fixed h ≥ 2 and m ≥ 3, we have
4h
t (x)h = c (m) xi
m i,h
i=0
X
satisfying
(14) 0 < c (m) ≤ (2mh)h i = 0,1,...,4h.
i,h
In fact, we have
(15) c (m) = c (m) = mh, c (m) = c (m) = hmh.
0,h 4h,h 1,h 4h−1,h
Proof. Adirect calculation shows thatt (x)2 andt (x)3 have property
m m
(14) provided m ≥ 3. Set h = 2h +3h with max(h ,h ) ≥ 1. Then
1 2 1 2
t (x)h = t (x)2...t (x)2·t (x)3...t (x)3.
m m m m m
h times h times
1 2
Since products of poly|nomia{lsz with }all|positiv{ez coeffic}ients have all
positive coefficients too, we get c (m) > 0 for all i = 0,1,...,4h. On
i,h
the other hand, the coefficients of t (x)h are clearly bounded by the
m
corresponding coefficients of the polynomial
h i j k
mh(1+x+x2 +x3 +x4)h = mh xi+j+k+l.
i j k l
0≤l≤k≤j≤i≤h(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)
X
Therefore, for all i with 0 ≤ i ≤ 4h, we have
h!
(16) c (m) ≤ mh
i,h
(h−i)!(i−j)!(j −k)!(k −l)!
0≤l≤k≤j≤i≤h
X
≤ mhh! exp(h−i+i−j +j −k +k −l)
≤ mhh!eh ≤ (2mh)h.
(cid:3)
Proof of Theorem 1.3. Now let k be such that qk > (2mh)h. By (14)
and (5) we then have
s (t (qk)h) = s (c (m))+s (c (m))+···+s (c (m))
q m q 0,h q 1,h q 4h,h
where s (c (m)) is bounded by a function which only depends on q,
q i,h
m and h. Together with (13) and letting k → ∞ we thus conclude for
fixed m ≥ 3,
lim s (t (qk)h)/s (t (qk)) = 0,
q m q m
k→∞
as wanted.
STOLARKSY’S CONJECTURE 9
Finally we consider the case with a general polynomial instead of xh.
Write
(17) p(t (x)) = a t (x)h +a t (x)h−1 +···+a t (x)+a
m h m h−1 m 1 m 0
where a > 0 and h ≥ 2. First suppose that all the coefficients are
h
nonnegative. Lemma 3.1 shows that for i with 2 ≤ i ≤ h all the
coefficients of t (x)i are positive. Also, the coefficient [x2] in p(t (x))
m m
is nonnegative if we choose m ≥ 3 sufficiently large. In fact, a sufficient
condition is a h mh −hmh−1 ≥ a which is true whenever
h 2 1
(cid:0)(cid:0) (cid:1) (cid:1) 1/(h−1)
2a
1
(18) m ≥ .
h(3h−5)a
(cid:18) h(cid:19)
If the polynomial p(x) has negative coefficients then there is a positive
integer b such that the polynomial p(x+b) has all positive coefficients.
A good choice for b is
λ λ
(19) b = 1+ = 1+ , λ = max{|a | : 0 ≤ i ≤ h}.
i
a a
(cid:24) h(cid:25) (cid:24) h(cid:25)
This is easy to see since both p(x+b)−(a (x+b)h −λ h−1(x+b)i)
h i=0
and
P
h−1
a (x+b)h−λ (x+b)i
h
i=0
X
λ λ
= a − (x+b)h +
h
x+b−1 x+b−1
(cid:18) (cid:19)
1
= (a x+(b−1)a −λ)(x+b)h +λ
h h
x+b−1
(cid:0) (cid:1)
have nonnegative coefficients when b ≥ 1+ λ . Thus if qk > m+b then
ah
s (t (qk) + b) = (q − 1)k + s (m − 1) + 2s (m) + s (m + b) and one
q m q q q
similarly obtains for fixed m,
lim s (p(t (qk)+b))/s (t (qk)+b) = 0.
q m q m
k→∞
(cid:3)
This completes the proof of Theorem 1.3.
4. Proof of Theorem 1.4
Theconstructionofanextremalsequence intheproofofTheorem1.3
gives a rough bound on the minimal n such that s (nh) < s (n). We
q q
first illustrate the method in the case q = 2, h = 3.
10 KEVIN G. HARE,SHANTALAISHRAM,AND THOMAS STOLL
Set m = 3. Then for all k with 2k > max c (m) = 225 we have
i,h
0≤i≤4h
s (t (2k)) = k +1+6 = k +7,
2 3
s (t (2k)3) = 2·(4+3+4+4+4+4)+4 = 50.
2 3
Therefore, by setting k = 44, we get
min{n : s (n3) < s (n)} < 2178.
2 2
Itispossibletoshowthattheminimalsuchntoben = 407182835067 ≈
239.
Proof of Theorem 1.4. Consider the general polynomial
p(x) = a xh +a xh−1 +···+a ∈ Z[x]
h h−1 0
with a > 0, h ≥ 2. Let λ = max|a |. Pick b such that p(x+b) has only
h i
nonnegative coefficients, as in (19). Pick m ≥ 3 such that p(t (x)+b)
m
has only nonnegative coefficients, as in (18). Our task is to bound the
coefficients of of p(t (x)+b) ∈ Z[x].
m
To begin with, we estimate the coefficient of xi,0 ≤ i ≤ h of p(x+b),
h h
j j
(20) a bj−i ≤ a bj−i ≤ λ(2b)h.
j j
i i
j=i (cid:18) (cid:19) j=i (cid:12) (cid:18) (cid:19)(cid:12)
X X(cid:12) (cid:12)
Combining (20)with(15), wefindth(cid:12)atthecons(cid:12)tanttermofp(t (x)+b)
(cid:12) (cid:12) m
is bounded by
h mh+1 −1
λ(2b)h mi = λ(2b)h ≤ λh(4mbh)h
m−1
i=0
X
since m ≥ 3 and h ≥ 2. Again from (20) and (14), we find that the
other coefficients of p(t (x)+b) are bounded by
m
h
(21) λ(2b)h (2mi)i ≤ λh(4mbh)h.
i=1
X
Therefore the coefficients of p(t (x) +b) are bounded by λh(4mbh)h.
m
Hence for qk > m+b, we have
log(λh(4mbh)h)
(22) s (p(t (qk))+b) ≤ (q −1)(4h+1) +1 .
q m
logq
(cid:18) (cid:19)
On the other hand, we clearly have s (t (qk) + b) > (q − 1)k for
q m
qk > m+b. Let
4h+1 log(λh(4mbh)h)
k = +1 +1.
ε logq
(cid:22) (cid:18) (cid:19)(cid:23)
