INTEGERS 15 (2015)
#A32
THE SUM OF DIGITS OF POLYNOMIAL VALUES
Shu-Yuan Mei
Department of Mathematics and Computer Science, Jiangsu Second Normal
University, Nanjing, P. R. China
shuyuan mei@sina.com
Received: 7/8/14, Revised: 3/5/15, Accepted: 6/9/15, Published: 7/17/15
Abstract
Let sq (n) denote the sum of the digits in the q-ary expansion of a nonnegative
integer n, and let p1 (x), p2 (x) be polynomials in Z[x] with distinct positive degrees.
If p1 (n) 1 and p2 (n) 1 for all positive integers n, then for any " > 0, we give
lower bounds of the number of n  N such that sq (p1 (n))/sq (p2 (n)) < ".
1. Introduction
For any integer q
2, let nonnegative integer
n=
k
X
i=0
Denote by sq (n) =
k
P
↵i (n)q i ,
↵i (n) 2 {0, 1, . . . , q
1}.
↵i (n) the sum of digits of n in base q. The study of the sum of
i=0
digits mainly focuses on the sum of digits of some special sequences of integers, the
average sum of the digits of integers, the asymptotic formula of the weighted sumof-digits function, and the ratio of the sum of digits of polynomial values. For the
study of the sum of digits of some special sequences of integers, several researchers
investigated the properties of sq of primes [14], polynomials [5, 7, 9, 11, 15, 16, 19],
Fibonacci numbers [21] and Bernoulli numbers [2]. For the study of the average sum
of the digits of integers, one may refer to [1, 4, 6, 17]. For the study of the asymptotic
formula of the weighted sum-of-digits function, one may refer to [12, 18]. For the
study of the ratio of the sum of digits of polynomial values, several researchers
investigated the problems and a lot of academic achievements have been achieved.
This work was supported by the Project of Graduate Education Innovation of Jiangsu
Province, CXZZ13 0385.
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In 1978, Stolarsky [20] showed that
lim inf
s2 (n2 )
=0
s2 (n)
lim inf
s2 (nh )
=0
s2 (n)
n!1
and conjectured that
n!1
for any integer h 2.
In 2011, Hare, Laishram and Stoll [10] proved that for any integer q
t
P
any polynomial p(x) =
ci xi 2 Z[x] with t 2 and ct > 0,
2 and for
i=0
lim inf
n!1
sq (p(n))
= 0.
sq (n)
In 2014, Madritsch and Stoll [13] proved that
✓
◆
sq (p1 (n))
sq (p2 (n)) n 1
is dense in R+ , where p1 (x), p2 (x) are polynomials in Z[x] of distinct positive
degrees with p1 (N), p2 (N) ✓ N.
In this paper, we always assume that
p1 (x) =
h
X
i=0
ai xi 2 Z[x],
p2 (x) =
l
X
i=0
bi xi 2 Z[x]
with h 1, l 1, ah > 0 and bl > 0.
By employing the methods in [10] and [20], the following theorems are proved.
Theorem 1. Let deg p1 > deg p2 . If p1 (n)
1 and p2 (n)
1 for any positive
integer n, then for any " > 0, there exists a positive constant C1 , dependent only
on ", q, p1 (x) and p2 (x), such that
⇢
sq (p1 (n))
nN :
<"
C1 N ↵
sq (p2 (n))
for all sufficiently large integers N , where ↵ = "(2h(l + 3)(h(l + 3) + 1) + ")
1
.
Theorem 2. Let deg p1 < deg p2 . If p1 (n)
1 and p2 (n)
1 for any positive
integer n, then for any " > 0, there exists a positive constant C2 , dependent only
on ", q, p1 (x) and p2 (x), such that
⇢
sq (p1 (n))
nN :
<"
C2 log N
sq (p2 (n))
for all sufficiently large integers N .
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2. Preliminary Lemmas
Let [a, b] denote the interval of integers n such that a  n  b. For convenience,
we write f (s) ⇣ g(s) (s 2 S) if f (s) and g(s) are positive for all s 2 S and
c1 g(s)  f (s)  c2 g(s) for all s 2 S, where c1 and c2 are two positive constants.
Lemma 1. (See [10, Proposition 2.1]) For any integers a, b, k with 1  b < q k
and a, k 1, we have
sq (aq k + b) = sq (a) + sq (b),
sq (aq k
b) = sq (a
1) + (q
1)k
sq (b
1).
Lemma 2. Let n be a positive integer. Then for any integer q
sq (n)  (q
2, we have
1)(1 + logq n).
Proof. Let n = ↵t q t + · · · + ↵1 q + ↵0 with
↵i 2 {0, 1, . . . , q
Then
sq (n) =
t
X
i=0
↵i  (q
1},
i = 0, 1, . . . , t,
1)(t + 1)  (q
↵t 6= 0.
1)(1 + logq n).
Lemma 3. (See [8, Bose-Chowla Theorem] or [3] ) Let d 2 be an integer, and
let M be a power of a prime. Then there exist integers y1 , y2 , . . . , yM +1 with 1 
y1 < y2 < · · · < yM +1 = M d such that all sums
yj1 + yj2 + · · · + yjd ,
1  j1  j2  · · ·  jd  M + 1
are distinct.
Lemma 4. Let l be a positive integer, and let
tm (x) = m + mx
x2
x3
xl+1 + mxl+2 + mxl+3
···
be a polynomial in Z[x]. For any positive integer i, let
(l+3)i
(tm (x))i =
X
(i,m) j
aj
x .
j=0
Then
(a) for all positive integers m and i, we have
(i,m)
|aj
|  (4m + l)i ,
0  j  (l + 3)i.
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(b) there exists a positive constant c0 , dependent only on l, such that for any
integer m with m c0 and any integer i with 1  i  l,
(i,m)
aj
⇣ mi if and only if j 2
[ ([0, i] + (l + 2)k),
0ki
and
(i,m)
aj
⇣ mi
1
if and only if j 2
[
([i + 1, l + 1] + (l + 2)k).
0ki 1
(c) for any integer i with i > l, there exists a positive constant c1 , dependent
only on i, such that for any integer m with m c1 ,
(i,m)
⇣ mi ,
aj
0  j  (l + 3)i.
Proof. (a) Let
fm (x) = m + mx + x2 + x3 + · · · + xl+1 + mxl+2 + mxl+3
and
(l+3)i
(fm (x))i =
X
(i,m) j
bj
x .
j=0
Since
(l+3)i
(i,m)
|aj
(i,m)
|  bj
it follows that
X
,
(i,m)
bj
= (4m + l)i ,
j=0
(i,m)
|aj
|  (4m + l)i .
(b) We will complete the proof by induction on i. It is easy to see that Lemma
4 (b) is true for i = 1, 2. Suppose that Lemma 4 (b) is true for an integer i with
2  i < l. Let
(l+3)i+1
X (i,m)
(tm (x))i (m + mx) =
cj
xj ,
j=0
(l+3)(i+1)
(tm (x))i (mxl+2 + mxl+3 ) =
X
(i,m) j
dj
x ,
j=l+2
and
(l+3)i+l+1
(tm (x))i ( x2
x3
···
xl+1 ) =
X
(i,m) j
ej
x .
j=2
By the induction hypothesis, for all sufficiently large integers m, it is easy to get
that
(i,m)
cj
⇣ mi+1 if and only if j 2 [ ([0, i + 1] + (l + 2)k),
0ki
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(i,m)
cj
(i,m)
dj
⇣ mi if and only if j 2
[
⇣ mi+1 if and only if j 2
(i,m)
dj
⇣ mi if and only if j 2
and
(i,m)
ej
⇣ mi ,
([i + 2, l + 1] + (l + 2)k),
0ki 1
[
([0, i + 1] + (l + 2)k),
1ki+1
[ ([i + 2, l + 1] + (l + 2)k),
1ki
2  j  (l + 3)i + l + 1.
Therefore, for all sufficiently large integers m, we have
(i+1,m)
aj
and
⇣ mi+1 if and only if j 2
(i+1,m)
aj
⇣ mi if and only if j 2
[
([0, i + 1] + (l + 2)k),
0ki+1
[ ([i + 2, l + 1] + (l + 2)k).
0ki
(c) From the proof of Lemma 4 (b), we see that Lemma 4 (c) is true for i = l + 1.
A proof is similar to the proof of Lemma 4 (b) by induction on i
l + 1. This
completes the proof of Lemma 4.
3. Proof of Theorem 1
(i,m)
Proof. Let tm (x) and aj
(i,m)
aj
(0,m)
(0  j  (l + 3)i) be as in Lemma 4, a0
= 1,
= 0 (j > (l + 3)i) and let
X
p1 (tm (x)) =
(m) i
fi
x,
0ih(l+3)
X
p2 (tm (x)) =
(m) i
gi
x,
0il(l+3)
and
= max{|a0 |, |a1 |, . . . , |ah |, |b0 |, |b1 |, . . . , |bl |}.
Then
(m)
fj
=
h
X
ai aj
l
X
bi aj
(i,m)
,
0  j  h(l + 3)
(1)
,
0  j  l(l + 3).
(2)
i=0
and
(m)
gj
=
(i,m)
i=0
Since l2 + 2l
1 > (l + 3)(l
1), we have
(i,m)
al2 +2l
1
= 0,
il
1.
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By (2) and Lemma 4 (b), noting that l2 + 2l 1 = l + 1 + (l + 2)(l
a positive constant m0 , dependent only on p2 (x), such that
(m)
(l,m)
gl2 +2l
= bl al2 +2l
1
1
1), there exists
<0
(3)
for all integers m m0 . Since h > l, it follows from Lemma 4 (b) and Lemma 4
0
(c) that there exists a positive constant m1 , dependent only on h, such that
(h,m)
aj
and
(i,m)
aj
= O(mh
1
0
for all integers m
p1 (x), such that
⇣ mh
) (i  h
1)
00
m1 . So there exists a positive constant m1 , dependent only on
(m)
fj
> 0,
0  j  h(l + 3)
00
for all integers m
m1 . Thus, by Lemma 4 (a) and (1), there exists a positive
constant m1 , dependent only on p1 (x) and p2 (x) with m1 l, such that
X
(m)
0 < fj 
(4m + l)i  2 (5m)h , 0  j  h(l + 3)
(4)
0ih
and
(m)
|gj
X
|
0il
(4m + l)i  2 (5m)l ,
0  j  l(l + 3)
(5)
for all integers m m1 . By Lemma 4 (b) and (2), there exists a positive constant
(m)
(m)
m2 , dependent only on p2 (x), such that g0 > 0 and g1 > 0 for all integers
m
m2 . For all integers m with m
m0 , by (3), at least one coefficient of
p2 (tm (x)) is negative. For m max{m0 , m2 }, let j be the least positive integer with
(m)
gj < 0. Then 2  j  l2 + 2l 1. If m max{m0 , m1 } and q k 2 > (2 (5m)l )2 ,
then, by Lemma 1 and (5), we have
sq (p2 (tm (q k )))
=
(m)
sq (g0
(m)
= sq (g0
+
(m)
) + sq (g1
(m)
g2 q 2k
+ ··· +
(m) k
+ g2
(m)
q + · · · + gl(l+3) q kl(l+3)
+ g2
(m)
+ gj+1 q k + · · · + gl(l+3) q kl(l+3)
sq (gj
(m) k
(m)
gl(l+3) q kl(l+3) )
(m)
sq (g1
···
+
(6)
(m)
g1 q k
(m)
(m)
(q
1)k
sq ( gj
(q
1)k
(q
(q 1)k (q
1
>
(q 1)k.
2
(m)
k
(m)
jk
q + · · · + gl(l+3) q kl(l+3)
1)
(m)
1)(logq ( gj
1) + 1)
1)(logq (2 (5m)l ) + 1)
)
)
k
)
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If q k > 2m, then, by the definition of tm (x), we have
mq (l+3)k < tm (q k ) < 2mq (l+3)k < q (l+4)k .
If q k > 2 (5m)l and m
(7)
m1 , then, by (4) and Lemma 1, we have
sq (p1 (tm (q k )))
=
(m)
sq (f0
(m)
) + sq (f1
(m)
) + sq (f1
= sq (f0
= ···
+
(8)
(m)
f1 q k
= sq (f0
(m)
(m)
 (h(l + 3) + 1)(q
k
+
(m)
f2 q 2k
+ ··· +
(m)
fh(l+3) q kh(l+3) )
(m) k
(m)
q + · · · + fh(l+3) q kh(l+3)
+ f2
k
)
(m)
) + · · · + sq (fh(l+3) )
1)(1 + logq (2 (5m)h )).
Let m3 = max{m0 , m1 , m2 , l}. For any integers m and k with m
[2 logq (2 (5m)l ) + 2], by (6) and (8), we have
m3 and
2(h(l + 3) + 1)(1 + logq (2 (5m)h ))
sq (p1 (n))

,
sq (p2 (n))
k
(9)
where n = tm (q k ).
Without loss of generality, we can assume that 0 < "  1. Let m be an integer
with m m3 ,
"
#
2(h(l + 3) + 1)(1 + logq (2 (5m)h )))
k(m) =
+ 1,
"
and n(m) = tm (q k(m) ). Then k(m)
[2 logq (2 (5m)l ) + 2]. By (9), we have
sq (p1 (n(m)))
< ".
sq (p2 (n(m)))
(10)
Now we prove that all n(m) (m m3 ) are distinct. Suppose that m00 > m0
Then k(m00 ) k(m0 ). If k(m00 ) = k(m0 ), then
n(m00 ) = n(m0 ) + (m00
0
0
0
m0 )(1 + q k(m ) + q (l+2)k(m ) + q (l+3)k(m ) ) > n(m0 ).
If k(m00 ) > k(m0 ), then
n(m00 )
n(m0 )
m3 .
00
00
m00 q (l+3)k(m )
q (l+3)(k(m
>
0
2
2m0 q (l+3)k(m )
) k(m0 ))
> 1.
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By the definitions of tm (x) and k(m), we have
n(m) = tm (q k(m) ) < 2mq (l+3)k(m)
 2mq (l+3)(2(h(l+3)+1)"
= 2mq (l+3)(2(h(l+3)+1)"
1
+1) 2(l+3)(h(l+3)+1)"
1
+1)
q
1
logq (2 (5m)h )
(2 (5m)h )2(l+3)(h(l+3)+1)"
1
< mq 3(l+3)(h(l+3)+1)" (10 m)2h(l+3)(h(l+3)+1)"
< (10 qm)2h(l+3)(h(l+3)+1)"
Hence, if
1
+1
1
1
.
0
m3  m  C1 N ↵ ,
0
then n(m)  N , where C1 = (10 q)
1
(11)
, and
↵ = "(2h(l + 3)(h(l + 3) + 1) + ")
1
.
Since m3 is positive constant dependent only on p1 (x) and p2 (x), by (11), there
exists a positive constant C1 , dependent only on ", q, p1 (x) and p2 (x) such that
⇢
sq (p1 (n))
nN :
<"
sq (p2 (n))
n
o
0
m : m3  m  C1 N ↵
0
C1 N ↵
> C1 N
m3
↵
for all sufficiently large integers N . This completes the proof of Theorem 1.
4. Proof of Theorem 2
Proof. We follow the proof of Hare, Laishram and Stoll [10]. Let b be a positive
h
l
P
P
integer, p1 (x + b) =
ui xi and p2 (x + b) =
vi xi . Then all
i=0
ui =
i=0
X
✓
◆
k+i k
ak+i
b ,
i
X
bk+i
0kh i
and
vi =
0kl i
✓
◆
k+i k
b ,
i
0ih
0il
are positive integers for all sufficiently large integers b. Without loss of generality,
we may assume that all the coefficients of p1 (x) and p2 (x) are positive integers. Let
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= max{a0 , a1 , . . . , ah , b0 , b1 , . . . , bl }. Let M be a prime, d = l and y1 , y2 , . . . , yM +1
be as in Lemma 3. Let
M
+1
X
TM (x) =
xyi ,
i=1
and
p2 (TM (x)) =
X
(M ) j
qj
x .
0jlM l
Then
p2 (TM (x)) =
X
0il
=
X
0il
bi
bi (xy1 + xy2 + · · · + xyM +1 )i
X
h1 +h2 +···+hM +1 =i
(12)
i!
xh1 y1 +h2 y2 +···+hM +1 yM +1 .
h1 !h2 ! · · · hM +1 !
By Lemma 3, for any fixed integer i with 0  i  l, we have all sums
h1 y1 + h2 y2 + · · · + hM +1 yM +1
with
h1 + h2 + · · · + hM +1 = i,
hj
0,
1j M +1
are distinct. Then for any nonnegative integer i  lM l , we have
X
(M )
0 < qi

bj j!  (l + 1)!.
(13)
0jl
By (12), we have


n
o
(M )
0  j  lM l : qj > 0
X
X
1
(14)
0il h1 +h2 +···+hM +1 =i
X ✓M + i◆ ✓M + l + 1◆
=
.
M
M +1
0il
Let n = TM (q k ) and k0 = [logq ( (l + 1)!)] + 1. Then for any integer k
have q k > (l + 1)!. Since all coefficients of
xh1 y1 +h2 y2 +···+hM +1 yM +1
with
h1 + h2 + · · · + hM +1 = l,
hj
0,
1j M +1
are positive integers and all sums
h1 y1 + h2 y2 + · · · + hM +1 yM +1
k0 , we
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with
h1 + h2 + · · · + hM +1 = l,
hj
0,
1j M +1
1=
✓
◆
M +l
.
M
are distinct, by Lemma 1, we have
X
sq (p2 (n))
h1 +h2 +···+hM +1 =l
By (13) and Lemma 2, for any nonnegative integer i  lM l , we have
⇣
⌘
(M )
sq qi
 (q 1)(1 + logq (l + 1)!).
(15)
(16)
By (14), (16), and Lemma 1, noting that q k > (l + 1)!, we have
⇣
⌘
X
(M )
sq (p2 (n)) =
sq qj
0jlM l
(M )
qj >0
X

⇣
⌘
(M )
1) log qj + 1
(q
0jlM l
(M )
qj >0
✓
◆
M +l+1
1)(1 + logq ( (l + 1)!))
.
M +1
 (q
As a similar argument for p1 (x), we have
sq (p1 (n))  (q
✓
◆
M +h+1
1)(1 + logq ( (h + 1)!))
.
M +1
(17)
By (15) and (17), we have
sq (p1 (n))
sq (p2 (n))


(q
1)(1 + logq ( (h + 1)!))
M +h+1
M +1
M +l
M
(q
(18)
1)l!(1 + logq ( (h + 1)!))
,
h!(M + 1)l h
where n = TM (q k ).
For any " > 0 and for any integer k k0 , by (18), there exists a prime M0 such
that
sq (p1 (n))
<"
sq (p2 (n))
l
for all integers n = TM0 (q k ). By n = TM0 (q k ) < q k(M0 +1) , we see that, if
k
(M0l
log N
,
+ 1) log q
INTEGERS: 15 (2015)
11
then n  N . Since TM0 (q k ) (k k0 ) are distinct, there exists a positive constant
C2 , dependent only on ", q, p1 (x) and p2 (x) such that
⇢
sq (p1 (n))
nN :
<"
sq (p2 (n))
⇢
log N
k : k0  k 
l
(M0 + 1) log q
C2 log N
for all sufficiently large integers N . This completes the proof of Theorem 2.
Acknowledgements The author would like to thank the anonymous referee for
helpful comments and suggestions.
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