Volume 9 (2008), Issue 1, Article 29, 8 pp.
PROPERTIES OF NON POWERFUL NUMBERS
VLAD COPIL AND LAURENŢIU PANAITOPOL
U NIVERSITY "S PIRU H ARET "
D EPARTMENT OF M ATHEMATICS
S TR . I ON G HICA , 13, 030045
B UCHAREST, ROMANIA
vladcopil@gmail.com
U NIVERSITY OF B UCHAREST
D EPARTMENT OF M ATHEMATICS
S TR . ACADEMIEI 14, 010014
B UCHAREST, ROMANIA
pan@al.math.unibuc.ro
Received 11 January, 2008; accepted 10 March, 2008
Communicated by L. Tóth
A BSTRACT. In this paper we study some properties of non powerful numbers. We evaluate the
n-th non powerful number and prove for the sequence of non powerful numbers some theorems
that are related to the sequence of primes: Landau, Mandl, Scherk. Related to the conjecture
of Goldbach, we prove that every positive integer ≥ 3 is the sum between a prime and a non
powerful number.
Key words and phrases: Powerful numbers, Sequences, Inequalities, Goldbach’s conjecture.
2000 Mathematics Subject Classification. 11B83, 11P32, 26D07.
1. I NTRODUCTION
A positive integer v is called non powerful if there exists a prime p such that p|v and p2 - v.
Otherwise, if v has the canonical decomposition v = q1α1 ·· · ··qrαr , there exists j ∈ {1, 2, . . . , r}
such that αj = 1.
It results that v can be written uniquely as v = f · u, where f is squarefree, u is powerful and
(f, u) = 1.
In this paper we use the following notations:
• K(x)= the number of powerful numbers less than or equal to x
• C(x)= the number of non powerful numbers less than or equal to x
• vn is the n-th non powerful number
We use a special case of a classical formula:
008-08
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V LAD C OPIL AND L AUREN ŢIU PANAITOPOL
Theorem A. If h ∈ C 1 , g is continuous, a is powerful and
X
G(x) =
g(v),
a≤v≤x
v non powerful
then
Z
X
x
h(v)g(v) = h(x)G(x) −
h0 (t)G(t)dt.
a
a≤v≤x
v non powerful
G. Mincu and L. Panaitopol proved [5] the following.
Theorem B.
and
√
√
K(x) ≥ c x − 1.83522 3 x
for x ≥ 961
√
√
K(x) ≤ c x − 1.207684 3 x
for x ≥ 4.
As C(x) = [x] − K(x) it results that
√
√
√
√
(1.1)
[x] − c x + 1.207684 3 x ≤ C(x) ≤ [x] − c x + 1.83522 3 x
the first inequality being true for x ≥ 4, while the second one is true for x ≥ 961.
We also use
Theorem C. We have the relation
1
3
ζ(3/2) √
ζ(2/3) √
− 15
3
5
6
K(x) =
x+
x + O x exp(−c1 log (log log x)
.
ζ(3)
ζ(2)
2. I NEQUALITIES FOR vn
Theorem 2.1. We have the relation
√
√
vn > n + c n − a 3 n
for n ≥ 88,
where a = 1.83522.
Proof. If we put x = vn in the second inequality from (1.1), it results that
√
√
n ≤ vn − c vn + a 3 vn
for n ≥ 4.
√
√
√
√
Let f (x) = x − c x + a 3 x − n and x0n = n + c n − k 3 n. As f (vn ) > 0, and f is
increasing, if we prove that f (x0n ) < 0, it results that vn > x0n .
Denote g(n) = f (x0n ). Proving that f (x0n ) < 0 is equivalent with proving that g(n) < 0.
Therefore we intend to prove that
q
q
√
√
√
√
√
√
3
3
3
g(n) = c n − k n − c n + c n − k n + a n + c n − k 3 n < 0.
We use the following relations for x > 0:
(2.1)
1+
x √
x x2
> 1+x>1+ −
2
2
8
1+
x √
x x2
> 31+x>1+ − .
3
3
9
and
(2.2)
J. Inequal. Pure and Appl. Math., 9(1) (2008), Art. 29, 8 pp.
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P ROPERTIES OF N ON P OWERFUL N UMBERS
3
Putting x = x0n in (2.1) gives
2
√ q
√
√
√
√
c
c
k
k
k
n
c
3
√ −√
n+ − √
> n+c n−k n> n+ − √
−
,
3
2 26n
2 26n
8
n
n2
while x = x0n gives from (2.2)
2
√
q
3
√
√
√
√
c
k
c
k
c
k
n
3
3
3
3
√ −√
.
n+ √
− √ > n+c n−k n> n+ √
− √ −
3
9
n
36n 33n
36n 33n
n2
Using the previous relations in the expression of g(n) yields
2
√ √
√
√
√
ac
ak
c
k
c2
ck
c n
3
3
√
√
√ .
√
+
a
n
+
−
−
g(n) < c n − k n − c n − + √
+
3
2
8
36n 33n
n
26n
n2
In order to prove g(n) > 0 it is enough to prove that
2
√ √
c2
ck ac
1
c
ak
c n
k
3
√
√ −√
(a − k) n − +
+
− √ +
< 0.
3
6
2
2
3
8
n 33n
n
n2
The best result is obtained by taking k = a, therefore
2
√ c2 5 ac
a2
c n
c
a
√ −√
− + √
− √ +
< 0.
3
2
6 6n 33n
8
n
n2
a
As √cn > √
3 2 for n ≥ 1, it is enough to prove that
n
√
5ac
c n c2
c2
a2
√
√ .
+
·
<
+
8
n
2
66n
33n
The last relation is true because
c3
a2
√ < √
⇔
8 n
33n
3c3
8a2
6
<n
that holds for n ≥ 816
and
6
5ac
5a
c2
√
⇔
< n that holds for n ≥ 8.
<
6
2
3c
6 n
In conclusion, we have
√
√
vn > n + c n − a 3 n
for n ≥ 816. Verifications done using the computer allow us to lower the bound to n ≥ 88.
Theorem 2.2. We have the relation
√
√
vn < n + c n − 3 n
for n ≥ 1.
Proof. If we put x = vn in the first inequality from (1.1), it results that
√
√
n > vn − c vn + α 3 vn ,
where α = 1.207684.√
√
√
√
Let f (x) = x − c x + α 3 x − n and x00n = n + c n − h 3 n. We have f (vn ) < 0, f is
increasing, so if we prove that f (x00n ) > 0, it results that vn < x00n .
Denote g(n) = f (x00n ). Proving that f (x00n ) > 0 is equivalent to proving that g(n) > 0.
Therefore we have to prove that
q
q
√
√
√
√
√
√
3
3
3
g(n) = n + c n − h n − c n + c n − h n + α n + c n − h 3 n − n < 0.
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4
V LAD C OPIL AND L AUREN ŢIU PANAITOPOL
Using the relations (2.1) and (2.2) as we did in the proof of Theorem 2.1, gives
2
√ √
√
√
√
ch
c2
αc
αh
α3n
c
h
3
3
√ −√
c n−h n−c n− + √
+α n+ √
− √ −
> 0.
3
2
9
26n
36n 33n
n
n2
The previous relation is equivalent to
2
√ √
αh
c
h
1
c2 α 3 n
ch αc
3
√ −√
√
+ √
.
>
+
n(α − h) +
+
3
6
2
9
2
2
3
33n
n
n
n
Thus, it is enough to prove that for h < α
√
√
h α
c2
αh
α 3 n c2
c
3
+
>
+ √
+
· .
n(α − h) + √
6
3
2
9
n
n 2
33n
√
2
We have 3 n(α − h) > c2 , if
3
c2
(2.3)
n>
.
2(α − h)
It remains to prove that
c
√
6
n
h α
+
2
3
αh
αc2
√ .
> √
+
3 3 n 9 3 n2
2
α
αc
√
√
Therefore it is enough to prove that c h2 > 3αh
6 n and that c 3 > 9 n ; both the relations are true for
n ≥ 1.
In conclusion, the condition (2.3) gives the lower bound for realizing the inequality from
Theorem 2.2: we take h = 1 so n > 1471. Verification using the computer allows us to take
n ≥ 1.
Theorem 2.3. There exists c2 > 0 such that
3
1
ζ(2/3) √
ζ(3/2) √
3
vn = n +
n+
n + O exp(−c2 log 5 n(log log n)− 5 .
ζ(3)
ζ(2)
Proof. We have C(x) = [x] − K(x), and put x = vn . It results that n = vn − K(vn ); we use
Theorem C to evaluate K, and obtain
1
√
√
3
6
n = vn − c vn − b vn + O n g(n) ,
−1
3
where c = ζ(3/2)/ζ(3), b = ζ(2/3)/ζ(2) and g(n) = exp −c2 (log n) 5 (log log n) 5 with
c2 > 0 and g(n) → ∞ as n → ∞.
So
1
√
√
(2.4)
−n + vn − c vn − b 3 vn = O n 6 g(n) .
From Theorem 2.1 and 2.2 we have
√
√
√
√
n + c n − 1.83522 3 n < vn < n + c n − 3 n,
therefore
(2.5)
It is known that
√
√
vn = n + c n − xn 3 n,
√
1+x=1+
J. Inequal. Pure and Appl. Math., 9(1) (2008), Art. 29, 8 pp.
with (xn )n≥1 bounded.
x x2
−
+ ···
2
8
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P ROPERTIES OF N ON P OWERFUL N UMBERS
and
√
3
1+x=1+
5
x x2
−
+ ··· .
3
9
Therefore
√
vn =
=
√
√
r
xn
c
1+ √ − √
3
n
n2
n
1
c
xn
n 1+ √ − √
−
3
2
8
2 n 2 n
so
√
(2.6)
vn =
√
c
xn
√ −√
3
n
n2
!
2
+ ···
,
2
√ c
xn
c
xn
n
√ −√
+ ··· .
n+ − √
−
3
8
2 26n
n
n2
In a similar manner, we get
2
√
3
c
n
c
xn
xn
√ −√
(2.7)
vn = n + √
− √ −
+ ··· .
3
9
36n 33n
n
n2
From (2.4), (2.6) and (2.7) it results that
2
√ √
√
√
c2
cxn
c
c n
xn
3
√ −√
c n − xn n − c n − + √
+
3
2
8
26n
n
n2
2
√
1
√
bc
bxn
b3n
c
xn
3
6
√ −√
+ √ +
+ · · · = O n g(n) .
−b n− √
3
9
36n 33n
n
n2
Therefore
1
√
− 3 n(xn + b) = O n 6 g(n) ,
√
3
√
3
which yields
xn = −b + O
(2.8)
g(n)
√
6
n
.
From (2.5) and (2.8) we obtain
√
√
√ vn = n + c n + b 3 n + O g(n) 3 n .
In conclusion, there exists c2 > 0 such that
√
√
3
1
3
5
5
vn = n + c n + b n + O exp(−c2 log n(log log n) ) .
3. S OME P ROPERTIES OF THE S EQUENCE OF N ON P OWERFUL N UMBERS
In relation to the prime number distribution function, E. Landau [4] proved in 1909 that
π(2x) < 2π(x)
for x ≥ x0 .
Afterwards J.B. Rosser and L. Schoenfeld proved [6] that
π(2x) < 2π(x) for all x > 2.
In relation to this problem we can state the following result.
Theorem 3.1. We have the relation
(3.1)
C(2x) ≥ 2C(x)
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for all integers x ≥ 7.
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6
V LAD C OPIL AND L AUREN ŢIU PANAITOPOL
Proof. Using Theorem B we obtain:
√
√
√
√
[x] − c x + 1.207684 3 x ≤ C(x) ≤ [x] − c x + 1.83522 3 x,
for x ≥ 961.
In order to prove (3.1) it is therefore sufficient to show that
√
√
√
√
3
[2x] − c 2x + 1.207864 2x ≥ 2[x] − 2c x + 3.67044 3 x.
As [2x] ≥ 2[x], it is sufficient to show that
√
√
√
c x(2 − 2) > 2.14885307 3 x,
√
which is true if 6 x ≥ 1.687939, more precisely for x ≥ 24. Verifications done using the
computer show that Theorem 3.1 is true for every integer 8 ≤ x ≤ 961, which concludes our
proof.
Remark 3.2. From Theorem 3.1 it follows that vn+1 < 2vn for every n ≥ 1.
The Mandl inequality [2] states that, for n ≥ 9
1
p1 + p2 + . . . + pn < npn ,
2
where pn is the n-the prime.
Related to this inequality, we prove that for non powerful numbers
Theorem 3.3. We have for n ≥ 7 that
1
v1 + v2 + . . . + vn > nvn .
2
(3.2)
Proof. Let n > C(961) + 1 = 912. In order to evaluate the sum
n
P
vi , we use Theorem A with
i=1
h(t) = t, g(t) = 1 and a = 961. It follows that G(x) = C(x) − C(961) and then we obtain
Z vn
n
X
vi = vn (n − C(961)) −
(C(t) − C(961))dt.
961
i=C(961)+1
Then
n
X
C(961)
vi =
X
i=1
Z
vn
vi + nvn − vn C(961) −
C(t)dt + C(961)(vn − 961).
961
i=1
Using Theorem B, we get a better upper bound for k 0 (x), namely
√
√
k 0 (x) ≤ x − c x + 1.83522 3 x for x ≥ 961.
Therefore, it is enough to prove that
C(961)
X
Z
vn
vi + nvn − 961C(961) −
961
i=1
√
√
nvn
3
t + c t + 1.83522 t dt >
.
2
Integrating and making some further numerical calculus (C(961) = 911,
911
P
vi = 445213)
i=1
lead us to
vn
√
n vn 2c √
3
−
+
vn − · 1.83522 3 vn
2
2
3
4
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P ROPERTIES OF N ON P OWERFUL N UMBERS
7
So, in order to prove (3.2), it is enough to prove that
√
n vn 2c √
3
−
+
vn − · 1.83522 3 vn > 0.
2
2
3
4
This is equivalent with proving that
√
4c √
3
vn < n +
vn − · 1.83522 3 vn .
3
2
Taking into account Theorem 2.2 and the fact that for n > C(961) + 1 we have n < vn < 2n,
it is enough to prove that
√
√
√
√
4c √
3
3
n+c n− 3n<n+
n − · 1.83522 · 2 · 3 n,
2
3
which is true for n ≥ 1565.
Verifications done with the computer, lead us to state that the theorem is true for every n ≥ 1,
excepting the case n = 7.
The well known conjecture of Goldbach states that every even number is the sum of two odd
primes. Related to this problem, Chen Jing-Run has shown [1] using the Large Sieve, that all
large enough even numbers are the sum of a prime and the product of at most two primes.
We present a weaker result, that has the advantage that is easily obtained and the proof is true
for every integer n ≥ 3.
Theorem 3.4. Every integer n ≥ 3 is the sum between a prime and a non powerful number.
Proof. Let n ≥ 3 and pi the largest prime that does not exceed n. Thus pi < n ≤ pi+1 and
(
π(n) − 1, if n is prime,
i=
π(n), otherwise
Then we consider the numbers n − p1 , n − p2 , . . . , n − pi . We prove that one of these i numbers
is non powerful.
Suppose that all these i numbers are powerful. It results that
√
c n − 2 ≥ k(n − 2) ≥ i ≥ π(n) − 1.
Taking into account that π(x) > logx x for x ≥ 59, we obtain
√
n
c n−2≥
− 1 for n ≥ 59.
log n
√
√
For n ≥ 4 we have c n − 2 > 2 n − 1, therefore it is enough to prove that
√
n
2 n≥
.
log n
√
But for n ≥ 75 we have 2 log n < n.
Therefore the supposition we made (that n − p1 , n − p2 , . . . , n − pi are all powerful) is
certainly false for n ≥ 75 and it results that every integer greater than 75 is the sum between a
prime and a non powerful number. Direct computation leads us to state that every integer n ≥ 3
is the sum between a prime and a non powerful number.
In 1830, H. F. Scherk found that
p2n = 1 ± p1 ± p2 ± . . . ± p2n−2 + p2n−1
and
p2n+1 = 1 ± p1 ± p2 ± · · · ± p2n−1 + 2p2n .
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8
V LAD C OPIL AND L AUREN ŢIU PANAITOPOL
The proof of these relations was first given by S. Pillai in 1928. W. Sierpinski gave a proof of
Scherk’s formulae in 1952, [7].
In relation to Scherk’s formulae, we have the following.
Theorem 3.5. For n ≥ 6, we have
vn = ±εn ± v1 ± v2 ± . . . ± vn−2 + vn−1
where εn is 0 or 1.
Proof. Following the method Sierpinski used in [7], we make an induction proof of this theorem.
If n = 6, we have v6 = 10 and
1 = −2 − 3 + 5 − 6 + 7,
2 = 1 − 2 − 3 + 5 − 6 + 7,
3 = −2 − 3 − 5 + 6 + 7,
4 = 1 − 2 − 3 − 5 + 6 + 7,
5=
2 − 3 + 5 − 6 + 7,
6 = 1 + 2 − 3 + 5 − 6 + 7,
7=
2 − 3 − 5 + 6 + 7,
8 = 1 + 2 − 3 − 5 + 6 + 7,
9 = −2 + 3 − 5 + 6 + 7,
10 = 1 − 2 + 3 − 5 + 6 + 7.
Therefore every natural number less than or equal to 10 can be expressed in the desired form.
We suppose the theorem is true for n and prove it for n + 1.
Let k be a positive integer less than or equal to vn+1 . Then, because vi+1 < 2vi for every
natural number i, we have
k ≤ vn+1 < 2vn ,
so
−vn < k − vn < vn .
It follows that 0 ≤ ±(k − vn ) < vn ; we can apply the induction hypothesis and write
±(k − vn ) = ±εn ± v1 ± v2 ± . . . ± vn−2 + vn−1 . It will immediately follow that there exist a
choice of the signs + and − such that
k = ±εn ± v1 ± v2 ± . . . ± vn−1 + vn .
As vn ≤ vn+1 , we get
vn = ±εn ± v1 ± v2 ± . . . ± vn−2 + vn−1 .
R EFERENCES
[1] JING-RUN CHEN, On the representation of a large even number as the sum of a prime and the
product of at most two primes, Sci. Sinica, 16 (1973), 157–176.
[2] P. DUSART, Sharper bounds for ψ, θ, π, pk , Rapport de Recherche, 1998.
[3] S.W. GOLOMB, Powerful numbers, Amer. Math. Monthly, 77 (1970), 848–852.
[4] E. LANDAU, Handbuch, Band I. Leipzig, (1909)
[5] G. MINCU AND L. PANAITOPOL, More about powerful numbers, in press.
[6] J.B. ROSSER AND L. SCHOENFELD, Abstracts of scientific communications, Intern. Congr. Math.
Moscow, (1966).
[7] W. SIERPINSKI, Elementary Theory of Numbers, Warszawa, P.W.N., (1964).
J. Inequal. Pure and Appl. Math., 9(1) (2008), Art. 29, 8 pp.
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