Int. J. Contemp. Math. Sciences, Vol. 8, 2013, no. 17, 815 - 819 HIKARI Ltd, www.m-hikari.com http://dx.doi.org/10.12988/ijcms.2013.3790
A Conjecture on the Difference between Consecutive Perfect Powers Rafael Jakimczuk Divisi´on Matem´atica, Universidad Nacional de Luj´an Buenos Aires, Argentina
[email protected] c 2013 Rafael Jakimczuk. This is an open access article distributed under Copyright the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Abstract Let Pn be the n-th perfect power and dn the difference Pn+1 − Pn . The Pillai’s conjecture establish the following limit limn→∞ dn = ∞. This is an unsolved problem. In this article we establish a more strong conjecture on dn .
Mathematics Subject Classification: 11A99, 11B99 Keywords: Perfect powers, conjectures
1
Introduction
A natural number of the form mn where m is a positive integer and n ≥ 2 is called a perfect power. The first few terms of the integer sequence of perfect powers are 1, 4, 8, 9, 16, 25, 27, 32, 36, 49, 64, 81, 100, 121, 125, 128 . . . In this article Pn denotes the n-th perfect power and dn denotes the difference Pn+1 − Pn . There exists an important conjecture on dn , the Pillai’s conjecture. This conjecture establish the following limit limn→∞ dn = ∞. This is an unsolved problem. In this article we shall establish a more strong conjecture.
R. Jakimczuk
816
A square-free number is a number without square factors, a product of different primes. The first few terms of the integer sequence of square-free numbers are 1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 17, 19, 21, 22, 23, 26, 29, 30, . . . On the other hand, the M¨obius function μ(n) is defined as follows: μ(1) = 1, if n is the product of r different primes, then μ(n) = (−1)r , if the square of a prime divides n, then μ(n) = 0. Jakimczuk [1] proved the following theorem. Theorem 1.1 Let ph be the h-th prime with h ≥ 3, where h is an arbitrary but fixed positive integer. We have 13 8/6 32 32/30 1+ p2 1+ 2q h + 2μ(q)n +o n Pn = n + n + n 3 15 2≤q≤ph 2
(1)
q=2,6,30
where q denotes a square-free number. Note that 2 = 1 + 22 , 32 2 = 1 + 30 . 30
8 6
= 1+
2 6
and
For example, if h = 4 then equation (1) becomes 5
7
Pn = n2 − 2n 3 − 2n 5 +
2
9 9 13 4 n 3 − 2n 7 + o n 7 . 3
The Conjecture
Let ph be the h-th prime with h ≥ 3, where h is an arbitrary but fixed positive integer. We establish the following conjecture. 2 52 2/6 512 2/30 2(q + 2) 2q dn = 2n + n + n n + o n ph + μ(q) 9 225 q 2≤q≤ph q=2,6,30
Note that is conjecture implies that dn ∼ 2n. For example, if h = 4 then equation (2) becomes dn = 2n −
2 10 2 14 2 52 2 18 2 n3 − n5 + n6 − n7 + o n7 . 3 5 9 7
The conjecture is possible since we have the following theorem. Theorem 2.1 Equation (2) implies equation (1).
(2)
A conjecture on perfect powers
817
Proof. Equation (2) can be written in the following schematic form di = 2i +
m
dj igj −
j=1
2 2(ph + 2) p2 i h + f (i)i ph ph
2 . ph
where f (i) → 0 and 1 > g1 > · · · > gm > n−1
n−1
m
n−1
(i ≥ 1),
Consequently
n−1 2 2 2(ph + 2) n−1 ph Pn = 1+ di = 1+2 i+ dj i − i + f (i)i ph . (3) ph i=1 i=1 j=1 i=1 i=1 i=1 gj
If α = 1, g1 , . . . , gm , p2h we have n−1
α
n
i =
i=1
0
nα+1 nα+1 1+ 2 + O (nα ) = + o n ph . x dx + O (n ) = α+1 α+1 α
α
(4)
Note that in equation (4) the left side is a sum of areas of rectangles of basis 1 and height iα . Therefore, the integral aproximates this sum with error O (nα ) since xα is a function strictly increasing. There exist n0 such that if i ≥ n0 we have |f (i)| < . Consequently (see (4)) n−1 2 f (i)i ph
≤
n−1
i=1
≤
n 0 −1
i=1
2
|f (i)| i ph +
i=1 1+ p2
+ o n
h
n 0 −1
2
|f (i)| i ph ≤
i=1
n−1
2
i ph =
i=1
2
|f (i)| i ph +
n 0 −1
n−1
2
i ph
i=n0 2
|f (i)| i ph +
i=1
1+ p2
n 1+
h
2 ph
.
Therefore, from a certain value of n we have 2 n−1 ph i=1 f (i)i
≤ .
1+ p2
n That is
n−1
f (i)i
2 ph
h
1+ p2
=o n
h
.
(5)
i=1
Finally, substituting equation (4) and equation (5) into (3) we obtain equation (1). The theorem is proved. We also have the following theorem
R. Jakimczuk
818 Theorem 2.2 The conjecture dn ∼ 2n.
(6)
is equivalent to the following limit
lim
n→∞
Pn+1 Pn
n2
= e.
(7)
Proof. Equation (1) implies that Pn ∼ n2
(8)
Equation (8) implies that Pn+1 ∼ Pn . Consequently lim
n→∞
Pn Pn = lim = lim n→∞ dn Pn+1 − Pn n→∞
1 Pn+1 Pn
−1
=∞
(9)
On the other hand, we have
Pn+1 Pn
n2
=
Pn + d n Pn
Now (see (9))
n 2
=
nd ⎛⎛ ⎞ Pn ⎞ 2Pnn dn 1 ⎜⎝ ⎟ ⎝ 1 + Pn ⎠ ⎠
(10)
dn
⎛
⎞ Pn
1 lim ⎝1 + Pn ⎠
n→∞
dn
=e
(11)
dn
If dn ∼ 2n then (see (8)) we have lim n→∞
ndn =1 2Pn
(12)
Equations (10), (11) and (12) give (7). Now, suppose that (7) holds. That is (see (10))
lim n→∞
Pn+1 Pn
n2
=
nd ⎛⎛ ⎞ Pn ⎞ 2Pnn dn 1 ⎜⎝ ⎟ lim ⎝ 1 + Pn ⎠ ⎠ n→∞
=e
(13)
dn
Therefore (see (11) and (13)) we have lim
n→∞
ndn =1 2Pn
Equations (14) and (8) give (6). The theorem is proved.
(14)
A conjecture on perfect powers
819
Theorem 2.3 The conjecture dn ∼ 2n is equivalent to the following limit
lim
n→∞
Pn+1 Pn
√
Pn 2
= e.
Proof. The proof is the same as theorem 2.2. Since
√
Pn ∼ n (see (8)).
ACKNOWLEDGEMENTS. The author is very grateful to Universidad Nacional de Luj´an.
References [1] R. Jakimczuk, Asymptotic formulae for the n-th perfect power, J. Integer Seq. 15 (2012), Article 12.5.5. Received: July 11, 2013
