A talk given at Dalian Maritime Univ. (Nov. 17, 2012)
Conjectures involving arithmetical sequences
Zhi-Wei Sun
Nanjing University
Nanjing 210093, P. R. China
[email protected]
http://math.nju.edu.cn/∼zwsun
Nov. 17, 2012
Abstract
A sequence (an )n>0 of natural numbers is said to be log-concave
2
2
(resp. log-convex) if an+1
> an an+2 (resp. an+1
6 an an+2 ) for all
n = 0, 1, 2, . . .. The log-concavity or log-convexity of combinatorial
sequences has been studied extensively by many authors.
During August-September 2012, the speaker formulated many
√
√
√
conjectures on monotonicity of ( n an )n>1 or ( n+1 an+1 / n an )n>1 for
various number-theoretic or combinatorial sequences (an )n>1 of
positive integers. In this talk we give an introduction to those
conjectures and related progress.
2 / 32
Part I. Conjectures on number-theoretic sequences
3 / 32
Firoozbakht’s Conjecture
In 1982 Faride Firoozbakht (from Iran) posed the following
challenging conjecture while he was studying a proof of the Prime
Number Theorem.
√
√
Firoozbakht’s Conjecture. n pn > n+1 pn+1 for all n = 1, 2, . . .,
√
i.e., the sequence ( n pn )n>1 is strictly decreasing
Remark. The conjecture implies that
pn+1 − pn < log2 pn − log pn + 1
for all n > 4, which is stronger than Cramer’s conjecture that
c := lim sup(pn+1 − pn )/ log2 pn coincides with 1 (A. Granville
thought that c should be 2/e γ ≈ 1.122918.)
Verification record. Verified for all primes less than 4 × 1018 .
√
Comments. Since log n pn ∼ (log n)/n, the conjecture seems
reasonable. I saw it few years ago but soon forgot the reference
and the proposer’s long name.
4 / 32
A refinement of Firoozbakht’s Conjecture
By the Prime Number Theorem, pn ∼ n log n. Note that
p
n+1
(n + 1) log(n + 1)
1
log n log log n
√
−
+
O
.
log
=
−
n
n2
n2
n2
n log n
This led me to make the following conjecture.
Conjecture (Sun, 2012-09-11) For any integer n > 4, we have the
inequality
√
n+1 p
log log n
n+1
<
1
−
.
√
n p
2n2
n
Remark. We have verified the conjecture for all n 6 3500000 and
all those n with pn < 4 × 1018 and pn+1 − pn 6= pk+1 − pk for all
1 6 k < n. If n = 49749629143526, then pn+1 − pn = 1132,
pn = 1693182318746371, and
√
√
(1 − n+1 pn+1 / n pn )n2 / log log n ≈ 0.5229.
5 / 32
Some easy facts
Easy things:
√
( n n)n>3 is strictly decreasing,
√
√
( n+1 n + 1/ n n)n>4 is strictly increasing.
Reason: For the function f (x) = (log x)/x on the interval
[4.5, +∞), we have
f 0 (x) =
1 − log x
2 log x − 3
< 0 and f 00 (x) =
> 0,
2
x
x3
and hence f (x) is strictly decreasing and strictly convex.
For
√
n
Pn = p1 p2 · · · pn and Sn = p1 + p2 + · · · + pn ,
( Pn )n>1 and (Sn /n)n>1 are strictly increasing (easy to prove).
√
n
Pn —— the geometric mean of p1 , p2 , . . . , pn .
Sn /n —— the arithmetic mean of p1 , p2 , . . . , pn .
6 / 32
Monotonicity related to Sn =
Pn
k=1 pk
p
√
Theorem (Sun). (i) (July 28-31) ( n Sn )n>2 and ( n Sn /n)n>1 are
strictly decreasing.
(ii) (Discovered on July 29 and proved on August 25)
!
!
p
p
n+1
n+1
Sn+1
Sn+1 /(n + 1)
√
p
and
n
n
Sn
Sn /n
n>5
n>10
are strictly increasing.
7 / 32
A general theorem
(α)
For Sn
=
Pn
α
k=1 pk ,
we have
Theorem (Sun, Bull. Aust. Math. Soc., in press). Let α > 1.
α
(i) If n > max{100, e 2×1.348 +1 }, then
s
s
(α)
(α)
n+1 S
n Sn
n+1
>
n
n+1
and hence
q
n
(α)
Sn >
q
(α)
Sn+1 .
n+1
(ii) The sequence
q
q
n
n+1
(α)
(α)
Sn+1 /(n + 1)
Sn /n
n>N(α)
is strictly increasing, where
o
n
2
2α+1 +(α+1)1.2α+1 )/α
N(α) = max 350000, de ((α+1) 1.2
e .
8 / 32
The cases α = 2, 3, 4
Corollary. The sequences
q
q
n
n+1
(2)
(2)
Sn+1
Sn
,
n>10
q
q
n
n+1
(3)
(3)
Sn+1
Sn
,
n>10
q
n+1
(4)
Sn+1
q
n
(4)
Sn
n>17
are all strictly increasing.
9 / 32
On squarefree numbers
A positive integer n is called squarefree if p 2 - n for any prime p.
Here is the list of all squarefree positive integers not exceeding 30
in alphabetical order:
1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 17, 19, 21, 22, 23, 26, 29, 30.
Conjecture (Sun, 2012-08-14) (i) For n = 1, 2, 3, . . . let sn be the
√
n-th squarefree positive integer. Then the sequence ( n sn )n>7 is
strictly decreasing.
(ii) For n = 1, 2, 3, . . . let S(n) be the sum
n squarefree
p of the first
p
n+1
n
positive integers. Then the sequence (
S(n + 1)/ S(n))n>7 is
strictly increasing.
Remark. I have checked both parts of the conjecture via
√
√
Mathematica; for example, n sn > n+1 sn+1 for all
p
n = 7, . . . , 500000. Note that limn→∞ n S(n) = 1 since S(n) does
not exceed the sum of the first n primes.
10 / 32
Conjecture on primitive roots modulo primes
Conjecture (Sun, 2012-08-17). Let a ∈ Z be not a perfect power
(i.e., there are no integers m > 1 and x with x m = a).
(i) Assume that a > 0. Then there are infinitely many primes p
having a as the smallest positive primitive root modulo p.
Moreover, if p1 (a), . . . , pn (a) are the first n such primes, then the
next such prime pn+1 (a) is smaller than pn (a)1+1/n .
(ii) Suppose that a < 0. Then there are infinitely many primes p
having a as the largest negative primitive root modulo p.
Moreover, if p1 (a), . . . , pn (a) are the first n such primes, then the
next such prime pn+1 (a) is smaller than pn (a)1+1/n with the only
exception a = −2 and n = 13.
p
p
(a)/ n Pn (a))n>3 is strictly increasing
(iii) The sequence ( n+1 Pn+1P
with limit 1, where Pn (a) = nk=1 pk (a).
Remark. The first 5 primes having 24 as the smallest positive
primitive root are
533821, 567631, 672181, 843781, 1035301.
11 / 32
Conjecture on twin primes
It is conjectured that there are infinitely many twin primes.
Conjecture (Sun, 2012-08-18) (i) If {t1 , t1 + 2}, . . . , {tn , tn + 2}
are the first n pairs of twin primes, then the first prime tn+1 in the
√
√
1+1/n
next pair of twin primes is smaller than tn
, i.e., n tn > n+1 tn+1 .
p
p
(ii) The sequence ( n+1 T (n P
+ 1)/ n T (n))n>9 is strictly increasing
with limit 1, where T (n) = nk=1 tk .
√
√
Remark. Via Mathematica I verified that n tn > n+1 tn+1 for all
n = 1, . . . , 500000, and
p
p
p
p
n+1
T (n + 1)/ n T (n) < n+2 T (n + 2)/ n+1 T (n + 1)
for all n = 9, 10, . . . , 500000. Note that t500000 = 115438667.
After I made the conjecture public, Marek Wolf verified the
√
√
inequality n tn > n+1 tn+1 for all the 44849427 pairs of twin primes
below 234 ≈ 1.718 × 1010 .
12 / 32
Conjecture on Sophie Germain primes
A prime p is called a Sophie Germain prime if 2p + 1 is also a
prime. It is conjectured that there are infinitely many Sophie
Germain primes, but this has not been proved yet.
Conjecture (Sun, 2012-08-18) (i) If g1 , . . . , gn are the first n
Sophie Germain primes, then the next Sophie Germain prime gn+1
√
√
1+1/n
is smaller than gn
(i.e., n gn > n+1 gn+1 ) with the only
exceptions n = 3, 4.
p
p
(ii) The sequence ( n+1 G (n P
+ 1)/ n G (n))n>13 is strictly increasing
with limit 1, where G (n) = nk=1 gk .
√
√
Remark. Via Mathematica I verified that n gn > n+1 gn+1 for all
n = 5, . . . , 200000, and
p
p
p
p
n+1
G (n + 1)/ n G (n) < n+2 G (n + 2)/ n+1 G (n + 1)
for all n = 13, 14, . . . , 200000. Note that g200000 = 42721961.
13 / 32
A general conjecture related to Hypothesis H
Schinzel’s Hypothesis H. If f1 (x), . . . , fk (x) are irreducible
polynomials with integer coefficients and positive leading
coefficients such that there is no prime dividing the product
f1 (q)f2 (q)...fk (q) for all q ∈ Z, then there are infinitely many
n ∈ Z+ such that f1 (n), f2 (n), . . . , fk (n) are all primes.
General Conjecture (Sun, 2012-09-08) Let f1 (x), . . . , fk (x) be
irreducible polynomials with integer coefficients and positive
Q
leading coefficients such that there is no prime dividing kj=1 fj (q)
for all q ∈ Z. Let q1 , q2 , . . . be the list (in ascending order) of
those q ∈ Z+ such that f1 (q), . . . , fk (q) are all primes. Then, for
all sufficiently large positive integers n, we have
1+1/n
qn+1 < qn
, i.e.,
√
n
qn >
√
n+1
qn+1 .
Also,
positive integer N such that the sequence
p there is a p
( n+1 Q(n + 1)/Pn Q(n))n>N is strictly increasing with limit 1,
where Q(n) = nk=1 qk .
14 / 32
Conjecture on Proth primes
Proth numbers: k × 2n + 1 with k odd and 0 < k < 2n .
F. Proth (1878): A Proth number p is a prime if (and only if)
a(p−1)/2 ≡ −1 (mod p) for some integer a.
A Proth prime is a Proth number which is also a prime number;
the Fermat primes are a special kind of Proth primes.
Conjecture (Sun, 2012-09-07) (i) The number of Proth primes
not exceeding a large integer x is asymptotically equivalent to
√
c x/ log x for a suitable constant c ∈ (3, 4).
(ii) If Pr(1), . . . , Pr(n) are thep
first n Prothp
primes, then
Pr(n + 1) < Pr(n)1+1/n (i.e., nPPr(n) > n+1 Pr(n + 1)) unless
n =p2, 4, 5. If we p
set PR(n) = nk=1 Pr(k), then the sequence
( n+1 PR(n + 1)/ n PR(n))n>34 is strictly increasing with limit 1.
Remark. I have checked the conjecture for the first 4000 Proth
primes.
15 / 32
On irreducible polynomials over finite fields
Let q > 1 be a prime power and let Fq denote the finite field of
order q. For n = 1, 2, 3, . . . let Nn (q) denote the number of monic
irreducible polynomials of degree n over Fq .
Theorem (Sun, October 2012)
(i) The sequence (Nn+1 (q)/Nn (q))n>1 is strictly increasing if
q > 9, and (Nn+1 (q)/Nn (q))n>19 is strictly increasing if q < 9.
p
(ii) The sequence ( n Nn (q))n>e 3+7/(q−1)2 is strictly increasing, and
the sequence
p
p
( n+1 Nn+1 (q)/ n Nn (q))n>5.835×1014
is strictly decreasing.
16 / 32
On partitions of integers
A partition of a positive integer n is a way of writing n as a sum of
positive integers with the order of addends ignored. Let p(n)
denote the number of partitions of n. It is known that
√
e π 2n/3
(Hardy and Ramanujan)
p(n) ∼ √
4 3n
p
and hence limn→∞ n p(n) = 1.
p
n
Conjecture (Sun, 2012-08-02) The
sequence
(
p(n))n>6 is
p
p
n+1
n
strictly decreasing. Moreover, (
p(n + 1)/ p(n))n>26 is
strictly increasing.
Remark. I have verified the conjecture for n up to 105 . The
log-concavity of (p(n))n>25 was conjectured by W.Y.C. Chen in
August 2010 and proved by J.E. Janoski in his PhD thesis who said
that he began the project in the summer of 2010.
17 / 32
On strict partitions of integers
A strict partition of n ∈ Z+ is a way of writing n as a sum of
distinct (or odd) positive integers with the order of addends
ignored. For n = 1, 2, 3, . . . we denote by p∗ (n) the number of
strict partitions of n. It is known that
√
e π n/3
as n → +∞
p∗ (n) ∼
4(3n3 )1/4
and hence
lim
n→∞
p
n
p∗ (n) = 1.
Conjecture
(Sun, 2012-08-02) (p∗ (n + 1)/p∗ (n))n>32 and
p
n
( p
p∗ (n))n>9 arepstrictly decreasing. Furthermore, the sequence
( n+1 p∗ (n + 1)/ n p∗ (n))n>45 is strictly increasing.
Remark. I have verified the conjecture for n up to 105 .
18 / 32
On harmonic numbers of order m
Harmonic numbers: Hn =
P
1/k (n = 0, 1, 2, . . .).
P
(m)
Harmonic numbers of order m: Hn = 0<k6n 1/k m .
q
n
(m)
It is easy to show that ( Hn )n>2 is strictly decreasing for any
positive integer m.
0<k6n
Conjecture (Sun, 2012-08-12) For any positive integer m, the
sequence
q
q
n+1
(m) n
(m)
Hn+1 / Hn
n>3
is strictly increasing.
This conjecture was confirmed in a joint paper with Qing-Hu Hou
and Haomin Wen (arXiv:1208.3903).
19 / 32
Bernoulli numbers, Euler numbers and tangent numbers
Bernoulli numbers are given by
∞
X xn
x
=
Bn
(|x| < 2π).
ex − 1
n!
n=0
It is known that (−1)n−1 B2n > 0 and B2n+1 = 0 for all n > 0.
Euler numbers are given by
∞
X xn
2
=
En .
e x + e −x
n!
n=0
(−1)n E
It is known that
2n > 0 and E2n+1 = 0 for all
n = 0, 1, 2, . . ..
Tangent numbers are given by
tan x =
∞
X
n=1
It is known that T (n) ∈
Z+
T (n)
x 2n−1
.
(2n − 1)!
for all n = 1, 2, 3, . . ..
20 / 32
Bernoulli numbers, Euler numbers and tangent numbers
Conjecture (Sun, 2012-08-02) The sequences
q
p
p
( n (−1)n−1 B2n )n>1 , ( n (−1)n E2n )n>1 , ( n T (n))n>1
are strictly increasing, and the sequences
q
p
n+2
n
(
(−1) B2n+2 / n (−1)n−1 B2n )n>2 ,
q
p
n+1
(
(−1)n+1 E2n+2 / n (−1)n E2n )n>1 ,
p
p
( n+1 T (n + 1)/ n T (n))n>2
are strictly decreasing.
This conjecture was confirmed in a preprint by Florian Luca and
Pantelimon Stǎnicǎ (arXiv:1208.5151).
21 / 32
Part II. Conjectures on combinatorial sequences
22 / 32
On the Fibonacci sequence
The Fibonacci sequence (Fn )n>0 is given by
F0 = 0, F1 = 1, and Fn+1 = Fn + Fn−1 (n = 1, 2, 3, . . .).
√
n
Conjecture (2012-08-11) The sequence ( p
Fn )n>2√is strictly
increasing, and moreover the sequence ( n+1 Fn+1 / n Fn )n>4 is
strictly decreasing.
This was confirmed in a joint paper with Qing-Hu Hou and
Haomin Wen (arXiv:1208.3903). Actually the paper contains a
result on general Lucas sequences.
23 / 32
On derangement numbers
The nth derangement number
Dn = |{σ ∈ Sn : σ(i) 6= i for all i = 1, . . . , k}|.
P
It is known that Dn /n! = nk=0 (−1)k /k!.
√
Conjecture (Sun, 2012-08-11) The sequence
( n Dn√)n>2 is strictly
p
increasing, and moreover the sequence ( n+1 Dn+1 / n Dn )n>3 is
strictly decreasing.
This was confirmed in a joint paper with Qing-Hu Hou and
Haomin Wen (arXiv:1208.3903).
24 / 32
On Bell numbers
The n-th Bell number Bn denotes the number of partitions of
{1, . . . , n} into disjoint nonempty subsets. It is known that
Bn+1
n X
n
=
Bk (with B0 = 1)
k
k=0
and
∞
Bn =
1 X kn
.
e
k!
k=0
√
Conjecture (Sun, 2012-08-11) The sequence
( n Bn√)n>1 is strictly
p
increasing, and moreover the sequence ( n+1 Bn+1 / n Bn )n>1 is
strictly decreasing with limit 1, where Bn is the n-th Bell number.
Remark. In 1994 K. Engel proved the log-convexity of (Bn )n>1 .
The above conjecture is still open and seems very challenging!
25 / 32
On Springer numbers
Springer numbers are given by
∞
X xn
1
=
Sn .
cos x − sin x
n!
n=0
It is known that Sn equals the numerator of |En (1/4)|, where
En (x) is the Euler polynomial of degree n.
Conjecture (Sun, 2012-08-05) The sequence
(S√
n+1 /Sn )n>0 is
p
strictly increasing, and the sequence ( n+1 Sn+1 / n Sn )n>1 is strictly
decreasing with limit 1, where Sn is the n-th Springer number.
26 / 32
On central trinomial coefficients
The n-th central trinomial coefficient Tn is the coefficient of x n in
the expansion of (x 2 + x + 1)n . Here is an explicit expression:
Tn =
n X
n
n−k
k=0
k
k
bn/2c =
X
k=0
n
2k
2k
.
k
In combinatorics, Tn is the number of lattice paths from the point
(0, 0) to (n, 0) with only allowed steps (1, 0), (1, 1) and (1, −1). It
is known that
(n + 1)Tn+1 = (2n + 1)Tn + 3nTn−1 .
√
Conjecture (2012-08-11) The sequence
(√n Tn )n>1 is strictly
p
increasing, and the sequence ( n+1 Tn+1 / n Tn )n>1 is strictly
decreasing.
F. Luca and P. Stǎnicǎ confirmed this for large n.
27 / 32
On Motzkin numbers
The n-th Motzkin number
bn/2c Mn =
X
k=0
n
2k
2k
1
k k +1
is the number of lattice paths from (0, 0) to (n, 0) which never dip
below the line y = 0 and are made up only of the allowed steps
(1, 0), (1, 1) and (1, −1). It is known that
(n + 3)Mn+1 = (2n + 3)Mn + 3nMn−1 .
√
Conjecture (Sun, 2012-08-11) The sequence
( n M√
n )n>1 is strictly
p
increasing, and moreover the sequence ( n+1 Mn+1 / n Mn )n>1 is
strictly decreasing.
Remark. The log-convexity of the sequence (Mn )n>1 was first
established by M. Aigner in 1998. F. Luca and P. Stǎnicǎ
confirmed the conjecture for large n.
28 / 32
On Schröder numbers
The n-th Schröder number
n n X
X
n
n+k
1
n+k
2k
1
Sn =
=
k
k
k +1
2k
k k +1
k=0
k=0
is the number of lattice paths from the point (0, 0) to (n, n) with
steps (1, 0), (0, 1) and (1, 1) that never rise above the line y = x
√
Conjecture (Sun, 2012-08-11) The sequence
( n S√
n )n>1 is strictly
p
increasing, and moreover the sequence ( n+1 Sn+1 / n Sn )n>1 is
strictly decreasing, where Sn stands for the n-th Schröder number.
29 / 32
On Domb numbers and Catalan-Larcombe-French numbers
Conjecture (Sun, 2012-08-13) For the Domb numbers
n 2 X
n
2k
2(n − k)
D(n) =
(n = 0, 1, 2, . . .),
k
k
n−k
k=0
p
the sequences (D(n + 1)/D(n))n>0 andp( n D(n))n>1
pare strictly
increasing. Moreover, the sequence ( n+1 D(n + 1)/ n D(n))n>1 is
strictly decreasing.
The Catalan-Larcombe-French numbers P0 , P1 , P2 , . . . are given by
bn/2c 2
2k 2 2(n−k) 2
n
X n
X
2k
k
n−k
n
=2
4n−2k ,
Pn =
n
2k
k
k
k=0
k=0
they arose from the theory of elliptic integrals. It is known that
(n + 1)Pn+1 = (24n(n + 1) + 8)Pn − 128n2 Pn−1 for all n ∈ Z+ .
Conjecture
(Sun, 2012-08-14) The sequences (Pn+1 /Pn )n>0 and
√
n
( p
Pn )n>1 are
√ strictly increasing. Moreover, the sequence
( n+1 Pn+1 / n Pn )n>1 is strictly decreasing.
30 / 32
On Franel numbers and Apéry numbers
Franel numbers:
fn :=
n 3
X
n
k
k=0
(n = 0, 1, 2, . . .).
Apéry numbers:
An :=
n 2 X
n
n+k 2
k=0
k
k
(n = 0, 1, 2, . . .).
Conjecture (Sun, 2012-08-11) Both
p p
n+1
fn+1 / n fn n>1
and
p p
An+1 / n An n>1
n+1
are strictly decreasing with limit 1.
F. Luca and P. Stǎnicǎ confirmed the conjecture for large n.
31 / 32
My above conjectures appeared in the following paper:
Z. W. Sun, Conjectures involving arithmetical sequences, in:
Arithmetic in Shangri-La (eds., S. Kanemitsu, H. Li and J. Liu),
Proc. 6th China-Japan Seminar (Shanghai, August 15-17, 2011),
World Sci., Singapore, 2013, pp. 244–258.
You are welcome to solve my
conjectures!
Thank you!
32 / 32