Various New Observations About Primes

A talk given at Univ. of Illinois at Urbana-Champaign (August 28, 2012) and the 6th National Number Theory Confer. (October 20, 2012) and the China-Korea Number Theory Seminar (October 27, 2012)

Various new observations about primes

Zhi-Wei Sun

Nanjing University Nanjing 210093, P. R. China [email protected] http://math.nju.edu.cn/∼zwsun

October 27, 2012 Abstract

This talk focuses on the speaker’s recent discoveries about primes. We will talk about the speaker’s new way to generate all primes or primes in certain arithmetic progressions, and his various conjectures for products of primes, sums of primes, recurrence for primes, representations of integers as alternating sums of consecutive primes. We will also mention his new observations (mainly made in August 2012 at UIUC) about twin primes, squarefree numbers, and primitive roots modulo primes.

2 / 42 Part I. On functions taking only prime values

3 / 42 Representing primes by polynomials

Euler: x2 − x + 41 is prime for every x = 1,..., 40. Theorem (Rabinowitz, 1913). Let p > 1 be an integer and let K √ be the imaginary quadratic ﬁeld Q( 1 − 4p). Let OK be the ring of algebraic integers in K. Then x2 − x + p is a prime for all x = 1,..., p − 1 if and only if K has class number one, i.e., OK is a principal ideal domain. Theorem (conjectured by Gauss and proved by H. Stark). The only√ imaginary quadratic ﬁeld having class number one are those Q( −d) with d ∈ {1, 2, 3, 7, 11, 19, 43, 67, 163}. Thus, for any p > 41, x2 − x + p cannot take prime values for all x = 1,..., p − 1.

In general, any non-constant polynomial P(x1,..., xn) with integer coeﬃcients cannot take primes for all x1,..., xn ∈ Z.

4 / 42 Mills’ Theorem

Theorem (Mills, 1947). There is a real number A such that M(n) = bA3n c takes only prime values. 5/8 Sketch of the Proof. Since pn+1 − pn = O(pn ) (A. E. Ingham, 1937), one can construct inﬁnitely many primes P0, P1, P2,... with

3 3 Pn < Pn+1 < (Pn + 1) − 1.

3−n Then the sequence un = Pn is increasing while the sequence 3−n vn = (Pn + 1) is decreasing. As un < vn, we see that A = limn→∞ un 6 B = limn→∞ vn, hence

3n 3n 3n Pn = un < A < Pn + 1 = vn .

3n So bA c = Pn is a prime for all n = 1, 2, 3,....

5 / 42 A problem on central binomial coeﬃcients Let p be an odd prime. For k = 0,..., (p − 1)/2 we have 2k (2k)! = 6≡ 0 (mod p); k k!2 but for k = (p + 1)/2,..., p − 1 we have 2k (2k)! = ≡ 0 (mod p). k k!2 Conjecture (Sun, Feb. 20, 2012). Let p > 5 be a prime. Then 2k p − 1 ± : k = 1,..., k 2 cannot be a reduced system of residues modulo p. Moreover, if 2k p > 11 then k (k = 1,..., (p − 3)/2) cannot be pairwise distinct modulo p. Remark. Later I realized that for any positive integer m the largest n such that 2k (k = 1,..., n) are pairwise distinct modulo m √k √ should be O( m) and probably less than 4.53 m. 6 / 42 A function taking only prime values

Conjecture (Z. W. Sun, Feb. 21, 2012). For n = 1, 2, 3,... deﬁne 2k s(n) as the least integer m > 1 such that k (k = 1,..., n) are pairwise distinct modulo m. Then s(n) is always a prime! I also guessed that s(n) < n2 for n = 2, 3, 4,.... I calculated s(n) for n = 1,..., 2065. For example,

s(1) = 2, s(2) = 3, s(3) = 5, s(4) = s(5) = s(6) = 11, s(7) = s(8) = s(9) = 23, s(10) = 31, s(11) = ··· = s(14) = 43, s(15) = s(16) = s(17) = s(18) = 59, s(19) = 107, s(20) = 149.

After I made the conjecture public via a message to Number Theory List, Laurent Bartholdi computed s(n) for n = 2001,..., 5000, and his computational result supports my conjecture.

7 / 42 Artin’s conjecture

Let p be an odd prime. If a ∈ Z is not divisible by p and ak 6≡ 1 (mod p) for k = 1,..., p − 2, then we say that a is a primitive root mod p (or the order of a mod p is p − 1).

Artin’s Conjecture (1927). If a ∈ Z is neither −1 nor a square, then there are inﬁnitely many primes p having a as a primitive root modulo p (moreover, the set of such primes has a positive asymptotic density inside the set of all primes). Progress: (1) C. Hooley (1967): The conjecture follows from the Generalized Riemann Hypothesis. (2) R. Gupta & M. R. Murty (1984): The conjecture holds for inﬁnitely many a (with help of sieve methods). (3) R. Heath-Brown (1986): There are at most two exceptional primes a for which Artin’s conjecture fails.

8 / 42 A conjecture implying Artin’s conjecture

Conjecture (Sun, Feb. 22-23, 2012). Let a ∈ Z with |a| > 1. For + n ∈ Z deﬁne fa(n) as the least integer m > 1 such that those ak (k = 1,..., n) are pairwise incongruent modulo m.

(i) fa(n) is a prime for all suﬃciently large n.

(ii) If a is not a square, then for any suﬃciently large n, fa(n) is the least prime p > n having a as a primitive root mod p;

(iii) If a is a square, then for any suﬃciently large n, fa(n) is just the least prime p > 2n such that a, a2,..., a(p−1)/2 are pairwise distinct modulo p.

+ Example. f−3(n) with n ∈ Z is the least prime p > n such that −3 is a primitive root mod p. √ 2k k Motivation. By Stirling’s formula, k ∼ 4 / kπ.

9 / 42 Conjectures on Lucas sequences Let A be an integer with |A| > 2. Deﬁne

u0(A) = 0, u1(A) = 1, and un+1(A) = Aun(A) − un−1(A);

v0(A) = 2, v1(A) = A, and vn+1(A) = Avn(A) − vn−1(A). Conjecture (Sun, Feb. 26, 2012). (i) If 2 + A is not a square, then there are infinitely many odd primes p such that those vk (A) mod p with k = 1, ..., (p − εp)/2 are pairwise distinct, A2−4 where εp = ( p ). (ii) If 2 − A is not a square, then there are infinitely many odd primes p such that those uk (A) mod p with 1 6 k 6 (p − εp)/2 are pairwise distinct. + Conjecture (Sun, Feb. 26, 2012). Let n ∈ Z be sufficiently large (n > 2|A| or n > 100 may suffice). Then tA(n) (the smallest integer m > 1 such that vk (A)(k = 1, ..., (p − εp)/2) are pairwise distinct modulo m) is prime. Moreover, if A + 2 is not a square, then tA(n) is the smallest odd prime p such that p − εp > 2n and those vk (A) mod p (k = 1, ..., (p − εp)/2) are pairwise distinct. 10 / 42 Generate all primes in a combinatorial manner

+ Theorem 1 (Sun, Feb. 29, 2012) For n ∈ Z let S(n) denote the smallest integer m > 1 such that those 2k(k − 1) mod m for k = 1,..., n are pairwise distinct. Then S(n) is the least prime greater than 2n − 2.

Remark. (a) The range of S(n) is exactly the set of all primes! + (b) I also showed that for any d ∈ Z whenever n > d + 2 the + least prime p > 2n + d is just the smallest m ∈ Z such that 2k(k + d)(k = 1,..., n) are pairwise distinct modulo m. (c) I proved that the least positive integer m such that those k 2 = k(k − 1)/2 (k = 1,..., n) are pairwise distinct modulo m, is just the least power of two not smaller than n.

11 / 42 Another theorem

+ Theorem 2 (Sun, March 2012) (i) Let d ∈ {2, 3} and n ∈ Z . Then the smallest positive integer m such that those k(dk − 1) (k = 1,..., n) are pairwise distinct modulo m, is the least power of d not smaller than n. (ii) Let n ∈ {4, 5,...}. Then the least positive integer m such that 18k(3k − 1) (k = 1,..., n) are pairwise distinct modulo m, is just the least prime p > 3n with p ≡ 1 (mod 3). Remark. We are also able to prove some other similar results including the following one: + For n > 5 the least m ∈ Z such that those 18k(3k + 1) (k = 1,..., n) are pairwise distinct modulo m, is just the ﬁrst prime p ≡ −1 (mod 3) after 3n.

12 / 42 One more theorem

+ Theorem (Sun, March 2012). (i) For d, n ∈ Z let λd (n) be the smallest integer m > 1 such that those (2k − 1)d (k = 1,..., n) are pairwise incongruent modulo m. Then λd (n) with d ∈ {4, 6, 12} and n > 2 is the least prime p > 2n − 1 with p ≡ −1 (mod d). (ii) Let q be an odd prime. Then the smallest integer m > 1 such that those kq(k − 1)q with k = 1,..., n are pairwise incongruent mod m, is just the least prime p > 2n − 1 with p 6≡ 1 (mod q).

Remark. In the proof I used the Brun-Titchmarsh theorem which + asserts that if a, q ∈ Z , gcd(a, q) = 1 and x > q then 2x |{p x : p ≡ a (mod q)}| . 6 6 ϕ(q) log(x/q)

13 / 42 Previous results by others Theorem 1 (L. K. Arnold, S. J. Benkoski and B. J. McCabe, 1985). For n > 4 the least positive integer m (denoted by D(n)) such that 12, 22,..., n2 are distinct modulo m, is

min{m > 2n : m = p or m = 2p with p an odd prime}. Remark. The range of D(n) does not contain those primes p = 2q + 1 with q an odd prime. Theorem 2 (P. S. Bremser, P. D. Schumer and L. C. Washington, 1990). Let k > 2 and n > 0 be integers, and let D(k, n) denote the the least positive integer m such that 1k , 2k ,..., nk are distinct modulo m. (i) If k is odd and n is suﬃciently large, then

D(k, n) = min{m > n : m is squarefree, and (k, ϕ(m)) = 1}. (ii) If k is even and n is suﬃciently large, then

D(k, n) = min{m > 2n : m = p or 2p with p a prime, and (k, ϕ(m)) = 2}.

14 / 42 A conjecture involving factorials Recall that 2n (2n)! = . n (n!)2 By Stirling’s formula, √ nn n! ∼ 2πn e as n → +∞. Conjecture (Sun, Feb. 27, 2012). For n = 1, 2, 3,... let t(n) be the least integer m > 1 such that 1!, 2!,..., n! are pairwise distinct mod m. Then t(n) is a prime with the only exception t(5) = 10. Examples. t(1) = t(2) = 2, t(3) = 3, t(4) = 7, t(5) = 10, t(6) = t(7) = t(8) = 13, t(9) = 31, t(10) = t(11) = t(12) = 37. Qing-Hu Hou (Nankai Univ.) has veriﬁed the conjecture for 4 n 6 10 . I noted that if we replace k! by (k + 1)! or (2k)! in the deﬁnition of t(n) then t(n) will take only prime values. 15 / 42 Another conjecture involving factorials

Conjecture (Sun, March 25, 2012). The least integer m > 1 such that n! 6≡ k!(mod m) for all 0 < k < n, is a prime in [n, 2n) unless n ∈ {4, 6}.

Remark. The Bertrand Postulate conﬁrmed by Chebyshev states that for any x > 1 the interval [x, 2x) contains a prime.

For Euler number E0, E1, E2,... , I have a similar conjecture. Conjecture (Sun, March, 2012). Let e(n) be the least integer m > 1 such that 2E2n 6≡ 2E2k (mod m) for all 0 < k < n. Then e(n) is a prime in the interval [2n, 3n] with the only exceptions as follows:

e(4) = 13, e(7) = 23, e(10) = 52, e(55) = 112.

16 / 42 A conjecture on products of consecutive primes + Conjecture (Sun, March 18, 2012). For k ∈ Z let Pk denote the product of the first k primes p1,..., pk . + (i) For n ∈ Z define w1(n) as the least integer m > 1 such that m divides none of those Pi − Pj with 1 6 i < j 6 n. Then w1(n) is always a prime. + (ii) For n ∈ Z define w2(n) as the least integer m > 1 such that m divides none of those Pi + Pj with 1 6 i < j 6 n. Then w2(n) is always a prime. 2 2 (iii) We have w1(n) < n and w2(n) < n for all n = 2, 3, 4,....

Remark. Clearly wi (n) 6 wi (n + 1) for i = 1, 2.

+ + W1 = {w1(n): n ∈ Z } and W2 = {w2(n): n ∈ Z }

are both inﬁnite. If wi (n) = pk , then k > n since pk | Pk ± Pk+1. I’m even unable to prove w2(n) > n.

17 / 42 Respondences from others

One of my students: “The conjecture might be wrong, it is not reasonable!” A Chinese professor in number theory: “This seems to be incorrect. Other number theorists kept silent and made no comments. In April, W. B. Hart (an English number theorist) veriﬁed the conjecture for n up to 105. The following comment comes from http://tech.groups.yahoo.com/group/ primenumbers/message/24181 “As would be expected when coming from Zhi-Wei Sun, if he presents it as a conjecture, you can be sure of two things ... It’s very likely true, and will be very hard to prove.” —Jack Brennen (March 21, 2012).

18 / 42 A conjecture on sums of consecutive prime + Conjecture (Sun, March 21, 2012). For k ∈ Z let Sk denote the sum of the first k primes p1,..., pk . + + (i) For n ∈ Z define S (n) as the least integer m > 1 such that + m divides none of Si ! + Sj ! with 1 6 i < j 6 n. Then S (n) is always a prime not exceeding Sn. + − (ii) For n ∈ Z define S (n) as the least integer m > 1 such that − m divides none of those Si ! − Sj ! with 1 6 i < j 6 n. Then S (n) is always a prime not exceeding Sn. Remark. When n > 1, clearly both S+(n) and S−(n) are greater than Sn−1. Thus, by the conjecture we should have + + − − S (n) 6 Sn < S (n + 1) and S (n) 6 Sn < S (n + 1) for all n = 1, 2, 3,.... It seems very challenging to prove that (Sn, Sn+1] + contains a prime for any n ∈ Z . Note that n X Z n n2 S ∼ k log k ∼ x log xdx ∼ log n. n 2 k=1 1

19 / 42 Alternating sums of primes

Let pn be the nth prime and deﬁne

n−1 sn = pn − pn−1 + ··· + (−1) p1.

Note that

n n X X s2n = (p2k − p2k−1) > 0, s2n+1 = (p2k+1 − p2k ) + p1 > 0. k=1 k=1

Here are values of s1,..., s16:

2, 1, 4, 3, 8, 5, 12, 7, 16, 13, 18, 19, 22, 21, 26, 27.

The sequence 0, s1, s2,... were ﬁrst introduced by N.J.A. Sloane and J.H. Conway (see A008347 at OEIS).

It is not diﬃcult to show that those sn (n = 1, 2, 3,...) are pairwise distinct.

20 / 42 An amazing recurrence for primes The following surprising conjecture on recurrence for primes allows us to compute pn+1 in terms of p1,..., pn. Conjecture (Sun, March 28, 2012). For any positive integer n 6= 1, 2, 4, 9, the (n + 1)-th prime pn+1 is the least positive integer m such that 2 2 2s1 ,..., 2sn are pairwise distinct modulo m. 5 Remark. I have veriﬁed the conjecture for n 6 10 , and proved 2 2 that 2s1 ,..., 2sn are indeed pairwise distinct modulo pn+1. No comments from number theorists though I made the conjecture public via a message to Number Theory List! They all keep silent on this mysterious recurrence. A Related Conjecture (Sun, March 21, 2012). The least integer 2 m > 1 such that 2Sk (k = 1,..., n) are pairwise distinct modulo 2 Pk m is a prime smaller than n unless n | 6, where Sk = j=1 pj . 21 / 42 Conjecture on alternating sums of consecutive primes Conjecture (Sun, April 2-3, 2012). For any positive integer m, there are consecutive primes p ,..., p (k n) not exceeding √ √ k n 6 2m + 2.2 m (or m + 4.6 m if 2 - m) such that n−k m = pn − pn−1 + ··· + (−1) pk . Examples. 10 = 17 − 13 + 11 − 7 + 5 − 3; 20 = 41 − 37 + 31 − 29 + 23 − 19 + 17 − 13 + 11 − 7 + 5 − 3;

303 = p76 − p75 + ··· + p52, √ p76 = 383 = b303 + 4.6 303c, p52 = 239;

2382 = p652 − p651 + ··· + p44 − p43, √ p652 = 4871 = b2 · 2382 + 2.2 2382c, p43 = 191. The conjecture has been veriﬁed for m up to 107. Most known results on primes are about local properties of primes, not about relations of primes.

Prize. I would like to oﬀer 1000 US dollars for the ﬁrst proof. 22 / 42 Part II. Inequalities involving primes

23 / 42 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 p > n+1 p for all n = 1, 2,..., √ n n+1 n i.e., the sequence ( pn)n>1 is strictly decreasing Remark. The conjecture implies that

2 pn+1 − pn < log pn − log pn + 1

for all n > 4, which is stronger than Cramer’s conjecture that 2 c := lim sup(pn+1 − pn)/ log pn coincides with 1 (A. Granville thought that c should be 2/eγ ≈ 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. 24 / 42 A refinement of Firoozbakht’s Conjecture

By the Prime Number Theorem, pn ∼ n log n. Note that n+1p(n + 1) log(n + 1) log n log log n 1 log √ = − − + O . n n log n n2 n2 n2 This led me to make the following conjecture. Conjecture (Sun, 2012-09-11) For any integer n > 4, we have the inequality √ n+1 pn+1 log log n √ < 1 − 2 . n pn 2n

Remark. We have veriﬁed the conjecture for all n 6 3500000 and 18 all those n with pn < 4 × 10 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 √ √ n 2 (1 − n+1 pn+1/ pn)n / log log n ≈ 0.5229.

25 / 42 Some easy facts Easy things: √ n ( n)n>3 is strictly decreasing, √ √ n+1 n ( n + 1/ n)n>4 is strictly increasing. Reason: For the function f (x) = (log x)/x on the interval [4.5, +∞), we have 1 − log x 2 log x − 3 f 0(x) = < 0 and f 00(x) = > 0, x2 x3 and hence f (x) is strictly decreasing and strictly convex. For Pn = p1p2 ··· pn and Sn = p1 + p2 + ··· + pn, √ n ( 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. 26 / 42 Pn Monotonicity related to Sn = k=1 pk

√ n pn Theorem (Sun). (i) (July 28-31) ( Sn)n>2 and ( Sn/n)n>1 are strictly decreasing. (ii) (Discovered on July 29 and proved on August 25) ! ! n+1pS n+1pS /(n + 1) √ n+1 and n+1 n pn Sn Sn/n n>5 n>10 are strictly increasing.

27 / 42 A general theorem (α) Pn α For Sn = k=1 pk , we have Theorem (Sun, Bull. Aust. Math. Soc., in press). Let α > 1. 2×1.348α+1 (i) If n > max{100, e }, then s s (α) (α) n S n+1 S n > n+1 n n + 1 and hence q q n (α) n+1 (α) Sn > Sn+1. (ii) The sequence q q n+1 (α) n (α) Sn+1/(n + 1) Sn /n n>N(α) is strictly increasing, where

n 2 2α+1 α+1 o N(α) = max 350000, de((α+1) 1.2 +(α+1)1.2 )/αe .

28 / 42 The cases α = 2, 3, 4

Corollary. The sequences q q n+1 (2) n (2) Sn+1 Sn , n>10 q q n+1 (3) n (3) Sn+1 Sn , n>10 q q n+1 (4) n (4) Sn+1 Sn n>17 are all strictly increasing.

29 / 42 On squarefree numbers 2 A positive integer n is called squarefree if p - 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 s be the √ n n n-th squarefree positive integer. Then the sequence ( sn)n>7 is strictly decreasing. (ii) For n = 1, 2, 3,... let S(n) be the sum of the ﬁrst n squarefree n+1p pn 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 pn n = 7,..., 500000. Note that limn→∞ S(n) = 1 since S(n) does not exceed the sum of the ﬁrst n primes.

30 / 42 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 xm = 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 1+1/n next such prime pn+1(a) is smaller than pn(a) . (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 1+1/n next such prime pn+1(a) is smaller than pn(a) with the only exception a = −2 and n = 13.

n+1p pn (iii) The sequence ( Pn+1(a)/ Pn(a))n>3 is strictly increasing Pn with limit 1, where Pn(a) = k=1 pk (a). Remark. The ﬁrst 5 primes having 24 as the smallest positive primitive root are 533821, 567631, 672181, 843781, 1035301. 31 / 42 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 t in the √ n+1√ 1+1/n n n+1 next pair of twin primes is smaller than tn , i.e., tn > tn+1.

n+1p pn (ii) The sequence ( T (n + 1)/ T (n))n>9 is strictly increasing Pn with limit 1, where T (n) = k=1 tk . √ √ n n+1 Remark. Via Mathematica I veriﬁed that tn > tn+1 for all n = 1,..., 500000, and

n+1pT (n + 1)/pn T (n) < n+2pT (n + 2)/ n+1pT (n + 1)

for all n = 9, 10,..., 500000. Note that t500000 = 115438667. After I made the conjecture public, Marek Wolf veriﬁed the √ √ n n+1 inequality tn > tn+1 for all the 44849427 pairs of twin primes below 234 ≈ 1.718 × 1010. 32 / 42 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.

n+1p pn (ii) The sequence ( G(n + 1)/ G(n))n>13 is strictly increasing Pn with limit 1, where G(n) = k=1 gk . √ √ Remark. Via Mathematica I veriﬁed that n gn > n+1 gn+1 for all n = 5,..., 200000, and

n+1pG(n + 1)/pn G(n) < n+2pG(n + 2)/ n+1pG(n + 1)

for all n = 13, 14,..., 200000. Note that g200000 = 42721961.

33 / 42 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 Qk leading coefficients such that there is no prime dividing j=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 n qn+1 < qn , i.e., qn > n+1 qn+1.

Also, there is a positive integer N such that the sequence n+1p pn ( Q(n + 1)/ Q(n))n>N is strictly increasing with limit 1, Pn where Q(n) = k=1 qk . 34 / 42 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 the ﬁrst n Proth primes, then Pr(n + 1) < Pr(n)1+1/n (i.e., pn Pr(n) > n+1pPr(n + 1)) unless Pn n = 2, 4, 5. If we set PR(n) = k=1 Pr(k), then the sequence n+1p pn ( PR(n + 1)/ PR(n))n>34 is strictly increasing with limit 1. Remark. I have checked the conjecture for the ﬁrst 4000 Proth primes.

35 / 42 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. pn (ii) The sequence ( Nn(q))n>e3+7/(q−1)2 is strictly increasing, and the sequence

n+1p pn ( N (q)/ N (q)) 14 n+1 n n>5.835×10 is strictly decreasing.

36 / 42 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 √ eπ 2n/3 p(n) ∼ √ (Hardy and Ramanujan) 4 3n

pn and hence limn→∞ p(n) = 1.

pn Conjecture (Sun, 2012-08-02) The sequence ( p(n))n>6 is strictly decreasing. Moreover, the sequence n+1p pn ( p(n + 1)/ p(n))n>26 is strictly increasing. Remark. I have veriﬁed the conjecture for n up to 105. √ I have many other conjectures on monotonicity of ( n a ) or √ √ n n>1 n+1 n ( an+1/ an)n>1 for various combinatorial sequences (an)n>1 of positive integers.

37 / 42 Part III. Selected conjectures on primes made before 2012

38 / 42 Selected conjectures made before 2012 Conjecture (Sun, 2009). Any natural number n 6= 216 can be written in the form p + Tx , where p is a prime or zero, and Tx = x(x + 1)/2 is a triangular number. Conjecture (Sun, 2010). For any prime p > 3 we have p−1 4k 2k X 2k+1 k ≡ 0 (mod p2). 48k k=0 Conjecture (Sun, 2010). For any odd prime p we have p−1 ( X 4x2 − 2p (mod p2) if p = x2 + 2y 2, A ≡ n 2 n=0 0 (mod p ) if p ≡ 5, 7 (mod 8),

where A0, A1,... are Ap´erynumbers deﬁned by n 2 2 X n n + k A = . n k k k=0 Remark. I have proved the congruence modulo p. 39 / 42 Selected conjectures made before 2012 Conjecture (Sun, 2011). For any odd prime p and n = 1, 2, 3,..., n−1 1 X (p − 1)k is always a p-adic integer. 2n k,..., k n n k=0

Conjecture (Sun, 2011). Let p > 3 be a prime and let Tk (b, c) denote the coeﬃcient of xk in (x2 + bx + c)k . Then

p−1 2k 2 X k Tk (3, −3) (−108)k k=0 4x2 − 2p (mod p2) if ( −1 ) = ( p ) = ( p ) = 1, p = x2 + 21y 2,  p 3 7 12x2 − 2p (mod p2) if ( −1 ) = ( p ) = −1, ( p ) = 1, p = 3x2 + 7y 2,  p 7 3  2 2 −1 p p 2 2 ≡ 2x − 2p (mod p ) if ( p ) = ( 3 ) = −1, ( 7 ) = 1, 2p = x + 21y ,  p p 6x2 − 2p (mod p2) if ( −1 ) = 1, ( ) = ( ) = −1, 2p = 3x2 + 7y 2,  p 3 7  2 −21 0 (mod p ) if ( p ) = −1. √ Remark. The quadratic ﬁeld Q( −21) has class number four. 40 / 42 Few of my 181 conjectural series Motivated by congruences modulo primes, I found: ∞ X (28k2 − 18k + 3)(−64)k 4 = −14ζ(3), 52k 3k k=1 k k k ∞ 2 n 2 X 40n2 + 26n + 52n X n 2k2(n − k) 24 = , (−256)n n k k n − k π2 n=0 k=0 ∞ √ X 66k + 17 540 2 T 3(10, 112) = , (21133)k k 11π k=0 ∞ √ X 126k + 31 880 5 T 3(22, 212) = , (−80)3k k 21π k=0 ∞ X 3990k + 1147 432 √ √ T 3(62, 952) = (195 14 + 94 2). (−288)3k k 95π k=0 Remark. I would like to oﬀer $300 for the person (not joint authors) who could prove the last three identities. 41 / 42 For sources of my conjectures, you may visit my homepage http://math.nju.edu.cn/∼zwsun You are welcome to solve my conjectures and win the prizes! Thank you!

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