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- Kostochka [8] showed that the sunflower bound for k = 3 is reduced from 2ss! to s log log log s cs! log log s for a constant c. The case k = 3 of the sunflower conjecture is especially emphasized by Erd¨os [4], which other researchers also believe includes some critical difficulty. − s 2 s - It has also been shown [9] that of cardinality greater than k 1+ csk− F + includes a k-sunflower for some cs R depending only on s. - With the sunflower bound (k 1)ss!,∈ Razborov proved an exponential lower bound on the monotone circuit complexity− of the clique problem [10]. Alon and Boppana strengthened the bound [1] by relaxing the condition to be a sunflower from U U ′ = V V to U U ′ V V for all U,U ′ ,U = U ′. - [2]∩ discusses the∈B sunflower∩ conjecture⊃ ∈B and its variants∈F in relation6 to fast matrix multiplicationT algorithms. Especially,T it is shown in the paper that if the sunflower conjecture is true, the Coppersmith-Winograd conjecture implying a faster matrix multiplication algorithm [3] does not hold. In this paper we show that the general sunflower bound can be indeed improved by an exponential factor. We prove the following theorem. Theorem 1.2. A family of sets, each of cardinality at most s, includes a k- sunflower if F 2 1 c s √10 2 k min , s!, |F| ≥ − · √10 2 log min(k,s) − for a constant c and any integers k 2 and s 2. ≥ ≥ s 1 This improves the sunflower bound by the factor of O log s whenever k exceeds sǫ for a given constant ǫ (0, 1). Also any h of cardinalityi at least 2 s ∈ F √10 2 k s! includes a k-sunflower. √10 2 − − We split its proof in two steps. We will show: 2 s Statement I: includes a k-sunflower if √10 2 k s! for any √10 2 F |F| ≥ − − positive integers s and k. s Statement II: includes a k-sunflower if ck s! for some constant F |F| ≥ log min(s,k) c and any integers s 2 and k 2. h i ≥ ≥ It is clear that the two statements mean Theorem 1.2. The rest of the paper is dedicated to the description of their proofs.
2. Terminology and Related Facts Denote an arbitrary set by S that is a subset of X. Given a family of sets of cardinality at most s, define F def (S) = U : U and U S = , F { ∈F ∩ 6 ∅} def j (S) = U : U and U S = j for positive integer j, F { ∈F | ∩ | } def sup(S) = U : U and U S , and F { ∈F ⊃ } def (S) = (v,U): v X,U and v U S . P { ∈ ∈F ∈ ∩ } ASYMPTOTIC IMPROVEMENT OF THE SUNFLOWER BOUND 3
Let ǫ (0, 1/8) be a constant and k Z+. Given such numbers, we use the following∈ two functions as lower bounds on∈ : |F| s def 2 k Φ (s) = √10 2 s!, and 1 − √ 10 2 s − def k s! def ǫ ln min (j, k) if j 2, Φ2(s) = , where pj = ≥ p p ps ǫ if j = 1. 1 2 ··· Here ln denotes the natural logarithm of a positive real number. We regardΦi(j)= 0 if j ·Z+ for each i =1, 2. 6∈ The following lemma shows that Statement II is proved if Φ2(s) means a k-sunflower in . |F| ≥ F s Lemma 2.1. There exists a constant c such that Φ (s) ck s! for any 2 ≤ ln min(k,s) s 2 and k 2. ≥ ≥ s It is shown by p1p2 ps (c′ ln min (k,s)) for another constant c′. Its exact proof is found in Appendix.··· ≥ We also have ks ks k(s 1) (2.1) Φ2 (s)= Φ2(s 1) = · − Φ2(s 2) = ps − psps 1 − ··· − kj s(s 1) (s j + 1) = − ··· − Φ2(s j) psps 1 ps j+1 − − ··· − kj s! = Φ2(s j) psps 1 ps j+1 · (s j)! − − ··· − − k j s! Φ2(s j), ≥ ps (s j)! − − for each positive integer j n n+ 1 n n+ 1 √2πn n e− 12n+1 3. The Improvement Method We will show Statements I and II by improving the original proof of the sunflower lemma in [5]. We review it in a way to introduce our proof method easily. The original proof shows by induction on s that includes a k-sunflower if > Φ0(s), where F |F| def Φ (s) = (k 1)ss!. 0 − The claim is clearly true in the induction basis s = 1; the family consists of more than k 1 different sets of cardinality at most 1 including a desiredF sunflower. To show− the induction step, let = B ,B ,...,Br B { 1 2 } be a sub-family of consisting of pairwise disjoint sets Bi, whose cardinality r is maximum. We sayF that such is a maximal coreless sunflower in for notational convenience in this paper. AlsoB write F def B = B B Br. 1 ∪ 2 ∪···∪ We show r k to complete the induction step. We have ≥ (3.1) sup (S) Φ (s S ) for any S X such that 1 S < s, |F |≤ 0 − | | ⊂ ≤ | | for sup (S), the sub-family of consisting of the sets including S as defined in Sec- tionF 2. Otherwise a k-sunflowerF exists in by induction hypothesis. Then the sub- family (B) consisting of the sets intersectingF with B meets (B) B Φ (s 1); F |F | ≤ | | 0 − because for every element v in B, the cardinality of ( v )= sup ( v ) is bounded F { } F { } by Φ (s 1). Also Φ (s 1) = Φ0(s) . Thus, 0 − 0 − ks (3.2) (B) B Φ (s 1) rsΦ (s 1) |F | ≤ | | 0 − ≤ 0 − Φ (s) r r = rs 0 = Φ (s) < . ks k 0 k |F| Now (B) is less than unless r k. In other words, if r < k, then would |F | |F| ≥ F have a set disjoint with any Bi , contradicting the maximality of r = . Hence r k, meaning includes∈ a Bk-sunflower with an empty core. This proves|B| the induction≥ step. B To improve this argument, we note that the proof works even if sup ( v ) for all F { } v B are disjoint. If so, each sup ( v ) includes elements U only intersecting ∈ F { } ∈F with v , and disjoint with B v . Let v Bi . If we replace Bi by any set { } −{ } ∈ ∈B U in sup ( v ) such that U (B v ) = , then is still a maximal coreless sunflowerF in{ }. ∩ −{ } ∅ B F On the other hand, if there are sufficiently many U sup ( v ) disjoint with ∈ F { } B v , we can find U among them such that U Bi is much smaller than s. (Here−{ we} assume both k are s are are large enough.)| ∩ This| gives us the following contradiction: Due to the maximality of r = , the family (U) must contain |B| F (Bi) (B Bi). So (U) (Bi) includes most sets in sup ( v ) being a not F −F − F ∩F F { } ASYMPTOTIC IMPROVEMENT OF THE SUNFLOWER BOUND 5 too small family. But by (3.1), (U) (Bi) is upper-bounded by |F ∩F | 2 s Φ (s 2)+ U Bi Φ (s 1) 0 − | ∩ | 0 − 2 Φ0(s) Φ0(s) = s + U Bi · k2s(s 1) | ∩ | · ks − 1 U Bi < + | ∩ | . |F| k2 1 1 ks − s ! As U Bi is much smaller than s, the cardinality (U) (Bi) is also small, | ∩ | |F ∩F | 1 U Bi i.e. bounded by times O 2 + | ∩ | . This contradiction on (U) (Bi) |F| k ks |F ∩F | essentially means that if r is around k, we can construct a larger coreless sunflower in . Hence r must be more than k with the cardinality lower bound Φ0(s). FOur proof of Theorem 1.2 in the next section generalizes the above observation. By finding Bi with sufficiently large (Bi) (B Bi) , we will show Statement ∈B 2 |F 2−B − | I that such that √10 2 k s! includes a k-sunflower. √10 2 F |F| ≥ − − We will further extend this argument to show Statement II. Instead of finding just one such Bi , we will find ′ such that a sub-family of ∈B B ⊂B H (3.3) S S′ F −F S ′ ! S′ ′ ! [∈B ∈B−B[ is sufficiently large. Then we show a maximal coreless sunflower in whose cardi- H nality is larger than ′ . This again contradicts the maximality of r = to prove the second statement.|B | |B| 4. Proof of Theorem 1.2 4.1. Statement I. We prove Statement I in this section. Put k k δ = √10 3=0.16227 ..., and x = = , − 1+ δ √10 2 − then 2 s k 2 s Φ1(s)= √10 2 s!=(1+ δ) x s!, − √10 2 − as defined in Section 2. We show that Φ1(s) means a (1+ δ) x-sunflower included in . |F| ≥ We proveF it by induction on s. Its basis occurs when s 2. The claim is true by 2 s ≤ s s 2 the sunflower lemma, since Φ1(s)=(1+ δ) x s! ((1 + δ)x) s!= k s! > (k 1) s! if s 2. Assume true for 1, 2,...,s 1 and prove≥ true for s 3. As in Section− 3, ≤ − ≥ let = B ,B ,...,Br be a maximal coreless sunflower of cardinality r in , and B { 1 2 } F B = B B Br. Contrarily to the claim, let us assume 1 ∪ 2 ∪···∪ (4.1) r< (1 + δ)x. We will find a contradiction caused by (4.1). Observe the following facts. For any nonempty set S X such that S < s, if sup(S) Φ (s S ), the • ⊂ | | |F |≥ 1 − | | family sup(S) contains a (1 + δ)x-sunflower by induction hypothesis. Thus we assumeF (4.2) sup(S) < Φ (s S ) for S X such that 1 S < s. |F | 1 − | | ⊂ ≤ | | 6 JUNICHIRO FUKUYAMA We also have k 3, because Φ1(s) > 0 for k = 1, and • ≥ 2 s Φ (s)= √10 2 2 s! >s! = (k 1)ss! if k = 2. 2 √10 2 − − − Since r has the maximum value, • (4.3) = (B), F F i.e., every set in intersects with B. (B), defined inF Section 2 as the family of pairs (v,U) such that U and •v P U B, has a cardinality bounded by ∈ F ∈ ∩ Φ (s) (4.4) (B) B Φ (s 1) rsΦ (s 1) < (1 + δ)xs 1 (1 + δ) , |P | ≤ | | 1 − ≤ 1 − · xs ≤ |F| due to (4.1) and (4.2). Here (B) B Φ1(s 1) because for each v B, there are at most Φ (s 1) pairs (|Pv,U) | ≤ |(B|). − ∈ 1 − ∈P We first see that many U intersect with B by cardinality 1, i.e., 1(B) is sufficiently large. Observe two∈F lemmas. |F | Lemma 4.1. (B) > (1 δ) . |F1 | − |F| Proof. Let 1(B) = (1 δ′) for some δ′ [0, 1]. By (4.3), there are δ′ elements U |F such| that−U |F|B 2, each of∈ which creates two or more pairs|F| in ∈F | ∩ |≥ (B). If δ′ δ, P ≥ (B) (1 δ′) +2δ′ =(1+ δ′) (1 + δ) , |P |≥ − |F| |F| |F| ≥ |F| contradicting (4.4). Thus δ′ <δ proving the lemma. Lemma 4.2. There exists Bi B1,B2,...,Br such that 1 (Bi) (B Bi) 1 δ Φ1(s) ∈{ } |F −F − |≥ − . 1+δ · x Proof. By Lemma 4.1, there exists Bi B ,B ,...,Br such that the number of ∈{ 1 2 } U intersecting with Bi by cardinality 1, and disjoint with B Bi, is at least ∈F − (1 δ) 1 δ 1 δ Φ (s) − |F| > − − 1 . r (1 + δ)x · |F| ≥ 1+ δ · x The family of such U is exactly 1(Bi) (B Bi), so its cardinality is no less 1 δ Φ1(s) F −F − than − . 1+δ · x Assume such Bi is B1 without loss of generality. Then 1 δ Φ (s) (B ) (B B ) (B ) (B B ) − 1 > 0. |F 1 −F − 1 | ≥ |F1 1 −F − 1 |≥ 1+ δ · x We choose any element B′ (B ) (B B ) that is not B . Switch B 1 ∈ F1 1 −F − 1 1 1 with B′ in . Since B′ is disjoint with any of B ,B ,...,Br, the obtained family 1 B 1 2 3 B1′ ,B2,...,Br is another maximal coreless sunflower in . We see the following inequality.{ } F ASYMPTOTIC IMPROVEMENT OF THE SUNFLOWER BOUND 7 - Lemma 4.3. Φ1(s) 1 1 (B ) (B′ ) < + . |F 1 ∩F 1 | x s x Proof. A set U F (B ) F (B′ ) intersects with B B′ of cardinality 1, or both ∈ 1 ∩ 1 1 ∩ 1 B B′ and B′ B of cardinality s 1. By (4.2), there are at most 1 − 1 1 − 1 − Φ (s) Φ (s) Φ (s 1)+(s 1)2Φ (s 2) = 1 + (s 1)2 1 1 − − 1 − sx − s(s 1)x2 − Φ (s) 1 1 < 1 + x s x such U F (B ) F (B′ ). The lemma follows. ∈ 1 ∩ 1 = (B′ B B Br) would be true if the cardinality r of the new F F 1 ∪ 2 ∪ 3 ∪···∪ coreless sunflower B′ ,B ,B ,...,Br were maximum. However, it means that { 1 2 3 } every element in (B1) (B B1) is included in (B1′ ). Thus, (B1) F (B1′ ) (B ) (B FB ), leading−F to− F F ∩ ⊃ F 1 −F − 1 1 δ Φ1(s) Φ1(s) 1 1 (B ) F (B′ ) (B ) (B B ) − + . |F 1 ∩ 1 | ≥ |F 1 −F − 1 |≥ 1+ δ · x ≥ x s x k 1 δ 1 1+δ Its last inequality is confirmed with s 3, k 3, x = , and − = + as ≥ ≥ 1+δ 1+δ 3 3 δ = √10 3. This contradicts Lemma 4.3, completing the proof of Statement I. − 4.2. Statement II. We prove the second statement by further developing the above method. We show that includes a k-sunflower if > Φ2(s) for a choice of sufficiently small constant ǫ F(0, 1/8). This suffices to prove|F| Statement II thanks to Lemma 2.1. The proof is∈ by induction on s. Its basis occurs when min(k,s) is smaller than a sufficiently large constant c1, i.e., when min(k,s) is smaller than a lower bound c1 on min(k,s) required by the proof below. To meet this case, we choose ǫ (0, 1/8) to be smaller than 1 ln c . Then Φ (s) kss! > (k 1)ss! by ∈ 1 2 ≥ − the definition of Φ2 in Section 2. The family thus includes a k-sunflower by the sunflower lemma in the basis. F Assume true for 1, 2,...,s 1 and prove true for s. The two integers k and s satisfy − (4.5) min(k,s) c , ≥ 1 i.e., they are sufficiently large. Put def 1 def k p = ln min(k,s), and x = . 8 p p is sufficiently large since both k and s are. Note. The lower bound c1 on min(s, k) is required in order to satisfy (4.8), (4.10), (4.11) and (4.12) below, which are inequalities with fixed coefficients and no ǫ. We choose c as the minimum positive integer such that min(k,s) c satisfies the 1 ≥ 1 inequalities, and also ǫ as min 1 , 1 . 2ln c1 9 As ps = ǫ ln min(k,s) Similarly to (4.2), we have sup (s S ) < Φ (s S ) for any S X with 1 S < s, |F − | | | 2 − | | ⊂ ≤ | | as induction hypothesis. We also keep denoting a maximal coreless sunflower in F by = B1,B2,...,Br , and B1 B2 Br by B. Also (4.3) holds due to the maximalityB { of r = . } ∪ ∪···∪ We prove r k |B|for the induction step. Suppose contrarily that ≥ (4.7) x r < k, ≤ and we will find a contradiction. Here r x is confirmed similarly to (3.2) in Section 3, i.e., by ≥ Φ (s) r (B) B Φ (s 1) rsΦ (s 1) rs 2 , |F | ≤ | | 2 − ≤ 2 − ≤ xs ≤ x |F| with (4.6), so r < x would contradict the maximality of r = . We start our proof by showing a claim seen similarly to Lemma|B| 4.1. Φ2(s) Lemma 4.4. There exists a positive integer j 2p such that j (B) . ≤ |F |≥ 4p 1 1 Proof. We first show 0 j 2p j (B) 2 . If not, there would be at least 2 ≤ ≤ F ≥ |F| |F| sets U such that U B > 2p by the definition of j given in Section 2. Each ∈F |P∩ | F such U creates at least 2p pairs (v,U ) (B), so ⌈ ⌉ ∈P (B) 2p |F| = p . |P |≥ · 2 |F| However, similarly to (4.4), pΦ (s) (B) B Φ (s 1) rsΦ (s 1) < ks 2 p , |P | ≤ | | 2 − ≤ 2 − · ks ≤ |F| by induction hypothesis and (4.6). By the contradictory two inequalities, 1 1 0 j 2p j (B) < 2 is false. Thus 0 j 2p j (B) 2 . ≤ ≤ F |F| ≤ ≤ F ≥ |F| PLet j be an integer in [1, 2p] with the P maximum cardinality of j (B). By the 1 Φ2(s) F above claim and (4.3), j (B) |F| , proving the lemma. |F |≥ 2 · 2p ≥ 4p Fix this integer j [1, 2p]. Next we find a small sub-family ′ of such that ∈ B B (3.3) is large enough. For each non-empty ′ , define B ⊂B def ( ′) = U : U j (B) , S ′,U S = and S′ ′,U S′ = . G B { ∈F ∀ ∈B ∩ 6 ∅ ∀ ∈B−B ∩ ∅} Observe a lemma regarding ( ′). G B Lemma 4.5. There exists a nonempty sub-family ′ of cardinality at most j B ⊂B such that ( ) Φ2(s) . ′ p r |G B |≥ 8 (j) Proof. By definition, the cardinality of ′ such that ( ′) = does not exceed j. Thus there are at most B G B 6 ∅ r r r r + + + + j j 1 j 2 ··· 1 − − r j j 2 j 3 r < 1+ + + + 2 j r j +1 r j +1 r j +1 ··· ≤ j − − − ! such possible ′ . Here its truth is confirmed by the following arguments. B ⊂B ASYMPTOTIC IMPROVEMENT OF THE SUNFLOWER BOUND 9 n m n Z+ r j r m 1 = n m+1 m for any n,m such that m n. So j 1 = r j+1 j , • − − 2 ∈ ≤ − 3 − r j 1 r j r r j 2 r j r j 2 = r −j+2 j 1 < r j+1 j , j 3 = r −j+3 j 2 < r j+1 j , . − − − − − − − − ··· The last inequality is due tor x = k = 8k > k by (4.7), and • ≥ p ln min( k,s) ln k j 2p< ln k. Thus ≤ j ln k 1 (4.8) < < , r j +1 k ln k +1 2 − ln k − by (4.5) where k is sufficiently large. So the last inequality holds in the above. Then by Lemma 4.4, there exists at least one nonempty ′ such that B ⊂B j (B) Φ2(s) Φ2(s) ( ′) > |F | = . |G B | 2 r ≥ 2 r 4p 8p r j j · j The lemma follows. We now construct a sub-family of (3.3) in which we will find a larger maximal H coreless sunflower. Fix a sub-family ′ decided by Lemma 4.5. Put B ⊂B def B′ = S, S ′ [∈B def r′ = ′ j 2p, |B |≤ ≤ def and = Fj (B′) F (B B′)= j S S′ . H − − F −F S ′ ! S′ ′ ! [∈B ∈B−B[ The family includes ( ′) by definition, so H G B Φ2(s) (4.9) r , |H| ≥ 8p j by Lemma 4.5. If we find a maximal coreless sunflower in whose cardinality H is larger than r′ = ′ , it means the existence of a coreless sunflower in with |B | F cardinality larger than r, since any U is disjoint with B B′. Extending the notation (S), write∈ H − F def (S) = U : U and U S = , H { ∈ H ∩ 6 ∅} for a nonempty set S X. Then ( v ) for an element v X is the family of U containing⊂v. H { } ∈ ∈H⊂FLet us show two lemmas on and ( v ). By them we will see that the latter is sufficiently smaller than the former.H H { } Lemma 4.6. >sj Φ (s j)exp( 4p). |H| 2 − − Proof. We have two facts on (4.9). The natural logarithm of the denominator 8p r is upper-bounded by • j r r k xp ln 8p < j ln + j + ln 8p < j ln + j + ln 8p = j ln + j + ln 8p, j j j j k due to (2.3), (4.7) and x = p . 10 JUNICHIRO FUKUYAMA Let d = ln Φ (s) ln Φ (s j), or Φ (s)=Φ (s j) exp(d). It satisfies • 2 − 2 − 2 2 − ks j2 ks j2 j2 d > j ln 1 > j ln 1= j ln sx 1 > j ln sx j 1, ps − s − p − s − − s − − − by Lemma 2.2, ps Φ2(s) xp > Φ2(s j)exp (j ln sx j 1) j ln + j + ln 8p |H| ≥ 8p r − − − − j j p = sj Φ (s j)exp j ln 2j ln 8p 1 . 2 − − j − − − p Find max1 j 2p j ln +2j regarding j as a real parameter. The maximum ≤ ≤ j value (4 2 ln 2)p is achieved when j =2p. Since p is sufficiently large by (4.5), − (4.10) 2p ln 2 > 1 + ln 8p. Hence, p > sj exp j ln 2j ln 8p 1 sj exp( (4 2 ln 2)p ln 8p 1) |H| − j − − − ≥ − − − − > sj Φ (s j)exp( 4p) , 2 − − completing the proof. Lemma 4.7. The following two statements hold true. 1 7p i) ( v ) s e for every v B′. |H { } |≤ ·1 · |H|7p ∈ ii) ( v ) (s j)x e for every v X B′. |H { } |≤ − · · |H| ∈ − Proof. i): Let U be any set in ( v ) for the given v B′, and write U ′ = U B′. H { } ∈ ∩ Since U ′ has cardinality j containing v, there are no more than B′ 1 r′s 1 j r′s j r′s | |− − = j 1 ≤ j 1 r′s j ≤ s j − −′ ′ r s r′s r s 1 choices of U ′. Here the identity j = j j −1 is used. By the induction hypothesis on s, the number of U ( v ) is upper-bounded− by ∈ H{ } j r′s Φ (s j) r′s ( v ) Φ (s j) 2 − exp ln j + j ln + j |H { } | ≤ s j · 2 − ≤ s j r′ exp j ln j + j + ln j exp(j + ln j) sj Φ (s j) sj Φ (s j) ≤ 2 − s ≤ 2 − s exp(2p + ln 2p) exp(3p) sj Φ (s j) def 1 k y y = min (s, k) , so that p = ln y, y = e8p, s y, and x = . 8 ≥ p ≥ p We have 3 p 1 (4.12) p 1 and 8p e− < , ≥ 2 due to (4.5). Fix each V . By Lemma 4.7, the family (V ) has a cardinality bounded by ∈ H H e7p e7p (V ) = ( v ) j + (s j) |H | |H { } |≤ · s − (s j)x |H| v V [∈ − 1 1 1 p 2p je7p + je7p + je7p ≤ s x |H| ≤ y y |H| ≤ · y · |H| 2 7p 2 7p 2 p 4p e |H| =4p e |H| =4p e− . ≤ · y · e8p · |H| Therefore, by ′ = r′ j 2p, |B | ≤ ≤ 3 p 1 ′ (V ) 2p (V ) 8p e− < . |B | · |H |≤ |H |≤ · |H| 2|H| The lemma follows. Hence each V intersects with at most 2|H|′ sets in . There exist more ∈ H |B | H than ′ pairwise disjoint sets in . By definition, every element in is disjoint |B | H H with a set in ′. The cardinality r of the coreless sunflower in is therefore not maximum.B−B This contradiction proves Statement II, completingB F the proof of Theorem 1.2. Appendix: Proofs of Lemmas 2.1 and 2.2 s Lemma 2.1. There exists a constant c> 0 such that Φ (s) ck s! 2 ≤ ln min(k,s) for any s 2 and k 2. ≥ ≥ Proof. We first show the lemma when s k. By the definition of Φ (s) in Section 2, ≤ 2 1 k s Φ (s)= s!. 2 ln 2 ln 3 ln s ǫ · ··· 12 JUNICHIRO FUKUYAMA We assume s 3 since the claim is trivially true for s = 2. It suffices to show s 1 k ≥s e2 k s ln 2 ln 3 ln s ǫ s! ln s ǫ s!, or · ··· ≤ ln s s (4.13) ln 2 ln 3 ln s · ··· ≥ e2 ln ln 2 + ln ln 3 + + ln ln s s ln ln s 2s. ⇔ ··· ≥ − ln ln x is a smooth, monotonically increasing function of x R+, so s ∈ ln ln 3 + ln ln 4 + + ln ln s> ln ln x dx ··· Z2 = (s ln ln s li(s)) (2 ln ln 2 li(2)). − − − s dx Here li(s) is the logarithmic integral 0 ln x . As ln ln 2 < 0, ln ln 2 + ln ln 3 + + ln ln s>s ln ln s li(s)+ li(2), ··· R − s dx where li(s) li(2) = 2 ln x is upper-bounded by (s 2)/ ln 2 < 2s. Hence, ln ln 2 + ln ln 3 + −+ ln ln s s ln ln s 2s, proving the lemma− when s k. If s >··· k, the lemma≥R is also proved− by (4.13): ≤ 1 k s Φ (s) = s! 2 s k ǫ ln 2 ln 3 ln k (ln k) − · ··· · s 1 k s e2k s! < s!. ≤ ln k k s k ǫ ǫ ln k 2 (ln k) − e · This completes the proof. ks j2 Lemma 2.2. Φ2(s) > Φ2(s j)exp j ln 1 for a positive integer − ps − s − j j 2 Proof. It suffices to show k s! > exp j ln ks j 1 due to (2.1), or ps (s j)! ps s − − − s! j2 > exp j ln s 1 . (s j)! − s − − s! By (2.2), ln (s j)! is at least − s ln s s + ln √2πs (s j) ln(s j) (s j) + ln s j +1 − − − − − − − √2πs p = s ln s (s j) ln(s j) j + ln 1 − − − − √s j − − j > s ln s (s j) ln s + ln 1 j 1 − − − s − − j = j ln s (s j) ln 1 j 1. − − − s − − j 1 j i j By the Taylor series of natural logarithm, ln 1 s = i∞=1 i s > s . 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