New inequalities for p ( n ) and log p ( n )

Let p ( n ) denote the number of partitions of n . A new inﬁnite family of inequalities for p ( n ) is presented. This generalizes a result by William Chen et al. From this inﬁnite family, another inﬁnite family of inequalities for log p ( n ) is derived. As an application of the latter family one, for instance obtains that for n ≥ 120,

The OEIS [12] for A000041 shows that a similarly refined asymptotic formula for p(n) was discovered by Jon E. Schoenfield in 2014, this reads Later Vaclav Kotesovec according to OEIS [12] for A000041 got the precise value of c 0 , c 1 , . . . , c 4 as follows: To the best of our knowledge, the details of the methods of Schoenfield and Kotesovec have not yet been published.
We summarize some of our main results: Theorem 1.1 For the usual partition function p(n) we have The proof of this theorem will be given in Sect. 6.

Theorem 1.3 For the partition numbers p(n) we have the inequalities
The proof of this is given in Sect. 3. This paper is organized as follows. In Sect. 2 we present the methods used in the mathematical experiments that led us Theorem 1.1 and Conjecture 1.2. In Sect. 3 we prove Theorem 1.3 by adapting methods used by Chen et al. to fit our purpose. In Sect. 4 we generalize an inequality by Chen et al. by extending it to an infinite family of inequalities for p(n). In Sect. 5 we introduce preparatory results required to prove Theorem 6.6. In Sect. 6 we prove our main result, Theorem 6.6, by using the main result from Sect. 4, Theorem 4.4. This gives an infinite family of inequalities for log p(n). Finally in Sect. 7 we give an application of the results in Sect. 5 which extends DeSalvo's and Pak's log concavity theorem for p(n). In Sect. 1 (the Appendix) we give additional information on the method used to discover the asymptotic formulas. We remark explicitly that to finalize the proof of Theorem 6.6, we use the Cylindrical Algebraic Decomposition in Mathematica; the details of this are also put to Sect. 1.

Mathematical experiments for better asymptotics for a(n) and p(n)
Before proving our theorems, in this section we briefly describe the experimental mathematics which led us to their discovery. Our strategy is as follows. If we have sufficiently many instances of a given sequence, how can we find an asymptotic formula for this sequence? Take the cubic partitions a(n) and the partition numbers p(n) as examples.
A plot of the two curves through the points (n, a(n)), resp. (n, p(n)), for n ∈ {1, . . . , 1000} is shown in the Fig. 1a and b. According to the Hardy-Ramanujan Theorem 1.1 and the asymptotic formula of Kotesovec (1.5), the curves are increasing with "sub-exponential" speeds. Thus, we may plot two curves using data points (n, log a(n)) and (n, log p(n)) as shown in Fig. 1c. One observes that the new curves look like parabolas y = √ x. This is also very natural in view of So if we modify further with ( √ n, log a(n)) and ( √ n, log p(n)) to plot the curves, we get two almost-straight lines as shown in the Fig. 1d.
This provides the starting point for finding the improved asymptotic formulas (1.6) for p(n) and (1.7) for a(n) from their data sets. We restrict our description to the latter case. Motivated by (2.1), we compute the differences of log a(n) with the estimation values a e (n) := e π √ n 8n 5/4 : Then we can plot curves from the data points (n, (n)) in Fig. 2a and b, and (n, n· (n)) and (n, √ n · (n)) in Fig. 2c and d, in order to confirm the next dominant term approximately. We can see in Fig. 2d that after multiplying (n) by √ n the curve is almost constant, confirming that the next term is C √ n . Also multiplying (n) by n, in Fig. 2c we see that the behavior is like √ n as expected. By using least square regression on the original data set (n, a(n)) for 1 ≤ n ≤ 10000, we aimed at finding the best constant C that minimizes 2 − log a(n) + α · √ n − β · log n − log γ + C √ n , 2 The fourth author of this paper told the result to V. Kotesovec in May 2016 and got a reply in January 2017 that the precise value of C could be Pi/16+15/(8*Pi)=0.7931... where we fixed α = π, β = 5/4, γ = 8 according to (1.5). As a result, we obtained that C ≈ 0.7925. In the Appendix, Sect. 1, we explain that the constants α, β, γ can also be found via regression analysis with Maple instead of getting them from (1.5) directly.

Proof of Theorem 1.3
We separate the proof into two lemmas. The first lemma is the upper bound for p(n) and second lemma is the lower bound. In order to prove these lemmas we will state several facts which are routine to prove. The curve in (a) is for (n, (n)) where 1 ≤ n ≤ 10, 000, b is for (n, (n)) where 1 ≤ n ≤ 100. The curve in (c) is for (n, n · (n)), and d is for (n, √ n · (n)) where 1 ≤ n ≤ 10, 000 Lemma 3.1 For all n ≥ 1, we have Proof By [1, (2.7)-(2.8)] and with A k (n) and R(n, N ) 3 as defined there, we have, We will exploit the case N = 2 together with A 1 (n) = 1 and A 2 (n) = (−1) n for any positive integer n. For N ≥ 1, Lehmer [11, (4.14), p. 294] gave the following error bound: We first estimate the absolute value of T 1 (n); for convenience we denote subexpressions by a 1 , b 1 , c 1 , and d 1 : The following facts are easily verified.

Now,
By Fact D and Fact E, we have Now, by Fact A, B, C, and Fact F we conclude that for all n ≥ 35, By (3.3), we have for all n ≥ 35 that Therefore from (3.2), (3.4), and Fact G, we have for all n ≥ 35, n for all n ≥ 23.
Therefore by Facts H, I, and (3.5), we have for all n ≥ 35, This completes the proof of the stated upper bound in Lemma 3.1.
Proof In the proof of [6,Prop 2.4], it is noted that for all n ≥ 1, where and R(n) is as in [6, (7)].
From the definition of T 2 (n) one verifies: Fact J T 2 (n) > 0 for all n ≥ 1.
n > 0, for all n ≥ 1. From all the above facts we can conclude that (3.6) holds for all n ≥ 631. Using Mathematica we checked that (3.6) also holds for all 1 ≤ n ≤ 630. This concludes the proof of Lemma 3.2.

A generalization of a result by Chen, Jia, and Wang
In this section, we have again that μ(n) = π 6 √ 24n − 1; this should not be confused with the real variable μ which we will use below. Eventually, we will set the real variable μ equal to μ(n). The main goal of this section is to generalize [1, Lem. 2.2] which says that for n ≥ 1206, we have Our improvement is Theorem 4.4 where we replace the 10 in this formula by k and the 1206 by a parametrized bound g(k). In order to achieve this, for a fixed k one needs to find an explicit constant ν(k) ∈ R such that 1 6 e μ/2 > μ k for all μ ∈ R with μ > ν(k). One can show that Theorem 4.4 is crucial for proving our main result, Theorem 6.6, presented in the next section. In Lemma 4.1 we find such a constant ν(k) for all k ≥ 7. In Lemma 4.2 we find a lower bound onν(k). In this way, we see that what is delivered by Lemma 4.1 is best possible in the sense that our ν(k) from Lemma 4.1 and the minimal possibleν(k) satisfies |ν(k) −ν(k)| < 3k log log k log k for all k ≥ 7.
It is also straight-forward to prove log log k > 6/5 for all k ≥ 28. For the remaining cases 7 ≤ k ≤ 27 the inequality (4.1) is verified by numerical computation, which completes the proof of Lemma 4.1.
Then we have Proof Let f defined as in Lemma 4.1, then the statement is equivalent to proving that we observe that In order to show f (κ(k)) < 0, it would be enough therefore to show that 2 κ(k) − log log k log k >κ 2 below. We have which is equivalent to the inequality 2 log k log 6 k + log 2 + log log k log k > (log log k) 2 log 6 k log log k + log 2 log log k it suffices to show which after division by (log log k) 2 gives the equivalent inequality Now note that log k (log log k) 2 is increasing and the right-hand side of the above inequality is decreasing for k ≥ e e 2 = 1619. Evaluating both sides at k = e e 2 gives 5 4 e 2 4 > 23 10 for the left, and 1 + 1 e 2 + log 2 2 + log 6 2e e 2 2 < 22 10 for the right side. This proves the inequality for k ≥ 1619. For 7 ≤ k ≤ 1618 the result follows by numerical evaluation.
where ν(k) is as in Lemma 4.1.
By (4.4) and (4.5) we obtain that |T (n)| < 1 μ(n) k for n > g(k) which proves that statement for k ≥ 7. To prove the statement for k ∈ {2, . . . , 6} we use the statement for k = 7 which says that for all n ≥ g(7) = 581 we have However, for k ∈ {2, . . . , 6} and n ≥ 581. To prove (4.6) for g(k) < n < 581 it is enough to do a numerical evaluation of (4.6) for these values of n with the exception n = 6 when k = 2. We did this using computer algebra. Analogously, we see that for k ∈ {2, . . . , 6} and n ≥ 581 we have (4.7) In the same way we prove (4.7) for g(k) < n < 581.

Preparing for the proof of Theorem 6.6
In this section we prepare for the proof of our main theorem, Theorem 6.6, which is presented in Sect. 6. To this end, we need to introduce a variety of lemmas.
, and its sequence of Taylor coefficients by ∞ u=1 g u y u := G(y).

Lemma 5.3 Let g(k) be as in Definition 4.3.
Then for all k ≥ 2 and n > g(k) with Proof Taking log of both sides of (4.3) gives where The quantity α := π 2 36 + π 2 will play an important role in this and the next section. and for n ≥ 0, Proof By using For n ≥ 0, Inputting this into the package Sigma developed by Carsten Schneider [13], we obtain (5.1) and (5.2).
We need various additional facts about the Taylor coefficients g u of G(y). .
This proves the upper bound. To prove the lower bound note that the first term of the sum is a 2 and the other terms are all positive.
Lemma 5.6 Let s n := (−1) n 1/2 n+1 . For n ≥ 0 we have s n ≥ 0 and s n is a decreasing sequence, that is s n > s n+1 for all n ≥ 0.
Proof From Lemmas 5.4, 5.5, and 5.6, we obtain Again by Lemmas 5.4, 5.5, and 5.6, we have The last line is because α n (1 + 2n) is a decreasing sequence of n for n ≥ 0.

Lemma 5.8
For n ≥ 1 we have Proof By Lemma 5.4 the statement follows from

Lemma 5.9
Define and μ 2 := Then for m ≥ 0 and 0 < y Proof First note that for 0 < y < √ 24 the function B(y) is increasing and also that Consequently, Lemma 5.14 For all k ≥ 2 and 0 < ≤ 1 √ 7 we have For all k ≥ 2 and 0 < ≤ 1 √ 7 , we have l s ≤ s < u s and l < ≤ u .
The following conventions for the letters l and u will be useful: l a denotes a lower bound for the quantity a, and u a will denote an upper bound for the quantity a. And again we use B(y) as defined in Definition 5.11. Then Let us define l B := 0 and u B := u π 6 l s . Then
Proof We obtain, using Lemma 5.14,
Proof We start with the first inequality: To prove the inequality in the rewritten form, define := 6048 6012+167π 2 and note that . For m, k ≥ 1 we define , g n y n − A −1,k (2m − 1)y 2m−1 ≤ G −1,k (y) and

Definition 6.4
For n, U ≥ 1 we define

Lemma 6.5 Let g(k) be as in Definition 4.3 and P n (U ) as in Definition
If m ≥ 2, k ≥ 2m − 1, and n > g(k), then Proof We start with the inequality from Lemma 5.3. Next we use Lemma 6.2 to bound G 1,k (y). Finally we set y = 1 √ n and obtain the desired result.
In order to prove (6.3) and (6.4) for the remaining values of m, firstly we will prove that if (6.3) holds for m ≥ 2 and all n ≥ y ≥ 1, then (6.3) holds for m − 1 and all n ≥ y. (6.5) In particular, if we subtract from the lower bound on log p(n) with parameter m in (6.3) the lower bound on log p(n) with parameter m − 1, we obtain f (2m, Similarly, if we subtract from the upper bound for m → m − 1 in (6.3) the upper bound for m, we obtain g(2m − 2, 2) − f (2m, 2). Hence in order to prove (6.5), it suffices to prove f (2m, −4) > g(2m − 2, −4) and f (2m, 2) < g(2m − 2, 2). (6.6) Analogously, in order to prove that if (6.4) holds for all m ≥ 3 and all n ≥ y ≥ 1, then (6.4) holds for m − 1 and all n ≥ y, it suffices to prove For proving (6.6) and (6.7), we shall prove and f (w, y 0 ) < g(w − 2, y 0 ) with y 0 > 0. (6.9) From Lemmas 5.7 and 5.8, we have where μ 1 and μ 2 are as in Lemma 5.9 and ν as in Definition 6.1. Consequently, In order to prove (6.8), it is enough to prove , (6.10) where Inequality (6.10) is equivalent to which is implied by since δ w = α w , x 0 < 0 and 1 √ n ≥ 1 n for all n ≥ 1. Inequality (6.11) is equivalent to We checked with Mathematica that N 1 (w, x 0 (w)) ≤ 1; see the Appendix, Sect. 3.
Similarly to above, for y 0 > 0 one has, In order to prove (6.9), it is enough to show This last inequality can be rewritten as the following equivalent inequality, which is implied by since y 0 > 0 and 1 √ n ≥ 1 n . Inequality (6.12) is equivalent to We checked using Mathematica that N 2 (w, y 0 ) ≤ 1 for all y 0 ≥ 1; see the Appendix, Sect. 3.
Finally, we are put into the position to prove Theorem 1.1.

An application to Chen-DeSalvo-Pak log concavity result
In 2010 at FPSAC [3], William Chen conjectured that { p(n)} n≥26 is log-concave and that for n ≥ 1, p(n) 2 < 1 + 1 n p(n − 1) p(n + 1). (7.1) DeSalvo and Pak [6] proved these two conjectures. Moreover, they refined (7.1) by proposing the following conjecture: Chen, Wang, and Xie [2] gave an affirmative answer to (7.2). In this section, using Theorem 6.6, we continue this research by obtaining the following inequality: for a more precise statement see Theorem 7.6. Note that the right inequality is just (7.2), but we give here our proof in order to show that, alternatively, one can obtain this from Theorem 6.6. In order to achieve our goal, we also need to prove the Lemmas 7.3 to 7.5 in this section. These lemmas deal with estimating the tail of an infinite series involving standard binomial coefficients.
Proof By simplifying quotients formed by taking each expression in k + 1 divided by the original expression in k.
Now we apply Lemma 7.2 to obtain, where the latter inequality is by n > 2s. This proves (7.4). Moreover, the bound we obtained also works for and geometric series summation implies (7.6). The proof of (7.7) is analogous.
give at least a brief sketch of what led us to the final formulas and how we were led to conjecture special cases of related asymptotics. The final asymptotic formulas can easily be derived from our main result, Theorem 6.6 presented in Sect. 6. In Sect. 3 we proved the inequality e π 2n which was found by mathematical experiments. Our proof uses methods similar to those used in [6] and [1]. In our attempt to prove the following formula for the asymptotics of log p(n), we first tried to prove the log-version of (7.24). However, we soon realized that this inequality is not sharp enough in order to prove (7.25). We noted that the inequality for p(n) in [1, Lemma 2.2] can be used instead. This formula says that for n ≥ 1206, where μ(n) := π 6 √ 24n − 1. We observed that after taking the log of both sides, with some extra work, (7.25) can be proven. When we saw the asymptotics (1.3), discovered by Schoenfield and Kotesovec, we naturally wondered whether these asymptotics can also be proven by taking the log of an appropriate inequality. We observed that (7.26) is enough also to prove these asymptotics, and we observed that (7.26) can be used to prove an even more refined asymptotic formula that takes the form log p(n) ∼ π 2n 3 − log n − log 4