MULTIPLE HARMONIC SUMS AND MULTIPLE HARMONIC STAR SUMS ARE (NEARLY) NEVER INTEGERS

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1 #A0 INTEGERS 7 (207) MULTIPLE HARMONIC SUMS AND MULTIPLE HARMONIC STAR SUMS ARE (NEARLY) NEVER INTEGERS Khodabakhsh Hessami Pilehrood The Fields Institute for Research in Mathematical Sciences, Toronto, Canada hessamik@gmail.com Tatiana Hessami Pilehrood The Fields Institute for Research in Mathematical Sciences, Toronto, Canada hessamit@gmail.com Roberto Tauraso Dipartimento di Matematica, Università di Roma Tor Vergata, Roma, Italy tauraso@mat.uniroma2.it Received: /4/6, Accepted: 3/5/7, Published: 4/4/7 Abstract It is well known that the harmonic sum H n () = P n k= k is never an integer for n >. In 946, Erdős and Niven proved that the nested multiple harmonic sum H n ({} r ) = P applek < <k rapplen k k r can take integer values only for a finite number of positive integers n. In 202, Chen and Tang refined this result by showing that H n ({} r ) is an integer only for (n, r) = (, ) and (n, r) = (3, 2). In 204, Hong and Wang extended these results to the general arithmetic progression case. In this paper, we consider the integrality problem for arbitrary multiple harmonic and multiple harmonic star sums and show that none of these sums is an integer with some natural exceptions like those mentioned above.. Introduction A classic result of elementary number theory is that even though the partial sum of the harmonic series P n k= /k increases to infinity, it is never an integer for n >. Apparently the first published proof goes back to Theisinger [9] in 95, and, since then, it has been proposed as a challenging problem in several textbooks; among all we mention [5, p.6, Problem 30], [7, p.33, Problem 37], [8, p.53, Problem 250]. In 946, Erdős and Niven [3] proved a stronger statement: there is only a finite number of integers n for which there is a positive integer r apple n such that the r-th

2 INTEGERS: 7 (207) 2 elementary symmetric function of, /2,..., /n, that is applek < <k rapplen k k r, is an integer. In 202, Chen and Tang [] refined this result and succeeded to show that the above sum is not an integer with the only two exceptions: either n = r = or n = 3 and r = 2. Recently, this theme has been further developed by investigating the case when the variables of the elementary symmetric functions are /f(), /f(2),..., /f(n) with f(x) being a polynomial of nonnegative integer coe cients: see [4] and [0] for f(x) = ax + b and see [6] for f of degree at least two. In this paper, we consider the integrality problem for sums which are not necessarily symmetric with respect to their variables. For an r-tuple of positive integers s = (s,..., s r ) and an integer n r, we define two classes of multiple harmonic sums: the ordinary multiple harmonic sum (MHS) H n (s,..., s r ) = k s applek < <k rapplen and the star version (MHS-star also denoted by S in the literature) H? n(s,..., s r ) = k s applek apple applek rapplen, () ksr r. (2) ksr r The number l(s) := r is called the length and s := P r j= s j is the weight of the multiple harmonic sum. Note that H n ({m} r ) is the r-th elementary symmetric function of /f(), /f(2),..., /f(n) with f(x) = x m. The multiple sums () and (2) are of a certain interest because by taking the limit as n goes to when s r > (otherwise the infinite sums diverge) we get the so-called multiple zeta value and the multiple zeta star value, lim H n(s,..., s r ) = (s,..., s r ) and lim n! n! H? n(s,..., s r ) =? (s,..., s r ), respectively. Note that the integrality of the MHS-star is quite simple to study. Theorem. Let n of H? (s ) =. r. Then H? n(s,..., s r ) is never an integer with the exception Proof. If n >, then by Bertrand s postulate, there is at least a prime p such that n/2 < p apple n. Then p apple n < 2p and H? (s) = applek apple applek rapplen 9i : k i6=p k s ksr r + applek apple applek rapplen 8i, k i=p k s ksr r = a bp t + p s

3 INTEGERS: 7 (207) 3 where gcd(b, p) = and t < s. Assume that H? (s) = m 2 N +. Then bmp s = ap s t + b, which is a contradiction because p divides the left-hand side and p does not divide the right-hand side. On the other hand, the case of the ordinary MHS is much more intricate. Our result is given below whereas the entire next section is dedicated to its proof. Theorem 2. Let n r. Then H n (s,..., s r ) is never an integer with the exceptions of H (s ) = and H 3 (, ) =. Summarizing and generalizing the results of Theorem, Theorem 2 and papers [, 4, 6, 0], we propose the following problem. Problem. Let n and r be positive integers such that r apple n. Let s = (s,..., s r ) be an r-tuple of positive integers and f(x) be a nonzero polynomial of nonnegative integer coe cients, f(x) 6= x. Let H n,f (s,..., s r ) = f(k ) s f(k r ) sr and H? n,f(s,..., s r ) = applek < <k rapplen applek apple applek rapplen f(k ) s f(k r ) sr. Find all values of n and s for which H n,f (s,..., s r ) and H? n,f (s,..., s r ) are integers. Note that when f(x) = x, the multiple sums H n,f (s,..., s r ) and H? n,f (s,..., s r ) coincide with multiple harmonic sums () and (2) and therefore, the integrality problem in this case is addressed in Theorem and Theorem 2, respectively. Throughout the paper, all the numerical computations were performed by using Maple TM. 2. Proof of Theorem 2 The main tool in the proof of Theorem is Bertrand s postulate. Unfortunately, such result is not strong enough to imply Theorem 2. The statement described in the following remark will replace it. Notice that the same argument had been used by Erdős and Niven in [3] and successively taken up in [, 4] and [0]. Remark. Let apple r apple n. If there is a prime p 2 ( n r+, n r ] and r < p, then H n(s) is not an integer when l(s) = r. Indeed, since < p < 2p < < rp apple n < (r +)p, it follows that H n (s) = applek < <k rapplen 9i : p-k i k s ksr r + applek < <k rapplen 8i, p k i k s ksr r = a bp t + cp s

4 INTEGERS: 7 (207) 4 where gcd(b, p) = gcd(c, p) = and t < s. Assume that H n (s) = m 2 N +. Then bcmp s = cap s t + b, which is a contradiction because p divides the left-hand side and p does not divide the right-hand side. For any integer r, let A r = [ p2p[rp, (r + )p) where P is the set of primes. Note that by Bertrand s postulate, A = [2, +). The crucial property of the set A r is that n 2 A r if and only if there exists a prime p such that p 2 ( n r+, n r ]. The next lemma is a variation of [4, Lemma 2.4]. Lemma. For any positive integer r, A r is cofinite, i.e., N \ A r is finite. Let m r = max(n \ A r ) +. Then the first few values of m r are as follows: r m r /r r m r /r and for any positive integer r 24, m r apple (r + ) exp( p.4 r). Proof. Let r and let n be such that x := n/(r + ) max(exp( p.4 r), 3275). Then ln 2 (x).4 r implies n r = x + x r x +.4 x ln 2 (x) x + x 2 ln 2 (x). Moreover, by [2, Theorem ], there is a prime p such that x n p 2 x, x + 2 ln 2 (x) r +, n r which implies that n 2 A r. This proves that N \ A r is finite. Note that if r 47, then exp( p.4 r) On the other hand, if 24 apple r apple 46 and n 2 [(r + ) exp( p.4 r), (r + )3275), one can verify directly that n 2 A r. In order to compare values of multiple harmonic sums of the same length, the following definition and lemma will be useful. Definition. Let s = (s,..., s r ) and t = (t,..., t r ) be two r-tuples of positive integers. We say that s t, if w(s) w(t), i.e., if s + + s r t + + t r, and s apple t,..., s l apple t l, s l+ t l+,..., s r t r for some 0 apple l apple r. In particular, s t if s i t i for i =,..., r.

5 INTEGERS: 7 (207) 5 Lemma 2. Let s = (s,..., s r ) and t = (t,..., t r ) be two r-tuples of positive integers, and s t. Then for any positive integer n, H n (s,..., s r ) apple H n (t,..., t r ). Proof. Let 0 apple l apple r be such that s apple t,..., s l apple t l and s l+ t l+,..., s r t r. Now we compare corresponding terms k s ksr r and, where k < < k r, ktr r k t of multiple sums H n (s) and H n (t). Since s + + s r k t s t2 s2 k and therefore, 2 k t l s l l apple k t+ +t l (s + +s l ) l = k (s l+ t l+ )+ +(s r t r) l k apple k s l+ t l+ l+ k k s apple ksr r which implies H n (s) apple H n (t), as required. Remark 2. Notice that H n () = + n k=2 k t sr tr r, ktr r Hence, if apple r apple n then, by the previous lemma, Z n k apple + dx = ln(n) +. x H n (s,..., s r ) apple H n ({} r ) apple (H n()) r r! apple t + + t r, we have t+ +tr (s+ +sr) l (ln(n) + )r r! where the second inequality holds because each term of H n ({} r ) is contained r! times in the expansion of (H n ()) r. Let p be a prime and let p (q) be the p-adic order of the rational number q, that is, if a, b are coprime with p and n 2 Z, then p (ap n /b) = n. It is known that the p-adic order satisfies the inequality p (a + b) min( p (a), p (b)) where the equality holds if p (a) 6= p (b). The following lemma will be our basic tool to find an upper bound for the index s. Lemma 3. Given 2 apple r apple n and (s 2,..., s r ), there exists an integer M (which depends on (s 2,..., s r ) and n) such that H n (s, s 2,..., s r ) is never an integer for any positive integer s > M.

6 INTEGERS: 7 (207) 6 Proof. If r = n, then H n (s, s 2,..., s r ) is trivially not an integer for all s > 0. If r < n, then we have that H n (s,..., s r ) = nr+ k= c k k s where c k := k<k 2< <k rapplen k s2 2 > 0. ksr r Let p be the largest prime in [2, n r + ]. Then 2p > n r + (otherwise by Bertrand s postulate there is a prime q such that p < q < 2p apple n r + ). Let 0 B M := p (c p ), p (c p ) min applekapplen r+ C ( p (c k )) A. We will show that p (H n (s)) < 0, which implies that H n (s) is not an integer. Assume that s > M, then Now i) s > p (c p ) and ii) s > p (c p ) min By ii), we have that 0 p applekapplen r+ and by i), Therefore c k k s C A H n (s) = c p p s + min applekapplen r+ 0 p (H n (s)) = min B applekapplen r+ c k ck p = min k s k s. applekapplen r+ cp p = p (c p ) s < 0. p s 0 applekapplen r+ c k k s C A, p cp p s applekapplen r+ p (c k ) > p (c p ) C A = p cp p s ( p (c k )). < 0. cp s = p, p s Remark 3. If, in addition to the inequalities 2 apple p apple n r +, the prime p satisfies n/2 < p, then ( p (c k ) min max 2appleiappler s i, if apple k < p, p k<k 2< <k rapplen k s2 2 = ksr r 0, if p apple k apple n r +.

7 INTEGERS: 7 (207) 7 Therefore This means that min applekapplen r+ ( p (c k )) max s i. 2appleiappler M 0 := p (c p ) + max 2appleiappler s i M. When the assumption n/2 < p is met, we will prefer to use the bound M 0 to M because the bound M 0 is computationally easier to determine. Proof of Theorem 2. Since H n (s ) = H n(s? ), by Theorem, the statement holds for r =. If e(ln(n) + ) apple r, then by Remark 2 and the fact that e r > r r /r!, we have r (ln(n) + )r ln(n) + H n (s) apple < apple, r! r/e which implies that H n (s) can not be an integer. Assume that 2 apple r < e(ln(n) + ). Then exp(r/e ) < n and, since it can be verified that (r + ) exp( p.4 r) apple exp(r/e ) for r 30, it follows that n 2 A r and we are done as soon as we use Remark. Remark can be applied successfully even when 2 apple r apple 29 and n 9599 because m r apple 33r apple = 9599 apple n. Hence the cases remained to consider are 2 apple r apple 29 and r apple n < m r (with n 62 A r ). For r = 25, 26, 27, 28, 29, by Remark 2, H n (s) < H mr (s) apple H mr ({} r ) apple (ln(m r) + ) r where the last inequality can be easily verified numerically. For r = 9, 20, 2, 22, 23, 24, by Lemma 2, r! < H n (s) < H mr (s) apple H mr ({} r ) and the non-integrality of H n (s) is implied by the following evaluations: H m24 ({} 24 ) < <, H m23 ({} 23 ) < <, H m22 ({} 22 ) < <, H m2 ({} 2 ) < <, H m20 ({} 20 ) < <, H m9 ({} 9 ) < <.

8 INTEGERS: 7 (207) 8 The strategy to handle the cases where 2 apple r apple 8 is fairly more complicated because H mr ({} r ) >. The analysis is based on the numerical values presented in Tables, 2, and 3. Here we give a detailed explanation of how the data in such tables are calculated and used for r = 5. The other cases can be treated in a similar way. What turns out at the end is that the only exception for r 2 is H 3 (, ) =. We first determine the optimal set of length 5, that is, a set of 5-tuples such that the multiple harmonic sums with n = m 5 = 45 are the largest sums less than with small weights (columns 2 and 3): H m5 ({} 4, 2) < <, H m5 (, 2, 2,, ) < <, H m5 (, 4, {} 3 ) < <. Then, thanks to Lemma 2, the size of the set of multiple harmonic sums less than (and therefore not integral) can be extended. Let 5 apple n < m 5. If s 2 4, then H n (s, s 2, s 3, s 4, s 5 ) apple H n (, 4, {} 3 ) < H m5 (, 4, {} 3 ) <. If s 2 2 {2, 3} and there is s j 2 with 3 apple j apple 5, then H n (s, s 2, s 3, s 4, s 5 ) apple H n (, 2, s 3, s 4, s 5 ) apple H n (, 2, 2,, ) < H m5 (, 2, 2,, ) <. If s 2 = and s 3 3, then H n (s,, s 3, s 4, s 5 ) apple H n (,, 3,, ) apple H n (, 2, 2,, ) < H m5 (, 2, 2,, ) <. If s 2 =, s 3 = 2, and s 4 2 or s 5 2, then H n (s,, 2, s 4, s 5 ) apple H n (,, 2, s 4, s 5 ) apple H n (,, 2, 2, ) apple H n (, 2, 2,, ) < H m5 (, 2, 2,, ) <. If s 2 =, s 3 =, and s 4 3, then H n (s,,, s 4, s 5 ) apple H n ({} 3, 3, ) apple H n (, 2, 2,, ) < H m5 (, 2, 2,, ) <. If s 2 =, s 3 =, s 4 = 2, and s 5 2, then H n (s,,, 2, s 5 ) apple H n ({} 3, 2, s 5 ) apple H n ({} 3, 2, 2) apple H n (, 2, 2, {} 2 ) < H m5 (, 2, 2, {} 2 ) <. If s 2 = s 3 = s 4 = and s 5 2, then H n (s, {} 3, s 5 ) apple H n ({} 4, s 5 ) apple H n ({} 4, 2) < H m5 ({} 4, 2) <.

9 INTEGERS: 7 (207) 9 The set of 5-tuples of positive integers which are excluded by the analysis above is what we call the exclusion set (column 4): (s, 3, {} 3 ), (s, 2, {} 3 ), (s,, 2,, ), (s,,, 2, ), (s, {} 4 ) with s. By using Lemma 3 and Remark 3, we are able to give an upper bound for s (column 5) and therefore to reduce the size of the exclusion set to a finite number. Finally, it su ces to compute the rational number H n (s) for 5 apple n < m 5 and for s in the exclusion set with the upper bound for s established in the last column. For r = 5, none of them is an integer. Acknowledgement. Kh. Hessami Pilehrood and T. Hessami Pilehrood gratefully acknowledge support from the Fields Institute Research Immersion Fellowships. References [] Y. Chen and M. Tang, On the elementary symmetric functions of, /2,..., /n, Amer. Math. Monthly 9 (202), [2] P. Dusart, Inégalités explicites pour (x), (x), (x) et les nombres premiers, C. R. Math. Acad. Sci. Soc. R. Can. 2 (999), [3] P. Erdős and I. Niven, Some properties of partial sums of the harmonic series, Bull. Amer. Math. Soc. 52 (946), [4] S. Hong and C. Wang, The elementary symmetric functions of reciprocals of elements of arithmetic progressions, Acta Math. Hungar. 44 (204), [5] K. Ireland and M. Rosen, A Classical Introduction to Modern Number Theory, Springer- Verlag, 990. [6] Y. Luo, S. Hong, G. Qian, C. Wang The elementary symmetric functions of a reciprocal polynomial sequence, C. R. Math. Acad. Sci. Paris 352 (204), [7] M. B. Nathanson, Elementary Methods in Number Theory, Springer-Verlag, [8] G. Pólya and G. Szegő, Problems and Theorems in Analysis II, Springer-Verlag, 976. [9] L. Theisinger, Bemerkung über die harmonische Reihe, Monatsh. Math. 26 (95), [0] C. Wang and S. Hong, On the integrality of the elementary symmetric functions of, /3,..., /(2n ), Math. Slovaca 65 (205),

10 INTEGERS: 7 (207) 0 Table : Optimal sets, exclusion sets and upper bounds for 2 apple r apple 8. Length Optimal set of Upper bound for Exclusion set Upper bound r (s, s 2,..., s r ) H mr (s, s 2,..., s r ) with s for s 2 (, 2) < (s, ) apple 3 3 (,, 2) < (s, 2, ) apple 3 (, 3, ) < (s,, ) apple 3 (,,, 2) < (s, 3,, ) apple 4 4 (, 2, 2, ) < (s, 2,, ) apple 3 (, 4,, ) < (s,, 2, ) apple 3 (s,,, ) apple 3 (s, 3, {} 3 ) apple 4 ({} 4, 2) < (s, 2, {} 3 ) apple 3 5 (, 2, 2,, ) < (s,, 2,, ) apple 3 (, 4, {} 3 ) < (s,,, 2, ) apple 3 (s, {} 4 ) apple 3 (s, 4, {} 4 ) apple 5 ({} 4, 2, ) < (s, 3, {} 4 ) apple 4 (,, 3, {} 3 ) < (s, 2, 2, {} 3 ) apple 3 6 (, 2,, 2,, ) < (s, 2, {} 4 ) apple 3 (, 3, 2, {} 3 ) < (s,, 2, {} 3 ) apple 3 (, 5, {} 4 ) < (s,,, 2,, ) apple 3 (s, {} 5 ) apple 2 (s, 4, {} 5 ) apple 5 ({} 4, 2,, ) < (s, 3, {} 5 ) apple 4 (,, 3, {} 4 ) < (s, 2, 2, {} 4 ) apple 3 7 (, 2,, 2, {} 3 ) < (s, 2, {} 5 ) apple 3 (, 3, 2, {} 4 ) < (s,, 2, {} 4 ) apple 3 (, 5, {} 5 ) < (s,,, 2, {} 3 ) apple 4 (s, {} 6 ) apple 2 (s, 4, {} 6 ) apple 5 ({} 5, 2,, ) < (s, 3, {} 6 ) apple 4 (,, 3, {} 5 ) < (s, 2, 2, {} 5 ) apple 5 8 (, 2,, 2, {} 4 ) < (s, 2, {} 6 ) apple 3 (, 3, 2, {} 5 ) < (s,, 2, {} 5 ) apple 4 (, 5, {} 6 ) < (s,,, 2, {} 4 ) apple 4 (s, {} 3, 2, {} 3 ) apple 2 (s, {} 6 ) apple 2

11 INTEGERS: 7 (207) Table 2: Optimal sets, exclusion sets and upper bounds for 9 apple r apple. Length Optimal set of Upper bound for Exclusion set Upper bound r (s, s 2,..., s r ) H mr (s, s 2,..., s r ) with s for s (s, 5, {} 7 ) apple 7 (s, 4, {} 7 ) apple 5 (s, 3, 2, {} 6 ) apple 6 ({} 6, 2,, ) < (s, 3, {} 7 ) apple 4 (, 2, {} 3, 2, {} 3 ) < (s, 2, 2, {} 6 ) apple 3 (,, 2,, 2, {} 4 ) < (s, 2,, 2, {} 5 ) apple 4 9 (,,, 3, {} 5 ) < (s, 2,,, 2, {} 4 ) apple 3 (, 3,, 2, {} 5 ) < (s, 2, {} 7 ) apple 4 (, 2, 3, {} 6 ) < (s,, 3, {} 6 ) apple 4 (, 4, 2, {} 6 ) < (s,, 2, 2, {} 5 ) apple 3 (, 6, {} 7 ) < (s,, 2, {} 6 ) apple 3 (s,,, 2, {} 5 ) apple 3 (s, {} 3, 2, {} 4 ) apple 3 (s, {} 4, 2, {} 3 ) apple 3 (s, {} 8 ) apple 2 (s, 5, {} 8 ) apple 6 (s, 4, {} 8 ) apple 6 (s, 3, 2, {} 7 ) apple 4 ({} 6, 2, {} 3 ) < (s, 3, {} 8 ) apple 4 (, 2, {} 3, 2, {} 4 ) < (s, 2, 2, {} 7 ) apple 4 (,, 2,, 2, {} 5 ) < (s, 2,, 2, {} 6 ) apple 4 0 (,,, 3, {} 6 ) < (s, 2,,, 2, {} 5 ) apple 3 (, 3,, 2, {} 6 ) < (s, 2, {} 8 ) apple 3 (, 2, 3, {} 7 ) < (s,, 3, {} 7 ) apple 4 (, 4, 2, {} 7 ) < (s,, 2, 2, {} 6 ) apple 3 (, 6, {} 8 ) < (s,, 2, {} 7 ) apple 4 (s,,, 2, {} 6 ) apple 3 (s, {} 3, 2, {} 5 ) apple 4 (s, {} 4, 2, {} 4 ) apple 3 (s, {} 9 ) apple 3 (s, 5, {} 9 ) apple 8 (s, 4, {} 9 ) apple 5 (s, 3, 2, {} 8 ) apple 6 ({} 6, 2, {} 4 ) < (s, 3,, {} 8 ) apple 5 (, 2,,, 2, {} 6 ) < (s, 2, 2, {} 8 ) apple 4 (,, 2, 2, {} 7 ) < (s, 2,, 2, {} 7 ) apple 3 (, 3,, 2, {} 7 ) < (s, 2,,, {} 7 ) apple 3 (, 2, 3, {} 8 ) < (s,, 3, {} 8 ) apple 5 (, 4, 2, {} 8 ) < (s,, 2, {} 8 ) apple 3 (, 6, {} 9 ) < (s,,, 2, {} 7 ) apple 3 (s, {} 3, 2, {} 6 ) apple 3 (s, {} 4, 2, {} 5 ) apple 3 (s, {} 0 ) apple 3

12 INTEGERS: 7 (207) 2 Table 3: Optimal sets, exclusion sets and upper bounds for 2 apple r apple 8. Length Optimal set of Upper bound for Exclusion set Upper bound r (s, s 2,..., s r ) H mr (s, s 2,..., s r ) with s for s (s, 4, {} 0 ) apple 5 (s, 3, {} 0 ) apple 4 ({} 6, 2, {} 5 ) < (s, 2, 2, {} 9 ) apple 3 (, 2,,, 2, {} 7 ) < (s, 2,, 2, {} 8 ) apple 4 2 (,, 2, 2, {} 8 ) < (s, 2, {} 0 ) apple 3 (, 3, 2, {} 9 ) < (s,, 3, {} 9 ) apple 4 (, 5, {} 0 ) < (s,, 2, {} 9 ) apple 4 (s,,, 2, {} 8 ) apple 3 (s, {} 3, 2, {} 7 ) apple 3 (s, {} 4, 2, {} 6 ) apple 3 (s, {} ) apple 3 (s, 4, {} ) apple 5 ({} 5, 2, {} 7 ) < (s, 3, {} ) apple 4 (, 2,, 2, {} 9 ) < (s, 2, 2, {} 0 ) apple 4 3 (,, 3, {} 0 ) < (s, 2, {} ) apple 3 (, 3, 2, {} 0 ) < (s,, 2, {} 0 ) apple 3 (, 5, {} ) < (s,,, 2, {} 9 ) apple 3 (s, {} 3, 2, {} 8 ) apple 3 (s, {} 2 ) apple 2 (s, 3, {} 2 ) apple 4 ({} 4, 2, {} 9 ) < (s, 2, {} 2 ) apple 3 4 (, 2, 2, {} ) < (s,, 2, {} ) apple 3 (, 4, {} 2 ) < (s,,, 2, {} 0 ) apple 3 (s, {} 3 ) apple 3 ({} 3, 2, {} ) < (s, 3, {} 3 ) apple 4 5 (, 2, 2, {} 2 ) < (s, 2, {} 3 ) apple 3 (, 4, {} 3 ) < (s,, 2, {} 2 ) apple 3 (s, {} 4 ) apple 2 6 (,, 2, {} 3 ) < (s, 2, {} 4 ) apple 3 (, 3, {} 4 ) < (s, {} 5 ) apple 3 7 (,, 2, {} 4 ) < (s, 2, {} 5 ) apple 3 (, 3, {} 5 ) < (s, {} 6 ) apple 3 8 (, 2, {} 6 ) < (s, {} 7 ) apple 3