CUBIC ALTERNATING HARMONIC NUMBER SUMS ANTHONY SOFO Received 12 September, 2018; Accepted 23 January, 2019; Published 27 May, 2019

The Australian Journal of Mathematical Analysis and Applications

AJMAA

Volume 16, Issue 1, Article 14, pp. 1-14, 2019

CUBIC ALTERNATING HARMONIC NUMBER SUMS ANTHONY SOFO Received 12 September, 2018; accepted 23 January, 2019; published 27 May, 2019.

VICTORIA UNIVERSITY,COLLEGEOF ENGINEERINGAND SCIENCE,MELBOURNE CITY,AUSTRALIA. [email protected]

ABSTRACT. We develop new closed form representations of sums of cubic alternating harmonic numbers and reciprocal binomial coefﬁcients. We also identify a new integral representation for the ζ (4) constant.

Key words and phrases: Polylogarithm function, Alternating cubic harmonic numbers, Combinatorial series identities, Sum- mation formulas, Partial fraction approach, Binomial coefﬁcients. 2000 Mathematics Subject Classiﬁcation. Primary 05A19, 33C20. Secondary 11M06, 11B83.

1. INTRODUCTIONAND PRELIMINARIES In this paper we will develop identities, closed form representations of alternating cubic harmonic numbers and reciprocal binomial coefﬁcients, including integral representations, of the form: ∞ n+1 X (−1) H3 (1.1) Ω(k, p) = n , n + k n =1 np k

th for p = 0, 1 and k ∈ N0. Here, the n harmonic number n X 1 Z 1 1 − tn (1.2) H = = γ + ψ (n + 1) = dt, H := 0 n r 1 − t 0 r=1 0 and as usual, γ denotes the Euler-Mascheroni constant and ψ(z) is the Psi (or Digamma) function defined by d Γ0(z) Z z ψ(z) := {log Γ(z)} = or log Γ(z) = ψ(t) dt. dz Γ(z) 1 For sums of harmonic numbers with positive terms [10], [27], [28] and [29] have given many results, including sums of the form ∞ X H3 n . n + k n =1 np k Other results are given by [5] and [17]. Let R and C denote, respectively the sets of real and complex numbers and let N := {1, 2, 3, ···} be the set of positive integers, and N0 := N ∪ {0} . λ A generalized binomial coefficient µ (λ, µ ∈ C) is defined, in terms of the familiar gamma function, by λ Γ(λ + 1) := , λ, µ ∈ . µ Γ(µ + 1) Γ (λ − µ + 1) C The Pochhammer symbol (λ)ν (λ, ν ∈ C) is also defined in terms of the gamma function, by  1 (ν = 0; λ ∈ \{0}) Γ(λ + ν)  C (λ) := = . ν Γ(λ)  λ(λ + 1) ··· (λ + n − 1) (ν = n ∈ N; λ ∈ C), (m) A generalized harmonic number Hn of order m is defined, for positive integers n and m, as follows: n X 1 (m) H(m) := , m, n ∈ and H := 0, m ∈ . n rm N 0 N r=1 In the case of non-integer values of n such as (for example) a value ρ ∈ R, the generalized (m+1) harmonic numbers Hρ may be defined, in terms of the polygamma functions dn dn+1 ψ(n)(z) := {ψ(z)} = {log Γ(z)}, n ∈ , dzn dzn+1 N0 by (−1)m (1.3) H(m+1) = ζ (m + 1) + ψ(m) (ρ + 1) ,H(m+1) = 0 ρ m! 0 (ρ ∈ R \ {−1, −2, −3, ···} ; m ∈ N) ,

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(α) r where ζ (z) is the Riemann zeta function. The evaluation of the polygamma function ψ a at rational values of the argument can be explicitly done via a formula as given by Kölbig [7], or Choi and Cvijovic [2] in terms of the polylogarithmic or other special functions. Some speciﬁc values are listed in the books[15], [21] and [22]. The polylogarithm or de-Jonquière function Lip (z), is deﬁned as, ∞ X zn Li (z) := , p ∈ when |z| < 1; < (p) > 1 when |z| = 1. p np C n =1 Some results for sums of alternating and non-alternating harmonic numbers may be seen in the works of [3], [4], [12], [13], [14], [16], [17], [18], [19], [20], [23], [24], [26] and references therein. Some explicit, and closely related results may also be seen in the well presented papers [9] and [25]. The following lemma will be useful in the development of the main theorems.

Lemma 1.1. Let r be a positive integer and p ∈ N. Then: r j X (−1) 1 (p) (p) (p) (1.4) F (p, r) = p = p H r + H r−1 − H r+1 j 2 [ 2 ] [ 2 ] 2[ 2 ]−1 j =1 where [x] is the integer part of x, and when p = 1,

r j X (−1) (1.5) F (1, r) = = H r − Hr. j [ 2 ] j =1 For p = 2,

r j X (−1) 1 (2) (2) 1 (1.6) F (2, r) = 2 = H r − H r+1 1 − ζ (2) . j 4 [ 2 ] [ 2 ]− 2 2 j =1 Proof. The proof is given in the paper [11]. Lemma 1.2. The following identity holds, ∞ n+1 X (−1) H3 5 3 1 9 (1.7) X (0) = n = ζ (4) + ζ (2) ln2 2 − ln4 2 − ζ (3) ln 2, n 8 4 4 8 n =1

1 2 Z log (1 − x) log (1 + x) + Li2 (−x) (1.8) = dx, 0 x (1 + x)

where Lip (·) is the polylogarithm function. Proof. Consider ∞ X tn (1.9) V (j, t) = n + j n =1 n j

∞ ∞ X tn Γ(n)Γ(j + 1) X = = tn B (n, j + 1) , Γ(n + j + 1) n =1 n =1

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where Γ(·) is the gamma function and B (·, ·) is the beta function. Now

Z 1 j ∞ Z 1 j (1 − x) X n (1 − x) V (j, t) = (tx) dx = t dx, x 1 − tx 0 n =1 0 and differentiating p times with respect to j and then letting j → 0 with t = −1, results in Z 1 p (p) log (1 − x) p 1 V (0, −1) = dx = (−1) p!Lip+1 . 0 1 + x 2

For p = 3 1 (1.10) V (0, −1)(3) = −6Li 4 2 From (1.9) we also have ∞ n (p) X (−1) σn V (0, −1) = n n =1 where " !# dp n + j −1 σn = lim , j→0 djp j when p = 3, ∞ n+1 (3) X (−1) φ 1 (1.11) V (0, −1) = n = 6Li n 4 2 n =1

3 (2) (3) and φn = Hn + 3HnHn + 2Hn . From the paper [8], we have the results

∞ n+1 (3) X (−1) Hn 19 3 = ζ (4) − ζ (3) ln 2, n 16 4 n =1

∞ n+1 (2) X (−1) HnHn 1 = 2Li − ζ (4) − 4ζ (2) ln2 2 n 4 2 n =1 1 7 + ln4 2 + ζ (3) ln 2, 4 8 and substituting into (1.11) we have the result (1.7). The representation of the integral (1.8) is obtained in the following way. From [11], we can express, for p ∈ N

p Z 1 Z 1 p n−1 p p Hn Y Y (−1) p = ··· xj ln (1 − xj) dxj n 0 0 j=1 j=1 R 1 R 1 where 0 ··· 0 is a p- fold integration procedure, for p = 3, 3 Z 1 Z 1 Z 1 n Hn (x1x2x3) ln (1 − x1) ln (1 − x2) ln (1 − x3) − 3 = dx1dx2dx3. n 0 0 0 x1x2x3 (1.12)

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Now,

∞ n+1 X (−1) H3 Z 1 Z 1 Z 1 n = − ln (1 − x) ln (1 − y) ln (1 − z) n n =1 0 0 0 ∞ X n−1 × n2 (−xyz) dxdydz n =1

Z 1 Z 1 Z 1 ln (1 − x) ln (1 − y) ln (1 − z) (1 − xyz) = − 2 dxdydz 0 0 0 (1 + xyz) Z 1 Z 1 ln (1 − x) ln (1 − y) (1 − ln (1 + xy)) = 2 dxdy 0 0 (1 + xy) Z 1 ln (1 − x) ln2 (1 + x) x = − Li2 dx 0 x (1 + x) 2 1 + x and applying Landen’s identity x ln2 (1 + x) Li = − − Li (−x) 2 1 + x 2 2 we obtain (1.8). By association we conclude Z 1 log (1 − x) log2 (1 + x) + Li (−x) 5 3 2 dx = ζ (4) + ζ (2) ln2 2 0 x (1 + x) 8 4 1 9 − ln4 2 − ζ (3) ln 2 4 8

∞ n+1 X (−1) H3 = X (0) = n . n n =1

An alternate manipulation of (1.7) leads to the following result and adds to the results on cubic sums some of which are published in [10]. Lemma 1.3. X H3 5 67 3 3 2n = ζ (4) + ζ (3) + 2ζ (2) + ζ (2) ln2 2 − ζ (2) ln 2 2n (2n − 1) 8 16 4 2 n≥1 9 1 3 − ζ (3) ln 2 + ln 2 − ln4 2 + ln3 2 − ln2 2. 8 4 2 (1.13) Proof. By re-arrangement

∞ 1 3 3 ! X H2n − H X (0) = 2n − 2n 2n − 1 2n n =1 3 2 X H − 3H X 3H2n X 1 = 2n 2n + − . 2n (2n − 1) 4n2 (2n − 1) 8n3 (2n − 1) n≥1 n≥1 n≥1

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Hence X H3 − 3H2 5 3 1 9 2n 2n = ζ (4) + ζ (2) ln2 2 − ln4 2 − ζ (3) ln 2 2n (2n − 1) 8 4 4 8 n≥1 33 3 3 + ln 2 + ζ (3) − ζ (2) + ln2 2 16 2 2 1 1 − ζ (3) − 3 ln 2 − ζ (2) . 8 4

Now we can evaluate X H2 5 1 1 3 2n = ζ (2) + ln 2 − ln2 2 − ζ (2) ln 2 + ln3 2 + ζ (3) , 2n (2n − 1) 4 2 3 4 n≥1 therefore (1.13) follows. Lemma 1.4. The following identity holds, ∞ n+1 X (−1) H3 X (1) = n n + 1 n =1

5 3 1 9 = − ζ (4) − ζ (2) ln2 2 + ln4 2 + ζ (3) ln 2 16 4 4 8 (1.14) 5 = −X (0) + ζ (4) , 16

1 2 5 Z log (1 − x) log (1 + x) + Li2 (−x) (1.15) = ζ (4) − dx. 16 0 x (1 + x) Proof. Consider X (1) and by a change of summation index ∞ n ∞ n+1 3 X (−1) H3 X (−1) 1 X (1) = n−1 = − H − n n n n n =1 n =1

∞ n+1 3 ∞ n+1 ∞ n+1 X (−1) H X (−1) HnHn−1 X (−1) = − n − 3 + . n n2 n4 n =1 n =1 n =1 Using lemma 1.2 9 7 X (1) = −X (0) − ζ (4) + ζ (4) 16 8 and (1.14) follows. The integral identity (1.15) follows directly from (1.8). We can also note that a similar calculation as that of lemma 1.3, yields the identity 3 X H 3H2nH2n−1 3 2n + = −X (0) + ζ (4) 2n (2n + 1) 4n2 (2n − 1) 8 n≥1

9 3 3 69 + ζ (2) − ln2 2 − ζ (2) ln 2 + ln3 2 + ζ (3) . 4 2 2 16

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Remark 1.1. It may be noticed from lemma 1.2 and lemma 1.4 that 5 X (0) + X (1) = ζ (4) 16 in which case manipulating the integral identities we obtain 8 Z 1 log (1 − x) log2 (1 + x) (1.16) ζ (4) = − dx. 3 0 x Further extraction from X (0) and X (1) gives us the results Z 1 2 log (1 − x) log (1 + x) 1 1 4 dx = 2ζ (3) ln 2 + 2Li4 + ln 2 0 1 + x 2 3

(1.17) −2ζ (4) − ζ (2) ln2 2,

Z 1 log (1 − x) Li2 (−x) 7 1 1 4 dx = ζ (3) ln 2 + 2Li4 + ln 2 0 x (1 + x) 8 2 12 (1.18) 1 −ζ (4) − ζ (2) ln2 2 4 and Z 1 log (1 − x) Li2 (−x) 7 1 1 4 dx = − ζ (3) ln 2 − 2Li4 − ln 2 0 x 4 2 12 11 1 + ζ (4) + ζ (2) ln2 2 4 2

∞ n+1 X (−1) Hn = . n3 n =1 The integrals (1.16), (1.17) and (1.18) cannot be analytically evaluated by "Mathematica". There are many integral representations of powers of Pi, for example

Z 1 √ Z 1 logp (1 − x) π = 4 1 − x2dx, (−1)p p!ζ (p + 1) = dx. 0 0 x

Guillera and Sondow [6] gave Z 1 Z 1 Z 1 Z 1 2 dydx 3 ln xy dydx π = 8 2 2 , π = −16 2 2 , 0 0 1 − x y 0 0 1 + x y

and amongst many other results [1] obtained the intriguing representation (after some manipulation) ∞ π X n+1 1 ln = (−1) ln 1 + , 2 n n =1 however the author believes (1.16) is a new representation.

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Lemma 1.5. Let r ≥ 2 be a positive integer, deﬁning

∞ n+1 X (−1) Hn (1.19) S (r) := n + r n =1 then S (r) has the recurrence relation 1 (−1)r+1 S (r) + S (r − 1) = (1 + (−1)r) ln 2 − F (1, r − 1) r − 1 r − 1 with representation

r+1 r S (r) = (−1) S (1) + (−1) 2Hr−1 − H r−1 ln 2 [ 2 ]

r−1 r X 1 + (−1) H j − Hj , j [ 2 ] j =1 (1.20) 1 2 1 2 where S (0) = 2 ζ (2) − ln 2 ,S (1) = 2 ln 2 and F (1, r − 1) is given by (1.5). Proof. The proof is detailed in [11]. Lemma 1.6. For a positive integer r ≥ 2, we deﬁne

∞ n+1 X (−1) H2 T (r) := n n + r n =1 then T (r) has the recurrence relation 2S ζ (2) ln2 2 ln 2 T (r) + T (r − 1) = − r−1 + − + r − 1 2 (r − 1) r − 1 (r − 1)2 1 − 2 H r−1 − H r−2 , 2 (r − 1) 2 2 with representation

1 (1.21) T (r) = (−1)r+1 T (1) + (−1)r+1 F (1, r − 1) ζ (2) 2

+ (−1)r F (1, r − 1) ln2 2 + (−1)r+1 F (2, r − 1) ln 2

r−1 ! j H j − H j−1 r X (−1) + (−1) 2S (j) + 2 2 , j 2j j =1

3 1 3 1 1 1 3 1 with T (0) = 4 ζ (3) + 3 ln 2 − 2 ζ (2) ln 2, T (1) = 2 ζ (2) ln 2 − 3 ln 2 − 4 ζ (3), S (j) is given by (1.19) and F (·, r − 1) is given by (1.4). Proof. The proof is detailed in [11].

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Lemma 1.7. Let r ∈ N\{1}, then ∞ n+1 X (−1) HnHn−1 Y (r − 1) = 3 n (n + r − 1) n =1

3 1 1 1 = ζ (3) + ln3 2 − ζ (2) ln 2 − T (r − 1) (r − 1) 8 3 2 3 (1.22) + ζ (2) − ln2 2 − 2S (r − 1) , 2 (r − 1)2 where S (r − 1) is given by (1.19) and T (r − 1) is given by (1.21). For r = 1 we have

∞ n+1 X (−1) HnHn−1 9 (1.23) Y (0) = 3 = − ζ (4) . n2 16 n =1 Proof. Consider ∞ n+1 X (−1) HnHn−1 Y (r − 1) = 3 n (n + r − 1) n =1 ∞ 3 X n+1 1 1 1 = (−1) H H − − r − 1 n n n n n + r − 1 n =1

∞ 2 2 3 X n+1 H Hn H Hn = (−1) n − − n + r − 1 n n2 n + r − 1 n (n + r − 1) n =1

3 3 1 1 5 = ζ (3) + ln3 2 − ζ (2) ln 2 − ζ (3) − T (r − 1) r − 1 4 3 2 8 ∞ 3 X n+1 Hn 1 1 + (−1) − r − 1 r − 1 n n + r − 1 n =1 3 3 1 1 5 = ζ (3) + ln3 2 − ζ (2) ln 2 − ζ (3) − T (r − 1) r − 1 4 3 2 8 3 1 1 + ζ (2) − ln2 2 − S (r − 1) (r − 1)2 2 2 and (1.22) follows. For the case r = 1, we have ∞ n+1 ∞ n+1 1 X (−1) HnHn−1 X (−1) Hn Hn − Y (0) = 3 = 3 n n2 n2 n =1 n =1

∞ 2 X n+1 H Hn = 3 (−1) n − n2 n3 n =1 and (1.23) follows.

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Lemma 1.8. Let r ∈ N, then ∞ n+1 X (−1) H3 X (r) = n n + r n =1 then 3ζ (3) ln3 2 ζ (2) ln 2 X (r) = − (−1)r X (1) + 3 (−1)1+r + − F (1, r − 1) 8 3 2

(−1)r+1 + 2ζ (2) − 3 ln2 2 F (2, r − 1) + (−1)1+r ln 2F (3, r − 1) 2 r−1 1+r X j 1 3 S (j) + (−1) (−1) H j−1 − H j − T (j) + 2j3 2 2 j j j=1 (1.24) where X (0) is given by (1.8), X (1) is given by (1.14), S (j) is given by (1.20), T (j) is given by (1.21) and F (·, r − 1) is given by (1.4). Proof. Consider ∞ n+1 X (−1) H3 X (r) = n n + r n =1 and by a change of summation index ∞ n 3 ∞ n+1 X (−1) H X (−1) HnHn−1 X (r) = n−1 = −X (r − 1) + 3 n + r − 1 n (n + r − 1) n =2 n =1 ∞ n+1 X (−1) + . n3 (n + r − 1) n =1 Now using lemma 1.5, 1.6 and 1.7

3ζ (3) ζ (2) ln 2 X (r) = −X (r − 1) + 3Y (r − 1) + − + 4 (r − 1) 2 (r − 1)2 (r − 1)3 1 + 3 H r−2 − H r−1 , 2 (r − 1) 2 2 hence we have the recurrence relation 3 3 1 1 3T (r − 1) X (r) + X (r − 1) = ζ (3) + ln3 2 − ζ (2) ln 2 − r − 1 8 3 2 r − 1 1 ln 2 + 2ζ (2) − 3 ln2 2 + 2 (r − 1)2 (r − 1)3 3S (r − 1) 1 − 2 + 3 H r−2 − H r−1 (r − 1) 2 (r − 1) 2 2 for r ≥ 2. The recurrence relation is solved by the subsequent reduction of the X (r) ,X (r − 1) , ..., X (1) terms and using lemma 1.1, ﬁnally arriving at the relation (1.24). The next few theorems relate the main results of this investigation, namely the closed form and integral representation of (1.1).

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2. CLOSED FORMAND INTEGRAL IDENTITIES We now prove the following theorems. Theorem 2.1. Let k ≥ 1 be real positive integer, then from (1.1) with p = 0 we have

∞ n+1 3 k X (−1) H X r+1 k (2.1) Ω(k, 0) = n = (−1) r X (r) n + k r n =1 r =1 k

" # Z 1 Z 1 Z 1 ln (1 − x) ln (1 − y) ln (1 − z) 2, 2, 2, 2

= 4F3 − xyz dxdydz 0 0 0 1 + k 1, 1, 2 + k where X (r) is given by (1.24). Proof. Consider the expansion

∞ n+1 ∞ n+1 X (−1) H3 X (−1) k! H3 Ω(k, 0) = n = n n + k (n + 1) n =1 n =1 1+k k

∞ k X n+1 X Φr = (−1) k! H3 n n + r n =1 r =1 where      n + r  (−1)r+1 r k Φ = lim = , r k n→−r  Q  k! r  n + r r=1 hence

k ∞ n+1 3 X r+1 k X (−1) H Ω(k, 0) = (−1) r n r n + r r =1 n =1 k X r+1 k = (−1) r X (r) , r r =1 and using lemma 1.8, gives k X r+1 k (2.2) Ω(k, 0) = (−1) r X (r) , r r =1 hence (2.1) follows. For the integral representation we employ the same technique as lemma 1.2.

The other case of Ω(k, 1) , can be evaluated in a similar fashion. We list the result in the next theorem.

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Theorem 2.2. Under the assumptions of Theorem 2.1, we have, ∞ n+1 X (−1) H3 Ω(k, 1) = n n + k n =1 n k

" # Z 1 Z 1 Z 1 ln (1 − x) ln (1 − y) ln (1 − z) 2, 2, 2

= 3F2 − xyz dxdydz, 0 0 0 1 + k 1, 2 + k

k X r+1 k (2.3) = X (0) − (−1) X (r) . r r =1 Proof. The proof follows directly from theorem 2.1 and using the same technique. Example 2.1. Some illustrative examples follow.

∞ n+1 X (−1) H3 255 21 261 15 Ω (3, 0) = n = − ζ (2) − ζ (3) − ζ (4) n + 3 16 2 16 4 n =1 3 63 +27ζ (2) ln 2 + 36 ln2 2 − 18 ln3 (2) − ln 2 2 27 −9ζ (2) ln2 2 + 3 ln4 2 + ζ (3) ln 2, 2

∞ n+1 X (−1) H3 32575 499 325 85 Ω (4, 1) = n = − + ζ (2) + ζ (3) + ζ (4) n + 4 1296 36 16 16 n =1 n 4 151 68 10636 −34ζ (2) ln 2 − ln2 2 + ln3 (2) + ln 2 3 3 216

+12ζ (2) ln2 2 − 4 ln4 2 − 18ζ (3) ln 2. Remark 2.1. Following a similar argument as that of Lemma 1.3, we note ﬁrst from (2.3), with k = 2 ∞ n+1 X (−1) H3 25 1 1 15 n = ζ (4) − + ζ (2) + ζ (3) + ln 2 n (n + 1) (n + 2) 32 2 2 16 n =1 3 3 1 3 9 − ζ (2) ln 2 − ln2 2 − ln4 2 + ln3 2 + ζ (2) ln2 2 − ζ (3) ln 2. 2 2 2 2 4 Also ∞ X H3 1 n = 5ζ (4) − 2ζ (3) − ζ (2) − , n (n + 1) (n + 2) 2 n =1 and hence, ∞ X H3 185 1 17 2n−1 = ζ (4) − 1 − ζ (2) − ζ (3) + ln 2 n (2n − 1) (2n + 1) 32 2 16 n =1

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3 3 1 3 9 − ζ (2) ln 2 − ln2 2 − ln4 2 + ln3 2 + ζ (2) ln2 2 − ζ (3) ln 2. 2 2 2 2 4 In the following remark we list, without proof, an extension of the results related to Lemma 1.2.

3 (2) (3) Remark 2.2. Let φn = Hn + 3HnHn + 2Hn , then ∞ n+1 X (−1) φ 3 21 1 n = ζ (2) ln2 2 − ζ (3) ln 2 + 6Li n (n + 1) 2 4 4 2 n =1 1 15 − ln4 2 + ζ (4) . 4 8

Z 1 3 log (1 − x) log (1 + x) 3 2 21 1 2 dx = ζ (2) ln 2 − ζ (3) ln 2 + 6Li4 0 x 2 4 2 1 33 − ln4 2 − ζ (4) . 4 8

∞ n+1 X (−1) φ Z 1 log3 (1 − x) log (1 + x) 1 n = − dx n + 1 1 + x x 1 + x n =1 0 21 3 1 15 = ζ (3) ln 2 − ζ (2) ln2 2 + ln4 2 − ζ (4) . 4 2 4 8

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