A new lower bound for k(3/2)kk Tome 19, no1 (2007), p. 311-323.
<http://jtnb.cedram.org/item?id=JTNB_2007__19_1_311_0>
© Université Bordeaux 1, 2007, tous droits réservés.
L’accès aux articles de la revue « Journal de Théorie des Nom-bres de Bordeaux » (http://jtnb.cedram.org/), implique l’accord avec les conditions générales d’utilisation (http://jtnb.cedram. org/legal/). Toute reproduction en tout ou partie cet article sous quelque forme que ce soit pour tout usage autre que l’utilisation à fin strictement personnelle du copiste est constitutive d’une infrac-tion pénale. Toute copie ou impression de ce fichier doit contenir la présente mention de copyright.
cedram
Article mis en ligne dans le cadre du
Centre de diffusion des revues académiques de mathématiques http://www.cedram.org/
de Bordeaux 19 (2007), 311-323
A new lower bound for k(3/2)
kk
par Wadim ZUDILIN
Résumé. Nous démontrons que pour tout entier k supérieur à une constante K effectivement calculable, la distance de (3/2)k à l’entier le plus proche est minorée par 0,5803k.
Abstract. We prove that, for all integers k exceeding some ef-fectively computable number K, the distance from (3/2)k to the nearest integer is greater than 0.5803k.
1. Historical overview of the problem
Let b · c and { · } denote the integer and fractional parts of a number, respectively. It is known [12] that the inequality {(3/2)k} ≤ 1 − (3/4)k for
k ≥ 6 implies the explicit formula g(k) = 2k+ b(3/2)kc − 2 for the least
integer g = g(k) such that every positive integer can be expressed as a sum of at most g positive kth powers (Waring’s problem). K. Mahler [10] used Ridout’s extension of Roth’s theorem to show that the inequality k(3/2)kk ≤ Ck, where kxk = min({x}, 1 − {x}) is the distance from x ∈ R
to the nearest integer, has finitely many solutions in integers k for any C < 1. The particular case C = 3/4 gives one the above value of g(k) for all k ≥ K, where K is a certain absolute but ineffective constant. The first non-trivial (i.e., C > 1/2) and effective (in terms of K) estimate of the form (1) 3 2 k > Ck for k ≥ K,
with C = 2−(1−10−64), was proved by A. Baker and J. Coates [1] by applying effective estimates of linear forms in logarithms in the p-adic case. F. Beuk-ers [4] improved on this result by showing that inequality (1) is valid with C = 2−0.9 = 0.5358 . . . for k ≥ K = 5000 (although his proof yielded the better choice C = 0.5637 . . . if one did not require an explicit evaluation of the effective bound for K). Beukers’ proof relied on explicit Padé approxi-mations to a tail of the binomial series (1 − z)m=Pm
n=0 m
n
(−z)nand was
later used by A. Dubickas [7] and L. Habsieger [8] to derive inequality (1) with C = 0.5769 and 0.5770, respectively. The latter work also includes
Manuscrit reçu le 8 juillet 2005.
The work is partially supported by grant no. 03-01-00359 of the Russian Foundation for Basic Research.
the estimate k(3/2)kk > 0.57434k for k ≥ 5 using computations from [6]
and [9].
By modifying Beukers’ construction, namely, considering Padé approxi-mations to a tail of the series
(2) 1 (1 − z)m+1 = ∞ X n=0 m + n m ! zn
and studying the explicit p-adic order of the binomial coefficients involved, we are able to prove
Theorem 1. The following estimate is valid:
3 2 k > 0.5803k= 2−k·0.78512916... for k ≥ K,
where K is a certain effective constant.
2. Hypergeometric background
The binomial series on the left-hand side of (2) is a special case of the generalized hypergeometric series
(3) q+1Fq A 0, A1, . . . , Aq B1, . . . , Bq z = ∞ X k=0 (A0)k(A1)k· · · (Aq)k k!(B1)k· · · (Bq)k zk, where (A)k= Γ(A + k) Γ(A) = ( A(A + 1) · · · (A + k − 1) if k ≥ 1, 1 if k = 0,
denotes the Pochhammer symbol (or shifted factorial). The series in (3) converges in the disc |z| < 1, and if one of the parameters A0, A1, . . . , Aq
is a non-positive integer (i.e., the series terminates) the definition of the hypergeometric series is valid for all z ∈ C.
In what follows we will often use the q+1Fq-notation. We will require
two classical facts from the theory of generalized hypergeometric series: the Pfaff–Saalschütz summation formula
(4) 3F2 −n, A, B C, 1 + A + B − C − n 1 = (C − A)n(C − B)n (C)n(C − A − B)n
(see, e.g., [11], p. 49, (2.3.1.3)) and the Euler–Pochhammer integral for the Gauss2F1-series (5) 2F1 A, B C z = Γ(C) Γ(B)Γ(C − B) Z 1 0 tB−1(1 − t)C−B−1(1 − zt)−Adt, provided Re C > Re B > 0 (see, e.g., [11], p. 20, (1.6.6)). Formula (5) is valid for |z| < 1 and also for any z ∈ C if A is a non-positive integer.
3. Padé approximations of the shifted binomial series
Fix two positive integers a and b satisfying 2a ≤ b. Formula (2) yields
3 2 3(b+1) = 27 8 b+1 = 3b+1 1 −1 9 −(b+1) (6) = 3b+1 ∞ X k=0 b + k b ! 1 9 k = 3b−2a+1 ∞ X k=0 b + k b ! 32(a−k) = an integer + 3b−2a+1 ∞ X k=a b + k b ! 32(a−k) ≡ 3b−2a+1 ∞ X ν=0 a + b + ν b ! 3−2ν (mod Z).
This motivates (cf. [4]) constructing Padé approximations to the function
(7) F (z) = F (a, b; z) = ∞ X ν=0 a + b + ν b ! zν = a + b b ! ∞ X ν=0 (a + b + 1)ν (a + 1)ν zν
and applying them with the choice z = 1/9.
Remark 1. The connection of F (a, b; z) with Beukers’ auxiliary series
H(ea,eb; z) from [4] is as follows: F (a, b; z) = z−a (1 − z)−b−1− a−1 X k=0 b + k b ! zk = H(−a − b − 1, a; z).
Although Beukers considers H(ea,eb; z) only for e
a,eb ∈ N, his construction
remains valid for anyea ∈ C,eb ∈ N and |z| < 1. However the diagonal Padé
approximations, used in [4] for H(z) and used below for F (z), are different. Taking an arbitrary integer n satisfying n ≤ b, we follow the general recipe of [5], [13]. Consider the polynomial
Qn(x) = a + b + n a + b ! 2F1 −n, a + n a + b + 1 x (8) = n X µ=0 a + n − 1 + µ µ ! a + b + n n − µ ! (−x)µ= n X µ=0 qµxµ∈ Z[x]
of degree n. Then Qn(z−1)F (z) = n X µ=0 qn−µzµ−n· ∞ X ν=0 a + b + ν b ! zν (9) = ∞ X l=0 zl−n n X µ=0 µ≤l qn−µ a + b + l − µ b ! = n−1 X l=0 rlzl−n+ ∞ X l=n rlzl−n= Pn(z−1) + Rn(z).
Here the polynomial (10) Pn(x) = n−1 X l=0 rlxn−l ∈ Z[x], where rl= l X µ=0 qn−µ a + b + l − µ b ! , has degree at most n, while the coefficients of the remainder
Rn(z) = ∞ X
l=n
rlzl−n
are of the following form: rl = n X µ=0 qn−µ a + b + l − µ b ! = n X µ=0 (−1)n−µ a + 2n − 1 − µ n − µ ! a + b + n µ ! a + b + l − µ b ! = (−1)n (a + b + n)! (a + n − 1)!n!b! n X µ=0 (−1)µ n µ ! (a + 2n − 1 − µ)!(a + b + l − µ)! (a + l − µ)!(a + b + n − µ)! = (−1)n (a + b + n)! (a + n − 1)!n!b! × (a + 2n − 1)!(a + b + l)! (a + l)!(a + b + n)! n X µ=0 (−n)µ(−a − l)µ(−a − b − n)µ µ!(−a − 2n + 1)µ(−a − b − l)µ = (−1)n(a + 2n − 1)!(a + b + l)! (a + n − 1)!(a + l)!n!b! ·3F2 −n, −a − l, −a − b − n −a − 2n + 1, −a − b − l 1 . If we apply (4) with the choice A = −a−l, B = −a−b−n and C = −a−b−l, we obtain rl= (−1)n (a + 2n − 1)!(a + b + l)! (a + n − 1)!(a + l)!n!b! · (−b)n(n − l)n (−a − b − l)n(a + n)n .
The assumed condition n ≤ b guarantees that the coefficients rl do not
vanish identically (otherwise (−b)n = 0). Moreover, (n − l)n = 0 for l
ranging over the set n ≤ l ≤ 2n − 1, therefore rl= 0 for those l, while
rl= (a + 2n − 1)!(a + b + l)! (a + n − 1)!(a + l)!n!b! × b!/(b − n)! · (l − n)!/(l − 2n)! (a + b + l)!/(a + b + l − n)! · (a + 2n − 1)!/(a + n − 1)! = (a + b + l − n)!(l − n)! n!(b − n)!(a + l)!(l − 2n)! for l ≥ 2n. Finally, Rn(z) = ∞ X l=2n rlzl−n= zn ∞ X ν=0 rν+2nzν (11) = zn 1 n!(b − n)! ∞ X ν=0 (a + b + n + ν)!(n + ν)! ν!(a + 2n + ν)! z ν = zn a + b + n b − n ! ·2F1 a + b + n + 1, n + 1 a + 2n + 1 z .
Using the integral (5) for the polynomial (8) and remainder (11) we arrive at
Lemma 1. The following representations are valid:
Qn(z−1) = (a + b + n)! (a + n − 1)!n!(b − n)! Z 1 0 ta+n−1(1 − t)b−n(1 − z−1t)ndt and Rn(z) = (a + b + n)! (a + n − 1)!n!(b − n)!z nZ 1 0 tn(1 − t)a+n−1(1 − zt)−(a+b+n+1)dt. We will also require linear independence of a pair of neighbouring Padé approximants, which is the subject of
Lemma 2. We have (12) Qn+1(x)Pn(x) − Qn(x)Pn+1(x) = (−1)n a + 2n + 1 a + n ! a + b + n b − n ! x.
Proof. Clearly, the left-hand side in (12) is a polynomial; its constant term
is 0 since Pn(0) = Pn+1(0) = 0 by (10). On the other hand,
Qn+1(z−1)Pn(z−1) − Qn(z−1)Pn+1(z−1)
= Qn+1(z−1) Qn(z−1)F (z) − Rn(z)
− Qn(z−1) Qn+1(z−1)F (z) − Rn+1(z)
and from (8), (11) we conclude that the only negative power of z originates with the last summand:
− Qn+1(z−1)Rn(z) = (−1)n a + 2n + 1 a + n ! z−n−1 1 + O(z)· a + b + n b − n ! zn 1 + O(z) = (−1)n a + 2n + 1 a + n ! a + b + n b − n ! 1 z + O(1) as z → 0. 4. Arithmetic constituents
We begin this section by noting that, for any prime p >√N , ordpN ! = N p and ordpN = λ N p , where λ(x) = 1 − {x} − {−x} = 1 + bxc + b−xc = ( 1 if x ∈ Z, 0 if x /∈ Z. For primes p >√a + b + n, let
ep = min µ∈Z − −a + n p + −a + n + µ p + µ p (13) − a + b + n p + a + b + µ p + n − µ p = min µ∈Z −a + n p − −a + n + µ p − µ p + a + b + n p − a + b + µ p − n − µ p ≤ min 0≤µ≤nordp a + n a + n + µ a + n + µ µ ! a + b + n n − µ ! = min 0≤µ≤nordp a + n − 1 + µ µ ! a + b + n n − µ ! and e0p= min µ∈Z − a + n + µ p + a + n p + µ p (14) − a + b + n p + a + b + µ p + n − µ p
= min µ∈Z a + n + µ p − a + n p − µ p + a + b + n p − a + b + µ p − n − µ p ≤ min 0≤µ≤nordp a + n + µ µ ! a + b + n n − µ ! . Set (15) Φ = Φ(a, b, n) = Y p>√a+b+n pep and Φ0= Φ0(a, b, n) = Y p>√a+b+n pe0p.
From (8), (10) and (13), (15) we deduce
Lemma 3. The following inclusions are valid:
Φ−1· a + n − 1 + µ µ ! a + b + n n − µ ! ∈ Z for µ = 0, 1, . . . , n, hence Φ−1Qn(x) ∈ Z[x] and Φ−1Pn(x) ∈ Z[x].
Supplementary arithmetic information for the case n replaced by n + 1 is given in
Lemma 4. The following inclusions are valid:
(16) (n + 1)Φ0−1· a + n + µ µ ! a + b + n + 1 n + 1 − µ ! ∈ Z for µ = 0, 1, . . . , n + 1, hence (n + 1)Φ0−1Qn+1(x) ∈ Z[x] and (n + 1)Φ0−1Pn+1(x) ∈ Z[x]. Proof. Write a + n + µ µ ! a + b + n + 1 n + 1 − µ ! = a + n + µ µ ! a + b + n n − µ ! ·a + b + n + 1 n + 1 − µ . Therefore, if p - n + 1 − µ then (17) ordp a + n + µ µ ! a + b + n + 1 n + 1 − µ ! ≥ ordp a + n + µ µ ! a + b + n n − µ ! ≥ e0p;
otherwise µ ≡ n + 1 (mod p), hence µ/p − (n + 1)/p ∈ Z yielding ordp a + n + µ µ ! a + b + n + 1 n + 1 − µ ! (18) = − a + n + µ p + a + n p + µ p − a + b + n + 1 p + a + b + µ p + n + 1 − µ p = − a + 2n + 1 p + a + n p + n + 1 p = ordp a + 2n + 1 n + 1 ! = ordp a + 2n n ! + ordp a + 2n + 1 n + 1 = ordp a + n + µ µ ! a + b + n n − µ ! µ=n + ordp a + 2n + 1 n + 1 ≥ e0p− ordp(n + 1).
Combination of (17) and (18) gives us the required inclusions (16).
5. Proof of theorem 1
The parameters a, b and n will now depend on an increasing parameter m ∈ N in the following way:
a = αm, b = βm, n = γm or n = γm + 1,
where the choice of the positive integers α, β and γ, satisfying 2α ≤ β and γ < β, is discussed later. Then Lemma 1 and Laplace’s method give us
C0(z) = lim m→∞ log |Rn(z)| m (19) = (α + β + γ) log(α + β + γ) − (α + γ) log(α + γ) − γ log γ − (β − γ) log(β − γ) + γ log |z|
+ max
0≤t≤1Re γ log t + (α + γ) log(1 − t) − (α + β + γ) log(1 − zt) and C1(z) = limm→∞ log |Qn(z−1)| m (20) = (α + β + γ) log(α + β + γ) − (α + γ) log(α + γ) − γ log γ − (β − γ) log(β − γ) + max
0≤t≤1Re (α + γ) log t + (β − γ) log(1 − t) + γ log(1 − z −1
t)
In addition, from (13)–(15) and the prime number theorem we deduce that (21) C2 = lim m→∞ log Φ(αm, βm, γm) m = Z 1 0 ϕ(x) dψ(x), C20 = lim m→∞ log Φ0(αm, βm, γm) m = Z 1 0 ϕ0(x) dψ(x),
where ψ(x) is the logarithmic derivative of the gamma function and the 1-periodic functions ϕ(x) and ϕ0(x) are defined as follows:
ϕ(x) = min 0≤y<1ϕ(x, y),b ϕ 0(x) = min 0≤y<1ϕb 0(x, y), b ϕ(x, y) = −{−(α + γ)x} + {−(α + γ)x − y} + {y} − {(α + β + γ)x} + {(α + β)x + y} + {γx − y}, b ϕ0(x, y) = −{(α + γ)x + y} + {(α + γ)x} + {y} − {(α + β + γ)x} + {(α + β)x + y} + {γx − y}.
Lemma 5. The functions ϕ(x) and ϕ0(x) differ on a set of measure 0,
hence
(22) C2 = C20.
Remark 2. The following proof is due to the anonymous referee. In an
earlier version of this article, we verify directly assumption (22) for our specific choice of the integer parameters α, β and γ.
Proof. First note that ϕ(x, 0) =b ϕb
0(x, 0) ∈ {0, 1}, which implies ϕ(x) ∈
{0, 1} and ϕ0(x) ∈ {0, 1}.
From the definition we see that ∆(x, y) =ϕ(x, y) −b ϕb
0(x, y) = λ((α + γ)x) − λ((α + γ)x + y).
Assume that (α+γ)x /∈ Z; we plainly obtain ∆(x, y) = −λ((α+γ)x+y) ≤ 0, hence ∆(x, y) = 0 unless y = −{(α + γ)x}. Therefore the only possibility to get ϕ(x) 6= ϕ0(x) is the following one:
(23) ϕ(x, y) ≥ 1b for y 6= −{(α + γ)x} and b ϕ(x, y) = 0 for y = −{(α + γ)x}. Sinceϕb 0(x, 0) ≥ 1, we have {(α + β)x} + {γx} = 1; furthermore, {(α + β)x + y} + {γx − y} − {(α + β + γ)x} = 1 if 0 ≤ y < 1 − {(α + β)x}, 0 if 1 − {(α + β)x} ≤ y < {γx}, 1 if {γx} ≤ y < 1,
and
{−(α + γ)x − y} + {y} − {−(α + γ)x} =
(
0 if 0 ≤ y ≤ {−(α + γ)x}, 1 if {−(α + γ)x} < y < 1. The above conditions (23) onϕ(x, y) imply 1 − {(α + β)x} = {−(α + γ)x}b
or, equivalently, (β − γ)x ∈ Z.
Finally, the sets {x ∈ R : (α + γ)x ∈ Z} and {x ∈ R : (β − γ)x ∈ Z} have measure 0, thus proving the required assertion. Our final aim is estimating the absolute value of εk from below, where
3 2 k = Mk+ εk, Mk∈ Z, 0 < |εk| < 1 2.
Write k ≥ 3 in the form k = 3(βm + 1) + j with non-negative integers m and j < 3β. Multiply both sides of (9) by Φe−13b−2a+j+1, where Φ =e
Φ(αm, βm, γm) if n = γm andΦ = Φe 0(αm, βm, γm)/(γm+1) if n = γm+1, and substitute z = 1/9: Qn(9)Φe−12j· 3 2 j 3b−2a+1F a, b;1 9 (24) = Pn(9)Φe−13b−2a+j+1+ Rn 1 9 e Φ−13b−2a+j+1. From (6), (7) we see that
3 2 j 3b−2a+1F a, b;1 9 ≡ 3 2 3(b+1)+j (mod Z) = 3 2 k ,
hence the left-hand side equals Mk0+ εkfor some Mk0 ∈ Z and we may write
equality (24) in the form
(25) Qn(9)Φe−12j· εk= Mk00+ Rn 1 9 e Φ−13b−2a+j+1, where Mk00 = Pn(9)Φe−13b−2a+j+1− Qn(9)Φe−12jMk0 ∈ Z
by Lemmas 3 and 4. Lemma 2 guarantees that, for at least one of n = γm or γm + 1, we have Mk00 6= 0; we make the corresponding choice of n. Assuming furthermore that (26) C0 1 9 − C2+ (β − 2α) log 3 < 0, from (19) and (21) we obtain
Rn 1 9 e Φ−13b−2a+j+1 < 1 2 for all m ≥ N1,
where N1 > 0 is an effective absolute constant. Therefore, by (25) and |Mk00| ≥ 1 we have |Qn(9)Φe−12j| · |εk| ≥ |Mk00| − Rn 1 9 e Φ−13b−2a+j+1 > 1 2, hence from (19), (20) we conclude that
|εk| > Φe 2j+1|Q n(9)| ≥ Φe 23β|Q n(9)| > e−m(C1(1/9)−C2+δ)
for any δ > 0 and m > N2(δ), provided that C1(1/9)−C2+δ > 0; here N2(δ)
depends effectively on δ. Finally, since k > 3βm, we obtain the estimate (27) |εk| > e−k(C1(1/9)−C2+δ)/(3β)
valid for all k ≥ K0(δ), where the constant K0(δ) may be determined in
terms of max(N1, N2(δ)).
Taking α = γ = 9 and β = 19 (which is the optimal choice of the integer parameters α, β, γ, at least under the restriction β ≤ 100) we find that
C0 1 9 = 3.28973907 . . . , C1 1 9 = 35.48665992 . . . , and ϕ(x) = 1 if {x} ∈2 37, 1 18 ∪3 37, 1 10 ∪4 37, 1 9 ∪6 37, 1 6 ∪7 37, 1 5 ∪8 37, 2 9 ∪10 37, 5 18 ∪11 37, 3 10 ∪12 37, 1 3 ∪14 37, 7 18 ∪16 37, 4 9 ∪18 37, 1 2 ∪20 37, 5 9 ∪22 37, 3 5 ∪24 37, 2 3 ∪28 37, 7 9 ∪32 37, 8 9 ∪36 37, 1 , 0 otherwise, hence C2= C20 = 4.46695926 . . . .
Using these computations we verify (26), C0
1
9
− C2+ (β − 2α) log 3 = −0.07860790 . . . ,
and find that with δ = 0.00027320432 . . .
e−(C1(1/9)−C2+δ)/(3β)= 0.5803.
6. Related results
The above construction allows us to prove similar results for the se-quences k(4/3)kk and k(5/4)kk using the representations
4 3 2(b+1) = 2b+1 1 +1 8 −(b+1) (28) ≡ (−1)a2b−3a+1F a, b; −1 8 (mod Z), where 3a ≤ b, and 5 4 7b+3 = 2 · 5b 1 + 3 125 −(2b+1) (29) ≡ (−1)a2 · 3a· 5b−3aF a, 2b; − 3 125 (mod Z), where 3a ≤ b.
Namely, taking a = 5m, b = 15m, n = 6m(+1) in case (28) and a = 3m, b = 9m, n = 7m(+1) in case (29) and repeating the arguments of Section 5, we arrive at
Theorem 2. The following estimates are valid:
4 3 k > 0.4914 k= 3−k·0.64672207... for k ≥ K1, 5 4 k > 0.5152 k= 4−k·0.47839775... for k ≥ K2, where K1, K2 are certain effective constants.
The general case of the sequence k(1 + 1/N )kk for an integer N ≥ 5 may be treated as in [4] and [2] by using the representation
N + 1 N b+1 = 1 − 1 N + 1 −(b+1) ≡ F 0, b; 1 N + 1 (mod Z). The best result in this direction belongs to M. Bennett [2]: k(1 + 1/N )kk > 3−k for 4 ≤ N ≤ k3k.
Remark 3. As mentioned by the anonymous referee, our result for k(4/3)kk is of special interest. It completes Bennett’s result [3] on the order of the additive basis {1, Nk, (N + 1)k, (N + 2)k, . . .} for N = 3 (case N = 2 corre-sponds to the classical Waring’s problem); to solve this problem one needs the bound k(4/3)kk > (4/9)k for k ≥ 6. Thus we remain verification of the
We would like to conclude this note by mentioning that a stronger argu-ment is required to obtain the effective estimate k(3/2)kk > (3/4)k and its
relatives.
Acknowledgements. I am very grateful to the anonymous referee for
his valuable comments (incorporated into the current presentation as Re-marks 1–3), providing the proof of Lemma 5 and indicating misprints. My profound gratitude goes to Anne de Roton, for supporting my talk at the
24th Journeés Arithmetiques as a muse. I thank kindly J. Sondow for several
suggestions.
References
[1] A. Baker, J. Coates, Fractional parts of powers of rationals. Math. Proc. Cambridge Philos. Soc. 77 (1975), 269–279.
[2] M. A. Bennett, Fractional parts of powers of rational numbers. Math. Proc. Cambridge Philos. Soc. 114 (1993), 191–201.
[3] M. A. Bennett, An ideal Waring problem with restricted summands. Acta Arith. 66 (1994), 125–132.
[4] F. Beukers, Fractional parts of powers of rationals. Math. Proc. Cambridge Philos. Soc.
90 (1981), 13–20.
[5] G. V. Chudnovsky, Padé approximations to the generalized hypergeometric functions. I. J. Math. Pures Appl. (9) 58 (1979), 445–476.
[6] F. Delmer, J.-M. Deshouillers, The computation of g(k) in Waring’s problem. Math. Comp. 54 (1990), 885–893.
[7] A. K. Dubickas, A lower bound for the quantity k(3/2)kk. Russian Math. Surveys 45 (1990),
163–164.
[8] L. Habsieger, Explicit lower bounds for k(3/2)kk. Acta Arith. 106 (2003), 299–309.
[9] J. Kubina, M. Wunderlich, Extending Waring’s conjecture up to 471600000. Math. Comp.
55 (1990), 815–820.
[10] K. Mahler, On the fractional parts of powers of real numbers. Mathematika 4 (1957), 122–124.
[11] L. J. Slater, Generalized hypergeometric functions. Cambridge University Press, 1966. [12] R. C. Vaughan, The Hardy–Littlewood method. Cambridge Tracts in Mathematics 125,
Cambridge University Press, 1997.
[13] W. Zudilin, Ramanujan-type formulae and irrationality measures of certain multiples of π. Mat. Sb. 196:7 (2005), 51–66.
Wadim Zudilin
Department of Mechanics and Mathematics Moscow Lomonosov State University Vorobiovy Gory, GSP-2
119992 Moscow, Russia
URL: http://wain.mi.ras.ru/ E-mail : wadim@ips.ras.ru
