Journal de Théorie des Nombres
de Bordeaux 19 (2007), 311–323
A new lower bound for k(3/2)k k
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 effectively 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)k c − 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)k k ≤ C k , 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
k 3 k
for k ≥ K,
(1)
2 >C
−64
with C = 2−(1−10 ) , was proved by A. Baker and J. Coates [1] by applying
effective estimates of linear forms in logarithms in the p-adic case. F. Beukers [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 approxiPm
m
n
mations to a tail of the binomial series (1 − z) = n=0 m
n (−z) and 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.
Wadim Zudilin
312
the estimate k(3/2)k k > 0.57434k for k ≥ 5 using computations from [6]
and [9].
By modifying Beukers’ construction, namely, considering Padé approximations to a tail of the series
∞ X
1
m+n n
(2)
=
z
(1 − z)m+1
m
n=0
and studying the explicit p-adic order of the binomial coefficients involved,
we are able to prove
Theorem 1. The following estimate is valid:
k 3 k
−k·0.78512916...
2 > 0.5803 = 2
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
∞
X
(A0 )k (A1 )k · · · (Aq )k k
A0 , A1 , . . . , Aq z ,
(3)
F
z
=
q+1 q
B 1 , . . . , Bq
k!(B1 )k · · · (Bq )k
k=0
where
Γ(A + k)
=
(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+1 Fq -notation. We will require
two classical facts from the theory of generalized hypergeometric series:
the Pfaff–Saalschütz summation formula
−n, A, B
1 = (C − A)n (C − B)n
(4)
3 F2
C, 1 + A + B − C − n (C)n (C − A − B)n
(see, e.g., [11], p. 49, (2.3.1.3)) and the Euler–Pochhammer integral for the
Gauss 2 F1 -series
Z 1
Γ(C)
A, B (5) 2 F1
z =
tB−1 (1 − t)C−B−1 (1 − zt)−A dt,
C Γ(B)Γ(C − B) 0
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.
A new lower bound for k(3/2)k k
313
3. Padé approximations of the shifted binomial series
Fix two positive integers a and b satisfying 2a ≤ b. Formula (2) yields
(6)
3(b+1) b+1
3
27
1 −(b+1)
b+1
=
=3
1−
2
8
9
∞ ∞
X
X b+k
b + k 2(a−k)
1 k
b−2a+1
b+1
3
=3
=3
b
9
b
k=0
k=0
∞ X
b + k 2(a−k)
b−2a+1
= an integer + 3
3
b
k=a
∞ X
a + b + ν −2ν
b−2a+1
≡3
3
(mod Z).
b
ν=0
This motivates (cf. [4]) constructing Padé approximations to the function
(7) F (z) = F (a, b; z) =
∞ X
a+b+ν
ν=0
b
ν
z =
∞
a + b X (a + b + 1)ν ν
z
b
(a + 1)ν
ν=0
and applying them with the choice z = 1/9.
Remark 1. The connection of F (a, b; z) with Beukers’ auxiliary series
H(e
a, eb; z) from [4] is as follows:
F (a, b; z) = z
−a
−b−1
(1 − z)
a−1 X
b+k k
−
z
= H(−a − b − 1, a; z).
b
k=0
Although Beukers considers H(e
a, eb; z) only for e
a, eb ∈ N, his construction
remains valid for any e
a ∈ 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
a+b+n
−n, a + n Qn (x) =
x
2 F1
a+b+1 a+b
n n
X
X
a+n−1+µ a+b+n
=
(−x)µ =
qµ xµ ∈ Z[x]
µ
n−µ
(8)
µ=0
µ=0
Wadim Zudilin
314
of degree n. Then
(9)
Qn (z −1 )F (z) =
n
X
qn−µ z µ−n ·
µ=0
=
∞
X
z
l−n
n
X
qn−µ
µ=0
µ≤l
n−1
X
b
ν=0
l=0
=
∞ X
a+b+ν
rl z l−n +
l=0
∞
X
zν
a+b+l−µ
b
rl z l−n = Pn (z −1 ) + Rn (z).
l=n
Here the polynomial
(10)
n−1
X
Pn (x) =
rl xn−l ∈ Z[x],
where
rl =
l
X
µ=0
l=0
qn−µ
a+b+l−µ
,
b
has degree at most n, while the coefficients of the remainder
Rn (z) =
∞
X
rl z l−n
l=n
are of the following form:
n
X
a+b+l−µ
rl =
qn−µ
b
µ=0
n
X
a+b+n a+b+l−µ
n−µ a + 2n − 1 − µ
=
(−1)
n−µ
µ
b
µ=0
n
(a + b + n)! X
n (a + 2n − 1 − µ)!(a + b + l − µ)!
= (−1)n
(−1)µ
(a + n − 1)!n!b!
µ
(a + l − µ)!(a + b + n − µ)!
µ=0
(a + b + n)!
= (−1)n
(a + n − 1)!n!b!
n
(a + 2n − 1)!(a + b + l)! X (−n)µ (−a − l)µ (−a − b − n)µ
(a + l)!(a + b + n)!
µ!(−a − 2n + 1)µ (−a − b − l)µ
µ=0
−n, −a − l, −a − b − n n (a + 2n − 1)!(a + b + l)!
1 .
= (−1)
· 3 F2
−a − 2n + 1, −a − b − l (a + n − 1)!(a + l)!n!b!
×
If we apply (4) with the choice A = −a−l, B = −a−b−n and C = −a−b−l,
we obtain
(a + 2n − 1)!(a + b + l)!
(−b)n (n − l)n
rl = (−1)n
·
.
(a + n − 1)!(a + l)!n!b! (−a − b − l)n (a + n)n
A new lower bound for k(3/2)k k
315
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
(a + 2n − 1)!(a + b + l)!
rl =
(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)!
=
for l ≥ 2n.
n!(b − n)!(a + l)!(l − 2n)!
Finally,
∞
∞
X
X
Rn (z) =
rl z l−n = z n
(11)
rν+2n z ν
l=2n
ν=0
∞
X
(a + b + n + ν)!(n + ν)! ν
1
z
n!(b − n)!
ν!(a + 2n + ν)!
ν=0
a
+
b
+
n
a + b + n + 1, n + 1 n
· 2 F1
=z
z .
a + 2n + 1
b−n
= zn
Using the integral (5) for the polynomial (8) and remainder (11) we arrive
at
Lemma 1. The following representations are valid:
Z 1
(a + b + n)!
−1
Qn (z ) =
ta+n−1 (1 − t)b−n (1 − z −1 t)n dt
(a + n − 1)!n!(b − n)! 0
and
Z 1
(a + b + n)!
n
Rn (z) =
z
tn (1 − t)a+n−1 (1 − zt)−(a+b+n+1) dt.
(a + n − 1)!n!(b − n)!
0
We will also require linear independence of a pair of neighbouring Padé
approximants, which is the subject of
Lemma 2. We have
n
(12) Qn+1 (x)Pn (x) − Qn (x)Pn+1 (x) = (−1)
a + 2n + 1 a + b + n
x.
a+n
b−n
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)
= Qn (z −1 )Rn+1 (z) − Qn+1 (z −1 )Rn (z),
Wadim Zudilin
316
and from (8), (11) we conclude that the only negative power of z originates
with the last summand:
− Qn+1 (z −1 )Rn (z)
a+b+n n
−n−1
n a + 2n + 1
z
1 + O(z) ·
z 1 + O(z)
= (−1)
a+n
b−n
a
+
2n
+
1
a
+
b
+
n
1
= (−1)n
+ O(1)
as z → 0.
a+n
b−n
z
4. Arithmetic constituents
√
We begin this section by noting that, for any prime p > N ,
N
N
and
ordp N = λ
,
ordp N ! =
p
p
where
(
1 if x ∈ Z,
λ(x) = 1 − {x} − {−x} = 1 + bxc + b−xc =
0 if x ∈
/ Z.
For primes p >
(13)
√
a + b + n, let
a+n+µ
µ
a+n
+ −
+
ep = min − −
µ∈Z
p
p
p
a+b+n
a+b+µ
n−µ
−
+
+
p
p
p
a+n
a+n+µ
µ
= min −
− −
−
µ∈Z
p
p
p
a+b+n
a+b+µ
n−µ
−
−
+
p
p
p
a+n
a+n+µ a+b+n
≤ min ordp
0≤µ≤n
a+n+µ
µ
n−µ
a+n−1+µ a+b+n
= min ordp
0≤µ≤n
µ
n−µ
and
(14)
e0p
a+n+µ
a+n
µ
= min −
+
+
µ∈Z
p
p
p
a+b+n
a+b+µ
n−µ
−
+
+
p
p
p
A new lower bound for k(3/2)k k
317
a+n+µ
a+n
µ
= min
−
−
µ∈Z
p
p
p
a+b+µ
n−µ
a+b+n
−
−
+
p
p
p
a+n+µ a+b+n
.
≤ min ordp
0≤µ≤n
µ
n−µ
Set
(15)
Φ = Φ(a, b, n) =
Y
pep
and
Y
Φ0 = Φ0 (a, b, n) =
√
p> a+b+n
0
pep .
√
p> a+b+n
From (8), (10) and (13), (15) we deduce
Lemma 3. The following inclusions are valid:
a+n−1+µ a+b+n
−1
∈Z
Φ ·
n−µ
µ
for
µ = 0, 1, . . . , n,
hence
Φ−1 Qn (x) ∈ Z[x]
and
Φ−1 Pn (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)
a+n+µ a+b+n+1
0 −1
∈Z
(n + 1)Φ ·
µ
n+1−µ
for
µ = 0, 1, . . . , n + 1,
hence
(n + 1)Φ0
−1
Qn+1 (x) ∈ Z[x]
and
(n + 1)Φ0
−1
Pn+1 (x) ∈ Z[x].
Proof. Write
a+n+µ a+b+n+1
a+n+µ a+b+n
a+b+n+1
=
·
.
µ
n+1−µ
µ
n−µ
n+1−µ
Therefore, if p - n + 1 − µ then
(17) a+n+µ a+b+n+1
a+n+µ a+b+n
ordp
≥ ordp
≥ e0p ;
µ
n+1−µ
µ
n−µ
Wadim Zudilin
318
otherwise µ ≡ n + 1 (mod p), hence µ/p − (n + 1)/p ∈ Z yielding
a+n+µ a+b+n+1
(18)
ordp
µ
n+1−µ
a+n+µ
a+n
µ
=−
+
+
p
p
p
a+b+n+1
a+b+µ
n+1−µ
−
+
+
p
p
p
a+n
n+1
a + 2n + 1
+
+
=−
p
p
p
a + 2n
a + 2n + 1
a + 2n + 1
+ ordp
= ordp
= ordp
n
n+1
n+1
a+n+µ a+b+n a + 2n + 1
+ ordp
= ordp
µ
n−µ
n+1
µ=n
≥ 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
(19)
log |Rn (z)|
m
= (α + β + γ) log(α + β + γ) − (α + γ) log(α + γ)
C0 (z) = lim
m→∞
− γ log γ − (β − γ) log(β − γ) + γ log |z|
+ max Re γ log t + (α + γ) log(1 − t) − (α + β + γ) log(1 − zt)
0≤t≤1
and
(20)
log |Qn (z −1 )|
m→∞
m
= (α + β + γ) log(α + β + γ) − (α + γ) log(α + γ)
C1 (z) = lim
− γ log γ − (β − γ) log(β − γ)
+ max Re (α + γ) log t + (β − γ) log(1 − t) + γ log(1 − z −1 t) .
0≤t≤1
A new lower bound for k(3/2)k k
319
In addition, from (13)–(15) and the prime number theorem we deduce that
Z 1
log Φ(αm, βm, γm)
C2 = lim
ϕ(x) dψ(x),
=
m→∞
m
0
(21)
Z 1
log Φ0 (αm, βm, γm)
C20 = lim
=
ϕ0 (x) dψ(x),
m→∞
m
0
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 ϕ(x,
b y),
0≤y<1
ϕ0 (x) = min ϕ
b0 (x, y),
0≤y<1
ϕ(x,
b y) = −{−(α + γ)x} + {−(α + γ)x − y} + {y}
− {(α + β + γ)x} + {(α + β)x + y} + {γx − y},
0
ϕ
b (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
C2 = C20 .
(22)
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,
b 0) = ϕ
b0 (x, 0) ∈ {0, 1}, which implies ϕ(x) ∈
0
{0, 1} and ϕ (x) ∈ {0, 1}.
From the definition we see that
∆(x, y) = ϕ(x,
b y) − ϕ
b0 (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,
b y) ≥ 1
for y 6= −{(α + γ)x}
ϕ(x,
b y) = 0
and
for y = −{(α + γ)x}.
Since ϕ
b0 (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,
Wadim Zudilin
320
and
(
0 if 0 ≤ y ≤ {−(α + γ)x},
{−(α + γ)x − y} + {y} − {−(α + γ)x} =
1 if {−(α + γ)x} < y < 1.
The above conditions (23) on ϕ(x,
b y) imply 1 − {(α + β)x} = {−(α + γ)x}
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
k
3
1
= Mk + ε k ,
Mk ∈ Z, 0 < |εk | < .
2
2
Write k ≥ 3 in the form k = 3(βm + 1) + j with non-negative integers
e −1 3b−2a+j+1 , where Φ
e =
m and j < 3β. Multiply both sides of (9) by Φ
0
e = Φ (αm, βm, γm)/(γm+1) if n = γm+1,
Φ(αm, βm, γm) if n = γm and Φ
and substitute z = 1/9:
j
1
3
b−2a+1
−1
j
e 2 ·
(24)
3
F a, b;
Qn (9)Φ
2
9
e −1 3b−2a+j+1 .
e −1 3b−2a+j+1 + Rn 1 Φ
= Pn (9)Φ
9
From (6), (7) we see that
j
3(b+1)+j
k
3
1
3
3
b−2a+1
3
F a, b;
≡
(mod Z) =
,
2
9
2
2
hence the left-hand side equals Mk0 + εk for some Mk0 ∈ Z and we may write
equality (24) in the form
1 e −1 b−2a+j+1
−1 j
00
e
(25)
Qn (9)Φ 2 · εk = Mk + Rn
Φ 3
,
9
where
e −1 3b−2a+j+1 − Qn (9)Φ
e −1 2j M 0 ∈ Z
Mk00 = Pn (9)Φ
k
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
1
(26)
C0
− C2 + (β − 2α) log 3 < 0,
9
from (19) and (21) we obtain
e −1 3b−2a+j+1 < 1
Rn 1 Φ
2
9
for all
m ≥ N1 ,
A new lower bound for k(3/2)k k
321
where N1 > 0 is an effective absolute constant. Therefore, by (25) and
|Mk00 | ≥ 1 we have
1 e −1 b−2a+j+1 1
−1 j
00
e
|Qn (9)Φ 2 | · |εk | ≥ |Mk | − Rn
Φ 3
> 2,
9
hence from (19), (20) we conclude that
|εk | >
e
Φ
2j+1 |Qn (9)|
≥
e
Φ
23β |Qn (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
|εk | > e−k(C1 (1/9)−C2 +δ)/(3β)
(27)
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
1
1
= 3.28973907 . . . ,
C1
= 35.48665992 . . . ,
C0
9
9
and
ϕ(x) =


1












0
if {x} ∈
2 1
,
∪ 3, 1 ∪ 4,1 ∪ 6,1 ∪ 7,1
37
818 2 3710 105 3711 9 3 37 126 1 37 145 7 ∪ 37 , 9 ∪ 37 , 18 ∪ 37 , 10 ∪ 37 , 3 ∪ 37 , 18
18 1 20 5 22 3 24 2 ∪ 16
, 49 ∪ 37
, ∪ 37 , 9 ∪ 37 , 5 ∪ 37 , 3
37
28 7 32 28 36
∪ 37 , 9 ∪ 37 , 9 ∪ 37 , 1 ,
otherwise,
hence
C2 = C20 = 4.46695926 . . . .
Using these computations we verify (26),
1
C0
− C2 + (β − 2α) log 3 = −0.07860790 . . . ,
9
and find that with δ = 0.00027320432 . . .
e−(C1 (1/9)−C2 +δ)/(3β) = 0.5803.
This result, in view of (27), completes the proof of Theorem 1.
322
Wadim Zudilin
6. Related results
The above construction allows us to prove similar results for the sequences k(4/3)k k and k(5/4)k k using the representations
2(b+1)
4
1 −(b+1)
(28)
= 2b+1 1 +
3
8
1
a b−3a+1
≡ (−1) 2
F a, b; −
(mod Z),
8
where 3a ≤ b,
and
(29)
7b+3
5
3 −(2b+1)
b
=2·5 1+
4
125
3
a
a
b−3a
(mod Z),
≡ (−1) 2 · 3 · 5
F a, 2b; −
125
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:
k 4 k
−k·0.64672207...
3 > 0.4914 = 3
k 5 k
−k·0.47839775...
4 > 0.5152 = 4
for
k ≥ K1 ,
for
k ≥ K2 ,
where K1 , K2 are certain effective constants.
The general case of the sequence k(1 + 1/N )k k for an integer N ≥ 5 may
be treated as in [4] and [2] by using the representation
−(b+1)
N + 1 b+1
1
1
= 1−
≡ F 0, b;
(mod Z).
N
N +1
N +1
The best result in this direction belongs to M. Bennett [2]: k(1 + 1/N )k k >
3−k for 4 ≤ N ≤ k3k .
Remark 3. As mentioned by the anonymous referee, our result for k(4/3)k k
is of special interest. It completes Bennett’s result [3] on the order of the
additive basis {1, N k , (N + 1)k , (N + 2)k , . . . } for N = 3 (case N = 2 corresponds to the classical Waring’s problem); to solve this problem one needs
the bound k(4/3)k k > (4/9)k for k ≥ 6. Thus we remain verification of the
bound in the range 6 ≤ k ≤ K1 .
A new lower bound for k(3/2)k k
323
We would like to conclude this note by mentioning that a stronger argument is required to obtain the effective estimate k(3/2)k k > (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 Remarks 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.
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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