1
Journal of Integer Sequences, Vol. 19 (2016),
Article 16.3.4
2
3
47
6
23 11
Two Properties of
Catalan-Larcombe-French Numbers
Xiao-Juan Ji
School of Mathematical Sciences
Soochow University
Suzhou
Jiangsu 215006
P. R. China
xiaojuanji2015@yahoo.com
Zhi-Hong Sun
School of Mathematical Sciences
Huaiyin Normal University
Huaian
Jiangsu 223001
P. R. China
zhihongsun@yahoo.com
Abstract
Let (Pn ) be the Catalan-Larcombe-French numbers. The numbers Pn occur in the
theory of elliptic integrals, and are related to the arithmetic-geometric-mean. In this
paper we investigate the properties of the related sequence Sn = Pn /2n instead, since
Sn is an Apéry-like sequence. We prove a congruence and an inequality for Pn .
1
Introduction
Let (Pn ) be the sequence given by
P0 = 1, P1 = 8 and (n + 1)2 Pn+1 = 8(3n2 + 3n + 1)Pn − 128n2 Pn−1 (n ≥ 1).
1
(1)
The numbers Pn are called Catalan-Larcombe-French numbers since Catalan first defined
Pn in [1], and Larcombe and French [6] proved that
⌊n/2⌋
Pn = 2
n
X
(−4)
k
k=0
2n − 2k
n−k
2 n−k
k
=
n
X
k=0
2k 2 2n−2k 2
n−k
k
,
n
k
where ⌊x⌋ is the greatest integer not exceeding x. The numbers Pn are related to the
arithmetic-geometric-mean. See [6] and A053175 in Sloane’s “On-Line Encyclopedia of Integer Sequences”.
Let (Sn ) be defined by
S0 = 1, S1 = 4 and (n + 1)2 Sn+1 = 4(3n2 + 3n + 1)Sn − 32n2 Sn−1 (n ≥ 1).
(2)
Comparing (2) with (1), we see that
Sn =
Pn
.
2n
Zagier noted that
Sn =
⌊n/2⌋ X
k=0
2k
k
2 n n−2k
4
.
2k
As observed by Jovović [7] in 2003 ,
n X
2n − 2k
n 2k
(n = 0, 1, 2, . . .).
Sn =
n
−
k
k
k
k=0
Recently Z. W. Sun stated that
2 n n X
1 X 2k
2k
k
2n − 2k
k
n−k
(−4)
=
Sn =
(−4)k .
n
k
n
−
k
k
n
−
k
n
−
k
(−2)
k=0
k=0
The first few values of Sn are shown below:
S0 = 1, S1 = 4, S2 = 20, S3 = 112, S4 = 676, S5 = 4304, S6 = 28496,
S7 = 194240, S8 = 1353508, S9 = 9593104, S10 = 68906320,
S11 = 500281280, S12 = 3664176400, S13 = 27033720640.
Let p be an odd prime. Jarvis, Larcombe, and French [3] proved that if n = ar pr + · · · +
a1 p + a0 with a0 , a1 , . . . , ar ∈ {0, 1, . . . , p − 1}, then
Pn ≡ Par · · · Pa1 Pa0
(mod p).
Jarvis and Verrill [5] showed that
Pn ≡ (−1)
p−1
2
128n Pp−1−n
(mod p) for n = 0, 1, . . . , p − 1
2
and
Pmpr ≡ Pmpr−1
(mod pr ) for m, r ∈ Z+ ,
where Z+ is the set of positive integers.
For a prime p let Zp denote the set of those rational numbers whose denominator is not
divisible by p. Let p be an odd prime, n ∈ Zp and n 6≡ 0, −16 (mod p). The second author
[11] proved that
p−1 X
2k
k=0
k
p−1
n(n + 16) X
Sk
≡
(n + 16)k
p
k=0
2k 2 4k
2k
k
n2k
(mod p),
where ( ap ) is the Legendre symbol.
In 1894 Franel [2] introduced the following Franel numbers (fn ):
fn =
n 3
X
n
k=0
k
(n = 0, 1, 2, . . .).
The first few Franel numbers are as below:
f0 = 1, f1 = 2, f3 = 10, f4 = 56, f5 = 346, f6 = 2252, f7 = 15184.
Franel [2] noted that the sequence (fn ) satisfies the recurrence relation:
(n + 1)2 fn+1 = (7n2 + 7n + 2)fn + 8n2 fn−1 (n ≥ 1).
Let r ∈ Z+ and p be a prime with p ≡ 5, 7 (mod 8). The second author [11] conjectured
that
(3)
S pr −1 ≡ 0 (mod pr ) and f pr −1 ≡ 0 (mod pr ).
2
2
In this paper we prove (3) in the case r = 2. We also prove the second author’s conjecture
[11]:
1+
2
2
1
S > Sm+1 Sm−1
m(m − 1) m
for m = 2, 3, . . . .
Basic lemmas
Lemma 1 (Lucas’ theorem [8]). Let p be an odd prime. Suppose a = ar pr + · · · + a1 p + a0
and b = br pr + · · · + b1 p + b0 , where ar , . . . , a0 , br , . . . , b0 ∈ {0, 1, . . . , p − 1}. Then
a0
a
ar
(mod p).
···
≡
b
b0
br
Lucas’ theorem is often formulated as follows.
3
Lemma 2 ([8]). Let p be an odd prime and a, b ∈ Z+ . Suppose a0 , b0 ∈ {0, 1, . . . , p − 1}.
Then
a a0
ap + a0
(mod p).
≡
b0
bp + b0
b
Lemma 3 ([4, Lemma 2.7]). For any positive integer n we have
n X
2n − 2k
n − 1 2k
.
Sn = 2
n−k
k
k−1
k=1
Lemma 4 ([9]). Let p be an odd prime. Suppose n = n1 p + n0 and k = k1 p + k0 with
k1 , n1 ∈ Z+ and k0 , n0 ∈ {0, 1, . . . , p − 1}. Then
n0 − p n0 − p
n0
n
n1
− k1
(mod p2 ).
− (n1 + k1 )
(1 + n1 )
≡
k0 + p
k0
k0
k
k1
Lemma 5. Let p be an odd prime. Then
(p−1)/2
X
t=0
p−1 + t p + p−1 + t− 1 2
2
2
2
(−1)
−
≡0
t
p+t
t
t
(mod p2 ).
Proof. For 0 ≤ t ≤ (p − 1)/2, from Lemma 2 we have
p+1
p−1
p+1
−p
+t−1
t+1 p + 2 + t − 1
t+1
2
2
= (−1)
≡ (−1)
p+t
p+t
t
p−1
−p
=− 2
(mod p)
t
and so
p−1
2
p−1
p−1 + t p + p−1 + t
−p
−p
t
2
2
2
+
= (−1)
−
≡0
t
p+t
t
p+t
(mod p).
We first assume p ≡ 1 (mod 4). Applying Lemma 4 we get
3(p−1) 3(p−1) 3(p−1) 3(p−1)
−p p + 3(p−1)
4
4
4
4
−
≡
− 2 p−1
−
p−1
p−1
p + p−1
4
2
2
2
3(p−1) 3(p−1) p−1
4
4
+ (−1) 2
= 0 (mod p2 )
= − p−1
p−1
4
p−1
4
2
2
4
and
p−1
p
+
p
−
1
−
t
p
+
p
−
1
−
t
+
t
+
t
2
2
−
−
+ p−1
p + p−1
−t
−t
t
p+t
2
2
p−1
p−1
+t
−p+t
−1 − t
p−1−t
2
2
≡−
+
+
−
p−1
p−1
p−1
p−1
p−1
2
2
2
2
p−1 + t p − 1 − t
p−1
p−1
2
+ (−1) 2 − 1
= (−1) 2 − 1
p−1
p−1
2
2
=0
(mod p2 ).
Also,
2
1 2
p−1 2 p−1 2 − 21
p−1
−2
−t
t
2
≡ (−1)
=0
− (−1) 2
(−1)
− p−12
p−1
t
t
−
t
−t
2
2
t
(mod p).
By the above four congruences, we have
(p−1)/2
p−1 + t p + p−1 + t− 1 2
2
2
2
(−1)
−
t
p
+
t
t
t=0
1 2 p−1
(p−5)/4
X
p + p−1
+t
+t t −2
2
2
−
(−1)
=
t
p+t
t
t=0
1 2 3(p−1) −2
p + 3(p−1)
p−1
4
4
+ (−1) 4 p−1
−
p−1
p + p−1
4
4
4
(p−5)/4
2 1
X
−2
p−1
p+p−1−t p−1−t
−t
(−1) 2
−
+
p−1
p−1
−t
−t
p + p−1
−t
2
2
2
t=0
1 2
2 p−1
(p−5)/4 X
− 12
p−1
p + p−1
+t
+t −t
t −2
2
2
−
(−1)
− (−1) 2
≡
p−1
t
p+t
t
−
t
2
t=0
1 2 3(p−1) −2
p + 3(p−1)
p−1
4
4
+ (−1) 4 p−1
−
≡ 0 (mod p2 ).
p−1
p−1
p
+
4
4
4
X
t
Thus the result is true for p ≡ 1 (mod 4).
Now we assume p ≡ 3 (mod 4). By Lemma 4,
p−1 + t p − 1 − t
+t
p + p−1
+t
2
2
−
≡−
(mod p2 ).
+
p−1
p−1
t
p+t
2
2
p−1
2
5
As
p−1
=
+t
2
p−1
2
p−1
+
2
p−1
2
+
t
p−1
p−1−t
+
+t
2
≡
p−1
p−1
2
2
t + p−1
p−1
2
(−1) 2
=
p−1
2
+
−1 − t
p−1
2
0
(mod p)
−
and
(−1)
t
− 12
t
2
+ (−1)
p−1
−t
2
− 12
p−1
−t
2
2
≡ (−1)
t
p−1 2
2
t
p−1
2
p−1
−
2
t
2 =0
(mod p),
we obtain
(p−1)/2
p−1 + t p + p−1 + t− 1 2
2
2
2
(−1)
−
t
p
+
t
t
t=0
(p−1)/2
p−1 + t p − 1 − t− 1 2
X
t
2
2
+
(−1)
≡−
p−1
p−1
t
2
2
t=0
1 2
2 (p−3)/4 p−1
X
+t
− 12
p−1
p − 1 − t t −2
−t
2
=−
(−1)
+
+ (−1) 2
p−1
p−1
p−1
t
−t
2
2
2
t=0
X
≡0
t
(mod p2 ).
Hence the result is also true in this case. The proof is now complete.
Lemma 6 ([10, Theorem 3.3]). Let p be a prime with p ≡ 5, 7 (mod 8). Then
p−1
2
X
k=0
2k 3
k
(−64)k
≡0
(mod p2 ).
Lemma 7 ([4, Lemma 2.8]). Let m ∈ Z and k, p ∈ Z+ . Then
k
r−1
r
mpr −1 Y mp − 1
k−⌊ kp ⌋ mp
1−
= (−1)
.
⌊k/p⌋
i
k
i=1, p∤i
3
Congruences for S p2−1 and f p2−1 (mod p2)
2
2
Theorem 8. Let p be a prime with p ≡ 5, 7 (mod 8). Then
S p2 −1 ≡ f p2 −1 ≡ 0
2
(mod p2 ).
2
Moreover,
S p2 −1 ≡ f p2 −1
2
2
6
(mod p3 ).
Proof. For
p−1
2
< t < p and 0 ≤ s ≤ p−1
, from Lemma 2 we see that
2
2
2
p −1
p−1
≡ 1 (mod p),
≡ p−1
p2 −1
2
2
p−1
2
p−1 p+ 2
sp + t
p−1 p−1 2
2
= 0 (mod p),
s
t
2s + 1 2t − p
(2s + 1)p + 2t − p
2sp + 2t
=0
≡
=
t
s
sp + t
sp + t
≡
(mod p)
and
p2 − 1 − 2sp − 2t
p2 −1−2sp−2t
2
(p − 2s − 2)p + 2p − 2t − 1
= p−1
(
− s − 1)p + p + p−1
−t
2
2
p − 2s − 2
2p − 2t − 1
≡ p−1
=0
p−1
−
s
−
1
p
+
−
t
2
2
Now we assert that
(p−1)/2 p−1
X
p + p−1
2
2
sp + t
t=0
3
≡0
(mod p).
(mod p2 ) for s = 0, 1, 2, . . . .
(4)
We prove the result by induction on s. For 0 ≤ t ≤ (p − 1)/2 we see that
p2 −1 1 2t
−
t
2
2
(mod p2 ).
≡
=
t
t
t
(−4)
From Lemma 6 we know that the result is true for s = 0. Suppose that (4) holds for s = k.
For s = k + 1, applying Lemma 4 we have
(p−1)/2 p−1
X
p
2
+ p−1
2
(k + 1)p + t
t=0
3
p−1
3 X
2 p−1
p + 1 p−1
−p
p−1
2
2
+k
≡
−
2
2
t
k+1
t
t=0
p−1
p−1
p−1
3
−p
−p
−p
2
2
−
+
(mod p2 ).
−k 2
t+p
t
t+p
p−1
2
P(p−1)/2 p−1 p+ p−1 3
2
2
≡ 0 (mod p2 ) for k ≥
Hence
t=0
(k+1)p+t
hypothesis and Lemma 4 we have
(p−1)/2 X
t=0
p+1
2
p−1 2
t
p−1
−
+k
2
p−1
2
p−1
.
2
For k <
p−1
,
2
p−1
3
−p
−
p
−k 2
≡0
t
t+p
7
by the inductive
(mod p2 ).
Also,
p−1
2
t
≡
p−1
−p
2
t
≡
− 21
t
(mod p) and
p−1
−p
2
t
0 (mod p) for t ∈ {0, 1, . . . , p−1
}. By Lemma 5,
2
+
p−1
−p
2
t+p
= (−1)t
p−1
+t
2
t
−
+t
p+ p−1
2
t+p
≡
(p−1)/2 p−1
X
p
2
3
+ p−1
2
(k + 1)p + t
t=0
p−1 3 (p−1)/2
p−1
X p + 1 p−1 p−1
p−1
−p
− p 3
2
2
2
2
≡
−
−k
+k
k+1
2
t
2
t
t+p
t=0
p−1
(p−1)/2 X p + 1 p−1 p−1
p−1
−p
−p 2
2
2
−
−k
+k
+3
2
t
2
t
t+p
t=0
p−1
p−1 − p
− p 2
2
2
+
×
t
t+p
p−1
(p−1)/2 X p + 1 p−1 p−1
−p
− p 2
p−1
2
2
2
−k
−3
−
+k
t
t+p
t
2
2
t=0
p−1
p−1 − p
−p 2
2
+
×
t
t+p
p−1
(p−1)/2 p−1
X
−p
− p 3 2
2
+
−
t
t+p
t=0
p−1 3 (p−1)/2
X − 1 2 p−1 − p p−1 − p
2
2
2
2
≡ −3
+
t
t+p
k+1
t
t=0
1 2 p−1
p−1 3 (p−1)/2
X
+t
p + p−1
+t t −2
2
2
2
−
(−1)
= −3
t
t+p
k+1
t
t=0
≡0
(mod p2 ).
Hence
f p2 −1 ≡
2
(p−1)/2 (p−1)/2 p−1
X X
p
2
s=0
t=0
+ p−1
2
sp + t
3
≡0
(mod p2 ).
Pk
P
Set H0 = H0 (1, 1) = 0, Hk = i=1 k1 and Hk (1, 1) = 1≤i<j≤k ij1 for k ∈ Z+ . For 0 ≤ s ≤
p−1
≡ 1 − pH2s + p2 H2s (1, 1) (mod p3 )
(p − 1)/2, it is easily seen that Hp−1 ≡ 0 (mod
p),
2s
1
2
≡ 1 + pH2s + p2 H2s
− H2s (1, 1) (mod p3 ). By Lemma 7, for 0 ≤ t ≤ (p − 1)/2
and so p−1
( 2s )
we see that
2sp+2t
2
2sp+2t X 1
p2 p−1 Y p −1
p−1 1−
=
1 − p2
≡
(mod p3 ).
2s
2sp + 2t
i
i
2s
i=1,p∤i
i=1,p∤i
8
Applying (4), Lemma 6 and the identity
a−b
b
a
c
a−c . a
=
c−d d
c
d b−d
b
we derive that
S p2 −1 ≡
2
(p−1)/2 (p−1)/2 p2 −1
X X
2
s=0
sp + t
t=0
=
2
(p−1)/2 (p−1)/2
p −1 X X
≡
≡
p2 −1
2
s=0
p2 − 1
p2 −1
2
p2 −1
2
+p
2
1
p−1
2s
s=0
p2 − 1
(p−1)/2
X
1
p−1
2s
s=0
p2 − 1
p2 −1
2
sp+t
p2 −1
2sp+2t
t=0
(p−1)/2
X
(p−1)/2
X
2sp + 2t p2 − 1 − 2sp − 2t
p2 −1
sp + t
− sp − t
2
(p2 −1)/2 3
(p−1)/2 p−1
X
p
2
3 2sp+2t X 1
2
1+p
i
(p−1)/2 p−1
X
p
2
3
t=0
t=0
2s 3
s
+ p−1
2
sp + t
i=1,p∤i
+ p−1
2
sp + t
(p−1)/2
2t 3
X
t
(−64)s t=0 (−64)t
s=0
2
(p−1)/2 (p−1)/2 p−1
3
p −1 X X
p + p−1
2
2
≡
p2 −1
sp + t
2
s=0
t=0
(p−1)/2
+p
X
H2s +
(p−1)/2 (p−1)/2 p−1
X X
p
2
s=0
≡ f p2 −1
t=0
3 (p−1)/2 p−1
X
p + p−1
2
2
− H2s (1, 1))
sp + t
t=0
p−1 3
2
p(H2s
s=0
≡
H2t
+ 2
sp + t
(mod p3 ).
2
Summarizing the above proves the theorem.
4
An inequality involving (Sm)
Theorem 9. For m = 2, 3, 4, . . . we have
1+
2
1
S > Sm+1 Sm−1 .
m(m − 1) m
Proof. It is easily seen that
1+
2
1
S
> Sm Sm−2
(m − 1)(m − 2) m−1
9
for m = 3, 4, . . . , 13.
Now suppose m ≥ 14 and 1 +
inductive hypothesis we have
1
(m−1)(m−2)
2
Sm−1
> Sm Sm−2 . By (2), Lemma 3 and the
2
1
S − Sm+1 Sm−1
m(m − 1) m
2
1
32m2
4(3m2 + 3m + 1)
= 1+
S
S
+
S2
Sm −
m m−1
m(m − 1)
(m + 1)2
(m + 1)2 m−1
m2 − m + 1
4(3m2 + 3m + 1)
32m2 (m − 1)(m − 2)
Sm −
S
+
S
>
m−1
m−2 Sm
m(m − 1)
(m + 1)2
(m + 1)2 (m2 − 3m + 3)
= (20m5 − 60m4 + 52m3 + 28m2 − 36m + 12)Sm−1
+ (−128m5 + 320m4 − 256m3 − 32m2 + 192m − 96)Sm−2
1+
×
=
(m +
Sm
− 3m + 3)m3 (m − 1)
m−2
X m − 22k 2m − 4 − 2k 16Sm
1)2 (m2
(m + 1)2 (m2 − 3m + 3)m3 (m − 1)
k=0
k
k
m−2−k
F (m, k),
where
2k + 1
k+1
− 8m5 + 20m4 − 16m3 − 2m2 + 12m − 6.
F (m, k) = (5m5 − 15m4 + 13m3 + 7m2 − 9m + 3)
< 4, 5m5 −15m4 +13m3 +7m2 −9m+3 > 0,
For m ≥ 14 it is easily seen that 3 < (2m−7)(2m−5)
(m−3)(m−2)
−8m5 +20m4 −16m3 −2m2 +12m−6 < 0, 6m7 −75m6 +223m5 −283m4 −61m3 +427m2 −87m−
42 > 0, and F (m, k + 1) > F (m, k) for k = 0, 1, . . . , m − 3. Thus, F (m, m − 3) + F (m, 1) >
F (m, 5) + F (m, 1) > 0 and
5
F (m, k) ≥ F (m, 2) = (5m5 − 15m4 + 13m3 + 7m2 − 9m + 3)
3
− 8m5 + 20m4 − 16m3 − 2m2 + 12m − 6 > 0 for k ≥ 2.
10
From the above we derive that
1+
2
1
S − Sm+1 Sm−1
m(m − 1) m
2 X
2m − 4 − 2k
m − 2 2k
16Sm
F (m, k)
>
m−2−k
k
k
(m + 1)2 (m2 − 3m + 3)m3 (m − 1) k=0
m−2
X m − 22k 2m − 4 − 2k F (m, k)
+
k
k
m−2−k
k=m−4
2m − 4
16Sm
=
F
(m,
m
−
2)
+
F
(m,
0)
m−2
(m + 1)2 (m2 − 3m + 3)m3 (m − 1)
2m − 8
+ 3(m − 2)(m − 3)
F (m, m − 4) + F (m, 2)
m−4
2m − 6
+ 2(m − 2)
F (m, 1) + F (m, m − 3)
m−3
4(2m − 7)(2m − 5)
> 3(m2 − 5m + 6)F (m, m − 4) +
F (m, 0)
(m − 3)(m − 2)
2m−8
16Sm m−4
×
(m + 1)2 (m2 − 3m + 3)m3 (m − 1)
Sm 2m−8
8(2m − 7)(2m − 5) m−4
6(m
−
2)(2m
−
7)
+
=
(m + 1)2 (m2 − 3m + 3)m3 (m − 1)
(m − 3)(m − 2)
5
4
3
2
× (40m − 120m + 104m + 56m − 72m + 24)
+ (−128m5 + 320m4 − 256m3 − 32m2 + 192m − 96)
4(2m − 7)(2m − 5) × 3(m − 2)(m − 3) +
(m − 3)(m − 2)
> 6(m − 2)(2m − 7) + 24 (40m5 − 120m4 + 104m3 + 56m2 − 72m + 24)
+ 3(m − 2)(m − 3) + 16 (−128m5 + 320m4 − 256m3 − 32m2 + 192m − 96)
Sm 2m−8
m−4
×
(m + 1)2 (m2 − 3m + 3)m3 (m − 1)
= (6m7 − 75m6 + 223m5 − 283m4 − 61m3 + 427m2 − 87m − 42)
16Sm 2m−8
m−4
×
(m + 1)2 (m2 − 3m + 3)m3 (m − 1)
> 0.
Hence the inequality is proved by induction.
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Corollary 10. For m = 2, 3, 4, . . . we have
1+
2
1
P > Pm+1 Pm−1 .
m(m − 1) m
Proof. Since Pm = 2m Sm , the result follows from Theorem 9.
References
[1] E. Catalan, Sur les nombres de Segner, Rend. Circ. Mat. Palermo 1 (1887), 190–201.
[2] J. Franel, On a question of Laisant, L’Intermédiaire des Mathématiciens 1 (1894), 45–
47.
[3] A. F. Jarvis, P. J. Larcombe, and D. R. French, On small prime divisibility of the
Catalan-Larcombe-French sequence, Indian J. Math. 47 (2005), 159–181.
[4] X. J. Ji and Z. H. Sun, Congruences for Catalan-Larcombe-French numbers, preprint,
2015, http://arxiv.org/abs/1505.00668v2.
[5] F. Jarvis and H. A. Verrill, Supercongruences for the Catalan-Larcombe-French numbers, Ramanujan J. 22 (2010), 171–186.
[6] P. Larcombe and D. French, On the “other” Catalan numbers: a historical formulation
re-examined, Congr. Numer. 143 (2000), 33–64.
[7] P. J. Larcombe and D. R. French, A new generating function for the Catalan-LarcombeFrench sequence: proof of a result by Jovovic, Congr. Numer. 166 (2004), 161–172.
[8] R. Meštrović, Lucas’ theorem: its generalizations, extensions and applications (1878–
2014), preprint, 2014, http://arxiv.org/abs/1409.3820.
[9] B. Sagan, Congruences via Abelian groups, J. Number Theory. 20 (1985), 210–237.
[10] Z. H. Sun, Congruences concerning Legendre polynomials II, J. Number Theory. 133
(2013), 1950–1976.
[11] Z. H. Sun, Congruences involving Franel and Catalan-Larcombe-French numbers,
preprint, 2015, http://arxiv.org/abs/1502.02499.
2010 Mathematics Subject Classification: Primary 11A07; Secondary 05A10, 05A19, 05A20.
Keywords: congruence, inequality, Catalan-Larcombe-French number.
(Concerned with sequence A053175.)
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Received November 21 2015; revised versions received November 24 2015; February 23 2016.
Published in Journal of Integer Sequences, April 6 2016.
Return to Journal of Integer Sequences home page.
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