Download Catalan Numbers Modulo a Prime Power

Catalan Numbers Modulo a Prime Power
Larry Shu-Chung Liu
National Hsinchu University of Education, Taiwan
a joint work with Jean C.-C. Yeh (Texas A&M)
In honor of Professor Doron Zeilberger’s birthday
Outline
1 Some results about m
n during the late 1800
2 Some recent works for a prime power modulus
3 Our technique
4 Catalan numbers modulus 2k
5 Catalan numbers modulus an odd prime power
Outline
1 Some results about m
n during the late 1800
2 Some recent works for a prime power modulus
3 Our technique
4 Catalan numbers modulus 2k
5 Catalan numbers modulus an odd prime power
Catalan Numbers Modulo a Prime Power
Some results about m
during the late 1800
n
Kummer’s Theorem
The p-adic order of an integer n is denoted and defined as
ωp (n) := max{α ∈ N : pα |n}.
Catalan Numbers Modulo a Prime Power
Some results about m
during the late 1800
n
Kummer’s Theorem
The p-adic order of an integer n is denoted and defined as
ωp (n) := max{α ∈ N : pα |n}.
Let C(m, n) be the number of total carries for operating
m + n.
Catalan Numbers Modulo a Prime Power
Some results about m
during the late 1800
n
Kummer’s Theorem
The p-adic order of an integer n is denoted and defined as
ωp (n) := max{α ∈ N : pα |n}.
Let C(m, n) be the number of total carries for operating
m + n.
Theorem (Kummer, 1852)
ωp
m+n
m
= C(m, n).
Catalan Numbers Modulo a Prime Power
Some results about m
during the late 1800
n
Lucas’ Theorem
Let p be a fixed prime and n be any integer. Define that
[n]p := h. . . n2 n1 n0 ip be the representation of n in the base-p
system.
Catalan Numbers Modulo a Prime Power
Some results about m
during the late 1800
n
Lucas’ Theorem
Let p be a fixed prime and n be any integer. Define that
[n]p := h. . . n2 n1 n0 ip be the representation of n in the base-p
system.
|h. . . n2 n1 n0 ip | := n0 + n1 p + n2 p2 + · · · = n.
Catalan Numbers Modulo a Prime Power
Some results about m
during the late 1800
n
Lucas’ Theorem
Let p be a fixed prime and n be any integer. Define that
[n]p := h. . . n2 n1 n0 ip be the representation of n in the base-p
system.
|h. . . n2 n1 n0 ip | := n0 + n1 p + n2 p2 + · · · = n.
Theorem (Lucas, 1877)
Y mi m
≡p
.
n
ni
i≥0
Catalan Numbers Modulo a Prime Power
Some results about m
during the late 1800
n
m
ω
/p (mod p)
n
Theorem (Anton, 1869; Stickelberger, 1890; Hensel, 1902; and etc)
Given non-negative integers m and n, let r = m − n. We have
Y mi !
(−1)ω m
≡p
,
ω
p
n
ni ! ri !
i≥0
where ω = ωp (
m
n
).
Catalan Numbers Modulo a Prime Power
Some results about m
during the late 1800
n
m
ω
/p (mod p)
n
Theorem (Anton, 1869; Stickelberger, 1890; Hensel, 1902; and etc)
Given non-negative integers m and n, let r = m − n. We have
Y mi !
(−1)ω m
≡p
,
ω
p
n
ni ! ri !
i≥0
where ω = ωp (
m
n
).
Notice that
ri ≡p mi − ni − κ(m, n, i − 1),
where κ(m, n, i − 1) is the possible borrow from the i-th place
to the (i − 1)-st place when operating m − n.
Catalan Numbers Modulo a Prime Power
Some results about m
during the late 1800
n
m
ω
/p (mod p)
n
Theorem (Anton, 1869; Stickelberger, 1890; Hensel, 1902; and etc)
Given non-negative integers m and n, let r = m − n. We have
Y mi !
(−1)ω m
≡p
,
ω
p
n
ni ! ri !
i≥0
where ω = ωp (
m
n
).
Notice that
ri ≡p mi − ni − κ(m, n, i − 1),
where κ(m, n, i − 1) is the possible borrow from the i-th place
to the (i − 1)-st place when operating m − n.
Analogous formulae w.r.t. modulus pk were recently given by
Granville and {Davis, Webb}.
Outline
1 Some results about m
n during the late 1800
2 Some recent works for a prime power modulus
3 Our technique
4 Catalan numbers modulus 2k
5 Catalan numbers modulus an odd prime power
Catalan Numbers Modulo a Prime Power
Some recent works for a prime power modulus
ω2 (cn )
The n-th Catalan number is defined as cn =
2n
1
n+1 n
.
Catalan Numbers Modulo a Prime Power
Some recent works for a prime power modulus
ω2 (cn )
The n-th Catalan number is defined as cn =
2n
1
n+1 n
Theorem (Deutsch and Sagan, 2006)
ω2 (cn ) = d(n + 1) − 1 = d(α),
where d(m) = m0 + m1 + . . . and
[n]2 = h[α]2 0 11
. . . 1}i2 .
| {z
This segment might be empty.
.
Catalan Numbers Modulo a Prime Power
Some recent works for a prime power modulus
ω2 (cn )
The n-th Catalan number is defined as cn =
2n
1
n+1 n
.
Theorem (Deutsch and Sagan, 2006)
ω2 (cn ) = d(n + 1) − 1 = d(α),
where d(m) = m0 + m1 + . . . and
[n]2 = h[α]2 0 11
. . . 1}i2 .
| {z
This segment might be empty.
Postnikov and Sagan generalized this formula for a weighted
Catalan number.
Catalan Numbers Modulo a Prime Power
Some recent works for a prime power modulus
ω2 (cn )
The n-th Catalan number is defined as cn =
2n
1
n+1 n
.
Theorem (Deutsch and Sagan, 2006)
ω2 (cn ) = d(n + 1) − 1 = d(α),
where d(m) = m0 + m1 + . . . and
[n]2 = h[α]2 0 11
. . . 1}i2 .
| {z
This segment might be empty.
Postnikov and Sagan generalized this formula for a weighted
Catalan number.
A general formula of ωp (cn ) for any prime p will be given later.
Catalan Numbers Modulo a Prime Power
Some recent works for a prime power modulus
cn (mod 8)
Let d(n) = n0 + n1 + · · · be the digit sum of [n]2 .
Catalan Numbers Modulo a Prime Power
Some recent works for a prime power modulus
cn (mod 8)
Let d(n) = n0 + n1 + · · · be the digit sum of [n]2 .
Proposition (Eu, Liu and Y.-N. Yeh, 2008)
First of all, cn 6≡8 3, 7

1


 5




2
cn ≡8
6





4


0
for any n. As for other congruences, we have
β = 0 or 1,
if d(α) = 0 and
β ≥ 2,
α = 1,
if d(α) = 1 and
α ≥ 2,
if d(α) = 2,
if d(α) ≥ 3,
where [n]2 = h[α]2 0 1β i2 .
Catalan Numbers Modulo a Prime Power
Some recent works for a prime power modulus
cn (mod 8)
Let d(n) = n0 + n1 + · · · be the digit sum of [n]2 .
Proposition (Eu, Liu and Y.-N. Yeh, 2008)
First of all, cn 6≡8 3, 7

1


 5




2
cn ≡8
6





4


0
for any n. As for other congruences, we have
β = 0 or 1,
if d(α) = 0 and
β ≥ 2,
α = 1,
if d(α) = 1 and
α ≥ 2,
if d(α) = 2,
if d(α) ≥ 3,
where [n]2 = h[α]2 0 1β i2 .
e.g. [83]2 = h1010011i2 has [α]2 = h1010i2 , d(α) = 2 and
β = 2.
Catalan Numbers Modulo a Prime Power
Some recent works for a prime power modulus
cn (mod 8)
Let d(n) = n0 + n1 + · · · be the digit sum of [n]2 .
Proposition (Eu, Liu and Y.-N. Yeh, 2008)
First of all, cn 6≡8 3, 7

1


 5




2
cn ≡8
6





4


0
for any n. As for other congruences, we have
β = 0 or 1,
if d(α) = 0 and
β ≥ 2,
α = 1,
if d(α) = 1 and
α ≥ 2,
if d(α) = 2,
if d(α) ≥ 3,
where [n]2 = h[α]2 0 1β i2 .
e.g. [83]2 = h1010011i2 has [α]2 = h1010i2 , d(α) = 2 and
β = 2.
Therefore, c83 ≡8 4.
Catalan Numbers Modulo a Prime Power
Some recent works for a prime power modulus
Mn (mod 4) with even value
The Motzkin number can be defined as Mn =
Pbn/2c
k=0
n
2k
Proposition (Eu, Liu and Y.-N. Yeh, 2008)
We have Mn ≡4 0 if and only if
n = (4i + 1)4j+1 − 1
or
n = (4i + 3)4j+1 − 2,
or
n = (4i + 3)4j+1 − 1,
and Mn ≡4 2 if and only if
n = (4i + 1)4j+1 − 2
where i, j ∈ N.
ck .
Catalan Numbers Modulo a Prime Power
Some recent works for a prime power modulus
Mn (mod 8) with even value
Proposition (Eu, Liu and Y.-N. Yeh, 2008)
The Motzkin number Mn is even if and only if
n = (4i + ε)4j+1 − δ for i, j ∈ N, ε = 1, 3 and δ = 1, 2. And we
never have Mn ≡8 0. Precisely, we have
(
4
if (ε, δ) = (1, 1) or (3, 2);
Mn ≡8
4y + 2 if (ε, δ) = (1, 2) or (3, 1),
where y is the number of digit 1’s in [4i + ε − 1]2 .
Outline
1 Some results about m
n during the late 1800
2 Some recent works for a prime power modulus
3 Our technique
4 Catalan numbers modulus 2k
5 Catalan numbers modulus an odd prime power
Catalan Numbers Modulo a Prime Power
Our technique
Main idea I
Main idea I: ωp , CFp and Eq,t
Problem 1: To evaluate
congruence for the combinatorial
Q
h
numbers of form
Qgi=1
j=1
Mi
Nj
(mod q := pk ).
Problem 2: To classify these combinatorial numbers (mod q)
according to their congruences.
Catalan Numbers Modulo a Prime Power
Our technique
Main idea I
Main idea I: ωp , CFp and Eq,t
Problem 1: To evaluate
congruence for the combinatorial
Q
h
numbers of form
Qgi=1
j=1
Mi
Nj
(mod q := pk ).
Problem 2: To classify these combinatorial numbers (mod q)
according to their congruences.
CFp (n) := pωpn(n) , the cofactor of n with respect to pωp (n) (or
non-p-cofactor).
Catalan Numbers Modulo a Prime Power
Our technique
Main idea I
Main idea I: ωp , CFp and Eq,t
Problem 1: To evaluate
congruence for the combinatorial
Q
h
numbers of form
Qgi=1
j=1
Mi
Nj
(mod q := pk ).
Problem 2: To classify these combinatorial numbers (mod q)
according to their congruences.
CFp (n) := pωpn(n) , the cofactor of n with respect to pωp (n) (or
non-p-cofactor).
Q
To evaluate a product M := hi=1 Mi modulo q = pk , let us
consider two cofactors of M , namely
Ph
pωp (M ) = p i=1 ωp (Mi ) and
h
Y
CFp (M ) =
CFp (Mi ).
i=1
Catalan Numbers Modulo a Prime Power
Our technique
Main idea I
Main idea I: ωp , CFp and Eq,t
We analyze further that
h
Y
i=1
CFp (Mi ) ≡q
Y
tEq,t (M ) ,
t∈Z∗q
where Z∗q = {1, 2, · · · , q − 1} − {mp | m ∈ N} and
P
Eq,t (M ) := hi=1 χ(CFp (Mi ) ≡q t).
Catalan Numbers Modulo a Prime Power
Our technique
Main idea I
Main idea I: ωp , CFp and Eq,t
We analyze further that
h
Y
i=1
CFp (Mi ) ≡q
Y
tEq,t (M ) ,
t∈Z∗q
where Z∗q = {1, 2, · · · , q − 1} − {mp | m ∈ N} and
P
Eq,t (M ) := hi=1 χ(CFp (Mi ) ≡q t).
We call Eq,t (M ) the t-encounter function of modulus q w.r.t.
Q
the product M := hi=1 Mi .
Catalan Numbers Modulo a Prime Power
Our technique
Main idea I
Main idea I: ωp , CFp and Eq,t
Q
Q
If ωp ( hi=1 Mi ) ≥ k, then hi=1 Mi ≡q 0; otherwise
h
Y
i=1
Qh
Mi ≡q pωp (
i=1
Mi )
Y
t∈Z∗q
t Eq,t (
Qh
i=1
Mi )
.
Catalan Numbers Modulo a Prime Power
Our technique
Main idea I
Main idea I: ωp , CFp and Eq,t
Q
Q
If ωp ( hi=1 Mi ) ≥ k, then hi=1 Mi ≡q 0; otherwise
h
Y
Qh
Mi ≡q pωp (
i=1
i=1
The idea can easily apply to
actually an integer.
Mi )
Y
t Eq,t (
Qh
i=1
Mi )
.
t∈Z∗q
Qh
M
Qgi=1 i ,
j=1 Nj
if this fraction is
Catalan Numbers Modulo a Prime Power
Our technique
Main idea I
Main idea I: ωp , CFp and Eq,t
Q
Q
If ωp ( hi=1 Mi ) ≥ k, then hi=1 Mi ≡q 0; otherwise
h
Y
Qh
Mi ≡q pωp (
i=1
Y
t Eq,t (
Qh
i=1
Mi )
.
t∈Z∗q
i=1
The idea can easily apply to
Mi )
Qh
M
Qgi=1 i ,
j=1 Nj
if this fraction is
actually an integer.
2n
1
Since cn = n+1
n , we have
cn ≡q p−ωp (n+1)+ωp ((2n)!)−2ωp (n!)
Y
1
tEq,t ((2n)!)−2Eq,t (n!) .
×
CFp (n + 1)
∗
t∈Zq
Catalan Numbers Modulo a Prime Power
Our technique
Main idea II
Main idea II
Let q = pk . We have a bijection Tq as follows
Tq : (Z∗2k , ×q ) → (C2 × C2k−2 , +)
Tq :
(Z∗pk , ×q )
for k ≥ 2;
→ (Cpk−1 (p−1) , +) for an odd prime p.
Catalan Numbers Modulo a Prime Power
Our technique
Main idea II
Main idea II
Let q = pk . We have a bijection Tq as follows
Tq : (Z∗2k , ×q ) → (C2 × C2k−2 , +)
Tq :
(Z∗pk , ×q )
for k ≥ 2;
→ (Cpk−1 (p−1) , +) for an odd prime p.
Let A = C2 × C2k−2 or A = Cpk−1 (p−1) . If we want to use A,
then we need to consider


h
h
X
Y
Y
CFp ( Mi ) ≡q Tq−1 
Tq (t)Eq,t ( Mi ) or
t∈Z∗q
i=1
Tq
h
Y
CFp (
i=1
!
Mi ) (mod q)
≡A
X
y∈A
i=1
h
Y
y Eq,T −1 (y) ( Mi ).
i=1
Catalan Numbers Modulo a Prime Power
Our technique
Main idea II
Main idea II
Let q = pk . We have a bijection Tq as follows
Tq : (Z∗2k , ×q ) → (C2 × C2k−2 , +)
Tq :
(Z∗pk , ×q )
for k ≥ 2;
→ (Cpk−1 (p−1) , +) for an odd prime p.
Let A = C2 × C2k−2 or A = Cpk−1 (p−1) . If we want to use A,
then we need to consider


h
h
X
Y
Y
CFp ( Mi ) ≡q Tq−1 
Tq (t)Eq,t ( Mi ) or
t∈Z∗q
i=1
Tq
h
Y
CFp (
i=1
!
Mi )
≡A
X
y∈A
i=1
h
Y
y Eq,T −1 (y) ( Mi ).
i=1
Catalan Numbers Modulo a Prime Power
Our technique
Evaluating ωp (n!) and CFp (n!) (mod pk )
Recall that d(n) = n0 + n1 + · · · , where [n]p = h. . . n1 n0 ip .
Define dk (n) = nk + nk+1 + · · · .
Lemma
Let q = pk , t ∈ Z∗q and [n]p = hnr nr−1 . . . n1 n0 ip . We have
n − d(n)
,
p−1
|hnr . . . nk−1 ip | − dk−1 (n)
Eq,t (n!) =
,
p−1
X
+
χ(|hni+k−1 . . . ni+1 ni ip | ≥ t).
ωp (n!) =
i≥0
Catalan Numbers Modulo a Prime Power
Our technique
Evaluating ωp (n!) and CFp (n!) (mod pk )
Recall that d(n) = n0 + n1 + · · · , where [n]p = h. . . n1 n0 ip .
Define dk (n) = nk + nk+1 + · · · .
Lemma
Let q = pk , t ∈ Z∗q and [n]p = hnr nr−1 . . . n1 n0 ip . We have
ωp (n!)
Eq0 (n!)
00
Eq,t
(n!)
n − d(n)
,
p−1
|hnr . . . nk−1 ip | − dk−1 (n)
=
,
p−1
X
=
χ(|hni+k−1 . . . ni+1 ni ip | ≥ t).
=
i≥0
Catalan Numbers Modulo a Prime Power
Our technique
Evaluating ωp (n!) and CFp (n!) (mod pk )
Recall that d(n) = n0 + n1 + · · · , where [n]p = h. . . n1 n0 ip .
Define dk (n) = nk + nk+1 + · · · .
Lemma
Let q = pk , t ∈ Z∗q and [n]p = hnr nr−1 . . . n1 n0 ip . We have
ωp (n!)
Eq0 (n!)
00
Eq,t
(n!)
n − d(n)
,
p−1
|hnr . . . nk−1 ip | − dk−1 (n)
=
,
p−1
X
=
χ(|hni+k−1 . . . ni+1 ni ip | ≥ t).
=
i≥0
where S(n) = {|hnk+i−1 . . . ni+1 ni ip | : except 0}i≥0 is a multi-set
and #(S, T ) is the number of elements (with multiplicity) in S
belonging to T .
Catalan Numbers Modulo a Prime Power
Our technique
Evaluating ωp (n!) and CFp (n!) (mod pk )
Recall that d(n) = n0 + n1 + · · · , where [n]p = h. . . n1 n0 ip .
Define dk (n) = nk + nk+1 + · · · .
Lemma
Let q = pk , t ∈ Z∗q and [n]p = hnr nr−1 . . . n1 n0 ip . We have
ωp (n!)
Eq0 (n!)
00
Eq,t
(n!)
n − d(n)
,
p−1
|hnr . . . nk−1 ip | − dk−1 (n)
=
,
p−1
= #(S(n), [t, q − 1]).
=
where S(n) = {|hnk+i−1 . . . ni+1 ni ip | : except 0}i≥0 is a multi-set
and #(S, T ) is the number of elements (with multiplicity) in S
belonging to T .
Catalan Numbers Modulo a Prime Power
Our technique
Evaluating ωp (n!) and CFp (n!) (mod pk )
Therefore, CFp (n!) ≡q
Q
t∈Z∗q t
Eq0 (n!)
×
Q
00
t∈Z∗q
tEq,t (n!) .
Catalan Numbers Modulo a Prime Power
Our technique
Evaluating ωp (n!) and CFp (n!) (mod pk )
Therefore, CFp (n!) ≡q
Q
t∈Z∗q t
Eq0 (n!)
×
Q
00
t∈Z∗q
tEq,t (n!) .
For p = 2 and k ≥ 3, we have
X
Y
t ≡q 1;
y ≡ (0, 0); equivalently,
y∈C2 ×C2k−2
t∈Z∗k
2
So Eq0 (n!) is useless for evaluating CFp (n!) (mod 2k ).
Catalan Numbers Modulo a Prime Power
Our technique
Evaluating ωp (n!) and CFp (n!) (mod pk )
Therefore, CFp (n!) ≡q
Q
t∈Z∗q t
Eq0 (n!)
×
Q
00
t∈Z∗q
tEq,t (n!) .
For p = 2 and k ≥ 3, we have
X
Y
t ≡q 1;
y ≡ (0, 0); equivalently,
y∈C2 ×C2k−2
t∈Z∗k
2
So Eq0 (n!) is useless for evaluating CFp (n!) (mod 2k ).
For an odd prime p, we have
X
Y
y ≡ pk−1 (p − 1)/2; equivalently,
t ≡q −1.
y∈Cpk−1 (p−1)
We only care about the parity of Eq0 (n!).
t∈Z∗k
p
Outline
1 Some results about m
n during the late 1800
2 Some recent works for a prime power modulus
3 Our technique
4 Catalan numbers modulus 2k
5 Catalan numbers modulus an odd prime power
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus 2k
Tq (CF2 (n!)) := (b(CF2 (n!)), uq (CF2 (n!)))
We use the bijection Tq : Z∗2k → C2 × Cq/4 to study cn (mod
2k ). Define Tq (t) := (b(t), uq (t)).
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus 2k
Tq (CF2 (n!)) := (b(CF2 (n!)), uq (CF2 (n!)))
We use the bijection Tq : Z∗2k → C2 × Cq/4 to study cn (mod
2k ). Define Tq (t) := (b(t), uq (t)).
0 if t ≡4 1;
b(t) =
1 if t ≡4 3.
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus 2k
Tq (CF2 (n!)) := (b(CF2 (n!)), uq (CF2 (n!)))
We use the bijection Tq : Z∗2k → C2 × Cq/4 to study cn (mod
2k ). Define Tq (t) := (b(t), uq (t)).
0 if t ≡4 1;
b(t) =
1 if t ≡4 3.
b(CF2 (n!)) ≡2 r(n) + n0 + n1 ≡2 zr(n) + n1 ,
where r(n) is the number of runs of 1 in [n]2 and
zr(n) is the number of runs of 0.
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus 2k
Tq (CF2 (n!)) := (b(CF2 (n!)), uq (CF2 (n!)))
We use the bijection Tq : Z∗2k → C2 × Cq/4 to study cn (mod
2k ). Define Tq (t) := (b(t), uq (t)).
0 if t ≡4 1;
b(t) =
1 if t ≡4 3.
b(CF2 (n!)) ≡2 r(n) + n0 + n1 ≡2 zr(n) + n1 ,
where r(n) is the number of runs of 1 in [n]2 and
zr(n) is the number of runs of 0.
X
uq (CF2 (n!)) ≡q/4
#(S(n), {s}) uq (t)
3≤t≤s≤q−1
t: odd
=
X
s∈[3,q−2]
#(S(n), {s})
X
t∈[3,s]odd
uq (t).
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus 2k
Tq (CF2 (n!)) := (b(CF2 (n!)), uq (CF2 (n!)))
We use the bijection Tq : Z∗2k → C2 × Cq/4 to study cn (mod
2k ). Define Tq (t) := (b(t), uq (t)).
0 if t ≡4 1;
b(t) =
1 if t ≡4 3.
b(CF2 (n!)) ≡2 r(n) + n0 + n1 ≡2 zr(n) + n1 ,
where r(n) is the number of runs of 1 in [n]2 and
zr(n) is the number of runs of 0.
X
uq (CF2 (n!)) ≡q/4
#(S(n), {s}) uq (t)
3≤t≤s≤q−1
t: odd
=
X
s∈[3,q−2]
#(S(n), {s})
X
uq (t).
t∈[3,s]odd
Q
P
We built a table for t∈[3,s]odd uq (t) or t∈[3,s]odd t according
to s ∈ [3, q − 2] to study cn (mod 16) and cn (mod 64).
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus 2k
CF2 (n!) ≡q CF2 (m!)
Lemma
Let m be an integer such that [m]2 is obtained by either extending
or truncating some runs of 0 or 1 of length ≥ k − 1 in [n]2 to be
different length but still ≥ k − 1. We have
b(CF2 (n!)) = b(CF2 (m!)) and uq (CF2 (n!)) = uq (CF2 (m!)),
and then
CF2 (n!) ≡q CF2 (m!).
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus 2k
CF2 (n!) ≡q CF2 (m!)
Lemma
Let m be an integer such that [m]2 is obtained by either extending
or truncating some runs of 0 or 1 of length ≥ k − 1 in [n]2 to be
different length but still ≥ k − 1. We have
b(CF2 (n!)) = b(CF2 (m!)) and uq (CF2 (n!)) = uq (CF2 (m!)),
and then
CF2 (n!) ≡q CF2 (m!).
Pf. b(CF2 (n!)) ≡2 zr(n) + n1 ≡2 zr(m) + m1 ≡2 b(CF2 (m!)).
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus 2k
CF2 (n!) ≡q CF2 (m!)
Lemma
Let m be an integer such that [m]2 is obtained by either extending
or truncating some runs of 0 or 1 of length ≥ k − 1 in [n]2 to be
different length but still ≥ k − 1. We have
b(CF2 (n!)) = b(CF2 (m!)) and uq (CF2 (n!)) = uq (CF2 (m!)),
and then
CF2 (n!) ≡q CF2 (m!).
Pf. b(CF2 (n!)) ≡2 zr(n) + n1 ≡2 zr(m) + m1 ≡2 b(CF2 (m!)).
uq (CF2 (n!)) ≡q/4
X
s∈[3,q−3]odd
#(S(n), {s, s + 1})
X
uq (t)
t∈[3,s]odd
is independent on the numbers of h0k−1 i2 and h1k−1 i2 in [n]2 .
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus 2k
cn ≡2k cn̄
Given n ∈ N, let n̄ be the integer such that [n̄]2 is obtained by the
following rules.
a. When the rightmost run of 0 in [n]2 is of length ≥ k + 1, let
us truncate it to be length k, otherwise keep it the same.
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus 2k
cn ≡2k cn̄
Given n ∈ N, let n̄ be the integer such that [n̄]2 is obtained by the
following rules.
a. When the rightmost run of 0 in [n]2 is of length ≥ k + 1, let
us truncate it to be length k, otherwise keep it the same.
b. For any other run of 0 or 1 of [n]2 with length ≥ k, truncate
them to be length k − 1.
Theorem (Liu and Yeh, 2010)
Let n, k ∈ N with k ≥ 3. We have
cn̄ for d(α) ≤ k − 1, and
cn ≡2k
0 for d(α) ≥ k.
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus 2k
cn ≡2k cn̄
Given n ∈ N, let n̄ be the integer such that [n̄]2 is obtained by the
following rules.
a. When the rightmost run of 0 in [n]2 is of length ≥ k + 1, let
us truncate it to be length k, otherwise keep it the same.
b. For any other run of 0 or 1 of [n]2 with length ≥ k, truncate
them to be length k − 1.
Theorem (Liu and Yeh, 2010)
Let n, k ∈ N with k ≥ 3. We have
cn̄ for d(α) ≤ k − 1, and
cn ≡2k
0 for d(α) ≥ k.
Example.
ch100001111i2 ≡8 6,
ch100011i2 ≡8 6,
ch1111111i2 ≡16 13,
ch111i2 ≡16 13.
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus 2k
A general formula for cn (mod 2k )
Theorem (Liu and Yeh, 2010)
Let n ∈ N and q = 2k with k ≥ 2. Then we have
cn ≡q (−1)zr(α) 2d(α) 5uq (CF2 (cn )) .
In particular, when k = 2 we have
cn ≡4 (−1)zr(α) 2d(α) .
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus 2k
cn (mod 16)
Proposition (Liu and Yeh, 2010)
Let cn be the n-th Catalan number. First of all,
cn 6≡16 3, 7, 9, 11, 15 for any n. As for the other congruences, we
have



1 
 β ≤ 1,



5
β = 2,
if d(α) = 0 and



 13 
β ≥3,



2
β = 0 or β ≥ 2,



 10 if d(α) = 1, α = 1 and β = 1,
6
(α = 2, β ≥ 2) or (α ≥ 3, β ≤ 1),
cn ≡16
if d(α) = 1, α ≥ 2 and


14
(α = 2, β ≤ 1) or (α ≥ 3, β ≥ 2),



4
zr(α) 6= 1,


if d(α) = 2 and

12
zr(α) = 1,




if d(α) = 3,
 8
0
if
d(α) ≥ 4.
where [n]2 = h[α]2 0 1β i2 .
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus 2k
cn (mod 64)
We also completely classified cn (mod 64). Here we only post the
classification for odd congruences.
Proposition (Liu and Yeh, 2010)
Let n ∈ N with d(α) = 0, i.e. n = 2β − 1. Then we have

1 if β = 0 or 1;




 5 if β = 2;
45 if β = 3;
cn ≡64


61 if β = 4;



29 if β ≥ 5.
Moreover, any number in [1, 63]odd − {1, 5, 29, 45, 61} can never be
the congruence of cn (mod 64).
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus 2k
Only k − 1 different cn (mod 2k ) with odd value
After observing all odd congruences from modulus 4 up to modulus
1024, once we conjectured the following property. This property
was proved recently.
Theorem (Lin, 2010)
Let k ≥ 2. Only k − 1 different odd congruences cn (mod 2k )
exist, and they are
c2m −1 (mod 2k )
for m = 1, 2, . . . , k − 1.
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus 2k
Only k − 1 different cn (mod 2k ) with odd value
After observing all odd congruences from modulus 4 up to modulus
1024, once we conjectured the following property. This property
was proved recently.
Theorem (Lin, 2010)
Let k ≥ 2. Only k − 1 different odd congruences cn (mod 2k )
exist, and they are
c2m −1 (mod 2k )
for m = 1, 2, . . . , k − 1.
Example
value:
There are only 6 congruences cn (mod 128) with odd
ch1i2 ≡128 1,
ch111i2 ≡128 45,
ch11111i2 ≡128 29,
ch11i2 ≡128 5,
ch1111i2 ≡128 125,
ch111111i2 ≡128 93.
Outline
1 Some results about m
n during the late 1800
2 Some recent works for a prime power modulus
3 Our technique
4 Catalan numbers modulus 2k
5 Catalan numbers modulus an odd prime power
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus an odd prime power
ωp (cn ) for any prime p
k
j
|hmi ...m0 ip |+|hni ...n0 ip |
(= 0 or 1) be the
Let κp (m, n; i) :=
i+1
p
possible carry from the i-th to the (i + 1)-st places for m + n
in the base-p system.
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus an odd prime power
ωp (cn ) for any prime p
k
j
|hmi ...m0 ip |+|hni ...n0 ip |
(= 0 or 1) be the
Let κp (m, n; i) :=
i+1
p
possible carry from the i-th to the (i + 1)-st places for m + n
in the base-p system.
P
Let Cp (m, n; i) := j≥i κp (m, n; j) and
Cp (m, n) = Cp (m, n; 0) which is the number of total carries.
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus an odd prime power
ωp (cn ) for any prime p
k
j
|hmi ...m0 ip |+|hni ...n0 ip |
(= 0 or 1) be the
Let κp (m, n; i) :=
i+1
p
possible carry from the i-th to the (i + 1)-st places for m + n
in the base-p system.
P
Let Cp (m, n; i) := j≥i κp (m, n; j) and
Cp (m, n) = Cp (m, n; 0) which is the number of total carries.
d(m + n) = d(m) + d(n) − (p − 1)Cp (m, n);
ωp ((m + n)!) = m+n−d(m)−d(n)
+ Cp (m, n).
p−1
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus an odd prime power
ωp (cn ) for any prime p
k
j
|hmi ...m0 ip |+|hni ...n0 ip |
(= 0 or 1) be the
Let κp (m, n; i) :=
i+1
p
possible carry from the i-th to the (i + 1)-st places for m + n
in the base-p system.
P
Let Cp (m, n; i) := j≥i κp (m, n; j) and
Cp (m, n) = Cp (m, n; 0) which is the number of total carries.
d(m + n) = d(m) + d(n) − (p − 1)Cp (m, n);
ωp ((m + n)!) = m+n−d(m)−d(n)
+ Cp (m, n).
p−1
Theorem
We have
ωp (cn ) = Cp (n, n) − β = Cp (n, n; β),
^
where [n]p = h. . . (p
− 1) (p − 1)β ip .
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus an odd prime power
ωp (cn ) for any prime p
k
j
|hmi ...m0 ip |+|hni ...n0 ip |
(= 0 or 1) be the
Let κp (m, n; i) :=
i+1
p
possible carry from the i-th to the (i + 1)-st places for m + n
in the base-p system.
P
Let Cp (m, n; i) := j≥i κp (m, n; j) and
Cp (m, n) = Cp (m, n; 0) which is the number of total carries.
d(m + n) = d(m) + d(n) − (p − 1)Cp (m, n);
ωp ((m + n)!) = m+n−d(m)−d(n)
+ Cp (m, n).
p−1
Theorem
We have
ωp (cn ) = Cp (n, n) − β = Cp (n, n; β),
^
where [n]p = h. . . (p
− 1) (p − 1)β ip .
If p = 2, then C2 (n, n; β) = d(α), where [n]2 = h[α]2 0 1β i2 .
⇒ Theorem of Deutsch and Sagan.
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus an odd prime power
CFq (cn ) for an odd prime power q = pk
Let ⊕q be the operator of addition over ring Zq , and
z(m, n; i) = |hmi+k−1 . . . mi ip| ⊕q |hni+k−1 . . . ni ip|. Then
h(m + n)i+k−1 . . . (m + n)i ip = [z(m, n; i) ⊕q κ(m, n; i − 1)]p .
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus an odd prime power
CFq (cn ) for an odd prime power q = pk
Let ⊕q be the operator of addition over ring Zq , and
z(m, n; i) = |hmi+k−1 . . . mi ip| ⊕q |hni+k−1 . . . ni ip|. Then
h(m + n)i+k−1 . . . (m + n)i ip = [z(m, n; i) ⊕q κ(m, n; i − 1)]p .
Lemma (From now on p is an odd prime.)
|h. . . mk−1 ip | + |h. . . nk−1 ip |−dk−1 (m) − dk−1 (n)
p−1
+C(m, n; k − 1);
X
χ(z(m, n; i) ≥ t)
00
Eq,t
((m + n)!) =
+ χ(z(m, n; i) = t − 1) κ(m, n; i − 1)
Eq0 ((m + n)!) =
i≥0
−σ(m, n),
where σ(m, n) be # of i st. z(m, n; i) = q − 1 and
κ(m, n; i − 1) = 1.
00 .
ecause σ(m, n) is independent on t, let’s modify Eq0 and Eq,t
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus an odd prime power
CFq (cn ) for an odd prime power q = pk
Let ⊕q be the operator of addition over ring Zq , and
z(m, n; i) = |hmi+k−1 . . . mi ip| ⊕q |hni+k−1 . . . ni ip|. Then
h(m + n)i+k−1 . . . (m + n)i ip = [z(m, n; i) ⊕q κ(m, n; i − 1)]p .
Lemma (From now on p is an odd prime.)
|h. . . mk−1 ip | + |h. . . nk−1 ip |−dk−1 (m) − dk−1 (n)
p−1
+C(m, n; k − 1) − σ(m, n);
X
χ(z(m, n; i) ≥ t)
00
Eq,t ((m + n)!) =
+ χ(z(m, n; i) = t − 1) κ(m, n; i − 1)
Eq0 ((m + n)!) =
i≥0
00 .
Because σ(m, n) is independent on t, let’s modify Eq0 and Eq,t
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus an odd prime power
CFq (cn ) for an odd prime power q = pk
Let ⊕q be the operator of addition over ring Zq , and
z(m, n; i) = |hmi+k−1 . . . mi ip| ⊕q |hni+k−1 . . . ni ip|. Then
h(m + n)i+k−1 . . . (m + n)i ip = [z(m, n; i) ⊕q κ(m, n; i − 1)]p .
Lemma (final version)
Eq0 ((m + n)!) ≡2
X
(m2j+k + n2j+k ) + C(m, n; k − 1) − σ(m, n);
j≥0
00
((m
Eq,t
+ n)!)
=
#(S + , [t, q − 1]) + #(S̄ + , {t − 1}),
where S + = {z(m, n; i) : z(m, n; i) 6= 0}i≥0 ,
S̄ + = {z ∈ S + : κ(m, n; i − 1) = 1 and z0 6= p − 1}, and
σ(m, n) is the number of i such that z(m, n; i) = q − 1 and
κ(m, n; i − 1) = 1
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus an odd prime power
CFq (cn ) for an odd prime power q = pk
CFq (cn ) ≡q
Eq0 ((n+n)!)−2Eq0 (n!)
1
CFq (n+1) (−1)
Q
00
00
× t∈Z∗q tEq,t ((n+n)!)−2Eq,t (n!) .
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus an odd prime power
CFq (cn ) for an odd prime power q = pk
CFq (cn ) ≡q
Eq0 ((n+n)!)−2Eq0 (n!)
1
CFq (n+1) (−1)
Q
00
00
× t∈Z∗q tEq,t ((n+n)!)−2Eq,t (n!) .
^
Let [n]p = h. . . (p
− 1) (p − 1)β ip = h. . . nβ (p − 1)β ip .
So [n + 1]p = h. . . nβ+1 (nβ + 1) 0β ip and then
CFq (n + 1) ≡q |hnβ+k−1 . . . nβ+1 (nβ + 1)ip |.
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus an odd prime power
CFq (cn ) for an odd prime power q = pk
CFq (cn ) ≡q
Eq0 ((n+n)!)−2Eq0 (n!)
1
CFq (n+1) (−1)
Q
00
00
× t∈Z∗q tEq,t ((n+n)!)−2Eq,t (n!) .
^
Let [n]p = h. . . (p
− 1) (p − 1)β ip = h. . . nβ (p − 1)β ip .
So [n + 1]p = h. . . nβ+1 (nβ + 1) 0β ip and then
CFq (n + 1) ≡q |hnβ+k−1 . . . nβ+1 (nβ + 1)ip |.
Eq0 ((n + n)!) ≡2 Cp (n, n; k − 1) − σ̄.
k−1 i
where σ̄ = |{i : hni+k−1 . . . ni i = h( p−1
2 )
and κ(n, n; i − 1) = 1}|.
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus an odd prime power
CFq (cn ) for an odd prime power q = pk
CFq (cn ) ≡q
Eq0 ((n+n)!)−2Eq0 (n!)
1
CFq (n+1) (−1)
Q
00
00
× t∈Z∗q tEq,t ((n+n)!)−2Eq,t (n!) .
^
Let [n]p = h. . . (p
− 1) (p − 1)β ip = h. . . nβ (p − 1)β ip .
So [n + 1]p = h. . . nβ+1 (nβ + 1) 0β ip and then
CFq (n + 1) ≡q |hnβ+k−1 . . . nβ+1 (nβ + 1)ip |.
Eq0 ((n + n)!) ≡2 Cp (n, n; k − 1) − σ̄.
k−1 i
where σ̄ = |{i : hni+k−1 . . . ni i = h( p−1
2 )
and κ(n, n; i − 1) = 1}|.
k−1 i with κ(n, n; i − 1) appears in [n] ,
However, once h( p−1
p
2 )
we have Cp (n, n; β) ≥ k; and then cn ≡q 0.
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus an odd prime power
CFq (cn ) for an odd prime power q = pk
CFq (cn ) ≡q
Eq0 ((n+n)!)−2Eq0 (n!)
1
CFq (n+1) (−1)
Q
00
00
× t∈Z∗q tEq,t ((n+n)!)−2Eq,t (n!) .
^
Let [n]p = h. . . (p
− 1) (p − 1)β ip = h. . . nβ (p − 1)β ip .
So [n + 1]p = h. . . nβ+1 (nβ + 1) 0β ip and then
CFq (n + 1) ≡q |hnβ+k−1 . . . nβ+1 (nβ + 1)ip |.
Eq0 ((n + n)!) ≡2 Cp (n, n; k − 1)−σ̄.
k−1 i
where σ̄ = |{i : hni+k−1 . . . ni i = h( p−1
2 )
and κ(n, n; i − 1) = 1}|.
k−1 i with κ(n, n; i − 1) appears in [n] ,
However, once h( p−1
p
2 )
we have Cp (n, n; β) ≥ k; and then cn ≡q 0.
Therefore, σ̄ is irrelevant for those c6 ≡q 0.
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus an odd prime power
CFq (cn ) for an odd prime power q = pk
CFq (cn ) ≡q
Eq0 ((n+n)!)−2Eq0 (n!)
1
CFq (n+1) (−1)
Q
00
00
× t∈Z∗q tEq,t ((n+n)!)−2Eq,t (n!) .
^
Let [n]p = h. . . (p
− 1) (p − 1)β ip = h. . . nβ (p − 1)β ip .
So [n + 1]p = h. . . nβ+1 (nβ + 1) 0β ip and then
CFq (n + 1) ≡q |hnβ+k−1 . . . nβ+1 (nβ + 1)ip |.
Eq0 ((n + n)!) ≡2 Cp (n, n; k − 1)−σ̄.
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus an odd prime power
CFq (cn ) for an odd prime power q = pk
CFq (cn ) ≡q
Eq0 ((n+n)!)−2Eq0 (n!)
1
CFq (n+1) (−1)
Q
00
00
× t∈Z∗q tEq,t ((n+n)!)−2Eq,t (n!) .
^
Let [n]p = h. . . (p
− 1) (p − 1)β ip = h. . . nβ (p − 1)β ip .
So [n + 1]p = h. . . nβ+1 (nβ + 1) 0β ip and then
CFq (n + 1) ≡q |hnβ+k−1 . . . nβ+1 (nβ + 1)ip |.
Eq0 ((n + n)!) ≡2 Cp (n, n; k − 1)−σ̄.
By last lemma, we need S + = {z(n, n; i) : z(n, n; i) 6= 0}i≥0 ,
S̄ + = {z ∈ S + : κ(n, n; i − 1) = 1 and z0 6= p − 1} and
S := {|hnk+i−1 . . . ni+1 ni ip | except 0}i≥0 .
Note that S + = {2s (mod q) : s ∈ S}.
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus an odd prime power
CFq (cn ) for an odd prime power q = pk
CFq (cn ) ≡q
Eq0 ((n+n)!)−2Eq0 (n!)
1
CFq (n+1) (−1)
Q
00
00
× t∈Z∗q tEq,t ((n+n)!)−2Eq,t (n!) .
^
Let [n]p = h. . . (p
− 1) (p − 1)β ip = h. . . nβ (p − 1)β ip .
So [n + 1]p = h. . . nβ+1 (nβ + 1) 0β ip and then
CFq (n + 1) ≡q |hnβ+k−1 . . . nβ+1 (nβ + 1)ip |.
Eq0 ((n + n)!) ≡2 Cp (n, n; k − 1)−σ̄.
By last lemma, we need S + = {z(n, n; i) : z(n, n; i) 6= 0}i≥0 ,
S̄ + = {z ∈ S + : κ(n, n; i − 1) = 1 and z0 6= p − 1} and
S := {|hnk+i−1 . . . ni+1 ni ip | except 0}i≥0 .
Note that S + = {2s (mod q) : s ∈ S}.
Q
00
00
To study t∈Z∗q tEq,t ((n+n)!)−2Eq,t (n!) , we need S + , S̄ + and S.
But S + = 2S and S̄ + ⊆ S + , we shall derive a formula using
only S.
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus an odd prime power
CFq (cn ) for an odd prime power q = pk
Finally, we get
Y
00
00
tEq,t ((n+n)!)−2Eq,t (n!)
t∈Z∗q
≡q
Y
Y
(
t)#(S,{u})
u∈[1,q−1] t∈[1,2u (mod q)]∩Z∗q
×
Y
(
Y
t2 )#(S,{u})
u∈[1,q−1] t∈[u+1,q−1]∩Z∗q
×
Y
(2u + 1 (mod q))#(S̄,{u}) ,
u∈[1,q−1]
2u+1∈Z∗
q
where S̄ = {u = |hnk+i−1 . . . ni+1 ni ip | : u 6= 0, u0 6=
(p − 1)/2 and κ(n, n; i − 1) = 1}i≥0 .
Note S̄ ⊆ S and S̄ + = {2u (mod q) : u ∈ S̄}.
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus an odd prime power
CFq (cn ) for an odd prime power q = pk
Q
00
00
The formula t∈Z∗q tEq,t ((n+n)!)−2Eq,t (n!) on the last page only
depends on S and S̄, and no more S + and S̄ + .
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus an odd prime power
CFq (cn ) for an odd prime power q = pk
Q
00
00
The formula t∈Z∗q tEq,t ((n+n)!)−2Eq,t (n!) on the last page only
depends on S and S̄, and no more S + and S̄ + .
Q
00
00
To evaluate t∈Z∗q tEq,t ((n+n)!)−2Eq,t (n!) for a particular prime
power q, we need to construct a table according to u ∈ Z∗q for
the following three values:
Y
A =
t,
t∈[1,2u (mod q)]∩Z∗q
B = (
Y
t)2
t∈[u+1,q−1]∩Z∗q
C = 2u + 1 (mod q).
and
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus an odd prime power
Example: c3212 (mod 27)
Example: c3212 (mod 27)
Let p = 3, k = 3 and n = 3212. What is c3212 (mod 27)?
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus an odd prime power
Example: c3212 (mod 27)
Example: c3212 (mod 27)
Let p = 3, k = 3 and n = 3212. What is c3212 (mod 27)?
[3212]3 = h11101222i3 . So β = 3.
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus an odd prime power
Example: c3212 (mod 27)
Example: c3212 (mod 27)
Let p = 3, k = 3 and n = 3212. What is c3212 (mod 27)?
[3212]3 = h11101222i3 . So β = 3.
Then ωp (c3212 ) = Cp (n, n; β) = 1.
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus an odd prime power
Example: c3212 (mod 27)
Example: c3212 (mod 27)
Let p = 3, k = 3 and n = 3212. What is c3212 (mod 27)?
[3212]3 = h11101222i3 . So β = 3.
Then ωp (c3212 ) = Cp (n, n; β) = 1.
CFq (cn ) ≡q
Eq0 ((n+n)!)
1
CFq (n+1) (−1)
Q
00
00
× t∈Z∗q tEq,t ((n+n)!)−2Eq,t (n!) .
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus an odd prime power
Example: c3212 (mod 27)
Example: c3212 (mod 27)
Let p = 3, k = 3 and n = 3212. What is c3212 (mod 27)?
[3212]3 = h11101222i3 . So β = 3.
Then ωp (c3212 ) = Cp (n, n; β) = 1.
CFq (cn ) ≡q
Eq0 ((n+n)!)
1
CFq (n+1) (−1)
Q
00
00
× t∈Z∗q tEq,t ((n+n)!)−2Eq,t (n!) .
In general, CFq (n + 1) ≡q |hnβ+k−1 . . . nβ+1 (nβ + 1)ip |.
In this case, CFq (3213) ≡27 |h102ip | = 11.
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus an odd prime power
Example: c3212 (mod 27)
Example: c3212 (mod 27)
Let p = 3, k = 3 and n = 3212. What is c3212 (mod 27)?
[3212]3 = h11101222i3 . So β = 3.
Then ωp (c3212 ) = Cp (n, n; β) = 1.
CFq (cn ) ≡q
Eq0 ((n+n)!)
1
CFq (n+1) (−1)
Q
00
00
× t∈Z∗q tEq,t ((n+n)!)−2Eq,t (n!) .
In general, CFq (n + 1) ≡q |hnβ+k−1 . . . nβ+1 (nβ + 1)ip |.
In this case, CFq (3213) ≡27 |h102ip | = 11.
Eq0 ((n + n)!) ≡2 Cp (n, n; k − 1) = 2 ≡2 0.
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus an odd prime power
Example: c3212 (mod 27)
Here
u
A
B
C
is the table
1 2 3
2 8 13
1 7 7
5 7
Q
00
00
for evaluating t∈Z∗ tE27,t ((2n)!)−2E27,t (n!) .
27
4
5
6
7
8
9 10 11 12
26 17 25 14 8
1
2 17 13
19 4
4 10 1
1 10 7
7
11 13
17 19
23 25
Y
A =
t∈[1,2u (mod
B = (
t,
q)]∩Z∗q
Y
t)2
t∈[u+1,q−1]∩Z∗q
C = 2u + 1 (mod q).
and
13 · · ·
26 · · ·
1 ···
···
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus an odd prime power
Example: c3212 (mod 27)
Here
u
A
B
C
Q
00
00
is the table for evaluating t∈Z∗ tE27,t ((2n)!)−2E27,t (n!) .
27
· · · 14 15 16 17 18 19 20 21 22 23 24
··· 2
8 13 26 17 25 14 8
1
2 17
··· 4
4 19 1
1 19 7
7 10 4
4
··· 2
4
8 10
14 16
20 22
Y
A =
t∈[1,2u (mod
B = (
t,
q)]∩Z∗q
Y
t)2
t∈[u+1,q−1]∩Z∗q
C = 2u + 1 (mod q).
and
25
13
1
26
26
1
26
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus an odd prime power
Example: c3212 (mod 27)
When κ = 1 and u0 =
6 1, we shall use A · B · C; otherwise A · B.
u
1 2
3 4 5
6 7 8 9 10 11 12
A·B
2 2 10 8 14 19 5 8 1 20 11 10
A·B·C
10 16
19 4
1 19
10 7
14
8
16
15
5
20
16
4
17
26
19
18
17
8
19
16
20
17
22
21
2
5
22
10
23
8
25
24
14
11
25
13
26
26
1
For our case, [3212]3 = h11101222i3 and
S = {001, 011, 111, 110, 101, 012, 122, 222}.
So u = |h001i3 | = 1 corresponding to 2 in the table.
Q
00
00
1
We have CF27 (c3212 ) ≡27 11
(−1)0 × t∈Z∗q tEq,t ((n+n)!)−2Eq,t (n!) .
13
26
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus an odd prime power
Example: c3212 (mod 27)
When κ = 1 and u0 =
6 1, we shall use A · B · C; otherwise A · B.
u
1 2
3 4 5
6 7 8 9 10 11 12
A·B
2 2 10 8 14 19 5 8 1 20 11 10
A·B·C
10 16
19 4
1 19
10 7
14
8
16
15
5
20
16
4
17
26
19
18
17
8
19
16
20
17
22
21
2
5
22
10
23
8
25
24
14
11
25
13
26
26
1
For our case, [3212]3 = h11101222i3 and
S = {001, 011, 111, 110, 101, 012, 122, 222}.
So u = |h001i3 | = 1 corresponding to 2 in the table.
Q
00
00
1
We have CF27 (c3212 ) ≡27 11
(−1)0 × t∈Z∗q tEq,t ((n+n)!)−2Eq,t (n!) .
13
26
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus an odd prime power
Example: c3212 (mod 27)
When κ = 1 and u0 =
6 1, we shall use A · B · C; otherwise A · B.
u
1 2
3 4 5
6 7 8 9 10 11 12
A·B
2 2 10 8 14 19 5 8 1 20 11 10
A·B·C
10 16
19 4
1 19
10 7
14
8
16
15
5
20
16
4
17
26
19
18
17
8
19
16
20
17
22
21
2
5
22
10
23
8
25
24
14
11
25
13
26
26
1
For our case, [3212]3 = h11101222i3 and
S = {001, 011, 111, 110, 101, 012, 122, 222}.
So u = |h001i3 | = 1 corresponding to 2 in the table.
Q
00
00
1
We have CF27 (c3212 ) ≡27 11
(−1)0 × t∈Z∗q tEq,t ((n+n)!)−2Eq,t (n!) .
13
26
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus an odd prime power
Example: c3212 (mod 27)
When κ = 1 and u0 =
6 1, we shall use A · B · C; otherwise A · B.
u
1 2
3 4 5
6 7 8 9 10 11 12
A·B
2 2 10 8 14 19 5 8 1 20 11 10
A·B·C
10 16
19 4
1 19
10 7
14
8
16
15
5
20
16
4
17
26
19
18
17
8
19
16
20
17
22
21
2
5
22
10
23
8
25
24
14
11
25
13
For our case, [3212]3 = h11101222i3 and
S = {001, 011, 111, 110, 101, 012, 122, 222}.
So u = |h001i3 | = 1 corresponding to 2 in the table.
Q
00
00
We have CF27 (c3212 ) ≡27 5 × t∈Z∗q tEq,t ((n+n)!)−2Eq,t (n!) .
26
26
1
13
26
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus an odd prime power
Example: c3212 (mod 27)
When κ = 1 and u0 =
6 1, we shall use A · B · C; otherwise A · B.
u
1 2
3 4 5
6 7 8 9 10 11 12
A·B
2 2 10 8 14 19 5 8 1 20 11 10
A·B·C
10 16
19 4
1 19
10 7
14
8
16
15
5
20
16
4
17
26
19
18
17
8
19
16
20
17
22
21
2
5
22
10
23
8
25
24
14
11
For our case, [3212]3 = h11101222i3 and
S = {001, 011, 111, 110, 101, 012, 122, 222}.
So u = |h001i3 | = 1 corresponding to 2 in the table.
We have CF27 (c3212 ) ≡27 5 × (2 × · · · ).
25
13
26
26
1
13
26
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus an odd prime power
Example: c3212 (mod 27)
When κ = 1 and u0 =
6 1, we shall use A · B · C; otherwise A · B.
u
1 2
3 4 5
6 7 8 9 10 11 12
A·B
2 2 10 8 14 19 5 8 1 20 11 10
A·B·C
10 16
19 4
1 19
10 7
14
8
16
15
5
20
16
4
17
26
19
18
17
8
19
16
20
17
22
21
2
5
22
10
For our case, [3212]3 = h11101222i3 and
S = {001, 011, 111, 110, 101, 012, 122, 222}.
So u = |h011i3 | = 4 corresponding to 8.
We have CF27 (c3212 ) ≡27 5 × (2 · 8 × · · · ).
23
8
25
24
14
11
25
13
26
26
1
13
26
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus an odd prime power
Example: c3212 (mod 27)
When κ = 1 and u0 =
6 1, we shall use A · B · C; otherwise A · B.
u
1 2
3 4 5
6 7 8 9 10 11 12
A·B
2 2 10 8 14 19 5 8 1 20 11 10
A·B·C
10 16
19 4
1 19
10 7
14
8
16
15
5
20
16
4
17
26
19
18
17
8
19
16
20
17
22
21
2
5
22
10
23
8
25
For our case, [3212]3 = h11101222i3 and
S = {001, 011, 111, 110, 101, 012, 122, 222}.
So u = |h111i3 | = 13 corresponding to 26.
We have CF27 (c3212 ) ≡27 5 × (2 · 8 · 26 × · · · ).
24
14
11
25
13
26
26
1
13
26
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus an odd prime power
Example: c3212 (mod 27)
When κ = 1 and u0 =
6 1, we shall use A · B · C; otherwise A · B.
u
1 2
3 4 5
6 7 8 9 10 11 12
A·B
2 2 10 8 14 19 5 8 1 20 11 10
A·B·C
10 16
19 4
1 19
10 7
14
8
16
15
5
20
16
4
17
26
19
18
17
8
19
16
20
17
22
21
2
5
22
10
23
8
25
24
14
11
For our case, [3212]3 = h11101222i3 and
S = {001, 011, 111, 110, 101, 012, 122, 222}.
So u = |h110i3 | = 12 with κ = 1 corresponding to 7.
We have CF27 (c3212 ) ≡27 5 × (2 · 8 · 26 · 7 × · · · ).
25
13
26
26
1
13
26
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus an odd prime power
Example: c3212 (mod 27)
When κ = 1 and u0 =
6 1, we shall use A · B · C; otherwise A · B.
u
1 2
3 4 5
6 7 8 9 10 11 12
A·B
2 2 10 8 14 19 5 8 1 20 11 10
A·B·C
10 16
19 4
1 19
10 7
14
8
16
15
5
20
16
4
17
26
19
18
17
8
19
16
20
17
22
21
2
5
22
10
23
8
25
24
14
11
25
13
26
26
1
For our case, [3212]3 = h11101222i3 and
S = {001, 011, 111, 110, 101, 012, 122, 222}.
So u = |h101i3 | = 10 corresponding to 20. (κ = 1 but u0 = 1)
We have CF27 (c3212 ) ≡27 5 × (2 · 8 · 26 · 7 · 20 × · · · ).
13
26
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus an odd prime power
Example: c3212 (mod 27)
When κ = 1 and u0 =
6 1, we shall use A · B · C; otherwise A · B.
u
1 2
3 4 5
6 7 8 9 10 11 12
A·B
2 2 10 8 14 19 5 8 1 20 11 10
A·B·C
10 16
19 4
1 19
10 7
14
8
16
15
5
20
16
4
17
26
19
18
17
8
19
16
20
17
22
21
2
5
22
10
23
8
25
24
14
11
25
13
For our case, [3212]3 = h11101222i3 and
S = {001, 011, 111, 110, 101, 012, 122, 222}.
So u = |h012i3 | = 5 with ka = 1 corresponding to 19.
We have CF27 (c3212 ) ≡27 5 × (2 · 8 · 26 · 7 · 20 · 19 × · · · ).
26
26
1
13
26
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus an odd prime power
Example: c3212 (mod 27)
When κ = 1 and u0 =
6 1, we shall use A · B · C; otherwise A · B.
u
1 2
3 4 5
6 7 8 9 10 11 12
A·B
2 2 10 8 14 19 5 8 1 20 11 10
A·B·C
10 16
19 4
1 19
10 7
14
8
16
15
5
20
16
4
17
26
19
18
17
8
19
16
20
17
22
21
2
5
22
10
23
8
25
24
14
11
25
13
26
26
1
For our case, [3212]3 = h11101222i3 and
S = {001, 011, 111, 110, 101, 012, 122, 222}.
So u = |h122i3 | = 17 with ka = 1 corresponding to 19.
We have CF27 (c3212 ) ≡27 5 × (2 · 8 · 26 · 7 · 20 · 19 · 19 × · · · ).
13
26
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus an odd prime power
Example: c3212 (mod 27)
When κ = 1 and u0 =
6 1, we shall use A · B · C; otherwise A · B.
u
1 2
3 4 5
6 7 8 9 10 11 12
A·B
2 2 10 8 14 19 5 8 1 20 11 10
A·B·C
10 16
19 4
1 19
10 7
14
8
16
15
5
20
16
4
17
26
19
18
17
8
19
16
20
17
22
21
2
5
22
10
23
8
25
24
14
11
25
13
For our case, [3212]3 = h11101222i3 and
S = {001, 011, 111, 110, 101, 012, 122, 222}.
So u = |h222i3 | = 26 corresponding to 26.
We have CF27 (c3212 ) ≡27 5 × (2 · 8 · 26 · 7 · 20 · 19 · 19 · 26).
26
26
1
13
26
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus an odd prime power
Example: c3212 (mod 27)
When κ = 1 and u0 =
6 1, we shall use A · B · C; otherwise A · B.
u
1 2
3 4 5
6 7 8 9 10 11 12
A·B
2 2 10 8 14 19 5 8 1 20 11 10
A·B·C
10 16
19 4
1 19
10 7
14
8
16
15
5
20
16
4
17
26
19
18
17
8
19
16
20
17
22
21
2
5
22
10
23
8
25
24
14
11
25
13
For our case, [3212]3 = h11101222i3 and
S = {001, 011, 111, 110, 101, 012, 122, 222}.
So u = |h222i3 | = 26 corresponding to 26.
We have CF27 (c3212 ) ≡27 4. Then c3212 ≡27 31 × 4 = 12.
26
26
1
13
26
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus an odd prime power
Example: c3212 (mod 27)
When κ = 1 and u0 =
6 1, we shall use A · B · C; otherwise A · B.
u
1 2
3 4 5
6 7 8 9 10 11 12
A·B
2 2 10 8 14 19 5 8 1 20 11 10
A·B·C
10 16
19 4
1 19
10 7
14
8
16
15
5
20
16
4
17
26
19
18
17
8
19
16
20
17
22
21
2
5
22
10
23
8
25
24
14
11
25
13
For our case, [3212]3 = h11101222i3 and
S = {001, 011, 111, 110, 101, 012, 122, 222}.
So u = |h222i3 | = 26 corresponding to 26.
We have CF27 (c3212 ) ≡27 4. Then c3212 ≡27 31 × 4 = 12.
26
26
1
13
26
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus an odd prime power
Example: c3212 (mod 27)
u
A·B
A·B·C
14
8
16
15
5
20
1
2
2
2
10
3
10
16
4
8
16
4
17
26
19
18
17
8
19
16
5
14
19
20
17
22
6
19
4
21
2
5
7
5
22
10
8
8
1
9
1
19
23
8
25
10
20
24
14
11
11
11
10
25
13
Lemma
A · B ≡q ±1 and A · B · C ± 1 for both u = (q − 1)/2 and
u = q − 1.
12
10
7
26
26
1
13
26
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus an odd prime power
Example: c3212 (mod 27)
Given n ∈ N, let n̄ be the integer such that [n̄]2 is obtained by the
following rules.
a. If the rightmost 0 run and p − 1 run of form the
0
[n]2 = h. . . 0β (p − 1)β ip with β, β 0 ≥ k + 1, let us truncate
their to length k, otherwise keep it the same.
b. For any other run of 0 and (p − 1)/2 of [n]2 with length
≥ k + 1, truncate them to be length of the same parity as k
or k − 1.
Theorem
Let n, k ∈ N with k ≥ 3. We have
cn̄ for d(α) ≤ k − 1, and
cn ≡2k
0 for d(α) ≥ k.
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus an odd prime power
Example: c3212 (mod 27)
Example.
ch3330006i7
≡49 43,
ch30006i7
≡49 43,
ch3006i7
≡49 43,
ch306i7
≡49 15,
ch3066i7
≡49 1,
ch30666i7
≡49 1,
ch33306i7
≡49 15,
ch3306i7
≡49 34.
Catalan Numbers Modulo a Prime Power
Catalan numbers modulus an odd prime power
Example: c3212 (mod 27)
Example.
ch3330006i7
≡49 43,
ch30006i7
≡49 43,
ch3006i7
≡49 43,
ch306i7
≡49 15,
ch3066i7
≡49 1,
ch30666i7
≡49 1,
ch33306i7
≡49 15,
ch3306i7
≡49 34.
The End!