on divisibility properties of some differences of …...on divisibility properties of some di...

MotivationOur Results

SummaryReferences

On divisibility properties of some differences ofMotzkin numbers [May 22, 2012]

Tamas Lengyel

Mathematics DepartmentOccidental College

The Fifteenth International Conference on Fibonacci Numbersand Their Applications, June 25-20, 2012

Lengyel Divisibility properties of differences of Motzkin numbers p1


SummaryReferences

Outline

1 MotivationThe p-adic Order of Certain Difference of Motzkin NumbersPrevious Work

2 Our ResultsMain ResultsPreparation: Binomial Coefficients and Catalan NumbersProofsFigures

3 Summary



SummaryReferences

Previous Work

Previous Work I

Previous Work

Deutsch and Sagan (2006): Deutsch and Sagan, Congruencesfor Catalan and Motzkin numbers and related sequences, JNT2006

Eu et al. (2008): Eu, Liu, and Yeh, Catalan and Motzkinnumbers modulo 4 and 8, EurJC 2008

Luca and Young (2010): Luca and Young, On the binaryexpansion of the odd Catalan numbers, Proc. XIVthInternational Conference on Fibonacci Numbers, 2010.

Xin and Xu (2011): Xin and Xu, A short approach to Catalannumbers modulo 2r , ElJC 2011



SummaryReferences

Main ResultsPreparation: Binomial Coefficients and Catalan NumbersProofsFigures

Notation: p-adic order: νp(n/k), sum of p-ary digits:dp(k)

Notation

n and k positive integers, p prime

νp(k): highest power of p dividing k

νp(n/k) = νp(n)− νp(k)

dp(k): sum of digits in the base p representation of k

Motzkin numbers (A001006): Mn =∑n

k=0

( n2k

)Ck , n ≥ 0

where Ck is the kth Catalan number (A000108)Ck = 1

k+1

(2kk

), k ≥ 0

# of walks withn 0 1 2 3 4 5 6 7 from to steps

Mn 1 1 2 4 9 21 51 127 (0,0) (n, 0) ULDCn 1 1 2 5 14 42 132 429 (0,0) (2n, 0) UD


http://oeis.org/A001006

http://oeis.org/A000108


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p = 2, a ≥ 1 odd, b = 0, 1 I

Theorem 1For p = 2, n ≥ 2, a ≥ 1 odd, and b = 0 or 1, we have

ν2(Ma2n+1+b −Ma2n+b) = n − 1, if n is even

and

ν2(Ma2n+1+b −Ma2n + b) ≥ n, if n is odd.

Theorem 2 provides us with a lower bound onν2(Ma2n+1+b −Ma2n+b) on a recursive fashion in b and potentially,it can give the exact order if a = 1.



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p = 2, a ≥ 1 odd, b ≥ 3 odd I

Theorem 2For a ≥ 1 odd and n ≥ n0 with some n0 = n0(a, b) ≥ 1, we getthat for b ≥ 2 even

ν2(Ma2n+1+b −Ma2n+b) =

= min{ν2(Ma2n+1+b−1 −Ma2n+b−1),

ν2(Ma2n+1+b−2 −Ma2n+b−2)} − ν2(b + 2)

if the two expressions under the minimum operation are not equal.However, it they are, then it is at least

ν2(Ma2n+1+b−1 −Ma2n+b−1) + 1− ν2(b + 2).



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p = 2, a ≥ 1 odd, b ≥ 3 odd II

On the other hand, if b ≥ 3 is odd then we have

ν2(Ma2n+1+b −Ma2n+b) = ν2(Ma2n+1+b−1 −Ma2n+b−1) (1)

if ν2(Ma2n+1+b−2 −Ma2n+b−2) + ν2(b − 1) >ν2(Ma2n+1+b−1 −Ma2n+b−1), and

ν2(Ma2n+1+b −Ma2n+b) = ν2(Ma2n+1+b−2 −Ma2n+b−2)

+ ν2(b − 1)

if ν2(Ma2n+1+b−2 −Ma2n+b−2) + ν2(b − 1) <ν2(Ma2n+1+b−1 −Ma2n+b−1)), and otherwise, it is at leastν2(Ma2n+1+b−1 −Ma2n+b−1). Note however the stipulation that in



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p = 2, a ≥ 1 odd, b ≥ 3 odd III

all equalities above, if the right hand side value is at leastn − 2ν2(b + 2) then the equality turns into the inequalityν2(Ma2n+1+b −Ma2n+b) ≥ n − 2ν2(b + 2).



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p = 2, a ≥ 1 odd, b ≥ 3 odd IV

Theorem 2 guarantees that as n grows, eventually more and morebinary digits of Ma2n+b and Ma2n+1+b agree, starting with the leastsignificant bit for every fixed a ≥ 1 and b ≥ 0.

Corollary 3 (a rather coarse lower bound).

For a ≥ 1 odd, b ≥ 0, and n ≥ n0 with some n0 = n0(a, b) ≥ 1, wehave

ν2(Ma2n+1+b −Ma2n+b) ≥ n − (b + 2) + d2(b + 2).



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p ≥ 3

Theorem 4 gives a lower bound on ν3(M3n+1+1 −M3n+1) andνp(Mpn+1+b −Mpn+b) for p ≥ 5 and 0 ≤ b ≤ p − 3.

Theorem 4For p ≥ 3 prime and n ≥ n0 with some integer n0 = n0(p) ≥ 0, wehave

νp(Mpn+1 −Mpn) ≥ n. (2)

Assuming that n ≥ n0, for p = 3, we have

ν3(M3n+1+1 −M3n+1) ≥ n − 1,

and for p ≥ 5, we have

νp(Mpn+1+b −Mpn+b) ≥ n

with 0 ≤ b ≤ p − 3.



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Congruences and Divisibility Properties of BinomialCoefficients I

Three theorems comprise the most basic facts re: divisibility andcongruence properties of binomial coefficients. (Assume that0 ≤ k ≤ n.)

Theorem 5 (Kummer, 1852).

The power of a prime p that divides the binomial coefficient(nk

)is

given by the number of carries when we add k and n−k in base p.

Theorem 6 (Legendre, 1830).

We haveνp((n

k

))=

n−dp(n)p−1 − k−dp(k)

p−1 − n−k−dp(n−k)p−1 =

dp(k)+dp(n−k)−dp(n)p−1 .

In particular, ν2((nk

)) = d2(k) + d2(n − k)− d2(n) represents the

carry count in the addition of k and n − k in base 2.



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Congruences and Divisibility Properties of BinomialCoefficients II

From now on M and N will denote integers such that 0 ≤ M ≤ N.

Theorem 7 (Lucas, 1877).

Let N = (nd , . . . , n1, n0)p = n0 + n1p + · · ·+ ndpd andM = m0 + m1p + · · ·+ mdpd with 0 ≤ ni ,mi ≤ p − 1 for each i ,be the base p representations of N and M, respectively.(

N

M

)≡(

n0

m0

)(n1

m1

)· · ·(

nd

md

)mod p.



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Congruences and Divisibility Properties of BinomialCoefficients III

Lucas’ theorem has some remarkable extensions.

Theorem 8 (Anton, 1869, Stickelberger, 1890, Hensel, 1902).

Let N = (nd , . . . , n1, n0)p = n0 + n1p + · · ·+ ndpd ,M =m0 + m1p + · · ·+ mdpd and R = N −M = r0 + r1p + · · ·+ rdpd

with 0 ≤ ni ,mi , ri ≤ p− 1 for each i , be the base p representationsof N,M, and R = N −M, respectively. Then with q = νp

((NM

)),

(−1)q1

pq

(N

M

)≡(

n0!

m0!r0!

)(n1!

m1!r1!

)· · ·(

nd !

md !rd !

)mod p.

Theorem 8 can be further extended.



SummaryReferences


Congruences and Divisibility Properties of BinomialCoefficients IV

Definition

For a given integer n and prime p, let (n!)p = n!/(pbn/pcbn/pc!)be the product of positive integers not exceeding n and notdivisible by p.

• closely related to the p-adic Morita gamma function.

Theorem 9 (Granville, 1995).

Let N = (nd , . . . , n1, n0)p = n0 + n1p + · · ·+ ndpd ,M =m0 + m1p + · · ·+ mdpd and R = N −M = r0 + r1p + · · ·+ rdpd

with 0 ≤ ni ,mi , ri ≤ p− 1 for each i , be the base p representationsof N,M, and R = N −M, respectively. LetNj = nj + nj+1p + · · ·+ nj+k−1pk−1 for each j ≥ 0, i.e., the least



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Congruences and Divisibility Properties of BinomialCoefficients V

positive residue of bN/pjc mod pk with some integer k ≥ 1; alsomake the corresponding definitions for Mj and Rj . Let εj be thenumber of carries when adding M and R on and beyond the jthdigit. Then with q = ε0 = νp

((NM

)),

1

pq

(N

M

)≡ (±1)εk−1

((N0!)p

(M0!)p(R0!)p

)· · ·(

(Nd !)p(Md !)p(Rd !)p

)mod pk

where ±1 is −1 except if p = 2 and k ≥ 3.

We also use the following generalization of theJacobstahl–Kazandzidis congruences.



SummaryReferences


Congruences and Divisibility Properties of BinomialCoefficients VI

Theorem 10 (Corollary 11.6.22 in Cohen (2007)).

Let M and N such that 0 ≤ M ≤ N and p prime. We setR = p4NM(N −M)

(NM

)and have

(pN

pM

)≡

(

1− Bp−3

3 p3NM(N −M)

)(NM

)modR, if p ≥ 5,

(1 + 45NM(N −M))(NM

)modR, if p = 3,

(−1)M(N−M)P(N,M)(NM

)modR, if p = 2,

where P(N,M) =1+6NM(N−M)−4NM(N−M)(N2−NM +M2)+2(NM(N−M))2.



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Congruences and Divisibility Properties of BinomialCoefficients VII

Remark 11It is well known that νp(Bn) ≥ −1 by the von Staudt–Clausentheorem. If the prime p divides the numerator of Bp−3, i.e.,νp(Bp−3) ≥ 1, or equivalently

(2pp

)≡ 2 mod p4, then it is

sometimes called a Wolstenholme prime. The only knownWolstenholme primes up to 109 are p = 16843 and 2124679.



SummaryReferences


Divisibility Properties for Catalan Numbers I

Main tools wrt differences of Catalan numbers in Lengyel (2012a).For the 2-adic order of Catalan numbers we obtain the followingtheorem.

Theorem 12For any prime p ≥ 2 and (a, p) = 1, we have

νp(Capn+1 − Capn) = n + νp

((2a

a

)), n ≥ 1.

We can introduce an extra additive term b ≥ 1 into Theorem 12.



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Divisibility Properties for Catalan Numbers II

Theorem 13For p = 2, a odd, and n ≥ n0 = 2 we have

ν2(Ca2n+1+1 − Ca2n+1) = n + ν2

((2a

a

))− 1,

and in general, for b ≥ 1 and n ≥ n0 = blog2 2bc+ 1

ν2(Ca2n+1+b − Ca2n+b) = n + ν2

((2a

a

))+ ν2(g(b))

= n + d2(a) + d2(b)−dlog2(b + 2)e − ν2(b + 1) + 1



SummaryReferences


Divisibility Properties for Catalan Numbers III

where g(b) = 2(2bb

)(b + 1)−1(H2b − Hb − 1/(2(b + 1))) =

2Cb(H2b − Hb − 1/(2(b + 1))) = (f (b)− Cb)/(b + 1) withHn =

∑nj=1 1/j being the nth harmonic number.

For any prime p ≥ 3, (a, p) = 1, and b ≥ 1 we have that

νp(Capn+1+b − Capn+b) = n + νp

((2a

a

))+ νp(g(b)),

with n ≥ n0 = max{νp(g(b)) + 2r − νp(Cb) + 1, r + 1} =max{νp(2(H2b − Hb − 1/(2(b + 1)))) + 2r + 1, r + 1} andr = blogp 2bc.



SummaryReferences


Divisibility Properties for Catalan Numbers IV

In general, for any prime p ≥ 2, (a, p) = 1, b ≥ 1, andn > blogp 2bc, we have

νp(Capn+1+b − Capn+b) ≥ n + νp

((2a

a

))+ νp

((2b

b

))−

blogp 2bc − νp(b + 1).

Note. Clearly, νp(g(b)) ≥ 0 for 1 ≤ b ≤ (p− 1)/2. We note thatin general, for b ≥ 1 we haveνp(g(b)) ≥ νp(

(2bb

))− blogp 2bc − νp(b + 1) if p ≥ 2 while

ν2(g(b)) = d2(b)− dlog2(b + 2)e − ν2(b + 1) + 1 =d2(b + 1)− dlog2(b + 2)e if p = 2.



SummaryReferences


Divisibility Properties for Catalan Numbers V

Remark 14The combination of Theorems 12 and 13 proves some kind ofgeneralization of the observation in Liu and Yeh (2010) onC2n+m−1−1 mod 2n, with n and m ≥ 0, being constant. The latterfact has been proven by Lin (2011) and Luca and Young (2010)and extended by Xin and Xu (2011) recently.

Our result focuses on Capm+b mod pn, with any large enough m,for non-negative values of b only.

Theorem 15For any prime p ≥ 2, (a, p) = 1, b ≥ 0, we have thatCapm+b mod pn is constant form ≥ n + νp(b + 1) + max{0, blogp 2bc}, n ≥ 1.



SummaryReferences


Divisibility Properties for Catalan Numbers VI

We also note that

ν2(Ck) = d2(k)− ν2(k + 1) = d2(k + 1)− 1 (3)

holds, i.e., ν2(C2n+1) = ν2(C2n) = 1. It follows that Ck is odd ifand only if k = 2q − 1 for some integer q ≥ 0.



SummaryReferences


Proof of Theorem 1: case a = 1 & b = 0

Prove for b = 0 (b = 1 is practically identical)First we deal with the case with a = 1.

1 M2n =∑2n

k=0

(2n

2k

)Ck

2 ν2(Ck) = d2(k + 1)− 1

3 select an infinite incongruent disjoint covering system(IIDCS): {2q (mod 2q+1)}q≥0, i.e., ∀k > 0 : k = 2q + 2q+1Kfor some unique q and K ≥ 0.

4

M2n+1−M2n =2n−1∑k=1

((2n+1

2(2k)

)C2k−

(2n

2k

)Ck

)+

2n∑k≡1 mod 2

(2n+1

2k

)Ck

(4)

5 break the 1st summation into parts according to the IIDCS:



SummaryReferences



2n−1∑k=1

((2n+1

2(2k)

)C2k −

(2n

2k

)Ck

)(5)

=n−1∑q=0

∑k=2q+2q+1K

0≤K≤ 2n−q−1−12

((2n+1

2(2k)

)C2k −

(2n

2k

)Ck

)

=n−2∑q=0

∑k=2q+2q+1K

0≤K≤2n−q−2−1

((2n+1

2(2k)

)C2k −

(2n

2k

)Ck

)

+ (C2n − C2n−1).

Clearly, νp(C2n − C2n−1) = n by Theorem 12.



SummaryReferences



Definition (an m-section of the recurrence for M2n+1)

M ′r (n,m) =∑

k≡r mod m1≤k≤2n

(2n+1

2k

)Ck

2nd summation in (4) is M ′1(n, 2)

its 2-adic order is at least n:

Theorem 16For integers n ≥ q ≥ 1, we have

ν2(M ′2q(n, 2q+1)) = n + 1− q.

If q = 0 then ν2(M ′1(n, 2)) = n if n is odd, otherwise the2-adic order is at least n + 1.



SummaryReferences



needs more insight into the 2-adic structure of the terms ofthe sum (5):

1 check how the 2-adic orders of the terms(

2n

2k

)Ck and

2(

2n+1

2(2k)

)C2k with k = 2q + 2q+1K behave in M ′2q (n − 1, 2q+1)

and M ′2q+1,a(n, 2q+2)

If 1 ≤ q ≤ n − 1 then both 2-adic orders are = n − q + d2(K ).More importantly, the difference

Aq,K =( 2n+1

2(2k)

)C2k−

(2n

2k

)Ck =

( 2n+1

2(2k)

)(C2k−Ck)+

(( 2n+1

2(2k)

)−(2n

2k

))Ck

has 2-adic order n + d2(K ) by Theorems 5 (Kummer), 12, and 10.BTW, ν2(Aq,K ) = ν2(first term in the last sum), and it is given by

combining ν2

(( 2n+1

2(2k)

))= n − 1− q and

ν2(C2k − Ck) = q + d2(1 + 2K ) = q + d2(K ) + 1 ⇒ν2(∑

K Aq,K ) = n,∀q ≥ 1 and comes from the term with K = 0.



SummaryReferences


Proof of Theorem 1: case a = 0 & b = 0⇒ a ≥ 1

...and now the 2-adic order will be determined:

If q = 0, i.e., k = 1 + 2K , then A0,K = M ′2(n, 4)−M ′1(n − 1, 2)and ν2(A0,K ) = n − 1 since

ν2

(( 2n+1

4(1+2K)

)C2+4K

)= n − 1 + d2(3 + 4K )− 1 = n + d2(K ) and

ν2

((2n

2(1 + 2K )

)C1+2K

)= n − 1 + d2(2 + 2K )− 1

= n − 2 + d2(1 + K ) ≥ n − 1;

min value is taken exactly for n − 1 values of K since in the range0 ≤ K ≤ 2n−2 − 1 there are exactly n − 1 terms withK = 2r − 1, r = 0, 1, . . . , n − 2, leading to d2(K + 1) = 1 ⇒ν2(∑

K A0,K ) = n − 1 if n is even and at least n if n is odd.

The proof with an arbitrary a ≥ 1 odd is very similar except itrequires a more detailed analysis of the terms than we had in (4).



SummaryReferences


Proof of Theorem 1: case a ≥ 1 & b = 0, 1

Definition

M ′r ,a(n,m) =∑

k≡r mod m1≤k≤a2n

(a2n+1

2k

)Ck

While in the case of a = 1 it trivially follows that ν2

((a2n+1

2k

))= n,

now we have to deal with the possibility that 2k > 2n+1. However,this case turns out to be identical to that of a = 1 and hence,ν2(M ′1,a(n, 2)) ≥ n with equality if and only if n is odd.

What if a ≥ 1 odd and b = 1?

switch from a2n and a2n+1 to a2n + 1 and a2n+1 + 1, → theproof is almost identical to that of the case with b = 0

2-adic orders of(a2n+1

2k

)and

(a2n

2k

)are equal



SummaryReferences


Proof of Theorem 1: case a ≥ 1 & b = 1, b ≥ 2

extra work needed for the second term((a2n+1+12(2k)

)−(a2n+1

2k

))Ck in the revised version of the definition

of Aq,K . In fact, its 2-adic order is at least n as it follows byTheorem 10:(

a2n+1 + 1

4k

)−(

a2n + 1

2k

)≡ a2n

(a2n − 1

2k − 1

)≡ 0 (mod 2n).

Cases with b ≥ 2 call for more refined methods.



SummaryReferences


Sketch of the proof of Theorem 4 ([a = 1!] b = 0 first)

b = 0 first

use the infinite incongruent disjoint covering system (IIDCS){ipq (mod pq+1)}i=1,2,...,p−1;q≥0 ⇐⇒∀k > 0 : k = ipq + Kpq+1 uniquely with some integers i , q,and K ≥ 0

∼ the proof of Theorem 1, the difference of the Motzkinnumbers can be rewritten as

Mpn+1 −Mpn =n−1∑q=0

p−1∑i=1

( ∑k=ipq+Kpq+1

0≤K≤ pn−q−2i2p

((pn+1

p(2k)

)Cpk −

(pn

2k

)Ck

))(6)

+

p−1∑i=1

( pn+1/2∑k≡i mod p

(pn+1

2k

)Ck

)(7)



SummaryReferences


Sketch of the proof of Theorem 4 (b = 0 first)

The first term (6) can be rewritten as

n−1∑q=0

p−1∑i=1

∑k=ipq+Kpq+1


((pn+1

p(2k)

)Cpk −

(pn

2k

)Ck

)

=n−1∑q=0

p−1∑i=1

∑k=ipq+Kpq+1


((pn+1

p(2k)

)(Cpk − Ck) +

((pn+1

p(2k)

)−(

pn

2k

))Ck

).



SummaryReferences


Sketch of the proof of Theorem 4 (b = 0, 1, 2, . . . , p − 3for p ≥ 3)

in the first summation: νp(( pn+1

p(2k)

)(Cpk − Ck)

)≥ n − q + q = n by

Theorem 12, and νp((( pn+1

p(2k)

)−(pn

2k

))Ck

)≥ n + 2 by Theorem 10

and Remark 11. Clearly, νp(every term) in (7) is at least n + 1.

Unfortunately, the above treatment cannot be easily extended tohigher values of b, however, the recurrence

Mm =3(m − 1)Mm−2 + (2m + 1)Mm−1

m + 2,m ≥ 0, (8)

comes to the rescue. If p = 3 and b = 1, or p ≥ 5 and1 ≤ b ≤ p − 3 then use (8) with m = pn+1 + b and pn + b. Innumerator only the 2 terms 3(b−1)(b + 2)

(Mpn+1+b−2−Mpn+b−2

)and (2b + 1)(b + 2)

(Mpn+1+b−1 −Mpn+b−1

)matter and

νp(denominator) = 2νp(b + 2).



SummaryReferences


Sketch of the proof of Theorem 4 (b = 0, 1, 2, . . . , p − 3for p ≥ 3)

adapt the proof of Theorem 2

prove the statement step by step for b = 1 ⇒ b = 2, ..., ⇒b = p − 3

in the initial case of b = 1, the multiplying factor m − 1 ofMm−2 in (8) is divisible by pn in both settings of m while theterms with Mm−1 are covered by the case of b = 0starting with b = 2, we can use the already proven statementwith b − 1 and b − 2.

proof cannot be directly extended beyond b = p − 3 since thecommon denominator in the recurrence has p-adic order2νp(b + 2) [also reason for the potential drop in the 3-adicorder when b = 1]



SummaryReferences


Figures

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2 4 6 8 10 12

2

4

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10

12

(a) p = 2, a = 1, b = 1

(which agrees with a = 1, 5, 9,

or 13, and b = 0 for n ≥ 1)

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(b) p = 2, a = 1, b = 4

(which agrees with a = 1, b = 5

for n ≥ 3)

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(c) p = 2, a = 3, b = 11

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8

(d) p = 3, a = 2, b = 0

Figure: The function νp(Mapn+1+b

− Mapn+b), 0 ≤ n ≤ 12 (with y = n and n − log2 b included for p = 2)



SummaryReferences

Summary

For different settings of a and b, we have found

more and more p-ary digits of Mapn+1+b and Mapn+b agree asn grows

lower bounds on the rate of growth in the number ofmatching p-ary digits



SummaryReferences

Further Study

Find more exact orders.

Conjecture 1

For p = 2, a ≡ 1 (mod 4), and n ≥ 2, we have

ν2(Ma2n+1 −Ma2n) = n, if n is odd.

For p = 3, (a, 3) = 1, and n ≥ n0 = n0(a) with some integern0(a) ≥ 0, we have

ν3(Ma3n+1 −Ma3n) = n + ν3

((2a

a

)).

(follow up)

Theorem 17For p = 2 prime and n ≥ 2 odd, we have ν2(M2n+1 −M2n) = n.



SummaryReferences

Further Study

Conjecture 2

For p ≥ 5 prime and n ≥ n0 with some integer n0 = n0(p) ≥ 0, wehave

νp(Mpn+1 −Mpn) = n.

In fact, recently we have succeeded in proving

Theorem 18For p ≥ 3 prime and n ≥ 2, we have

νp(Mpn+1 −Mpn) = n.



SummaryReferences

References I

H. Cohen, Number Theory, vol 2: Analytic and Modern Tools,Graduate Texts in Mathematics, Springer, 2007

Emeric Deutsch and Bruce E. Sagan, Congruences forCatalan and Motzkin numbers and related sequences, Journal ofNumber Theory Vol. 117 (2006), 191–215.

S.-P. Eu, S.-C. Liu, and Y.-N. Yeh, Catalan and Motzkinnumbers modulo 4 and 8, European J. Combin. Vol. 29(2008),1449–1466.



SummaryReferences

References II

A. Granville, Arithmetic properties of binomial coefficients. I.Binomial coefficients modulo prime powers, in Organicmathematics (Burnaby, BC, 1995), volume 20 of CMS Conf.Proc., 253–276, Amer. Math. Soc., Providence, RI, 1997.Electronic version (a dynamic survey):http://www.dms.umontreal.ca/~andrew/Binomial/

T. Lengyel, On divisibility properties of some differences of thecentral binomial coefficients and Catalan numbers, manuscript,2012

T. Lengyel, Exact p-adic orders for differences of Motzkinnumbers, manuscript, 2012

H.-Y. Lin, Odd Catalan numbers modulo 2k ,INTEGERS,mElectronic Journal of Combinatorial NumberTheory Vol. 11 (2011), #A55, 1–5.


http://www.dms.umontreal.ca/~andrew/Binomial/


SummaryReferences

References III

Shu-Chung Liu and Jean C.-C. Yeh, Catalan NumbersModulo 2k , J. Integer Sequences Vol. 13 (2010), Article 10.5.4.

F. Luca and P. T. Young, On the binary expansion of the oddCatalan numbers, Proceedings of the XIVth InternationalConference on Fibonacci Numbers, Morelia, Mexico, 2010. eds.F. Luca and P. Stanica, to appear.

G. Xin and J-F. Xu, A short approach to Catalan numbersmodulo 2r , Electronic Journal of Combinatorics Vol. 18 (2011),#P177, 1–12.


on divisibility properties of some differences of …...on divisibility properties of some di...

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