On the periodicity problem of residual r-Fubini sequences

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by  Amir Abbas Asgari

TRANSACTIONSOFTHE
AMERICANMATHEMATICALSOCIETY
Volume00,Number0,Pages000–000
S0002-9947(XX)0000-0
ON THE PERIODICITY PROBLEM OF
RESIDUAL r-FUBINI SEQUENCES
AMIRABBASASGARIANDMAJIDJAHANGIRI
7
Abstract. For any positive integer r, the r-Fubini number with parameter n, de-
1
0 notedbyFn,r,isequal tothenumberofwaysthat theelements ofasetwithn+r
elements can be weak ordered such that the r least elements are in distinct orders.
2
Inthisarticlewefocus onthesequence of residuesofther-Fubininumbers modulo

n apositiveintegersandshowthatthissequenceisperiodicandthen,exhibithowto
a calculate its periodlength. As an extra result, an explicit formulafor the r-Stirling
J numbersisobtained whichisfrequentlyusedincalculations.

5
2

]
T 1. Introduction
N
The Fubini numbers (also known as the Ordered Bell numbers) form an integer se-
.
h quence in which the nth term counts the number of weak orderings of a set with n
t elements. Weak orderingmeans that the elements canbe ordered, allowingties. A. Cay-
a
m ley studied the Fubini numbers as the number of a certain kind of trees with n + 1
terminal nodes [2]. The Fubini numbers can also be defined as the summation of the
[
Stirling numbers of the second kind. The Stirling number of the second kind which is

1 denotedby n ,countsthenumberofpartitionsofnelementsintok non-emptysubsets.
v k
The sequence of residues of the Fubini numbers modulo a positive integer s is pointed
6 (cid:8) (cid:9)
outbyBjornPoonen.Heshowedthatthis sequenceisperiodicandcalculatedtheperiod
7
2 length for each positive integer s [5].
7 The r-Stirling numbers of the second kind are defined as an extension to the Stirling
0 numbers of the second kind, and similarly,an r-Fubini number is defined as the number
.
1 ofwayswhichtheelementsofasetwithn+relementscanbeweakorderedsuchthatthe
0 first r elements are in distinct places. One can study the same problem of periodicity of
7
residual sequence in case of the r-Fubini numbers. I. Mezo investigated this problem for
1
s=10[4].Inthisarticle,ω(A ),theperiodofther-Fubininumbersmoduloanypositive
: r,s
v integer s ∈ N is computed. Based on the Fundamental Theorem of Arithmetic, ω(A )
r,p
Xi is calculated for powers of odd primes pm. The cases s = 2m are studied separately.
Therefore if s = 2mpm1pm1...pmk is the prime factorization, then the ω(A ) is equal
r 1 1 k r,s
a to the least common multiple (LCM) of ω(Ar,pmi)’s and ω(Ar,2m), for i=1,2,...,k.
i
ThepreliminariesarepresentedinSection2andinSections3and4thelengthofthe
periods are computed in the case of odd prime powers and the 2 powers, respectively.
The last section contains the final theorem which presents the conclusion of this article.
2. Basic Concepts
The Stirling number of the second kind with the parameters n and k counts the
number of ways that the set A = {1,2,...,n} with n elements can be partitioned into
k non-empty subsets. If we want that the first r elements of A are in distinct subsets,
the numberofwaysto dosois the r-Stirling number ofthe secondkindwith parameters
2010 Mathematics Subject Classification. Primary11B50, 11B75,05A10;Secondary11B73,11Y55.
Keywords andphrases. residuesmoduloprimepowerfactors,r-Fubininumbers,periodicsequences,
r-Stirlingnumbersofthesecondkind.
(cid:13)cXXXXAmericanMathematicalSociety
1
2 AMIRABBASASGARIANDMAJIDJAHANGIRI
n and k, which is denoted by n (so it is clear that n ≥ k ≥ r). Fubini numbers are
k r
defined as follows [4]
(cid:8) (cid:9)
n
n
F = k! .
n k
k=0 (cid:26) (cid:27)
X
In a similar way we can define the r-Fubini numbers as the number of ways which the
elements of A can be weak ordered such that the elements {1,2,...,r} are in distinct
ranks. These numbers are denoted by F and are evaluated by
n,r
n
n+r
F = (k+r)! .
n,r k+r
k=0 (cid:26) (cid:27)r
X
Therearesimplerelationsandformulaeabout n whicharelistedbelow.Onecanfind
k r
a proof of them in [4], [1] and [3, §4].
(cid:8) (cid:9)
n n n−1
(1) = −(r−1) ,1≤r ≤n
m m m
(cid:26) (cid:27)r (cid:26) (cid:27)r−1 (cid:26) (cid:27)r−1
n n
(2) =
m m
(cid:26) (cid:27)1 (cid:26) (cid:27)
n+r
(3) =rn
r
(cid:26) (cid:27)r
n+r
(4) =(r+1)n−rn
r+1
(cid:26) (cid:27)r
m
n 1 m
(5) = (−1)m−j jn.
m m! j
(cid:26) (cid:27) j=1 (cid:18) (cid:19)
X
Inadditiontoaboverecurrencerelationsofther-Stirlingnumbersofthesecondkind,
a direct way to compute these numbers is given in the next theorem.
Theorem 2.1. For n,m ∈ N and r ≤ m ≤ n, the r-Stirling number of the second kind
with the parameters n and m is
m
n 1 m (j−1)!
= (−1)m−j jn−(r−1) .
m m! j (j−r)!
(cid:26) (cid:27)r j=r (cid:18) (cid:19) (cid:18) (cid:19)
X
Proof. We prove it by the induction on r. For r =2, the relations (1) and (5) result
n n n−1
= −
m m m
(cid:26) (cid:27)2 (cid:26) (cid:27) (cid:26) (cid:27)
m m
1 m m
= (−1)m−j jn− (−1)m−j jn−1
m! j j 
j=1 (cid:18) (cid:19) j=1 (cid:18) (cid:19)
X X
m 
1 m
= (−1)m−j jn−1(j−1)
m! j
j=2 (cid:18) (cid:19)
X
Assume that
m
n 1 m (j−1)!
(6) = (−1)m−j jn−(r−1) .
m m! j (j−r)!
(cid:26) (cid:27)r j=r (cid:18) (cid:19) (cid:18) (cid:19)
X
For n , use (1) to conclude that
m r+1
(cid:8) (cid:9)
PERIODICITY OF RESIDUAL r-FUBINI SEQUENCES 3
n n n−1
= −r
m m m
(cid:26) (cid:27)r+1 (cid:26) (cid:27)r (cid:26) (cid:27)r
m
1 m (j−1)!
= (−1)m−j jn−(r−1)
m! j (j−r)!
j=r (cid:18) (cid:19) (cid:18) (cid:19)
X
m
1 m (j−1)!
−r (−1)m−j jn−1−(r−1)
m! j (j−r)!
(cid:18) (cid:19)j=r (cid:18) (cid:19) (cid:18) (cid:19)
X
m
1 m (j−1)!
= (−1)m−j jn−r(j−r)
m! j (j−r)!
j=r (cid:18) (cid:19) (cid:18) (cid:19)
X
m
1 m (j−1)!
= (−1)m−j jn−r .
m! j (j−(r+1))!
j=r+1 (cid:18) (cid:19) (cid:18) (cid:19)
X
(cid:3)
Byϕ(n)weindicatethe numberofintegernumberslessthannandco-primetoit.It
isknownasEuler’stotientfunction.Thevalueofϕ(n)canbecomputedviathefollowing
relation [3]
1
ϕ(n)=n (1− )
p
Yp|n
3. The r-Fubini residues modulo prime powers
Letpbeaprimenumbergreaterthan2andmbeapositiveinteger.If{F }denotes
n,r
the r-Fubini numbers for a fixed positive integer r, we indicate by A = {F (mod
r,q n,r
q)}, for n ∈ N, the sequence of residues of the r-Fubini numbers modulo the positive
integer q. In this section we try to compute the period length of the sequence A when
r,q
q =pm. We denote this length by ω(A ).
r,q
Proposition 3.1. Let p be an odd prime and let q = pm, m ∈ N. If q ≤ r, then
ω(A )=1
r,q
Proof. Theproofisverysimple.Sincep≤r,wecandeducethatp|(k+r)!,fork ≥0,and
by the relationF = n (k+r)! n+r ,we have p|F .Thereforeω(A )=1. (cid:3)
n,r k=0 k+r r n,r r,p
P (cid:8) (cid:9)
As pointed out in above proposition, the cases in which the prime power factor of s
is less than or equal to r have the period length 1, so it is sufficient to investigate the
period length in the cases of q >r.
Lemma 3.2. Let p be an odd prime and r,m∈N with p≥r+1. Then
pm−r ≥m.
Proof. For m = 1 the result is obvious. Suppose the inequality holds for any m ≥ 2.
Since p(p+m)>2(p+m)>2p+m, we have
(7) p2+pm−p≥p+m.
Since p − 1 ≥ r, the induction hypothesis can be reformulated to pm ≥ p − 1 + m.
Multiplication by p results pm+1 ≥p2+pm−p. By (7) we havepm+1 ≥p+(m+1)−1,
Q.E.D. (cid:3)
Theorem 3.3. Let p bean odd prime and q =pm. After the (m−1)thterm thesequence
A has a period with length ω(A )=ϕ(q).
r,q r,q
4 AMIRABBASASGARIANDMAJIDJAHANGIRI
Proof. If n≥q−r−1 we can write
n+ϕ(q) n
n+ϕ(q)+r n+r
F −F = (k+r)! − (k+r)!
n+ϕ(q),r n,r k+r k+r
k=0 (cid:26) (cid:27)r k=0 (cid:26) (cid:27)r
X X
q−r−1
n+ϕ(q)+r n+r
≡ (k+r)! − (mod q)
k+r k+r
k=0 (cid:18)(cid:26) (cid:27)r (cid:26) (cid:27)r(cid:19)
X
q−r−1k+r
k+r (j−1)!
≡ (−1)k+r−j jn+1 (jϕ(q)−1) (mod q).
j (j−r)!
k=0 j=r (cid:18) (cid:19) (cid:18) (cid:19)
X X
If j = cp, c ∈ N, then jn+1 = (cp)q−r+h, for some h ≥ 0, so from Lemma 3.2 it follows
that jn+1 ≡ 0 (mod q). If (j,q) = 1, by Euler’s Theorem jϕ(q) −1 ≡ 0(mod q), so the
right hand side of the above congruence relation vanished and we have
(8) F ≡F (mod q), for n≥q−r−1.
n+ϕ(q),r n,r
If m−1≤n<q−r−1 then
q−r−1
n+ϕ(q)+r n+r
F −F ≡ (k+r)! −
n+ϕ(q),r n,r k+r k+r
k=0 (cid:18)(cid:26) (cid:27)r (cid:26) (cid:27)r(cid:19)
X
q−r−1
n+ϕ(q)+r
− (k+r)!
k+r
k=n+Xϕ(q)+1 (cid:26) (cid:27)r
q−r−1
n+r
+ (k+r)! (mod q)
k+r
k=n+1 (cid:26) (cid:27)r
X
q−r−1k+r
k+r (j−1)!
≡ (−1)k+r−j jn+1 (jϕ(q)−1)
j (j−r)!
k=0 j=r (cid:18) (cid:19) (cid:18) (cid:19)
X X
q−r−1
n+ϕ(q)+r
− (k+r)!
k+r
k=n+Xϕ(q)+1 (cid:26) (cid:27)r
q−r−1
n+r
+ (k+r)! (mod q).
k+r
k=n+1 (cid:26) (cid:27)r
X
Since n ≥ m−1, in the indices where j = cp, c ∈ N, we have jn+1 = (cp)m+h, for
some h ≥ 0, and it deduced that jn+1 ≡ 0 (mod q). When (j,q) = 1, again by Euler’s
Theoremjϕ(q)−1≡0(modq).Inthesums q−r−1(k+r)! n+r and q−r−1 (k+
k=n+1 k+r r k=n+ϕ(q)+1
r)! n+ϕ(q)+r the upper parameter of the r-Stirling number is less than the lower one
k+r r P (cid:8) (cid:9) P
and therefore these two sums are equal to zero. So
(cid:8) (cid:9)
q−r−1k+r
k+r (j−1)!
F −F ≡ (−1)k+r−j jn+1
n+ϕ(q),r n,r j (j−r)!
k=0 j=r (cid:18) (cid:19) (cid:18) (cid:19)
X X
×(jϕ(q)−1)≡0(mod q),
and therefore
(9) F ≡F (mod q), for m−1≤n<q−r−1.
n+ϕ(q),r n,r
Combining results (8) and (9) gives F ≡F (mod q), for n≥m−1. (cid:3)
n+ϕ(q),r n,r
PERIODICITY OF RESIDUAL r-FUBINI SEQUENCES 5
4. The r-Fubini residues modulo powers of 2
Similar to many computations in number theory, the case of p = 2 has its own
difficulties which needs special manipulations. In the case of powers of 2, initially we
calculate the residues of 2-Fubini numbers and then use the results in the case of the r-
Fubininumbers.Weclassifythesequenceofremaindersof2-Fubininumbersmodulo2m,
m ≥ 7, in Theorem 4.4 and then, work on remainders of the r-Fubini numbers modulo
2m, m ≥ 7 in Theorem 4.7. The special cases will be proved in Theorems 4.1, 4.5 and
4.6. The trivial cases in which 2m ≤r with period length 1 are omitted.
Theorem 4.1. If 3 ≤ m ≤ 6, then after the (m−1)th term the sequence A has a
2,2m
period with length ω(A )=2.
2,2m
Proof. ByusingtheformulaF = n (k+2)! n+2 weprovethatF −F ≡0
n,2 k=0 k+2 2 n+2,2 n,2
(mod 26) then it is concluded that F −F ≡0 (mod 2m), 3≤m≤5.
Pn+2,2 n,2(cid:8) (cid:9)
n+2 n
n+4 n+2
F −F = (k+2)! − (k+2)!
n+2,2 n,2 k+2 k+2
k=0 (cid:26) (cid:27)2 k=0 (cid:26) (cid:27)2
X X
5
n+4 n+2
≡ (k+2)! − (mod 26)
k+2 k+2
k=0 (cid:18)(cid:26) (cid:27)2 (cid:26) (cid:27)2(cid:19)
X
5 k+2
k+2
≡ (−1)k+2−j jn+1(j2−1)(j−1) (mod 26).
j
k=0j=2 (cid:18) (cid:19)
XX
m = 6 implies that n ≥ 5, so if j is even, then jn+1 = (2c)6+h, for some h ≥ 0 and
therefore64|jn+1.For odd j’s,(j,64)=1 so by Euler’s Theoremwe havej32 ≡1 (mod
64) and therefore jn+1+32 ≡jn+1 (mod 64). This implies
5 ⌊(k+1)/2⌋
k+2
F −F ≡ (−1)k+2−(2l+1) (2l+1)n+1
n+2,2 n,2 2l+1
k=0 l=1 (cid:18) (cid:19)
X X
× (2l+1)2−1) ×2l (mod 64)
5 ⌊(cid:0)(k+1)/2⌋ k+2(cid:1) l(l+1)
≡16 (−1)k+1 (2l+1)n+1 l .
 2l+1 2 
k=0 l=1 (cid:18) (cid:19) (cid:18) (cid:19)
X X
 
Enumerating the last summation for 2 ≤ n ≤ 33 shows that it is divisible by 64 and
because of periodicity of remainders of jn+1 modulo 64, the result follows. (cid:3)
Now we present the following lemma analogues to Lemma 3.2 in previous section.
Lemma 4.2. If m∈N and m>1, then 2m−2≥m.
Proof. For m=2 the result is obvious. If we assume that
(10) 2m−2≥m
then multiplication by 2 gives 2m+1−4 ≥ 2m. Since 2m+1 ≥ 2m+4 > m+3, we have
2m+1 ≥m+3 and so 2m+1−2≥m+1. (cid:3)
Thefollowinglemmaprovidesasimplebutessentialrelationusedinthenexttheorem.
Its proof is provided in Appendix A.
Lemma 4.3. For m≥7 and 5≤i≤2m−6 we have 2m−6−i|2i−5 2m−6−1 .
i
Theorem 4.4. If m≥7, after the (m−1)th term, the sequence A (cid:0) has a(cid:1)period with
2,2m
length ω(A )=2m−6.
2,2m
6 AMIRABBASASGARIANDMAJIDJAHANGIRI
Proof. Inthecaseofn≥2m−3,fromLemma4.2wecandeducethatn≥2m−3≥m−1.
So we have
n+2m−6 n+2m−6+2 n n+2
Fn+2m−6,2−Fn,2 ≡ (k+2)! k+2 − (k+2)! k+2
k=0 (cid:26) (cid:27)2 k=0 (cid:26) (cid:27)2
X X
2m−3 n+2m−6+2 n+2
≡ (k+2)! − (mod 2m)
k+2 k+2
k=0 (cid:18)(cid:26) (cid:27)2 (cid:26) (cid:27)2(cid:19)
X
2m−3k+2
≡ (−1)k+2−j k+2 jn+1(j2m−6 −1)(j−1) (mod 2m).
j
k=0 j=2 (cid:18) (cid:19)
X X
When j is even, then jn+1 =(2c)2m−2+h, for some h ≥0. So by Lemma 4.2, 2m | jn+1.
For odd j’s we have
2m−3⌊(k+1)/2⌋
k+2
Fn+2m−6,2−Fn,2 ≡ (−1)k+2−(2l+1) 2l+1 (2l+1)n+1
k=0 l=1 (cid:18) (cid:19)
X X
×((2l+1)2m−6 −1)×2l (mod 2m)
2m−3 ⌊(k+1)/2⌋
k+2
≡2m−4 (−1)k+1 (2l+1)n+1
2l+1
k=0 l=1 (cid:18) (cid:19)
X X
(2l+1)2m−6 −1
× l (mod 2m)
2m−5
!
2m−3 ⌊(k+1)/2⌋
k+2
≡2m−4 (−1)k+1 (2l+1)n+1
2l+1
k=0 l=1 (cid:18) (cid:19)
X X
2m−6 (2m−6−1)!
× li2i−1 l (mod 2m).
i!(2m−6−i)!
i=1 (cid:18) (cid:19)
X
Last expression contains m− 4 factors of 2, so it is sufficient to prove the last sum-
mation is divisible by 16. We denote this summation by S. Simplify the summation
2m−6li2i−1 (2m−6−1)! and use the Lemma 4.3 gives
i=1 i!(2m−6−i)!
P 2m−6 (2m−6−1)! 4 (2m−6−1)!
li2i−1 ≡ li2i−1 (mod 16)
i!(2m−6−i)! i!(2m−6−i)!
i=1 (cid:18) (cid:19) i=1 (cid:18) (cid:19)
X X
l3×2(2m−6−1)(2m−6−2)
≡l+l2(2m−6−1)+
3
l4(2m−6−1)(2m−6−2)(2m−6−3)
+ (mod 16).
3
Assume m ≥ 10 (the case 7 ≤ m ≤ 9 is proceeded at last). So 16 | 2m−6. Let 3a =
2(2m−6−1)(2m−6−2) and3b=(2m−6−1)(2m−6−2)(2m−6−3).Then 3a≡4(mod 16)
and 3b ≡ −6(mod 16). Therefore a ≡ −4(mod 16) and b ≡ −2(mod 16). So the proof
continues as follows
2m−3 ⌊(k+1)/2⌋
S 1≡6 (−1)k+1 k+2 (2l+1)n+1(l−l2−4l3−2l4)l
 2l+1 
k=0 l=1 (cid:18) (cid:19)
X X
2m−3 ⌊(k+1)/2⌋ 
S ≡8 (−1)k+1 k+2 (2l+1)n+1 l(l+1) (−2l2−2l+1)l.
2l+1 2
k=0 l=1 (cid:18) (cid:19) (cid:18) (cid:19)
X X
PERIODICITY OF RESIDUAL r-FUBINI SEQUENCES 7
Let P(l) and A(k,r,n) be the remainder of 1(2l+1)n+1(l(l+1))(−2l2−2l+1)l and
2
∞ k+2 P(l) divided by 8, respectively. By Pascal identity, we have k+2 =
l=−∞ 2l+r 2l+r
k+1 + k+1 and therefore
P2l+r (cid:0)2l+r(cid:1)−1 (cid:0) (cid:1)
(cid:0) (cid:1) (cid:0) ∞ (cid:1) ∞ ∞
k+2 k+1 k+1
P(l)= P(l)+ P(l),
2l+r 2l+r 2l+r−1
l=−∞(cid:18) (cid:19) l=−∞(cid:18) (cid:19) l=−∞(cid:18) (cid:19)
X X X
so
(11) A(k,r,n)=A(k−1,r,n)+A(k−1,r−1,n).
We can write
∞
k+2
A(k,r+32,n)≡ P(l). (mod 8)
2l+r+32
l=−∞(cid:18) (cid:19)
X
The sequence (P(l))∞ has period 16, so P(l+16)=P(l). Set l′ =l+16, then
l=−∞
∞
k+2
(12) A(k,r+32,n)≡ P(l′)≡A(k,r,n) (mod 8).
2l′+r
l′=−∞(cid:18) (cid:19)
X
Since (2l+1,16)=1, the Euler’s Theorem implies (2l+1)8 ≡1 (mod 16) and therefore
(2l+1)n+1+8 ≡(2l+1)n+1 (mod 16). A(6,r,n)vanishes for 1≤r ≤32and 9≤n≤24,
by enumeration, then by (11) and (12), we deduce that
(13) A(k,r,n)=0, for k ≥6.
Therefore
∞
k+2 l(l+1)
A(k,1,n)≡ (2l+1)n+1 (−2l2−2l+1)l (mod 8)
2l+1 2
l=−∞(cid:18) (cid:19) (cid:18) (cid:19)
X
⌊(k+1)/2⌋
k+2 l(l+1)
≡ (2l+1)n+1 (−2l2−2l+1)l
2l+1 2
l=1 (cid:18) (cid:19) (cid:18) (cid:19)
X
≡0 (mod 8),
for k ≥6. If 1≤k ≤5, 9≤n≤24 and 1≤r ≤32 we have 5 (−1)k+1A(k,r,n)≡0
k=1
(mod 8). The period length of A(k,r,n) with respect to r and n implies that
5 (−1)k+1A(k,1,n)≡0 (mod 8), for n≥9. Combining tPhis with (13) we have
k=1
P 2m−3
S ≡ (−1)k+1A(k,1,n)≡0 (mod 8), for n≥0.
k=1
X
So the result follows in the case of n≥2m−3. If m−1≤n<2m−3 we can write
n+2m−6 n+2m−6+2 n n+2
Fn+2m−6,2−Fn,2 = (k+2)! k+2 − (k+2)! k+2
k=0 (cid:26) (cid:27)2 k=0 (cid:26) (cid:27)2
X X
2m−3 n+2m−6+2 n+2
= (k+2)! −
k+2 k+2
k=0 (cid:18)(cid:26) (cid:27)2 (cid:26) (cid:27)2(cid:19)
X
2m−3 n+2m−6+2 2m−3 n+2
− (k+2)! + (k+2)!
k+2 k+2
k=n+X2m−6+1 (cid:26) (cid:27)2 k=Xn+1 (cid:26) (cid:27)2
2m−3k+2
≡ (−1)k+2−j k+2 jn+1(j2m−6 −1)(j−1) (mod 2m)
j
k=0 j=1 (cid:18) (cid:19)
X X
8 AMIRABBASASGARIANDMAJIDJAHANGIRI
When j is even, then jn+1 = (2c)m+h, for some h ≥ 0, so 2m | jn+1. Since m ≥ 10, for
odd j’s we have
2m−3k+2
(−1)k+2−j k+2 jn+1(j2m−6 −1)(j−1)
j
k=0 j=1 (cid:18) (cid:19)
X X
2m−3 ⌊(k+1)/2⌋
k+2
≡2m−4 (−1)k+1 (2l+1)n+1
2l+1
k=0 l=1 (cid:18) (cid:19)
X X
×(l−l2−4l3−2l4)l (mod 2m).
The last summation is exactly the S and the proof will be similar as above. Combine
with the previous case we have the following congruence relation
(14) Fn+2m−6,2 ≡Fn,2 (mod 2m), for m≥10.
In the case where 7≤m≤9, the remainder of the sum 2m−3(−1)k+1
k=0
× ⌊(k+1)/2⌋ k+2 (2l+1)n+1 4 li2i−1 (2m−6−1)! l modulo16iscomputedform−
l=1 2l+1 i=1 i!(2m−6−i)! P
1 ≤Pn ≤ m+(cid:0) 14.(cid:1)Divisibility(cid:16)oPf all these values by (cid:17)16 implies that the recent sum is
divisible by 16 therefore
(15) Fn+2m−6,2 ≡Fn,2 (mod 2m), for 7≤m≤9.
Summing up the congruence relations (14) and (15) gives
ω(A )=2m−6, for m≥7.
2,2m
(cid:3)
Theorem 4.5. For m=1 and m=2, the sequence A is periodic from the first term
r,2m
and the period length is ω(A )=1.
r,2m
Proof. The proof of this theorem is divided into three cases. For r =2 we have
n+1 n
n+3 n+2
F −F = (k+2)! − (k+2)!
n+1,2 n,2 k+2 k+2
k=0 (cid:26) (cid:27)2 k=0 (cid:26) (cid:27)2
X X
n+3 n+2 n+3 n+2
≡2 − +6 − (mod 4)
2 2 3 3
(cid:18)(cid:26) (cid:27)2 (cid:26) (cid:27)2(cid:19) (cid:18)(cid:26) (cid:27)2 (cid:26) (cid:27)2(cid:19)
=2 2n+1−2n +6 3n+1−2n+1−(3n−2n)
=2n(cid:0)+1+6(2×(cid:1)3n−(cid:0)2n)=4(2n−1+3n+1−3(cid:1)×2n−1)
=4(3n+1−2n)≡0(mod 4).
So we can deduce that ω(A )=1 and obviously ω(A )=1.
2,4 2,2
For r =3 we can write
n+1 n
n+4 n+3
F −F = (k+3)! − (k+3)!
n+1,3 n,3 k+3 k+3
k=0 (cid:26) (cid:27)3 k=0 (cid:26) (cid:27)3
X X
n+4 n+3
≡6 − (mod 4)
3 3
(cid:18)(cid:26) (cid:27)3 (cid:26) (cid:27)3(cid:19)
=6 3n+1−3n =6×2×3n =4×3n+1 ≡0(mod 4).
Therefore we have ω(A3,4)=(cid:0) 1 and ω(A(cid:1)3,2)=1.
Finally if r ≥4, let r =4+h, for some h≥0, then
n+1 n
n+1+r n+r
F −F = (k+r)! − (k+r)! .
n+1,r n,r k+r k+r
k=0 (cid:26) (cid:27)r k=0 (cid:26) (cid:27)r
X X
Since 4 | (k +r)!, for all k ≥ 0, we can write F −F ≡ 0 (mod 4). Therefore
n+1,r n,r
ω(A )=1 and ω(A )=1. (cid:3)
r,4 r,2
PERIODICITY OF RESIDUAL r-FUBINI SEQUENCES 9
Theorem 4.6. If 3≤m≤6, after the (m−1)th term, the sequence Ar,2m has a period
with length ω(A )=2.
r,2m
Proof. The proof of this theorem is similar to the proof of Theorem 4.1. Proving for
m = 6 deduces the result for m = 3,4,5. Let m = 6, so n ≥ 5; because n ≥ m−1. For
3≤r ≤7 we have
n+2 n
n+2+r n+r
F −F = (k+r)! − (k+r)!
n+2,r n,r k+r k+r
k=0 (cid:26) (cid:27)r k=0 (cid:26) (cid:27)r
X X
7−rk+r
k+r (j−1)!
≡ (−1)k+r−j jn+1(j2−1) (mod 64)
j (j−r)!
k=0j=r (cid:18) (cid:19) (cid:18) (cid:19)
XX
7−rk+r
k+r (j−2)!
≡ (−1)k+r−j jn+1(j2−1)(j−1) (mod 64)
j (j−r)!
k=0j=r (cid:18) (cid:19) (cid:18) (cid:19)
XX
When j is even, then jn+1 = (2c)6+h, for some h ≥ 0, and so 64 | jn+1. For odd
j’s we have (j,64) = 1 and the Euler’s Theorem gives j32 ≡ 1 (mod 64). Therefore
jn+1+32 ≡jn+1 (mod 64) and we can write
7−rk+r
k+r (j−2)!
(−1)k+r−j jn+1(j2−1)(j−1)
j (j−r)!
k=0j=r (cid:18) (cid:19)
XX
7−r⌊(k+r−1)/2⌋
k+r
≡ (−1)k+r−(2l+1) (2l+1)n+1((2l+1)2−1)
2l+1
kX=0 l=X⌊r/2⌋ (cid:18) (cid:19)
(2l−1)!
×2l (mod 64)
(2l+1−r)!
(cid:18) (cid:19)
7−r ⌊(k+r−1)/2⌋
k+r l(l+1)
≡16 (−1)k+r−1 (2l+1)n+1
2l+1 2
Xk=0 l=X⌊r/2⌋ (cid:18) (cid:19) (cid:18) (cid:19)
(2l−1)!
×l (mod 64)
(2l+1−r)!
(cid:18) (cid:19)
By computation we see that the recent summation is divisible by 4, for 2 ≤ n ≤ 33. So
the proof for 3≤r ≤7 is completed.
If r ≥8, since 64|8!, then 64|(k+r)!, and
n+2 n
n+2+r n+r
F −F = (k+r)! − (k+r)!
n+2,r n,r k+r k+r
k=0 (cid:26) (cid:27)r k=0 (cid:26) (cid:27)r
X X
≡0 (mod 64),
so ω(Ar,26)=2, for r ≥8, and the proof is completed. (cid:3)
Theorem 4.7. If m≥7, after the (m−1)th term, the sequence A has a period with
r,2m
length ω(A )=2m−6.
r,2m
10 AMIRABBASASGARIANDMAJIDJAHANGIRI
Proof. The proof of this theorem is similar to the proof of Theorem 4.4. In the case of
n≥2m−r−1 and r ≥8 we have
n+2m−6 n+2m−6+r n n+r
Fn+2m−6,r−Fn,r = (k+r)! k+r − (k+r)! k+r
k=0 (cid:26) (cid:27)r k=0 (cid:26) (cid:27)r
X X
2m−r−1 n+2m−6+r n+r
≡ (k+r)! − (mod 2m)
k+r k+r
k=0 (cid:18)(cid:26) (cid:27)r (cid:26) (cid:27)r(cid:19)
X
2m−r−1k+r
≡ (−1)k+r−j k+r jn+1(j2m−6 −1) (j−1)! (mod 2m)
j (j−r)!
k=0 j=r (cid:18) (cid:19) (cid:18) (cid:19)
X X
In the case of 2m >r >2m−m, since m≥7 this implies that r >2m−m≥2m−1 and
so
2m |(2m−1)!|(k+r)!, for each k≥0.
Therefore both summations in the above first equation are zero modulo 2m and in this
caseω(A )=2m−6. When r ≤2m−m,if j is eventhenjn+1 =(2c)2m−r+h, for some
r,2m
h≥0.So2m |jn+1.Foroddj’s,(j,2m−5)=1andbyEuler’sTheorem2m−5 |j2m−6−1.
Since r ≥ 8 we can write (j−1)! = (j−8)! 7 (j − i). Therefore 32 | (j−1)! and
(j−r)! (j−r)! i=1 (j−r)!
2m |(j2m−6 −1) (j−1)! . (cid:16) (cid:17)Q
(j−r)!
In the case of(cid:16)m−1(cid:17)≤n<2m−r−1 and r ≥8 we have
n+2m−6 n+2m−6+r n n+r
Fn+2m−6,r−Fn,r = (k+r)! k+r − (k+r)! k+r
k=0 (cid:26) (cid:27)r k=0 (cid:26) (cid:27)r
X X
2m−r−1 n+2m−6+r n+r
≡ (k+r)! −
k+r k+r
k=0 (cid:18)(cid:26) (cid:27)r (cid:26) (cid:27)r(cid:19)
X
2m−r−1 n+2m−6+r
− (k+r)!
k+r
k=n+X2m−6+1 (cid:26) (cid:27)
2m−r−1
n+r
+ (k+r)! (mod 2m)
k+r
k=n+1 (cid:26) (cid:27)
X
2m−r−1 n+2m−6+r n+r
= (k+r)! − +0,
k+r k+r
k=0 (cid:18)(cid:26) (cid:27)r (cid:26) (cid:27)r(cid:19)
X
and the proof proceeds as the previous case. In the case of 3 ≤ r ≤ 7 one can deduce
similar to the proof of Theorem 4.4 that
2m−r−1k+r
Fn+2m−6,r−Fn,r ≡ (−1)k+r−j k+j r jn+1(j2m−6 −1)
k=0 j=r (cid:18) (cid:19)
X X
(j−1)!
× (mod 2m).
(j−r)!
(cid:18) (cid:19)