Congruences of Finite Summations of the Coefficients in certain ...
Congruences of Finite Summations of the Coefficients in certain Generating Functions Po-Yi Huang∗
Shu-Chung Liu†
Department of Mathematics National Cheng Kung University Tainan City, Taiwan [email protected]
Department of Applied Mathematics National Hsinchu University of Education Hsinchu City, Taiwan [email protected]
Yeong-Nan Yeh‡ Institute of Mathematics Academia Sinica Taipei, Taiwan [email protected] Submitted: Sep 4, 2013; Accepted: May 25, 2014; Published: Jun 9, 2014 Mathematics Subject Classifications: 11B50, 05A15, 05A10
Abstract In this paper we develop a general to enumerate the congruences of finite P Pmethod p−1−h ak ak+h ak summations p−1 (mod p) and (mod p) for the infinite sequence k=0 mk k=0 Bk N
{an }n>0 with generating functions (1+xf (x)) 2 , where f (x) is an integer polynomial and N is an odd integer with |N | < p. We also enumerate the √ congruences of some similar finite summations involving generating functions √ 1−αx− 1−2αx+(α2 −4β)x2 . 2βx2
1
1−αx−
1−2(α+β)x+Bx2 βx
and
Introduction
Let p be an odd prime number and m be an integer with m 6≡ 0 (mod p). The initial topic of this article is to enumerate p−1 X ak mk k=0
(mod p)
(1)
∗
Partially supported by NSC 100-2115-M-006-008. Corresponding author. Partially supported by NSC 101-2115-M-134-134-006. ‡ Partially supported by NSC 101-2115-M-001-013-MY3. †
the electronic journal of combinatorics 21(2) (2014), #P2.45
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the congruence of the pth partial sum of the infinite sequence {an }∞ n=0 . The study of the congruence of a single term, an (mod p), has a long history extending form the most famous and edge-old problem of Pascal’s fractal, which is originally formed by the parities of binomial coefficients nk [5, 6, 13, 14, 15, 16, 19, 25], to the most recent works about Ap´ery numbers [2, 4, 9, 11, 20], central Delannoy numbers [9], Catalan numbers [1, 7, 10, 17, 21], Motzkin numbers [7, 9, 18] and etc. [12]. This article focuses on the sequence {an }∞ n=0 with generating functions (GF’s) (1 + N xf (x)) 2 , where f (x) is an integer polynomial and N is an odd integer with |N | < p. ak Neither an nor m k in (1) is necessarily an integer if we deal with the field of rational ak numbers; therefore, we shall consider both an and m k the congruences in the modular arithmetic field with modulus p. For instance, usually 1/2 is a fraction; however, in the k modular arithmetic field Zp = {0, 1, . . . , p − 1}, we have 1 p+1 2 2 ≡ (mod p) for k = 0, 1, . . . , p − 1, (2) k k 1 which is always an integer. For example let p = 5, and then 22 = 12 · −1 /2! = 2 (mod 5) = 32 . Z.-W. Sun [24] applied (2) and some other tools to verify the equivalence p−1 X m−4 m(m − 4) ck ≡ 1− k m 2 p k=1 where ck is the kth Catalan number and 0 a = 1 p −1
· p
(mod p),
−1 8
≡3
(3)
denotes the Legendre symbol defined by
if p | a; if a is a quadratic residue modulo p; if a is a quadratic nonresidue modulo p.
One has the following well-known and useful congruence: p−1 a a ≡a 2 (mod p), with ∈ {−1, 0, 1}. (4) p p P ak In [24], Sun also derived p−1 k=1 mk (mod p) for ak being the kth central Delannoy number or Schr¨oder number respectively. The results by Sun initiate the motivation of our study. In this article, we wish to provide a systematic method to deal with the problems of this kind. The article is organized as follows. In Section 2, we exam the problem in (1) by generating functions. √ As applications, similar problems of (1) involving generating functions modified from 1P+ Ax + Bx2 are given in Section 3. In Section 4, we focus on another ak ak+h type of problem p−1−h (mod p). And then some applications are demonstrated k=0 (±B)k in Sections 5 and 6. the electronic journal of combinatorics 21(2) (2014), #P2.45
2
2
N
GF’s of the form (1 + xf (x)) 2
Let {ak }k>0 be the sequence associating with the generating function N
(1 + xf (x)) 2 , where Pp−1 f (x)k is an integer polynomial Pp−1 ak and N is an odd integer with |N | < p. To formulate k=0 ak m (mod p) and k=0 mk (mod p), which are same problem in different forms, we need the following facts: p+N N (i) k2 ≡ k2 (mod p) for k = 0, 1, . . . , p − 1; (ii) (iii)
N 2
k N 2
p
≡
p+N 2
k
p+N +2 ,...,p 2
≡ 0 (mod p) for k =
≡ −2−1 (mod p); moreover
N 2
p+k
≡ −2−1
− 1; p+N 2
k
(mod p) for k = 0, 1, . . . , p − 1
The reason for (i) is trivial and the congruence 0 in (ii) is due to that p+N is an integer 2 p+N and k > 2 . The reason for the first equivalence in (iii) is because of the bijection between {N, N − 2, · · · , N − 2p + 2} and {0, 1, . . . p − 1} under modulo p; however, the element N − 2i such that N − 2i ≡ 0 (mod p) is −p, and additionally 2−p ≡ 2−1 (mod p). The second equivalence in (iii) directly follows the first one. The advantage of (i) is that p+N p+N p+N 2 2 2 is a normal binomial coefficient; therefore, we can apply = . p+N k k −k 2
In the following, the equivalence f (x) ≡ g(x) (mod h(x)) means that f (x) and g(x) share same residue with respect to divisor h(x). Moreover, f (x) ≡ g(x) (mod xq , p) indicates that the coefficients of f (x) and g(x) for the term xk , with k = 0, 1, . . . , q − 1, are congruent modulo p. Theorem 1. Given an odd prime p and an integer m with m 6≡ 0 (mod p). Let {ak }k>0 N be a sequence of integers whose generating function is (1 + xf (x)) 2 , where f (x) is an integer polynomial and N is an odd integer with |N | < p. We have p+N
N
(1 + xf (x)) 2 ≡ (1 + xf (x)) 2 (mod xp , p), and p−1 X N +1 1 + mf (m) k ak m ≡ (1 + mf (m)) 2 − E(m) p k=0
(5) (mod p),
(6)
where E(x) is the polynomial consisting of each terms xt with t > p in the expansion of p+N (1 + xf (x)) 2 . Proof. Actually, (5) and (6) are equivalent. The following can prove both at the same time. p−1 X
k
ak m
=
h
(1 + xf (x))
N 2
i (mod x ) p
k=0 the electronic journal of combinatorics 21(2) (2014), #P2.45
(7) x=m
3
=
≡ = ≡
" p−1 # X N 2 (xf (x))k (mod xp ) k k=0 x=m p+N 2 p+N X 2 (xf (x))k (mod xp , p) k k=0 x=m i h p+N (1 + xf (x)) 2 − E(x) x=m N +1 1 + mf (m) − E(m) (1 + mf (m)) 2 p
(8) (9) (mod p).
Notice that the equivalence in (8) responses to modulus p by applying (i) and (ii), and the is an exponent of positive integer. The equation in (9) yields by modulus xp and that p+N 2 last equivalence is because of (4). Referring to (7) and (8) without substituting x = m, we then verify (5). It takes time to enumerate E(x) unless f (x) is simple enough. In the rest of this section we assume f (x) = A + Bx. Let [xn ]g(x) denote the coefficient of xn in the power series g(x). We derive the following rules: (r1) If B ≡ 0 (mod p) then E(x) ≡ 0 (mod p). Moreover, we have t X
k
ak m ≡ (1 + mf (m))
N +1 2
k=0
for t =
p+N p+N +2 , 2 ,...,p 2
1 + mf (m) p
(mod p),
− 1 due to a p+N +2 ≡ · · · ≡ ap−1 ≡ 0 (mod p) by (ii). 2
(r2) If N 6 −1 then E(x) = 0 again. (r3) If N > 1 and B 6≡ 0 (mod p), then E(x) 6= 0. Let (1 + Ax + Bx2 )
p+N 2
= q0 + q1 x + · · · + qp+N xp+N .
For the precise value of qm , we can simply apply the following formula: [xe ](1 + αx + βx2 )M (Pmin{e,M } =
M α2k−e β e−k k=d 2e e M −k, 2k−e, e−k 2k−e e−k Pe −M +k−1 k β k=d 2e e (−1) −M −1, 2k−e, e−k α
if M > 0 and e 6 2M ; if M < 0.
Therefore, min{m, p+N } 2
qm =
p+N 2
X
k=d m e 2
p+N 2
− k, 2k − m, m − k
the electronic journal of combinatorics 21(2) (2014), #P2.45
A2k−m B m−k .
4
Particularly, if N = 1 then E(x) ≡ A2 Bp xp + B Bp xp+1 (mod p); if p > 3 and N = 3 then 3 1 3 −1 B 3 B 3 2 B B p E(x) ≡ − 4A B x + B + 3A xp+1 A 22 p 2 p 2 p 2 p 3 B B p+2 2 + AB x +B xp+3 (mod p). 2 p p In case that E(x) = 0, we reach a particular case as follows. Corollary 2. If f (x) is a constant function or f (x) = A + Bx with N 6 −1, then we have E(x) = 0 and 2 p−1 X ak m + Am + B ≡ mk m2 k=0
N +1 2
m2 + Am + B p
(mod p).
Here we recall some sequences as applications of the last corollary:
sequence name
first few terms
Pp−1
ak k=0 mk
GF
√ 1 1−4x
central binomial coeff.
1, 2, 6, 20, 70, . . .
central trinomial coeff.
1, 1, 3, 7, 19, 51, . . .
√
central Delannoy numbers
1, 3, 13, 63, 321, . . .
√
1 1−6x+x2 −3
(1 − 4x) 2 √ 1, −2, −2, −4, −10, . . . 1 − 4x
A002457, in [22]
1, 6, 30, 140, 630, . . .
A002420, in [22]
1 1−2x−3x2
(mod p)
m2 −4m p
m2 −2m−3 p
m2 −6m+1 p
(see [24]) 2
m2 m2 −4m
m −4m p
m2 −4m m2
m2 −4m p
Table 1: Some direct applications of Corollary 2 P Pp+1 ak ak Let us take advance to enumerate pk=0 m k and k=0 mk , which will be used in the ap+1 ap nest section. We need to enumerate two additional terms, m p and mp+1 . Using (4), (i), (ii) and (iii), we calculate ap and ap+1 as follows:
p
2
[x ](1 + Ax + Bx ))
N 2
p X
N = [x ] 2 (x(A + Bx))k k k=0 N p X 2 [xp−k ](A + Bx)k = k p+1 k=
≡
p
2
N p−1 2 2 ](A p+1 [x 2
+ Bx)
the electronic journal of combinatorics 21(2) (2014), #P2.45
p+1 2
N + 2 [x0 ](A + Bx)p (mod p) p 5
p+N
≡
[x
p+1
2
](1 + Ax + Bx ))
p−1 p+1 1 AB 2 − Ap (mod p) 2 2 ! p+N B 2 −1 (mod p), p+1 p 2
2 p+1 2
≡ A 2
N 2
=
p+1 N X 2
k= p+1 2
≡
k
[xp−k+1 ](A + Bx)k
N N p+1 p+1 2 2 ](A + Bx) 2 + 2 [x](A + Bx)p p+1 [x p 2 N 2 + [x0 ](A + Bx)p+1 (mod p)
p+1 p+N p+1 1 p+N 2 2 2 ≡ Ap+1 (mod p) +0− p+1 B 2 1 2 p+N N B 2 B − 2 A2 (mod p). ≡ p+1 p 2 2 P Pp+1 ak ak If B ≡ 0 (mod p) or N 6 −1, then pk=0 m k and k=0 mk are ready because we have ap+1 ap A N A2 ≡ − 2m (mod p), mp+1 ≡ − 4m2 (mod p), E(x) = 0 and referring (6). When B 6≡ 0 mp (mod p) and N > 1, then the larger N is, the more complicate E(x) is. However the two summations for N = 1 are also ready as follows. √ Corollary 3. Let {an }n>0 be the sequence whose generating function is 1 + Ax + Bx2 . We have p X ak m2 + Am + B m2 + Am + B B B A ≡ − − (mod p), mk m2 p 2m m2 p k=0 p+1 X ak m2 + Am + B m2 + Am + B A A2 ≡ − − (mod p). k 2 2 m m p 2m 4m k=0
3
GF’s modified from
√
1 + Ax + Bx2
Many well-known sequences are associated with generating functions modified from (1 + 1 Ax + Bx2 ) 2 . In this section, we consider two types: p p 1 − αx − 1 − 2(α + β)x + Bx2 1 − αx − 1 − 2αx + (α2 − 4β)x2 and . βx 2βx2 √ 1−αx− 1−2(α+β)x+Bx2 Let be the generating function of the sequence {bn }n>0 , and let βx A = −2(α + β) for convenience. For examples, (α, β, B) = (0, 2, 0) yields the Catalan the electronic journal of combinatorics 21(2) (2014), #P2.45
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number, (1, 2, 1) and (−1, 4, 1) provide the large and the little Schr¨oder numbers respectively, and (1, 2, 5) is associated with the number of restricted hexagonal polyominoes [22, A002212]. Theorem 4. We have 2 p−1 X m + Am + B B B bk m2 + Am + B 1− + −1 (mod p). ≡ 1+ mk βm p βm p k=0 √ Proof. We still let {an }n>0 associate with 1 + Ax + Bx2 . Notice that bk = − ak+1 for β A k > 1. Also it is clear that a0 = 1, a1 = 2 and b0 = 1. p−1 X bk mk k=0
p
m X ak = 1− β k=2 mk p
m A m X ak + − = 1+ β 2β β k=0 mk m A m m2 + Am + B m2 + Am + B A ≡ 1+ + − − 2 β 2β β m p 2m B B − 2 (mod p), m p 2 m2 + Am + B m + Am + B B B = 1+ 1− + −1 . βm p βm p
In particular, the last term in the last theorem is zero if B ≡ 0 (mod p) or Bp = 1. This particular case happens for the Catalan numbers (see (3) and also [24, Lemma 2.1]), the large Schr¨oder√numbers (see [24, Theorem 1.2]) and the little Schr¨oder numbers. 1−αx−
1−2αx+(α2 −4β)x2
Now let be the generating function of the sequence {dn }n>0 . 2βx2 For convenient, let A = −2α and B = α2 − 4β. Lots of famous sequences have GF’s of this form. For examples, (α, β) = (1, 1) yields the Motzkin number, and (3, 9) as well as (5, 5) are associated with the numbers of Motzkin paths with multiple colors for level steps. Also (α, β) = (2, 1) creates a shifted Catalan number, (3, 2) a shifted little Schr¨oder number [22, A001003], and (4, 1) the number of walks on cubic lattice starting and finishing on the horizontal plane but never going below it [22, A005572]. For more examples, please refer to [3]. The next result can be verified by a process similar to the proof of the last theorem. Theorem 5. We have 2 p−1 X m + Am + B m2 + Am + B dk ≡ 1− + 2 (mod p). k m 2β p k=0 the electronic journal of combinatorics 21(2) (2014), #P2.45
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4
Enumerating
Pp−1−h ak ak+h (±B)k
k=0
(mod p) N
Let {an }n>0 still associate with a generating functions of form (1 + Ax + Bx2 ) 2 . In this and the next sections we assume B 6≡ 0 (mod p). Given nonnegative integer h with Pp−1−h (mod p) and (S2) h 6 p − 1, we are interested in two summations: (S1) k=0 akBak+h k Pp−1−h ak ak+h (mod p). For convenience, we use ±B to denote B and −B simultaneously. k=0 (−B)k N
Let G(x) = a0 + a1 x + · · · ap−1 xp−1 ≡ (1 + Ax + Bx2 ) 2 (mod xp ) and Q(x) = (1 + Ax + p+N Bx2 ) 2 = q0 + q1 x + · · · +P qp+N xp+N . (See (r3) for the formula of qm .) Now we apply p [x−h ]G(x)G( x1 ) to evaluate p−1 k=0 ak ak+h and also use the fact G(x) ≡ Q(x) (mod p, x ) (see Theorem 1). p−h−1
X ak ak+h 1 x )G( ) = [x−h ]G( k (±B) ±B x k=0 X qk qk+h x 1 ≡ [x−h ]Q( )Q( ) − (mod p) ±B x (±B)k k>p−h = [x
−h
x2 Ax + ] 1+ ±B B
p+N 2
A B 1+ + 2 x x
p+N 2 −
X qk qk+h (±B)k k>p−h
p+N p+N X qk qk+h Ax x2 2 Ax x2 2 [x ] 1+ + + = B − 1+ ±B B B B (±B)k k>p−h ( A X qk qk+h p+N (1 + B x + B1 x2 )p+N for (S1); p+N −h 2 = − + B × [x ] (10) p+N 2 (±B)k (1 + 2B−A for (S2). x2 + B12 x4 ) 2 B2 k>p−h P k qk+h The first common term k>p−h q(±B) k can be avoided if and only if N 6 −1 (see (r2) and (r3)). Now the result of (S2) is ready. P ak ak+h (mod p) is equivalent to Theorem 6. The summation p−h−1 k=0 (−B)k p+N 2
B
N +1 2
p+N −h
p+N X qk qk+h B 2B − A2 2 1 4 2 p+N −h [x ] 1+ x + x − p B2 B2 (−B)k k>p−h
(mod p).
Particularly, if h is odd then the first term is 0, and if N 6 −1 then the last term is 0. 1 Example 7. Let N = −1 and B = 1. For instances, √1−6x+x 2 is the GF of the central De1 1 √ lannoy numbers, 1−10x+x2 is associated with the colored Delannoy paths, and √1−14x+x 2 P k is associated with the central coefficients of (1 + 7x + 12x2 ). So p−h−1 (−1) a a ≡ 0 k=0 Pp−h−1 k kk+h p−1 (mod p) if h is odd; otherwise let p−h−1 = 2e and 2 = M , and then k=0 (−1) ak ak+h (mod p) is equivalent to e X M e 2 2 M [x ](1 + (2 − A )x + x ) = (2 − A2 )2k−e . M − k, 2k − e, e − k e k=d 2 e
the electronic journal of combinatorics 21(2) (2014), #P2.45
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As for (S1), we continuously simplify the last term in (10) as [x
p+N −h
N N A N −h A 1 2 A 1 2 + [x , ] 1+ x+ x ] 1+ x+ x B B B B B
A p A x+ B1 x2 )p ≡ 1+ B x + B1 x2p (mod p) in which x2p is ignored for its exponent because (1+ B being too large. P Pp+N p+N N Theorem 8. Let (1 + Ax + Bx2 ) 2 = n>0 an xn and (1 + Ax + Bx2 ) 2 = n=0 q n xn . Pp−h−1 ak ak+h Then k=0 (mod p) is equivalent to Bk
B
N N N −1 B A 1 2 B A 1 2 p+N −h N −h [x ] 1+ x+ x + AB 2 [x ] 1+ x+ x p B B p B B X qk qk+h − . Bk k>p−h
N +1 2
Particularly, only the first term remains when N 6 −1, the first term is zero when N > 1 and p > N + h, and the second term is zero when N < h. P ak ak+h (mod p) is equivalent to Example 9. Again, let N = −1 and then p−h−1 k=0 Bk e X B k A 1 2 −1 k e (−1) A2k−e B −k [x ](1 + x + x ) = 2k − e, e − k p B B e k=d 2 e
where e = p − h − 1. A Since x/(1 + B x + B1 x2 ) is the generating function of the Lucas sequence un that satisfies the second order recurrence relation
1 A un = − un−1 − un−2 , B B with the initial u0 = 0 and u1 = 1, we also have p−h−1
X ak ak+h B ≡ up−h k B p k=0 αp−h − β p−h = α−β
(mod p) B , p
A where α and β are roots of x2 + B x + B1 .
the electronic journal of combinatorics 21(2) (2014), #P2.45
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5
Summations
Pp−2−h bk bk+h (±B)k √
k=0
1−αx−
and
Pp−3−h dk dk+h Bk
k=0
1−2αx+(α2 −4β)x2
We observe some applications here. Let be the generating function 2βx2 2 of {dn }n>0 . For convenience let A = −2α and B = α − 4β. ak (mod p) for k = 2, 3, . . . , p − 1, we have Since dk−2 = − 2β p−3−h
p−1−h X dk dk+h B 2 X ak ak+h ≡ (mod p) (±B)k 4β 2 k=2 (±B)k k=0
B2 = 4β 2
p−1−h
X ak ak+h a1 ah+1 − a0 ah − k (±B) ±B k=0
! .
Pp−1−h ak ak+h given in the last section, let us recall the Before we apply the formula of k=0 (±B)k P P qk qk+h k qk+h term k>p−h (±B)k in both Theorems 6 and 8. With N = 1 we have k>p−h q(±B) k = qp−h qp qp−h+1 qp+1 + (±B)p−h+1 . Notice that, the coefficients of (1 + Ax + Bx2 )M = q0 + q1 x + · · · + (±B)p−h q2M x2M have a sort of symmetric property, i.e., B M −k qk = q2M −k for k = 1, 2, . . . , 2M , and we get which can be easily proved by induction on M . Let M = p+1 2 p2M −k−h p2M −k (±B)2M −k−h
pk pk+h (±B)k ak ak+h ≡ (±1)h (±B)k = (±1)h
pp+1−h pp+1 (±B)p+1−h
p
(mod p).
(11)
p
p−h p h a1 ah+1 (mod p). More≡ (±1)h a0 ah (mod p) and (±B) p−h ≡ (±1) ±B N +1 B A over, for N = 1 and h 6 p−3, the term B 2 [xp+N −h ](1+ B x+ B1 x2 )N in Theorem 8 p must be 0. Now we derive conclusion and show some examples as follows. Pp−3−h dk dk+h (mod p) is equivalent to Theorem 10. The congruence k=0 Bk B2 A 1 2 B 2a1 a1+h 1−h [x ] 1 + x + x − 2a0 ah − . A 4β 2 p B B B
Particularly,
Corollary 11. Let 2 6 h 6 p − 3. We have p−3−h
X dk dk+h a1 a1+h B2 ≡ − a a + 0 h Bk 2β 2 B k=0
(mod p),
which is a fixed number modulo any prime p > 2. 2
3
+ A16 and a4 = Example 12. Notice that a0 = 1, a1 = A2 , a2 = B2 − A8 , a3 = − AB 4 2 2 4 − B8 + 3A16B − 5A . For h = 0, 1, 2, 3 and p > h + 3 we have following examples: 128 p−3 X dk 2 k=0
Bk
B ≡ 8β 2
B 2 2 2A − 4B − A p
the electronic journal of combinatorics 21(2) (2014), #P2.45
(mod p),
10
p−4 X dk dk+1 k=0 p−5
Bk
X dk dk+2 k=0 p−6
Bk
X dk dk+3 k=0
Bk
AB ≡ 32β 2 ≡ − ≡
B 2 8B − 12B + A p
B 16B 2 − 8A2 B + A4 2 64β
5AB 2 2 4 16B − 8A B + A 512β 2
(mod p),
(mod p),
(mod p)
Particularly, given (A, B, β) = (−2, −3, 1) (so we need p 6= 3) that yields the Motzkin numbers, the four summations above are respectively equivalent to 9 −3 15 −3 − 3, − + , 12 and 15 (mod p). −3 p 2 p 2 Pp−3−h dk dk+h Theorem 13. The congruence k=0 (−B)k (mod p) is equivalent to 0 if h is odd, and equivalent to ! p+1 2 2 B2 B 2B − A 1 a a 1 1+h B [xp+1−h ] 1 + , x2 + 2 x4 − 2a0 ah − 2 4β 2 p B2 B B if h is even. p+1 2 2 1 4 2 One can refer to (r3) for the precise value of [xp+1−h ] 1 + 2B−A . x + x B2 B2 √ 1−αx− 1−2(α+β)x+Bx2 Let be the generating function of the sequence {bn }n>0 , and βx let A = −2(α + β) for convenience. Since b0 = 1 and bk−1 ≡ − aβk (mod p) for k = 2, 3, . . . , p − 1, we have p−2−h
p−1−h X bk bk+h ±B X ak ak+h ≡ b0 b h + 2 (±B)k β k=2 (±B)k k=0
B = bh ± 2 β
p−1−h
(mod p)
X ak ak+h a1 a1+h − a0 ah − k (±B) ±B k=0
! .
This is quick similar to (11), so we can directly get Pp−2−h bk bk+h Theorem 14. The congruence k=0 (mod p) is equivalent to Bk B A 1 2 2a1 a1+h B 1−h [x ](1 + x + x ) − 2a0 ah − . bh + 2 A β p B B B P bk bk+h The congruence p−2−h (mod p) is equivalent to bh if h is odd, and equivalent k=0 (−B)k to ! p+1 2 2 B B 1 2B − A a a 1 1+h bh − 2 B x2 + 2 x4 , [xp+1−h ] 1 + − 2a0 ah − 2 β p B2 B B if h is even. the electronic journal of combinatorics 21(2) (2014), #P2.45
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6
More applications
Let an = 2n be the central binomial coefficient whose generating function is G(x) = n √ 1 . We have 1−4x p−1−h
X
16k ak ak+h = 4h [x−h ]G(x/4)G(1/4x)
k=0
p−1 −1 [x 2 −h ](1 − x)p−1 (mod p) ≡ 4 p p−1 −1 ≡ 4h [x 2 −h ](1 − x)−1 (mod p) p −1 = 4h . p h
Pp−1−h −2k Almost same procedure can be applied on the problem k=0 A ak ak+h for some other N an whose generating function is G(x) = (1 − Ax) 2 for odd integer N with |N | < p. In the following final example, we show an application involving two different sequences by the same technique. c, . . . , p−3 , p−1 and 0 6 h 6 b 3p+1 c − 2, we have Lemma 15. Given integers r = b p−1 4 2 2 4 ( 0 (mod p) 4k 64 ck+h ≡ 2k 22h+1 −1 p k=0
r X
−k
if 0 6 h < p−3 or p−1 < h 6 b 3p+1 c − 2; 4 2 4 p−3 p−1 (mod p) if 4 6 h 6 2 . p−1−2h p+1 2
1 that Proof. The sequence of the central binomial coefficients 2k (k > 0) has GF √1−4x k 2k p−1 p+1 k −1/2 comes from the identity k = 4 k . Given t = 2 , 2 , . . . , p − 1, we have t X 2k xk k=0
k
4k
t X −1/2 k = x k k=0 p−1 X −1/2 k ≡ x k k=0 1
≡ (1 − x)− 2 ≡ (1 − x)
p−1 2
(mod p)
(mod xp ) (mod xp , p) or (mod xp−1 , p),
where the equivalence on the second line dues to (ii) in Section 1. By the same reason, we can write (mod xp−1 , p) additionally on the last line. And then we get t
b2c X 4k x2k k=0
2k 16k
≡
p−1 p−1 1 (1 − x) 2 + (1 + x) 2 2
the electronic journal of combinatorics 21(2) (2014), #P2.45
(mod xp−1 , p).
12
For convenience, let t = p−1 in the rest of the proof. 2 On the other hand, the sequence of the Catalan numbers ck has GF C(x) = Again, ck ≡ 0 (mod p) for k = p+1 , . . . p − 2 by (ii). So we have 2
√ 1− 1−4x . 2x
p−1
p−2 X x−2k x−2k ck k ≡ ck k (mod p) 4 4 k=0 k=0 1 (mod xp−1 ) ≡ C 2 4x
2 X
= 2x2 − 2x−(p−1) (x2 − 1)
p+1 2
.
Given 0 6 h 6 b 3p+1 c − 2 (for b p−1 c + h 6 p − 2 by considering the term ck+h ), finally we 4 4 derive that b p−1 c 4k 4 X p−1 p−1 p+1 ck+h −2h −(p−1) 2 2k 2 2 2 (1 − x) + (1 + x) (mod p) · ≡ [x ] − x (x − 1) 16k 4k+h k=0 p−1 p+1 p−1 −1 p−1−2h ≡ [x ] (1 − x2 ) 2 (1 − x) 2 + (1 + x) 2 (mod p) p p+1 p+1 −1 p−1−2h (1 + x) 2 (1 − x)p + (1 + x)p (1 − x) 2 = [x ] p p+1 p+1 −1 ≡ [xp−1−2h ] (1 + x) 2 + (1 − x) 2 (mod p) p ( 0 if 0 6 h < p−3 or p−1 < h 6 b 3p+1 c − 2; 4 2 4 p+1 = 2 2 −1 if p−3 6 h 6 p−1 . p 4 2 p−1−2h c by any of b p−1 c, . . . , p−1 . We complete the proof by replacing the upper limit b p−1 4 4 2
References [1] R. Alter and K. Kubota. Prime and prime power divisibility of Catalan numbers. J. Combin. Theory Ser. A, 15:243–256, 1973. [2] F. Beukers. Another congruence for the Ap´ery numbers. J. Number Theory, 25:201– 210, 1987. [3] D. Callan. On generating functions involving the square root of a quadratic polynomial. J. Integer Seq., 10:Article 07.5.2, 2007. [4] S. Chowla, J. Cowles and M. Cowles. Congruence properties of Ap´ery numbers. J. Number Theory, 12:188–190, 1980. [5] K.S. Davis and W.A. Webb. Lucas’ theorem for prime powers. Europ. J. Combin., 11:229–233, 1990. [6] K.S. Davis and W.A. Webb. Pascal’s triangle modulo 4. Fibonacci Quart., 29:79–83, 1991. the electronic journal of combinatorics 21(2) (2014), #P2.45
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[7] E. Deutsch and B. Sagan. Congruences for Catalan and Motzkin numbers and related sequences. J. Number Theory, 117:191–215, 2006. [8] R. Donaghey and L.W. Shapiro. Motzkin Numbers. J. Combin. Theory Ser. A, 23:291–301, 1977. [9] S.-P. Eu, S.-C. Liu, and Y.-N. Yeh. On the congruences of some Combinatorial numbers. Studies in Applied Mathematics, 116:135–144, 2006. [10] S.-P. Eu, S.-C. Liu, and Y.-N. Yeh. Catalan and Motzkin numbers modulo 4 and 8. European J. Combin. 29:1449–1466, 2008. [11] I.M. Gessel. Some congruences for Ap´ery numbers. J. Number Theory, 14:362–368, 1982. [12] I.M. Gessel. Some congruences for generalized Euler numbers. Can. J. Math., 35:687– 709, 1983. [13] A. Granville. Arithmetic Properties of Binomial Coefficients I: Binomial Coefficients modulo prime powers. http://www.cecm.sfu.ca/organics/papers/granville/Binomial/toppage.html [14] A. Granville. Zaphod Beeblebrox’s brain and the fifty-ninth row of Pascal’s Triangle. Amer. Math. Monthly 99:318–381, 1992. [15] J.G. Huard, B.K. Spearman, and K.S. Williams. Pascal’s triangle modulo 4. Europ. J. Combin., 19:45–62, 1998. ¨ [16] E.E. Kummer. Uber die Erg¨anzungss¨atze zu den allgemeinen Reciprocit¨atsgesetzen. J. Reine Angew. Math., 44:93–146, 1852. [17] S.-C. Liu and J. C.-C. Yeh. Catalan numbers modulo 2k , J. Integer Seq., 13:Article 10.5.4, 2010. [18] F. Luca and M. Klazar. On integrality and periodicity of the Motzkin numbers. Aequationes Mathematicae, 69:68–75, 2005. [19] F. Lucas. Sur les congruences des numbres eul´eriens et des coefficients diff´erentiels des fonctions trigonom´etriques suivant un modulo premier. Bull. Soc. Math. France, 6:49–54, 1877–1878. [20] Y. Mimura. Congruence properties of Apery numbers. J. Number Theory, 16:138– 146. 1983. [21] A. Postnikov and B. Sagan. What power of two divides a weighted Catalan number. J. Combin. Theory Ser. A, 114:970–977, 2007. [22] N.J.A. Sloane. The On-Line Encyclopedia of Integer Sequences. http://oeis.org [23] Z.-W. Sun. Binomial coefficients, Catalan numbers and Lucas quotients. Sci. China Math., 53:2473–2488, 2010. [24] Z.-W. Sun. On Delannoy numbers and Schr¨oder numbers. J. Number Theory, 131:2387–2397, 2011. [25] S. Wolfram. A New Kind of Science, Champaign, IL: Wolfram Media. 870 and 931– 932, 2002.
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14
Shu-Chung Liu†
Department of Mathematics National Cheng Kung University Tainan City, Taiwan [email protected]
Department of Applied Mathematics National Hsinchu University of Education Hsinchu City, Taiwan [email protected]
Yeong-Nan Yeh‡ Institute of Mathematics Academia Sinica Taipei, Taiwan [email protected] Submitted: Sep 4, 2013; Accepted: May 25, 2014; Published: Jun 9, 2014 Mathematics Subject Classifications: 11B50, 05A15, 05A10
Abstract In this paper we develop a general to enumerate the congruences of finite P Pmethod p−1−h ak ak+h ak summations p−1 (mod p) and (mod p) for the infinite sequence k=0 mk k=0 Bk N
{an }n>0 with generating functions (1+xf (x)) 2 , where f (x) is an integer polynomial and N is an odd integer with |N | < p. We also enumerate the √ congruences of some similar finite summations involving generating functions √ 1−αx− 1−2αx+(α2 −4β)x2 . 2βx2
1
1−αx−
1−2(α+β)x+Bx2 βx
and
Introduction
Let p be an odd prime number and m be an integer with m 6≡ 0 (mod p). The initial topic of this article is to enumerate p−1 X ak mk k=0
(mod p)
(1)
∗
Partially supported by NSC 100-2115-M-006-008. Corresponding author. Partially supported by NSC 101-2115-M-134-134-006. ‡ Partially supported by NSC 101-2115-M-001-013-MY3. †
the electronic journal of combinatorics 21(2) (2014), #P2.45
1
the congruence of the pth partial sum of the infinite sequence {an }∞ n=0 . The study of the congruence of a single term, an (mod p), has a long history extending form the most famous and edge-old problem of Pascal’s fractal, which is originally formed by the parities of binomial coefficients nk [5, 6, 13, 14, 15, 16, 19, 25], to the most recent works about Ap´ery numbers [2, 4, 9, 11, 20], central Delannoy numbers [9], Catalan numbers [1, 7, 10, 17, 21], Motzkin numbers [7, 9, 18] and etc. [12]. This article focuses on the sequence {an }∞ n=0 with generating functions (GF’s) (1 + N xf (x)) 2 , where f (x) is an integer polynomial and N is an odd integer with |N | < p. ak Neither an nor m k in (1) is necessarily an integer if we deal with the field of rational ak numbers; therefore, we shall consider both an and m k the congruences in the modular arithmetic field with modulus p. For instance, usually 1/2 is a fraction; however, in the k modular arithmetic field Zp = {0, 1, . . . , p − 1}, we have 1 p+1 2 2 ≡ (mod p) for k = 0, 1, . . . , p − 1, (2) k k 1 which is always an integer. For example let p = 5, and then 22 = 12 · −1 /2! = 2 (mod 5) = 32 . Z.-W. Sun [24] applied (2) and some other tools to verify the equivalence p−1 X m−4 m(m − 4) ck ≡ 1− k m 2 p k=1 where ck is the kth Catalan number and 0 a = 1 p −1
· p
(mod p),
−1 8
≡3
(3)
denotes the Legendre symbol defined by
if p | a; if a is a quadratic residue modulo p; if a is a quadratic nonresidue modulo p.
One has the following well-known and useful congruence: p−1 a a ≡a 2 (mod p), with ∈ {−1, 0, 1}. (4) p p P ak In [24], Sun also derived p−1 k=1 mk (mod p) for ak being the kth central Delannoy number or Schr¨oder number respectively. The results by Sun initiate the motivation of our study. In this article, we wish to provide a systematic method to deal with the problems of this kind. The article is organized as follows. In Section 2, we exam the problem in (1) by generating functions. √ As applications, similar problems of (1) involving generating functions modified from 1P+ Ax + Bx2 are given in Section 3. In Section 4, we focus on another ak ak+h type of problem p−1−h (mod p). And then some applications are demonstrated k=0 (±B)k in Sections 5 and 6. the electronic journal of combinatorics 21(2) (2014), #P2.45
2
2
N
GF’s of the form (1 + xf (x)) 2
Let {ak }k>0 be the sequence associating with the generating function N
(1 + xf (x)) 2 , where Pp−1 f (x)k is an integer polynomial Pp−1 ak and N is an odd integer with |N | < p. To formulate k=0 ak m (mod p) and k=0 mk (mod p), which are same problem in different forms, we need the following facts: p+N N (i) k2 ≡ k2 (mod p) for k = 0, 1, . . . , p − 1; (ii) (iii)
N 2
k N 2
p
≡
p+N 2
k
p+N +2 ,...,p 2
≡ 0 (mod p) for k =
≡ −2−1 (mod p); moreover
N 2
p+k
≡ −2−1
− 1; p+N 2
k
(mod p) for k = 0, 1, . . . , p − 1
The reason for (i) is trivial and the congruence 0 in (ii) is due to that p+N is an integer 2 p+N and k > 2 . The reason for the first equivalence in (iii) is because of the bijection between {N, N − 2, · · · , N − 2p + 2} and {0, 1, . . . p − 1} under modulo p; however, the element N − 2i such that N − 2i ≡ 0 (mod p) is −p, and additionally 2−p ≡ 2−1 (mod p). The second equivalence in (iii) directly follows the first one. The advantage of (i) is that p+N p+N p+N 2 2 2 is a normal binomial coefficient; therefore, we can apply = . p+N k k −k 2
In the following, the equivalence f (x) ≡ g(x) (mod h(x)) means that f (x) and g(x) share same residue with respect to divisor h(x). Moreover, f (x) ≡ g(x) (mod xq , p) indicates that the coefficients of f (x) and g(x) for the term xk , with k = 0, 1, . . . , q − 1, are congruent modulo p. Theorem 1. Given an odd prime p and an integer m with m 6≡ 0 (mod p). Let {ak }k>0 N be a sequence of integers whose generating function is (1 + xf (x)) 2 , where f (x) is an integer polynomial and N is an odd integer with |N | < p. We have p+N
N
(1 + xf (x)) 2 ≡ (1 + xf (x)) 2 (mod xp , p), and p−1 X N +1 1 + mf (m) k ak m ≡ (1 + mf (m)) 2 − E(m) p k=0
(5) (mod p),
(6)
where E(x) is the polynomial consisting of each terms xt with t > p in the expansion of p+N (1 + xf (x)) 2 . Proof. Actually, (5) and (6) are equivalent. The following can prove both at the same time. p−1 X
k
ak m
=
h
(1 + xf (x))
N 2
i (mod x ) p
k=0 the electronic journal of combinatorics 21(2) (2014), #P2.45
(7) x=m
3
=
≡ = ≡
" p−1 # X N 2 (xf (x))k (mod xp ) k k=0 x=m p+N 2 p+N X 2 (xf (x))k (mod xp , p) k k=0 x=m i h p+N (1 + xf (x)) 2 − E(x) x=m N +1 1 + mf (m) − E(m) (1 + mf (m)) 2 p
(8) (9) (mod p).
Notice that the equivalence in (8) responses to modulus p by applying (i) and (ii), and the is an exponent of positive integer. The equation in (9) yields by modulus xp and that p+N 2 last equivalence is because of (4). Referring to (7) and (8) without substituting x = m, we then verify (5). It takes time to enumerate E(x) unless f (x) is simple enough. In the rest of this section we assume f (x) = A + Bx. Let [xn ]g(x) denote the coefficient of xn in the power series g(x). We derive the following rules: (r1) If B ≡ 0 (mod p) then E(x) ≡ 0 (mod p). Moreover, we have t X
k
ak m ≡ (1 + mf (m))
N +1 2
k=0
for t =
p+N p+N +2 , 2 ,...,p 2
1 + mf (m) p
(mod p),
− 1 due to a p+N +2 ≡ · · · ≡ ap−1 ≡ 0 (mod p) by (ii). 2
(r2) If N 6 −1 then E(x) = 0 again. (r3) If N > 1 and B 6≡ 0 (mod p), then E(x) 6= 0. Let (1 + Ax + Bx2 )
p+N 2
= q0 + q1 x + · · · + qp+N xp+N .
For the precise value of qm , we can simply apply the following formula: [xe ](1 + αx + βx2 )M (Pmin{e,M } =
M α2k−e β e−k k=d 2e e M −k, 2k−e, e−k 2k−e e−k Pe −M +k−1 k β k=d 2e e (−1) −M −1, 2k−e, e−k α
if M > 0 and e 6 2M ; if M < 0.
Therefore, min{m, p+N } 2
qm =
p+N 2
X
k=d m e 2
p+N 2
− k, 2k − m, m − k
the electronic journal of combinatorics 21(2) (2014), #P2.45
A2k−m B m−k .
4
Particularly, if N = 1 then E(x) ≡ A2 Bp xp + B Bp xp+1 (mod p); if p > 3 and N = 3 then 3 1 3 −1 B 3 B 3 2 B B p E(x) ≡ − 4A B x + B + 3A xp+1 A 22 p 2 p 2 p 2 p 3 B B p+2 2 + AB x +B xp+3 (mod p). 2 p p In case that E(x) = 0, we reach a particular case as follows. Corollary 2. If f (x) is a constant function or f (x) = A + Bx with N 6 −1, then we have E(x) = 0 and 2 p−1 X ak m + Am + B ≡ mk m2 k=0
N +1 2
m2 + Am + B p
(mod p).
Here we recall some sequences as applications of the last corollary:
sequence name
first few terms
Pp−1
ak k=0 mk
GF
√ 1 1−4x
central binomial coeff.
1, 2, 6, 20, 70, . . .
central trinomial coeff.
1, 1, 3, 7, 19, 51, . . .
√
central Delannoy numbers
1, 3, 13, 63, 321, . . .
√
1 1−6x+x2 −3
(1 − 4x) 2 √ 1, −2, −2, −4, −10, . . . 1 − 4x
A002457, in [22]
1, 6, 30, 140, 630, . . .
A002420, in [22]
1 1−2x−3x2
(mod p)
m2 −4m p
m2 −2m−3 p
m2 −6m+1 p
(see [24]) 2
m2 m2 −4m
m −4m p
m2 −4m m2
m2 −4m p
Table 1: Some direct applications of Corollary 2 P Pp+1 ak ak Let us take advance to enumerate pk=0 m k and k=0 mk , which will be used in the ap+1 ap nest section. We need to enumerate two additional terms, m p and mp+1 . Using (4), (i), (ii) and (iii), we calculate ap and ap+1 as follows:
p
2
[x ](1 + Ax + Bx ))
N 2
p X
N = [x ] 2 (x(A + Bx))k k k=0 N p X 2 [xp−k ](A + Bx)k = k p+1 k=
≡
p
2
N p−1 2 2 ](A p+1 [x 2
+ Bx)
the electronic journal of combinatorics 21(2) (2014), #P2.45
p+1 2
N + 2 [x0 ](A + Bx)p (mod p) p 5
p+N
≡
[x
p+1
2
](1 + Ax + Bx ))
p−1 p+1 1 AB 2 − Ap (mod p) 2 2 ! p+N B 2 −1 (mod p), p+1 p 2
2 p+1 2
≡ A 2
N 2
=
p+1 N X 2
k= p+1 2
≡
k
[xp−k+1 ](A + Bx)k
N N p+1 p+1 2 2 ](A + Bx) 2 + 2 [x](A + Bx)p p+1 [x p 2 N 2 + [x0 ](A + Bx)p+1 (mod p)
p+1 p+N p+1 1 p+N 2 2 2 ≡ Ap+1 (mod p) +0− p+1 B 2 1 2 p+N N B 2 B − 2 A2 (mod p). ≡ p+1 p 2 2 P Pp+1 ak ak If B ≡ 0 (mod p) or N 6 −1, then pk=0 m k and k=0 mk are ready because we have ap+1 ap A N A2 ≡ − 2m (mod p), mp+1 ≡ − 4m2 (mod p), E(x) = 0 and referring (6). When B 6≡ 0 mp (mod p) and N > 1, then the larger N is, the more complicate E(x) is. However the two summations for N = 1 are also ready as follows. √ Corollary 3. Let {an }n>0 be the sequence whose generating function is 1 + Ax + Bx2 . We have p X ak m2 + Am + B m2 + Am + B B B A ≡ − − (mod p), mk m2 p 2m m2 p k=0 p+1 X ak m2 + Am + B m2 + Am + B A A2 ≡ − − (mod p). k 2 2 m m p 2m 4m k=0
3
GF’s modified from
√
1 + Ax + Bx2
Many well-known sequences are associated with generating functions modified from (1 + 1 Ax + Bx2 ) 2 . In this section, we consider two types: p p 1 − αx − 1 − 2(α + β)x + Bx2 1 − αx − 1 − 2αx + (α2 − 4β)x2 and . βx 2βx2 √ 1−αx− 1−2(α+β)x+Bx2 Let be the generating function of the sequence {bn }n>0 , and let βx A = −2(α + β) for convenience. For examples, (α, β, B) = (0, 2, 0) yields the Catalan the electronic journal of combinatorics 21(2) (2014), #P2.45
6
number, (1, 2, 1) and (−1, 4, 1) provide the large and the little Schr¨oder numbers respectively, and (1, 2, 5) is associated with the number of restricted hexagonal polyominoes [22, A002212]. Theorem 4. We have 2 p−1 X m + Am + B B B bk m2 + Am + B 1− + −1 (mod p). ≡ 1+ mk βm p βm p k=0 √ Proof. We still let {an }n>0 associate with 1 + Ax + Bx2 . Notice that bk = − ak+1 for β A k > 1. Also it is clear that a0 = 1, a1 = 2 and b0 = 1. p−1 X bk mk k=0
p
m X ak = 1− β k=2 mk p
m A m X ak + − = 1+ β 2β β k=0 mk m A m m2 + Am + B m2 + Am + B A ≡ 1+ + − − 2 β 2β β m p 2m B B − 2 (mod p), m p 2 m2 + Am + B m + Am + B B B = 1+ 1− + −1 . βm p βm p
In particular, the last term in the last theorem is zero if B ≡ 0 (mod p) or Bp = 1. This particular case happens for the Catalan numbers (see (3) and also [24, Lemma 2.1]), the large Schr¨oder√numbers (see [24, Theorem 1.2]) and the little Schr¨oder numbers. 1−αx−
1−2αx+(α2 −4β)x2
Now let be the generating function of the sequence {dn }n>0 . 2βx2 For convenient, let A = −2α and B = α2 − 4β. Lots of famous sequences have GF’s of this form. For examples, (α, β) = (1, 1) yields the Motzkin number, and (3, 9) as well as (5, 5) are associated with the numbers of Motzkin paths with multiple colors for level steps. Also (α, β) = (2, 1) creates a shifted Catalan number, (3, 2) a shifted little Schr¨oder number [22, A001003], and (4, 1) the number of walks on cubic lattice starting and finishing on the horizontal plane but never going below it [22, A005572]. For more examples, please refer to [3]. The next result can be verified by a process similar to the proof of the last theorem. Theorem 5. We have 2 p−1 X m + Am + B m2 + Am + B dk ≡ 1− + 2 (mod p). k m 2β p k=0 the electronic journal of combinatorics 21(2) (2014), #P2.45
7
4
Enumerating
Pp−1−h ak ak+h (±B)k
k=0
(mod p) N
Let {an }n>0 still associate with a generating functions of form (1 + Ax + Bx2 ) 2 . In this and the next sections we assume B 6≡ 0 (mod p). Given nonnegative integer h with Pp−1−h (mod p) and (S2) h 6 p − 1, we are interested in two summations: (S1) k=0 akBak+h k Pp−1−h ak ak+h (mod p). For convenience, we use ±B to denote B and −B simultaneously. k=0 (−B)k N
Let G(x) = a0 + a1 x + · · · ap−1 xp−1 ≡ (1 + Ax + Bx2 ) 2 (mod xp ) and Q(x) = (1 + Ax + p+N Bx2 ) 2 = q0 + q1 x + · · · +P qp+N xp+N . (See (r3) for the formula of qm .) Now we apply p [x−h ]G(x)G( x1 ) to evaluate p−1 k=0 ak ak+h and also use the fact G(x) ≡ Q(x) (mod p, x ) (see Theorem 1). p−h−1
X ak ak+h 1 x )G( ) = [x−h ]G( k (±B) ±B x k=0 X qk qk+h x 1 ≡ [x−h ]Q( )Q( ) − (mod p) ±B x (±B)k k>p−h = [x
−h
x2 Ax + ] 1+ ±B B
p+N 2
A B 1+ + 2 x x
p+N 2 −
X qk qk+h (±B)k k>p−h
p+N p+N X qk qk+h Ax x2 2 Ax x2 2 [x ] 1+ + + = B − 1+ ±B B B B (±B)k k>p−h ( A X qk qk+h p+N (1 + B x + B1 x2 )p+N for (S1); p+N −h 2 = − + B × [x ] (10) p+N 2 (±B)k (1 + 2B−A for (S2). x2 + B12 x4 ) 2 B2 k>p−h P k qk+h The first common term k>p−h q(±B) k can be avoided if and only if N 6 −1 (see (r2) and (r3)). Now the result of (S2) is ready. P ak ak+h (mod p) is equivalent to Theorem 6. The summation p−h−1 k=0 (−B)k p+N 2
B
N +1 2
p+N −h
p+N X qk qk+h B 2B − A2 2 1 4 2 p+N −h [x ] 1+ x + x − p B2 B2 (−B)k k>p−h
(mod p).
Particularly, if h is odd then the first term is 0, and if N 6 −1 then the last term is 0. 1 Example 7. Let N = −1 and B = 1. For instances, √1−6x+x 2 is the GF of the central De1 1 √ lannoy numbers, 1−10x+x2 is associated with the colored Delannoy paths, and √1−14x+x 2 P k is associated with the central coefficients of (1 + 7x + 12x2 ). So p−h−1 (−1) a a ≡ 0 k=0 Pp−h−1 k kk+h p−1 (mod p) if h is odd; otherwise let p−h−1 = 2e and 2 = M , and then k=0 (−1) ak ak+h (mod p) is equivalent to e X M e 2 2 M [x ](1 + (2 − A )x + x ) = (2 − A2 )2k−e . M − k, 2k − e, e − k e k=d 2 e
the electronic journal of combinatorics 21(2) (2014), #P2.45
8
As for (S1), we continuously simplify the last term in (10) as [x
p+N −h
N N A N −h A 1 2 A 1 2 + [x , ] 1+ x+ x ] 1+ x+ x B B B B B
A p A x+ B1 x2 )p ≡ 1+ B x + B1 x2p (mod p) in which x2p is ignored for its exponent because (1+ B being too large. P Pp+N p+N N Theorem 8. Let (1 + Ax + Bx2 ) 2 = n>0 an xn and (1 + Ax + Bx2 ) 2 = n=0 q n xn . Pp−h−1 ak ak+h Then k=0 (mod p) is equivalent to Bk
B
N N N −1 B A 1 2 B A 1 2 p+N −h N −h [x ] 1+ x+ x + AB 2 [x ] 1+ x+ x p B B p B B X qk qk+h − . Bk k>p−h
N +1 2
Particularly, only the first term remains when N 6 −1, the first term is zero when N > 1 and p > N + h, and the second term is zero when N < h. P ak ak+h (mod p) is equivalent to Example 9. Again, let N = −1 and then p−h−1 k=0 Bk e X B k A 1 2 −1 k e (−1) A2k−e B −k [x ](1 + x + x ) = 2k − e, e − k p B B e k=d 2 e
where e = p − h − 1. A Since x/(1 + B x + B1 x2 ) is the generating function of the Lucas sequence un that satisfies the second order recurrence relation
1 A un = − un−1 − un−2 , B B with the initial u0 = 0 and u1 = 1, we also have p−h−1
X ak ak+h B ≡ up−h k B p k=0 αp−h − β p−h = α−β
(mod p) B , p
A where α and β are roots of x2 + B x + B1 .
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5
Summations
Pp−2−h bk bk+h (±B)k √
k=0
1−αx−
and
Pp−3−h dk dk+h Bk
k=0
1−2αx+(α2 −4β)x2
We observe some applications here. Let be the generating function 2βx2 2 of {dn }n>0 . For convenience let A = −2α and B = α − 4β. ak (mod p) for k = 2, 3, . . . , p − 1, we have Since dk−2 = − 2β p−3−h
p−1−h X dk dk+h B 2 X ak ak+h ≡ (mod p) (±B)k 4β 2 k=2 (±B)k k=0
B2 = 4β 2
p−1−h
X ak ak+h a1 ah+1 − a0 ah − k (±B) ±B k=0
! .
Pp−1−h ak ak+h given in the last section, let us recall the Before we apply the formula of k=0 (±B)k P P qk qk+h k qk+h term k>p−h (±B)k in both Theorems 6 and 8. With N = 1 we have k>p−h q(±B) k = qp−h qp qp−h+1 qp+1 + (±B)p−h+1 . Notice that, the coefficients of (1 + Ax + Bx2 )M = q0 + q1 x + · · · + (±B)p−h q2M x2M have a sort of symmetric property, i.e., B M −k qk = q2M −k for k = 1, 2, . . . , 2M , and we get which can be easily proved by induction on M . Let M = p+1 2 p2M −k−h p2M −k (±B)2M −k−h
pk pk+h (±B)k ak ak+h ≡ (±1)h (±B)k = (±1)h
pp+1−h pp+1 (±B)p+1−h
p
(mod p).
(11)
p
p−h p h a1 ah+1 (mod p). More≡ (±1)h a0 ah (mod p) and (±B) p−h ≡ (±1) ±B N +1 B A over, for N = 1 and h 6 p−3, the term B 2 [xp+N −h ](1+ B x+ B1 x2 )N in Theorem 8 p must be 0. Now we derive conclusion and show some examples as follows. Pp−3−h dk dk+h (mod p) is equivalent to Theorem 10. The congruence k=0 Bk B2 A 1 2 B 2a1 a1+h 1−h [x ] 1 + x + x − 2a0 ah − . A 4β 2 p B B B
Particularly,
Corollary 11. Let 2 6 h 6 p − 3. We have p−3−h
X dk dk+h a1 a1+h B2 ≡ − a a + 0 h Bk 2β 2 B k=0
(mod p),
which is a fixed number modulo any prime p > 2. 2
3
+ A16 and a4 = Example 12. Notice that a0 = 1, a1 = A2 , a2 = B2 − A8 , a3 = − AB 4 2 2 4 − B8 + 3A16B − 5A . For h = 0, 1, 2, 3 and p > h + 3 we have following examples: 128 p−3 X dk 2 k=0
Bk
B ≡ 8β 2
B 2 2 2A − 4B − A p
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(mod p),
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p−4 X dk dk+1 k=0 p−5
Bk
X dk dk+2 k=0 p−6
Bk
X dk dk+3 k=0
Bk
AB ≡ 32β 2 ≡ − ≡
B 2 8B − 12B + A p
B 16B 2 − 8A2 B + A4 2 64β
5AB 2 2 4 16B − 8A B + A 512β 2
(mod p),
(mod p),
(mod p)
Particularly, given (A, B, β) = (−2, −3, 1) (so we need p 6= 3) that yields the Motzkin numbers, the four summations above are respectively equivalent to 9 −3 15 −3 − 3, − + , 12 and 15 (mod p). −3 p 2 p 2 Pp−3−h dk dk+h Theorem 13. The congruence k=0 (−B)k (mod p) is equivalent to 0 if h is odd, and equivalent to ! p+1 2 2 B2 B 2B − A 1 a a 1 1+h B [xp+1−h ] 1 + , x2 + 2 x4 − 2a0 ah − 2 4β 2 p B2 B B if h is even. p+1 2 2 1 4 2 One can refer to (r3) for the precise value of [xp+1−h ] 1 + 2B−A . x + x B2 B2 √ 1−αx− 1−2(α+β)x+Bx2 Let be the generating function of the sequence {bn }n>0 , and βx let A = −2(α + β) for convenience. Since b0 = 1 and bk−1 ≡ − aβk (mod p) for k = 2, 3, . . . , p − 1, we have p−2−h
p−1−h X bk bk+h ±B X ak ak+h ≡ b0 b h + 2 (±B)k β k=2 (±B)k k=0
B = bh ± 2 β
p−1−h
(mod p)
X ak ak+h a1 a1+h − a0 ah − k (±B) ±B k=0
! .
This is quick similar to (11), so we can directly get Pp−2−h bk bk+h Theorem 14. The congruence k=0 (mod p) is equivalent to Bk B A 1 2 2a1 a1+h B 1−h [x ](1 + x + x ) − 2a0 ah − . bh + 2 A β p B B B P bk bk+h The congruence p−2−h (mod p) is equivalent to bh if h is odd, and equivalent k=0 (−B)k to ! p+1 2 2 B B 1 2B − A a a 1 1+h bh − 2 B x2 + 2 x4 , [xp+1−h ] 1 + − 2a0 ah − 2 β p B2 B B if h is even. the electronic journal of combinatorics 21(2) (2014), #P2.45
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6
More applications
Let an = 2n be the central binomial coefficient whose generating function is G(x) = n √ 1 . We have 1−4x p−1−h
X
16k ak ak+h = 4h [x−h ]G(x/4)G(1/4x)
k=0
p−1 −1 [x 2 −h ](1 − x)p−1 (mod p) ≡ 4 p p−1 −1 ≡ 4h [x 2 −h ](1 − x)−1 (mod p) p −1 = 4h . p h
Pp−1−h −2k Almost same procedure can be applied on the problem k=0 A ak ak+h for some other N an whose generating function is G(x) = (1 − Ax) 2 for odd integer N with |N | < p. In the following final example, we show an application involving two different sequences by the same technique. c, . . . , p−3 , p−1 and 0 6 h 6 b 3p+1 c − 2, we have Lemma 15. Given integers r = b p−1 4 2 2 4 ( 0 (mod p) 4k 64 ck+h ≡ 2k 22h+1 −1 p k=0
r X
−k
if 0 6 h < p−3 or p−1 < h 6 b 3p+1 c − 2; 4 2 4 p−3 p−1 (mod p) if 4 6 h 6 2 . p−1−2h p+1 2
1 that Proof. The sequence of the central binomial coefficients 2k (k > 0) has GF √1−4x k 2k p−1 p+1 k −1/2 comes from the identity k = 4 k . Given t = 2 , 2 , . . . , p − 1, we have t X 2k xk k=0
k
4k
t X −1/2 k = x k k=0 p−1 X −1/2 k ≡ x k k=0 1
≡ (1 − x)− 2 ≡ (1 − x)
p−1 2
(mod p)
(mod xp ) (mod xp , p) or (mod xp−1 , p),
where the equivalence on the second line dues to (ii) in Section 1. By the same reason, we can write (mod xp−1 , p) additionally on the last line. And then we get t
b2c X 4k x2k k=0
2k 16k
≡
p−1 p−1 1 (1 − x) 2 + (1 + x) 2 2
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(mod xp−1 , p).
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For convenience, let t = p−1 in the rest of the proof. 2 On the other hand, the sequence of the Catalan numbers ck has GF C(x) = Again, ck ≡ 0 (mod p) for k = p+1 , . . . p − 2 by (ii). So we have 2
√ 1− 1−4x . 2x
p−1
p−2 X x−2k x−2k ck k ≡ ck k (mod p) 4 4 k=0 k=0 1 (mod xp−1 ) ≡ C 2 4x
2 X
= 2x2 − 2x−(p−1) (x2 − 1)
p+1 2
.
Given 0 6 h 6 b 3p+1 c − 2 (for b p−1 c + h 6 p − 2 by considering the term ck+h ), finally we 4 4 derive that b p−1 c 4k 4 X p−1 p−1 p+1 ck+h −2h −(p−1) 2 2k 2 2 2 (1 − x) + (1 + x) (mod p) · ≡ [x ] − x (x − 1) 16k 4k+h k=0 p−1 p+1 p−1 −1 p−1−2h ≡ [x ] (1 − x2 ) 2 (1 − x) 2 + (1 + x) 2 (mod p) p p+1 p+1 −1 p−1−2h (1 + x) 2 (1 − x)p + (1 + x)p (1 − x) 2 = [x ] p p+1 p+1 −1 ≡ [xp−1−2h ] (1 + x) 2 + (1 − x) 2 (mod p) p ( 0 if 0 6 h < p−3 or p−1 < h 6 b 3p+1 c − 2; 4 2 4 p+1 = 2 2 −1 if p−3 6 h 6 p−1 . p 4 2 p−1−2h c by any of b p−1 c, . . . , p−1 . We complete the proof by replacing the upper limit b p−1 4 4 2
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