1. Introduction. = (2n)! (n N)

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1 Included in: Combinatorial Additive Number Theory: CANT (edited by M.B. Nathanson, Sringer Proc. in Math. & Stat., Vol. 101, Sringer, New Yor, 2014, ON SUMS RELATED TO CENTRAL BINOMIAL AND TRINOMIAL COEFFICIENTS ZHI-WEI SUN Abstract. A generalized central trinomial coefficient T n (b, c is the coefficient of x n in the exansion of (x 2 +bx+c n with b, c Z. In this aer we investigate congruences series for sums of terms related to central binomial coefficients generalized central trinomial coefficients. The aer contains many conjectures on congruences related to reresentations of rimes by certain binary quadratic forms, 62 roosed new series for 1/π motivated by congruences related dualities. 1. Introduction Let N = {0, 1, 2,...}. The central binomial coefficients ( n = (2n! (n N n (n! lay imortant roles in combinatorics number theory. In this section we first review some nown results on sums involving roducts of at most three central binomial coefficients. Let Z + = {1, 2, 3,...}. Recall that for given numbers A B the Lucas sequence u n = u n (A, B (n N its comanion v n = v n (A, B (n N are defined by u 0 = 0, u 1 = 1, u n+1 = Au n Bu n 1 (n Z +, v 0 = 2, v 1 = A, v n+1 = Av n Bv n 1 (n Z +. It is well nown that (α βu n = α n β n v n = α n + β n for all n N, where α = (A + /2 β = (A /2 with = A 2 4B. Key words hrases. Central binomial coefficients, central trinomial coefficients, congruences, reresentations of rimes by binary quadratic forms, series for 1/π 2010 Mathematics Subject Classification. Primary 11B65, 11E25; Secondary 05A10, 11A07, 33F05. 1

2 2 ZHI-WEI SUN Let be an odd rime let m be any integer not divisible by. The author [Su1] roved that ( m 2 m 4m + u ( m 2 (m 2, 1 (mod 4m, where ( denotes the Jacobi symbol. Let 1 (mod 4 be a rime. Write = x 2 +y 2 with x 1 (mod 4 y 0 (mod 2. Gauss showed in 1828 the famous congruence ( ( 1/2 2x (mod, ( 1/4 this was further refined by S. Chowla, B. Dwor R. J. Evans [CDE] in 1986 who used Gauss Jacobi sums to show that ( ( 1/ ( 2x (mod 2. ( 1/4 2 2x For more such congruences involving roducts of one or more binomial coefficients, the reader may consult the excellent survey [HW] by R. H. Hudson K. S. Williams. In 2009 the author (cf. [Su2, Conjecture 5.5] conjectured that ( 16 ( 2 32 ( ( 2x 2x (mod 2, this was later confirmed by Z.-H. Sun [S1]. Recently the author [Su5] determined x mod 2 via the congruence ( ( 1/2 2 x ( ( 1/ ( 16 Note that ( 2 for all = ( + 1/2,..., 1. Let be an odd rime. By [I, vh], 1 ( 2 (mod 2. 3 { 4x 2 2 (mod 2 if = x 2 + y 2 (2 x & 2 y, 64 0 (mod 2 if 3 (mod 4. In [Su2, Su4] the author made conjectures on 1 3/m mod 2 for m = 1, 8, 16, 64, 256, 512, 4096; for examle, Conjecture 5.3 of [Su2] states that 1 3 { 4x 2 2 (mod 2 if ( = 1 & = 7 x2 + 7y 2, 0 (mod 2 if ( = 1. 7

3 CENTRAL BINOMIAL AND TRINOMIAL COEFFICIENTS 3 (Throughout this aer, when we write a multile of a rime in the form ax 2 + by 2, we always assume that x y are nonzero integers. To attac such conjectures, Z.-H. Sun [S2] deduced the useful combinatorial identity n ( ( n + 2 x = P n ( 1 + 4x (1 2 where P n (x is the Legendre olynomial of degree n given by n ( ( ( n n + x 1 P n (x :=. 2 Actually (1 is just a secial case of the well-nown Clausen formula for hyergeometric series. We can rewrite (1 in the form n ( ( ( n n + 2 (x(x + 1 = D n (x (2 where D n (x is the Delannoy olynomial of degree n given by n ( ( n n + D n (x := x. Note that those D n = D n (1 (n = 0, 1, 2,... are central Delannoy numbers (see, e.g., [CHV], [Su3] [St,. 178]. It is well nown that P n ( x = ( 1 n P n (x, i.e., ( 1 n D n (x = D n ( x 1 (cf. [Su3, Remar 1.2]. As observed by Z.-H. Sun [S1, Lemma 2.2], if 0 n = ( 1/2 then (n + 2 ( 16 (mod 2 hence (n ( ( ( n + n + 2 = 2 ( 16 (mod 2. This simle tric was also realized by van Hamme [vh,. 231]. Combining this useful tric with the identity (2, we see that 3 ( ( 16 (x(x + 1 ( 16 x (mod 2. (3 To study the author s conjectures on ( 3 1 m, 1 ( 4 2 m, 1 ( 3 ( 6 3 m

4 4 ZHI-WEI SUN modulo 2 (with m suitable integers not divisible by given in [Su4, Su7], Z.-H. Sun [S2, S3, S4] managed to rove the following congruences similar to (3: 1 1 ( 4 2 ( 64 ( 3 ( 27 ( 1 (x(x + 1 (x(x + 1 ( 3 ( 6 3 (x(x + 1 ( 432 ( 4 2 2( ( 64 x (mod 2, (4 ( 3 ( 27 x (mod 2 ( > 3, (5 ( 6 3 3( ( 432 x (mod 2 ( > 3. ( 1 ( 1 In 1859 G. Bauer roved that ( 2 3 (4 + 1 ( 64 = 2 π. In 1914 S. Ramanujan [R] found 16 new series for 1/π which are quite similar to Bauer s series. The (rational Ramanujan-tye series for 1/π (cf. B. C. Berndt [Be, ], also [BB] [ChCh] have the following form: (a + d f( = C m π, (7 where f( refers to one of ( 3 ( ( 2 2 3,, ( 4 2, ( 3 ( 6 a, d, m are integers with dm 0, C 2 is rational. U to now, 36 such series have been established via the theory of modular forms. The reader may consult [CCL], [CC], S. Cooer [C] for some other series for 1/π. Let be an odd rime. Note that Γ(1/2 = π while Γ (1/2 = ( 1 (+1/2, where Γ (x denotes the -adic Γ-function. In view of this, in 1997 van Hamme [vh] studied -adic suercongruences for artial sums of some hyergeometric series involving the Gamma function. (If a -adic congruence haens to be true modulo a higher ower of, then it is called a suercongruence. For examle, Bauer s series led, (6

5 CENTRAL BINOMIAL AND TRINOMIAL COEFFICIENTS 5 him to conjecture that 2 3 ( 1 (4 + 1 ( 64 (mod 3, which was later confirmed by E. Mortenson [M2] in More suercongruences motivated by Ramanujan-tye series have been investigated by some followers of van Hamme, see, e.g., L. Long [L] Zudilin s wor stated there. The author [Su6] refined the congruence by van Hamme Mortenson to the following congruence mod 4 : 2 3 (4+1 ( 64 ( 1/2 3 ( 1 (4+1 ( E 3 (mod 4, where E 0, E 1,... are the Euler numbers defined by E 0 = 1, n 2 ( n E n = 0 (n Z +. For more conjectural connections between Ramanujan-tye congruences Euler numbers or Euler olynomials, the reader may consult [Su4]. Goser announced in 1974 that (25 3/(2( 3 = π/2 (see G. Almvist, C. Krattenthaler J. Petersson [AKP] for a simle roof. Though this is not a Ramanujan-tye series, the author conjectures that for any rime > 3 we have 1 ( ( 3 + (E (mod 3, ( 1/2 ( ( ( 3 2 (mod 2, 1 ( ( 3 1 ( (mod 2. The author [Su4] found some new series for owers of π motivated by corresonding -adic congruences. Here is a new examle: Immediately

6 6 ZHI-WEI SUN after the author discovered the conjectural congruence 1 ( n 28n + 5 2n 5 ( 2 2 ( 2(n n ( 576 n n n n=0 ( ( ( (mod 2 for any rime > 3, he conjectured (on Jan. 14, 2012 that ( n 28n + 5 2n 5 ( 2 2 ( 2(n n ( 576 n n n = 9 π ( (8 n=0 The author [Su4, Su7] found that an identity lie (7 usually corresonds to a congruence for 1 f(/m modulo 2 in terms of arameters in reresentations of a rime or its multile by certain binary quadratic forms. This was the main starting oint of the author s discoveries of many new series for 1/π. Let n N. Clearly ( 2n n is the coefficient of x n in the exansion of (x 2 + 2x + 1 n = (x + 1n. The nth central trinomial coefficient T n = [x n ](x 2 + x + 1 n is the coefficient of x n in the exansion of (x 2 + x + 1 n. Since T n is the constant term of (1 + x + x 1 n, by the multi-nomial theorem we see that T n = n/2 n!!!(n 2! = n/2 ( n 2 ( = n ( n ( n Central trinomial coefficients arise naturally in enumerative combinatorics (cf. [Sl], e.g., T n is the number of lattice aths from the oint (0, 0 to (n, 0 with only allowed stes (1, 1, (1, 1 (1, 0. Given b, c Z, we define the generalized central trinomial coefficients T n (b, c :=[x n ](x 2 + bx + c n = [x 0 ](b + x + cx 1 n n/2 ( ( n/2 n 2 ( ( n n = b n 2 c = b n 2 c. 2 Clearly T n (2, 1 = ( 2n n Tn (1, 1 = T n. An efficient way to comute T n (b, c is to use the initial values T 0 (b, c = 1 T 1 (b, c = b, the recursion (n+1t n+1 (b, c = (2n+1bT n (b, c n(b 2 4cT n 1 (b, c (n = 1, 2,.... Note that the recursion is rather simle if b 2 4c = 0.. (9

7 CENTRAL BINOMIAL AND TRINOMIAL COEFFICIENTS 7 Let b, c Z d = b 2 4c. It is nown that T n (b, c = d n P n (b/ d if d 0 (see, e.g., [N] [Su9]. Thus n ( ( ( n + 2 b d d n T n (b, c =. ( (In the case d = 0, (10 holds trivially since x 2 +bx+c = (x+b/2. By the Lalace-Heine formula (cf. [Sz,. 194], for any comlex number x [ 1, 1] we have P n (x (x + x 2 1 n+1/2 2nπ 4 x 2 1 It follows that if b > 0 c > 0 then T n (b, c f n (b, c := (b + 2 c n+1/2 2 4 c nπ as n +. as n +. (11 Note that T n ( b, c = ( 1 n T n (b, c. The generalized central trinomial coefficients seem to be natural extensions of the central binomial coefficients. To see this, in the next section we study congruences for 1 T (b, c m 1 T2 (b, c modulo an odd rime, where b, c, m Z m 0 (mod. One may comare them with congruences for 1 /m mod 2 with m = 8, 16, 32. Since ( ( ( T (2, 1 =, T 2 (2, 1 = T 3 (2, 1 =, 2 3 in Section 3 we are going to investigate general sums m T ( (b, c, T m (b, c, 2( T m (b, c 1 m T 2(b, c, 1 ( 3 m m T 3 (b, c modulo 2, where is an odd rime, b, c, m Z m 0 (mod. For this urose, we need to extend those congruences (3-(6 in Section 3. Section 4 contains many conjectural congruences involving generalized central binomial coefficients they offer bacgrounds for those

8 8 ZHI-WEI SUN conjectural series for 1/π in Sect. 5. In the fifth section we first show a theorem on dualities then roose 61 new conjectural series for 1/π based on our investigation of congruences. 2. On 1 T (b, c/m 1 T2 (b, c/m modulo Lemma 2.1. Let = 2n + 1 be an odd rime let {0,..., n}. Then ( ( 2 2(n ( 1 n 16 (mod. (12 n Given b, c Z with b 2 4c (mod, we also have ( b 2 4c T2 (b, c T 2(n (b, c (mod. (13 (b 2 4c 2 Proof. (12 holds because ( 1/2 ( 4 = ( ( n n = n ( 1/2 = n ( 2(n n (mod. ( 4 n For b, c Z with d = b 2 4c 0 (mod, we get (13 from the nown result d j T 1 j (b, c ( dt j(b, c for j = 0,..., 1 (see [N, (14] or [Su9, Lemma 2.2]. Theorem 2.1. Let be an odd rime let m, b, c Z with m 0 (mod. If m 4b (mod, then 1 m T (b, c { ( m 2xc( 1/4 (mod if = x 2 + y 2 (4 x 1, 0 (mod if 3 (mod 4. (14 If m 4b (mod, then ( m(m 4b 4 2 m T (b, c 2( c (m 4b 2 (mod. (15 Also, rovided that d = b 2 4c 0 (mod, for any h Z + we have 1 ht2 (b, c m ( ( 1 h dm ht2 (b, c (mod. (16 (16 h d 2 /m

9 CENTRAL BINOMIAL AND TRINOMIAL COEFFICIENTS 9 Proof. Set n = ( 1/2. As ( ( n 1/2 ( = 2 /( 4 (mod for all = 0,..., 1, we have 1 n ( ( n m T (b, c 4 [x 0 ](x + b + cx 1 m ( =[x 0 ] 1 4 n m x2 + bx + c x ( m [x n ](mx 4(x 2 + bx + c n ( m ( 1 n ( m = ( m ( [x n ] ( m 4b T n, c 4 x 2 m 4b n x + c 4 ( m Tn (m 4b, 16c = 2 2n T n (m 4b, 16c (mod. Observe that T n (m 4b, 16c = n/2 n/2 = n/2 ( ( n 2 2 ( 4 2 ( 4 ( 4 2 (m 4b n 2 (16c ( (m 4b n 2 (16c (m 4b n 2 c (mod. Thus (15 holds when m 4b (mod. If m 4b (mod, then T n (m 4b, 16c {( 2n ( n n n/2 c n/2 (mod if 2 n, 0 (mod if 2 n. Clearly ( ( 2n n = 1 n ( 1 n (mod. If = 2n (mod 4 = x 2 + y 2 with x 1 (mod 4, then ( n n/2 2x (mod as observed by Gauss. Thus, (14 holds when m 4b (mod.

10 10 ZHI-WEI SUN Now suose that d = b 2 4c 0 (mod h Z +. In view of Lemma 2.1, we have ( n ht2 (b, c n (( 1 n 16 ( 2(n n h ( d d 2 T m m 2(n (b, c ( n ( d 16 =( 1 hn h d 2 n j ( h 2j T 2j(b, c m j j=0 ( n ( ( 1 h 2 ht2 dm (b, c (mod. (16 h d 2 /m Recall that for each = n + 1,..., 1. So (16 follows. Corollary 2.1. Let be an odd rime. Then 1 4 T (1, 2 ( 1 (x 1/2+y/4 2x (mod if 8 1 & = x 2 + y 2 (2 x, ( 1 (y 2/4 2y (mod if 8 5 & = x 2 + y 2 (2 y, 0 (mod if 3 (mod 4. Proof. If 3 (mod 4, then 1 T (1, 2/4 0 (mod by (14 with m = 4, b = 1 c = 2. Now assume that 1 (mod 4 write = x 2 + y 2 with x 1 (mod 4 y 0 (mod 2. Alying (14 with m = 4, b = 1 c = 2, we get 1 4 T (1, 2 2x 2 ( 1/4 (mod. By Exercise 27 of [IR,. 64] (an observation of Dirichlet, ( y xy/2 2 ( 1/4 (mod. x Note that ( y x = y 2 So we have x 2 1 (mod hence ( y x ( y y/2 2 ( 1/4 (mod. x 4 1 (mod 4.

11 CENTRAL BINOMIAL AND TRINOMIAL COEFFICIENTS 11 If 1 (mod 8, then 4 y hence 2 ( 1/4 ( 1 y/4 (mod. If 5 (mod 8, then y 2 (mod 4 hence ( y (y 2/4 2 ( 1/4 y x x y ( 1(y 2/4 x (mod. Combining the above, we obtain the desired result. Corollary 2.2. For any rime > 3 we have 1 T ( 4 ( 1 (mod 1 T 12 ( 3 (mod. Proof. Alying (15 with b = c = 1 m { 4, 12} we obtain 1 T ( 4 ( ( 4( 8 1 ( 4 2 2( (mod 64 1 T 12 ( ( 4 2 2( (mod. 64 It is nown that 1 ( 4 2 2( 64 ( 2 (mod 2, which was conjectured in [RV] roved in [M1]. So the two congruences in Corollary 2.2 are valid. Theorem 2.2. Let be an odd rime. Then 1 T2 4 ( 2 (mod.

12 12 ZHI-WEI SUN Proof. Set n = ( 1/2. Then 1 4 T 2 n ( n ( 1 [x 0 ](1 + x + x 1 =[x 0 ] ( 1 (1 + x + x 1 n =[x 0 ]( 1 n ( x x (x + 12 x n =( 1 n [x 2n ](x n (x + 1n = ( 1 n n ( n ( 1 n = ( 2 n ( 2 n (mod. ( ( n 2n 2 This concludes the roof. Remar 2.1. For any rime > 3 we observe the following congruences: T 2(5, 4 1 (mod, ( 16 T 2(4, 9 (mod, 3 4 T 2(3, 1 1 Conjecture 2.1. Let > 3 be a rime. Then ( 1/2 ( 1/ T ( 3 ( 16 T 2(8, T 2(4, 1 1 (mod 2, (( 3 ( T 2(4, ( 1/2 ( 1/2 4 T 2(3, 4 16 T 2(8, 9 4 ( ( 3 (mod 2, (mod 2, ( 1 (mod 2, (mod, ( 5 (mod. (mod 2,

13 CENTRAL BINOMIAL AND TRINOMIAL COEFFICIENTS 13 1 ( T 3(6, 1 1 (mod. Conjecture 2.2. Let > 3 be a rime. Then Also, ( 1/2 ( 1/2 ( 1/2 ( 1/2 16 T 2(2, 3 { 16 T 2(4, 3 ( 1( 3 2x 2 if = x 2 + 3y 2, 0 (mod if 2 (mod 3. 4 T 2(1, 3 { ( 1 xy/2 ( x2x (mod 3 if = x2 + 3y 2, 0 (mod if 2 (mod 3; 16 T 2(4, 3 ( 1 x/6 +y/2 2x (mod if 12 1 & = x 2 + y 2 (2 x, ( 1 (x+y+1/2 ( xy 2y (mod if 12 5 & = 3 x2 + y 2 (2 x, 0 (mod if 3 (mod 4. Conjecture 2.3. Let be an odd rime. Then ( 1/2 16 T 2(12, 7 { 2x( x (mod if ( = 1 & = 7 7 x2 + 7y 2, 0 (mod if ( = 1, i.e., 3, 5, 6 (mod Extensions of (2 (6 with alications to sums involving generalized central trinomial coefficients Our following theorem is a natural generalization of (2. Theorem 3.1. For any n N we have n ( ( ( ( n j 2j D n (xd n (y = (xy+y j (x y j, 2 2j j j=0 (17

14 14 ZHI-WEI SUN Proof. Let a n denote the left h side or the right-h side of (17. It is easy to see that a 0 = 1, a 1 = (2x + 1(2y + 1, a 2 = (6x 2 + 6x + 1(6y 2 + 6y + 1 a 3 = (20x x x + 1(20y y y + 1. Alying the Zeilberger algorithm (cf. [PWZ, ] via Mathematica we find the recursion for n 3: (n + 1 (2n 3a n+1 (2n 3(2n + 1 (2x + 1(2y + 1a n + (2n 1A(n, x, ya n 1 (2n 3 (2n + 1(2x + 1(2y + 1a n 2 + (n 2 (2n + 1a n 3 = 0, where A(n, x, y := 6n 2 6n 5 + (16n 2 16n 12(x + y x 2 y 2. Thus (17 holds by induction. Now we give our extensions of (3 (6. Theorem 3.2. Let be a rime let a Z +. Let h be a -adic integer set w (h = ( ( h h+ for N. Then ( a 1 ( 1 w (hx w (hy w (h a 1 ( + j j=0 2j In articular, if 2 then ( a 1 ( ( a 1 ( 16 x a 1 ( a 1 a 1 ( 4 ( 16 ( + j j=0 2 2( ( 64 x ( 4 2 2( ( 64 2j ( a 1 ( + j j=0 2j ( 2j j (xy + y j (x y j (mod 2. ( 16 y ( 2j j ( 4 (xy + y j (x y j (mod 2 2 2( ( 64 y ( 2j j (xy + y j (x y j (mod 2 ; (18 (19 (20

15 CENTRAL BINOMIAL AND TRINOMIAL COEFFICIENTS 15 rovided > 3 we have ( a 1 ( ( 3 ( a 1 ( 27 x a 1 ( a 1 a 1 ( 6 ( 3 ( 27 ( + j j=0 ( 6 3 3( ( 432 x 3 3( ( 432 2j ( a 1 ( + j j=0 2j ( 3 ( 27 y ( 2j j (xy + y j (x y j (mod 2 ( 6 3 3( ( 432 y ( 2j j Remar 3.1. Note that ( w 1 = 2 ( 16, ( w 1 ( 3 =( 3 ( 27, (xy + y j (x y j (mod 2. w w ( 1 ( 4 2 = 2( 4 ( 64, ( 1 ( 6 3 = 3( 6 ( 432. Also, (19-(22 in the case x = y a = 1 yield (3-(6 resectively. (21 (22 The reader may wonder how we found Theorem 3.2. In fact, (17 is our main clue to the congruence (19. By refining our roof of (19-(22 we found (18. To rove Theorem 3.2 we need two lemmas. Lemma 3.1. For m, n N we have n ( ( n + m w +m (h = w m(hw n (h, (23 n where w (h = ( h ( h+ ( m+n n as defined in Theorem 3.2. Proof. Let u n denote the left-h side of (23. By alying the Zeilberger algorithm via Mathematica, we find the recursion: (n + 1(m + n + 1u n+1 = (h n(h + n + 1u n (n = 0, 1, 2,.... Thus (23 can be easily roved by induction on n.

16 16 ZHI-WEI SUN Lemma 3.2. For, m, n N we have the combinatorial identity m ( ( ( ( m + j 2j j + + m j ( 1 m j 2j j n ( ( ( + m + n + m m =. m m n j=0 (24 Proof. If m < n then both sides of (24 vanish. (24 in the case m = n can be directly verified. Let s m denote the left-h side of (24. By the Zeilberger algorithm we find the recursion (m+1(m n+1s m+1 = (+m+1(+m+n+1s m (m = n, n+1,.... So we can show (24 by induction. Proof of Theorem 3.2. In view of Remar 3.1, it suffices to rove (18. Note that both sides of (18 are olynomials in x y the degrees with resect to x or y are all smaller than a. Fix m, n {0,..., a 1} let c(m, n denote the coefficient of x n y m in the right-h side of (18. Define ( z = 0 for = 1, 2, 3,.... Then c(m, n coincides with ( + j [x n ] w (h 2j 0 j < a w (h a 1 = =m a 1 m = a 1 m = m j=0 w +m (h w +m (h ( 2j j ( 1 m j ( + j 2j m j=0 m j=0 Alying Lemma 3.2 we get ( m + n c(m, n = m ( m + n = m ( j (x + 1 j ( 1 m j x m m j ( ( ( 2j j j m j ( 1 m j ( + m + j 2j ( 1 m j ( m + j 2j a 1 m a 1 m ( 2j j ( 2j j j m + n ( + m j ( + m + j ( ( + m m w +m (h m n ( ( + m n w +m (h. n ( j n ( j. n

17 By Lemma 3.1, CENTRAL BINOMIAL AND TRINOMIAL COEFFICIENTS 17 = ( ( + m n w +m (h n n ( ( + m n w +m (h = w m(hw n (h. n a 1 So, it remains to show ( m + n m a 1 = a m To rove (25 we only need to show ( m+n m ( ( + m n w +m (h 0 (mod 2. (25 n ( ( m + n + m 0 (mod m n under the suosition n a m. Note that m + n + m a 0 < a n + m n m < a. As the addition of m n in base has at least one carry, we have ( m+n m by Kummer s theorem (cf. [Ri,. 24]. Similarly, ( +m n. So far we have comleted the roof of Theorem 3.2. Theorem 3.2 imlies the following useful result on congruences for sums of central binomial coefficients generalized central trinomial coefficients. Theorem 3.3. Let be an odd rime let x be a -adic integer. Let a Z +, b, c Z d = b 2 4c. Set D := 1 + 2bx + dx 2. Then we have a 1 ( a 1 ( 16 T (b, cx ( 32 (1 D + d x ( a 1 32 (1 D d x P ( a 1/2( D + d xp ( a 1/2( D d x (mod 2 (26

18 18 ZHI-WEI SUN a 1 ( 4 2 ( 2( a 1 ( 64 T (b, cx If > 3, then a 1 a 1 ( 4 a 1 ( 4 ( 3 ( a 1 ( 27 T (b, cx ( 6 3 3( ( 432 T (b, cx a 1 ( ( a 1 2 2( (1 D + d x ( (1 D d x (mod ( 6 a 1 ( 6 ( 3 (1 D + d x 54 ( 3 (27 54 (1 D d x (mod 2 3 3( (1 D + d x 864 ( 3 ( (1 D d x (mod 2. (29 Remar 3.2. Note that d D in Theorem 3.3 are viewed as algebraic -adic integers. Proof of Theorem 3.3. Let n = ( a 1/2. For = 0,..., n we have ( n + = 2 ( 16 0<j ( 2a 1 (2j 1 ( 16 (mod 2 hence ( ( ( ( n n + n + 2 = 2 ( 16 (mod 2.

19 CENTRAL BINOMIAL AND TRINOMIAL COEFFICIENTS 19 Note also that ( 2 for = n + 1,..., a 1 by Kummer s theorem. Thus n ( ( ( n n + t 1 P n (t = 2 ( n ( t 1 a 1 ( ( (1 t (mod 2, hence the second congruence in (26 follows. Set D + d x 1 D d x 1 u = v =. 2 2 Then uv + v = D ( d x + 1 = b d x 4 2 u v = d x. In view of (10, for any N we have ( ( + j 2j (uv + v j (u v j 2j j j=0 ( ( ( + j 2j =x b j d d j = x T (b, c. 2j j 2 j=0 So the first congruence in (26 follows from (19. Similarly, (27-(29 are consequences of (20-(22 resectively. For d {2, 3, 4, 7}, it is well nown that an odd rime can be written in the form x 2 + dy 2 with x, y Z if only if ( d = 1 (see, e.g., [BEW] [Co]. Alying (26 we get the following new results. Theorem 3.4. Let be an odd rime. Then T (1, 2 32 Also, 1 1 { ( 2 (4x2 2 (mod 2 if = x 2 + 4y 2, 0 (mod 2 if 3 (mod 4. (30 T { (2, 1 ( 1 4x2 (mod if = x 2 + 2y 2, 8 0 (mod 2 if 5, 7 (mod 8; T (4, 1 ( 4 { 4x 2 (mod if = x 2 + 3y 2, 0 (mod 2 if 2 (mod 3; (31 (32

20 20 ZHI-WEI SUN 1 ( 2 ( 1 T (16, 1 { ( 1 T (1, 16 ( 256 4x 2 (mod if ( = 1 = 7 x2 + 7y 2, 0 (mod 2 if ( = 1, i.e., 3, 5, 6 (mod 7. 7 (33 Remar 3.3. Let be an odd rime. We guess that 4x 2 (mod in (31-(33 can be relaced by 4x 2 2 (mod 2. To rove Theorem 3.4 we need a lemma. Lemma 3.3. Let be an odd rime. Then 1 32 x ( ( 2 x n P n 1 4 x ( 3 x P n (x (2x + 2 n P n 1 + x Proof. With the hel of Lemma 2.1, we get ( n n 256 ( 2(n 32 x 32 x n x = 32 ( 8 ( 2 ( 2 = x n x n (mod (34 (mod. (35 n ( n ( 2 ( 16 x n ( ( ( n n + (1 4/x 1 ( x n P n 1 4 x (mod. 2 ( 2 (8x n This roves (34. (35 follows from [S1, Theorem 2.6] its roof. Proof of Theorem 3.4. For convenience we set n = ( 1/2. (i Alying (26 with b = 1, c = 2 x = 1/2, we obtain that ( ( 32 T (1, 2 (mod ( 16

21 CENTRAL BINOMIAL AND TRINOMIAL COEFFICIENTS 21 The author [Su2, Conjecture 5.5] conjectured that 1 /32 0 (mod 2 if 3 (mod 4, ( 2 32 ( 16 2x 2x (mod 2 if = x 2 + y 2 with x 1 (mod 4 y 0 (mod 2. This was confirmed by Z.-H. Sun [S1]. So the desired (30 follows. (ii Alying (26 with b = 2, c = 1 x = 2 we get T ( (2, 1 n ( n 8 32 α 32 β (mod 2. where α = 4(1 + 2 β = 4(1 2. Clearly αβ = 16. By Lemma 3.3, ( n ( 2 32 α α n P n ( 2 (mod ( n 32 β ( β n P n ( 2 = ( 2 β n P n ( 2 (mod. By [S2, Theorem 2.7], P n ( 2 0 (mod if ( 2 = 1, P n ( 2 ( 1 4x2 (mod if ( 2 = 1 = x2 + 2y 2 (x, y Z. So (31 holds. (iii (26 with b = x = 4 c = 1 yields that 1 ( 4 T (4, 1 P n ( P n ( (mod 2. By Lemma 3.3, ( ( 3 3 (±1 n P n = P n ± 2 2 (2 ± ( 3 n 3 3/2 P n 1 ± = (2 ± 3 n P n ( (mod. 3/2 By [S2, Theorem 2.8], P n ( 3/2 0 (mod if 2 (mod 3, P n ( 3/2 ( 1 n 4x 2 (mod if 1 (mod 3 = x 2 + 3y 2 (x, y Z. Therefore (32 is valid.

22 22 ZHI-WEI SUN (iv Alying (26 with b = 1, c = 16 x = 1/16 we obtain that T ( n ( (1, 16 2 ( n ( α 32 β (mod 2, where α = (1+3 7/16 β = (1 3 7/16. Note that αβ = 1/4. By Lemma 3.3, ( 2 32 α α n P n ( 63 (mod 1 32 β ( β n P n ( 63 = (26 with b = 16, c = 1 x = 16 yields that ( 2 β n P n ( 63 (mod. 1 ( 2 ( 1 T (16, 1 P n ( P n ( (mod 2. By Lemma 3.3, ( (±1 n 3 7 P n (8 ± 3 7 n P n ( (mod. 8 Therefore 1 ( ( 2 2 ( n ( 1 T (16, 1 ( 1 n 3 7 P n (mod. 8 ( 7 By [S2, Theorem 2.5], P n ( 63 P n (3 7/8 0 (mod = 1, P n ( ( 63 ( 1 n 3 7 P n 4x 2 (mod 8 if ( 7 = 1 = x2 + 7y 2 (x, y Z. Therefore (33 holds. Motivated by Theorem 3.4 the congruence 1 ( 3 2 ( B 3 (mod 5 if

23 CENTRAL BINOMIAL AND TRINOMIAL COEFFICIENTS 23 roved in [Su4] (where B 0, B 1,... are Bernoulli numbers, we conjecture that 1 2 T ( (1, (3 + 1( 32 3 ( 1 (mod 2, 1 2 T ( (2, 1 1 (5 + 2( + (mod 2, T (4, 1 (5 + 2( 2 ( ( 1 ( (mod 2, 1 ( ( ( 2 1 ( ( 1 T (16, (mod 2, 1 2 T ( (1, 16 1 (30 + 7( 7 (mod 2. ( 256 The last congruence led the author to find the conjectural identity ( T ( 256 (1, 16 = 24 π in Jan which was the starting oint of the discovery of many series for 1/π of new tyes given in Section Conjectural congruences related to reresentations of rimes by binary quadratic forms In view of (26, our following conjecture imlies that for any rime = x 2 + 7y 2 with x, y Z + we have T ( (1, 16 1 (4x 2 2 (mod 2. ( 256 Conjecture 4.1. Let be an odd rime with ( = 1. Write = 7 x 2 + 7y 2 with x, y Z such that x 1 (mod 4 if 1 (mod 4, y 1 (mod 4 if 3 (mod 4. Then u (1, v (1, 16 { 0 (mod 2 if 1 (mod 4, 1 3 7y 2 if 3 (mod 4; { 2( 2(2x 2x 2 if 1 (mod 4, 0 (mod 2 if 3 (mod 4.

24 24 ZHI-WEI SUN When 1 (mod 4, we have 1 16 u (1, u (1, 16 ( 2 42 ( x 2x 1 2 (4 + 3( 16 v (1, ( 2 3 (4 + 1( 256 v (1, 16 6 x (mod 2. When 3 (mod way: ( u 16 (1, 16 (mod 2 4, we can determine y mod 2 in the following ( u (1, 16 ( 2 2 v 16 (1, 16 ( v (1, 16 ( y 2 (mod 2 ( y 2 (mod 2. Just lie Q( 7, the imaginary quadratic field Q( 11 also has class number one. Let be an odd rime. Whenever ( = 1 we can 11 write in the form (x y 2 /4 with x, y Z. We guess that ( 3 { T (46, 1 x 2 2 (mod 2 if 4 = x y 2, (mod 2 if ( = To attac this we note that (28 with b = 46, c = 1 x = 27/512 yields ( 3 ( ( 3 ( 3 T 512 (46, 1 ( 64 α ( 64 β (mod 2, where α = (1 + 33/2 β = (1 33/2. Observe that 2α = v (1, 8 + (α βu (1, 8 2β = v (1, 8 (α βu (1, 8. So we have ( 3 T (46, ( 1 ( 3 ( 64 v (1, 8 33( 1 ( 3 ( 64 v (1, 8 (mod 2.

25 CENTRAL BINOMIAL AND TRINOMIAL COEFFICIENTS 25 This, together with the author s conjecture on 1 ( 3 /64 mod 2 (cf. [Su2, Conjecture 5.4] leads us to raise the following conjecture. Conjecture 4.2. Let > 3 be a rime. If ( = 1, then 11 1 ( 3 ( 64 u (1, 8 1 ( 3 ( 64 v (1, 8 0 (mod. When ( 11 = 1, 1 (mod 3, 4 = x2 + 11y 2 with x 1 (mod 3, we have ( 3 ( 64 u (1, 8 0 (mod 2, ( ( 2 3 ( 64 u (1, ( 11 x x (mod 2, ( ( 2 3 u 216 (8, 27 4 ( 99 x x (mod 2, ( 3 ( 64 v ( 3 ( (1, 8 v 216 (8, 27 2 x x (mod 2, 1 2 ( 3 ( + 60( ( 64 v (1, 8 60x (mod 2, 1 2 ( 3 (9 + 2( v 216 (8, 27 2x (mod 2. When ( 11 = 1, 2 (mod 3, 4 = x2 + 11y 2 with y 1 (mod 3, we have ( 3 11 ( 64 u (1, 8 ( 3 3 ( 64 v (1, ( y 11y (mod 2,

26 26 ZHI-WEI SUN ( 3 (2 155( ( 64 u (1, y (mod 2, 1 2 ( 3 (2 243( ( 64 v (1, y (mod 2, ( 3 1 ( u (8, u (8, 27 y 11y (mod 2, 1 ( v (8, 27 y 9 (mod 2. Motivated by the author s investigation of 1 ( 3 T (3, 1/27 mod 2 (with > 3 a rime the congruence (28, we ose the following conjecture which involves the well-nown Fibonacci numbers F = u (1, 1 ( N Lucas numbers L = v (1, 1 ( N. Note that the imaginary quadratic field Q( 15 has class number 2. Conjecture 4.3. Let > 5 be a rime. If 1, 4 (mod = x y 2 (x, y Z with x 1 (mod 3, then 1 ( ( 2 3 F ( 3 ( x 2x 27 L 4x x (mod 2 (mod 2, 1 2 ( 3 (3 + 2( L 27 4x (mod If 2, 8 (mod 15 = 3x 2 +5y 2 (x, y Z with y 1 (mod 3, then ( 3 F 27 5y 4y (mod 2 1 ( ( ( ( 2 3 F 27 L y (mod 2.

27 CENTRAL BINOMIAL AND TRINOMIAL COEFFICIENTS 27 Remar 4.1. By [Su8, Theorem 1.6], for any rime > 3 we have ( F 0 (mod 2 if 1 (mod 3, ( 3 27 L 0 (mod 2 if 2 (mod 3. In fact, we have many other conjectures similar to Conjectures ; for the sae of brevity we don t include them in this aer. Conjecture 4.4. Let > 3 be a rime. (i If 1, 4 (mod 15 = x y 2 with x, y Z, then P ( 1/2 (7 15 ± 16 ( 15 ( x ( 3 2x (mod x (ii Suose that ( = ( = 1 write 4 = 5 7 x2 + 35y 2 with x, y Z. If 1 (mod 3, then ( ( ± ( x ( 35 3 x x (mod 2. If 2 (mod 3, then ( ( ± 35 ± 35 ( y 3 ( y 35y (mod 2. (iii If ( 2 = ( 3 = ( 5 = 1 = x2 + 30y 2 with x, y Z, then 1 ( ( ± 5 ( x 3 ( 2x 2x (mod 2. (iv If ( 2 = ( = ( = 1, = 3 7 x2 + 42y 2 with x, y Z, then ( ( ± 2 ( x ( 14 2x (mod x (v If ( 2 = ( 3 = ( 13 = 1 = x2 + 78y 2 with x, y Z, then 1 ( ( ± 5 26 ( x 3 ( 2x 2x (mod 2.

28 28 ZHI-WEI SUN 1 (vi If ( 2 = ( 3 = ( 17 = 1 = x y 2 with x, y Z, then ( ( ±35 2 ( x 3 ( 2x 2x (mod 2. (vii If ( 1 = ( = ( = 1 = 3 11 x2 + 33y 2 with x, y Z, then 4 2 2( ( ( ± 65 ( x ( 3 3 2x 2x (mod 2. Remar 4.2. Let 1, 4 (mod 15 be a rime with = x y 2 (x, y Z. Alying (26 we see that 1 ( 2 ( 1 T P ( 1/2 ( P ( 1/2 ( (mod 2. Thus art (i of Conjecture 4.4 imlies that 1 ( 2 ( 1 T ( 2x 4x 2 2 (mod 2. 2x We omit here similar comments on arts (ii-(vii of Conjecture 4.4. We also have many other conjectures similar to Conjecture 4.4. Conjecture 4.5. Let > 5 be a rime. Then ( 1 T2 (62, 1 ( T2 (62, 1 ( ( x 2 2 (mod 2 if 1, 9 (mod 20 & = x 2 + 5y 2, 2x 2 2 (mod 2 if 3, 7 (mod 20 & 2 = x 2 + 5y 2, 0 (mod 2 if 11, 13, 17, 19 (mod 20. And 1 2 T2 ( (62, 1 1 ( ( 3 ( T2 (62, 1 ( ( ( ( ( ( 1 ( ( (mod 2, (mod 2.

29 CENTRAL BINOMIAL AND TRINOMIAL COEFFICIENTS 29 Conjecture 4.6. Let > 3 be a rime. Then 1 T 2 ( T (10, 1 ( ( 64 3 T 2 (4, T2 (6, 1 T2 (6, x 2 2 (mod 2 if 1, 7 (mod 24, = x 2 + 6y 2, 8x 2 2 (mod 2 if 5, 11 (mod 24, = 2x 2 + 3y 2, 0 (mod 2 if ( 6 = 1; 1 T 2 (6, ( 1 (4x2 2 (mod 2 if 1, 7 (mod 24, = x 2 + 6y 2, 8x 2 2 (mod 2 if 5, 11 (mod 24, = 2x 2 + 3y 2, 0 (mod 2 if ( 6 = 1. Also, 1 2 T (10, 1 (3 + 1( ( T 2 (4 + 1( (6, Conjecture 4.7. Let > 5 be a rime. (i We have ( ( ( ( (mod 2, ( (mod 2. T 2 ( (3, 1 T (34, 1 T2 (18, 1 36 ( x 2 2 (mod 2 if 1, 9, 11, 19 (mod 40, = x y 2, 8x 2 2 (mod 2 if 7, 13, 23, 37 (mod 40, = 2x 2 + 5y 2, 0 (mod 2 if ( 10 = 1.

30 30 ZHI-WEI SUN Also, 1 2 T (3, 1 (16 + 5( 2 5 (mod 2, T ( ( ( (34, ( ( ( 64 Conjecture 4.8. Let > 7 be a rime. Then (mod 2. 1 ( 3 T (18, ( 10 1 ( 3 T3 (6, 1 ( 512 x 2 2 (mod 2 if ( = ( = 1 & 4 = 5 7 x2 + 35y 2, 2 5x 2 (mod 2 if ( = ( = 1 & 4 = 5 7 5x2 + 7y 2, 0 (mod 2 if ( = And 1 2 ( 3 T (18, 1 (35 + 9( T3 (6, 1 (35 + 9( 9 ( ( ( (mod 2, ( ( ( (mod 2. Conjecture 4.9. Let 2, 29 be a rime. When 5, 7, we have 1 2T2 (19602, x 2 2 (mod 2 if ( 2 = ( 29 = 1 & = x2 + 58y 2, 8x 2 2 (mod 2 if ( 2 = ( 29 = 1 & = 2x2 + 29y 2, 0 (mod 2 if ( 58 = 1. Provided 13 we have 1 2T2 (19602, x 2 2 (mod 2 if ( 2 = ( 29 = 1 & = x2 + 58y 2, 2 8x 2 (mod 2 if ( 2 = ( 29 = 1 & = 2x2 + 29y 2, 0 (mod 2 if ( 58 = 1.

31 CENTRAL BINOMIAL AND TRINOMIAL COEFFICIENTS 31 Conjecture Let > 5 be a rime. Then ( 6 ( 3 ( T3 (26, 1 15 ( 3 T3 (62, 1 ( 24 3 ( x 2 2 (mod 2 if ( = ( = 1 & 4 = 7 13 x2 + 91y 2, 2 7x 2 (mod 2 if ( = ( = 1 & 4 = x2 + 13y 2, 0 (mod 2 if ( = And 1 2 ( 3 T3 (26, 1 ( ( ( 24 3 ( 6 ( ( (mod , 1 2 ( 3 T3 (62, 1 ( ( ( ( 105 ( ( (mod Remar 4.3. Note that the imaginary quadratic field Q( d has class number two for d = 5, 6, 10, 15, 35, 58, 91. Conjecture Let > 3 be a rime. We have ( ( T (110, 1 ( x 2 2 (mod 2 if ( 1 = ( = ( = 1, = 3 7 x2 + 21y 2, 12x 2 2 (mod 2 if ( 1 = ( = 1, ( = 1, = 7 3 3x2 + 7y 2, 2x 2 2 (mod 2 if ( 1 = ( = 1, ( = 1, 2 = 3 7 x2 + 21y 2, 6x 2 2 (mod 2 if ( 1 = 1, ( = ( = 1, 2 = 3 7 3x2 + 7y 2, 0 (mod 2 if ( 21 = 1, 1 ( ( 4 2 T (110, 1 ( ( 6 ( ( (mod 2.

32 32 ZHI-WEI SUN Conjecture Let > 3 be a rime. Then 1 T2 (18, x 2 2 (mod 2 if ( 2 = ( = ( = 1, = 3 5 x2 + 30y 2, 12x 2 2 (mod 2 if ( = 1, ( 2 = ( = 1, = 3 5 3x2 + 10y 2, 2 8x 2 (mod 2 if ( 2 = 1, ( = ( = 1, = 3 5 2x2 + 15y 2, 2 6x 2 (mod 2 if ( 5 = 1, ( 2 = ( = 1, 2 = 3 3x2 + 10y 2, 0 (mod 2 if ( 30 = 1. And 1 T2 (30, x 2 2 (mod 2 if ( 2 = ( = ( = 1, = 3 7 x2 + 42y 2, 12x 2 2 (mod 2 if ( 2 = 1, ( = ( = 1, = 3 7 3x2 + 14y 2, 2 8x 2 (mod 2 if ( 2 = 1, ( = ( = 1, = 7 3 2x2 + 21y 2, 2 6x 2 (mod 2 if ( 2 = 1, ( 3 = ( = 1, 2 = 7 3x2 + 14y 2, 0 (mod 2 if ( 42 = 1. Conjecture Let > 3 be a rime. When 13, 17, we have 1 ( 3 T (102, x 2 2 (mod 2 if ( 2 = ( = ( = 1, = 3 13 x2 + 78y 2, 2 8x 2 (mod 2 if ( 2 = 1, ( = ( = 1, = x2 + 39y 2, 12x 2 2 (mod 2 if ( = 1, ( 2 = ( = 1, = x2 + 26y 2, 2 24x 2 (mod 2 if ( 3 = 1, ( 2 = ( = 1, = 13 6x2 + 13y 2, 0 (mod 2 if ( 78 = 1.

33 CENTRAL BINOMIAL AND TRINOMIAL COEFFICIENTS 33 Provided 11, 17, we have 1 T (198, 1 ( x 2 2 (mod 2 if ( 2 = ( = ( = 1, = 3 17 x y 2, 2 8x 2 (mod 2 if ( = 1, ( 2 = ( = 1, = x2 + 51y 2, 12x 2 2 (mod 2 if ( = 1, ( 2 = ( = 1, = x2 + 34y 2, 2 24x 2 (mod 2 if ( 2 = 1, ( = ( = 1, = x2 + 17y 2, 0 (mod 2 if ( 102 = 1. Conjecture Let be an odd rime let m belong to the set {2, 3, 6, 10, 18, 30, 102, 198}. If m, then ( 3 T (m, 1 T2 (m, 1 (mod 2. (36 m If m 2 12 (mod, then T2 ( (m, 1 m ( 4 2 T (m 2 2, 1 (m (mod 2. (37 Remar 4.4. We note that (36 holds mod for any integer m 0 (mod, (37 holds mod for any m Z with m 2 12 (mod. Conjecture Let 2, 5, 19 be a rime. We have 1 2T2 (5778, x 2 2 (mod 2 if ( 2 = ( = ( = 1, = 5 19 x y 2, 8x 2 2 (mod 2 if ( 2 = 1, ( = ( = 1, = x2 + 95y 2, 2 20x 2 (mod 2 if ( 2 = ( = 1, ( = 1, = x2 + 38y 2, 2 40x 2 (mod 2 if ( 2 = ( = 1, ( = 1, = x2 + 19y 2, 0 (mod 2 if ( 190 = 1, 1 2 T2 (5778, 1 ( ( ( ( (mod 95 2.

34 34 ZHI-WEI SUN Provided 17 we have 1 T2 (5778, 1 ( T2 (5778, (mod 2, 1 2 T2 (5778, 1 ( ( ( ( ( (mod Conjecture Let > 5 be a rime. Then T2 (198, 1 ( T2 (322, x 2 2 (mod 2 if ( 2 = ( = ( = 1, = 5 7 x2 + 70y 2, 8x 2 2 (mod 2 if ( = 1, ( 2 = ( = 1, = 7 5 2x2 + 35y 2, 2 20x 2 (mod 2 if ( = 1, ( 2 = ( = 1, = 5 7 5x2 + 14y 2, 28x 2 2 (mod 2 if ( 2 = 1, ( = ( = 1, = 5 7 7x2 + 10y 2, 0 (mod 2 if ( 70 = 1. Also, 1 2T2 (322, 1 ( x 2 2 (mod 2 if ( 1 = ( = ( = 1, = 5 17 x2 + 85y 2, 2 2x 2 (mod 2 if ( 1 = 1, ( = ( = 1, 2 = 17 5 x2 + 85y 2, 2 20x 2 (mod 2 if ( 1 = 1, ( = ( = 1, = x2 + 17y 2, 10x 2 2 (mod 2 if ( 1 = 1, ( 5 = ( = 1, 2 = 17 5x2 + 17y 2, 0 (mod 2 if ( 85 = 1. And 1 2T2 (1298, x 2 2 (mod 2 if ( 2 = ( = ( = 1, = 5 13 x y 2, 8x 2 2 (mod 2 if ( 2 = 1, ( = ( = 1, = x2 + 65y 2, 2 20x 2 (mod 2 if ( 2 = 1, ( = ( = 1, = x2 + 26y 2, 2 40x 2 (mod 2 if ( 2 = 1, ( 13 = ( = 1, = 5 10x2 + 13y 2, 0 (mod 2 if ( 130 = 1.

35 CENTRAL BINOMIAL AND TRINOMIAL COEFFICIENTS 35 Remar The imaginary quadratic field Q( d has class number four for d = 21, 30, 42, 70, 78, 85, 102, 130, 190. Conjecture Let > 3 be a rime. Then 1 1 ( 3 T3 ( 3 ( 3 ( 27 If 13, then 1 ( 3 If 73, then 1 { ( 3 (4x2 2 (mod 2 if = x 2 + 7y 2, 0 (mod 2 if ( = 1; 7 { T3 (26, 81 ( 6 (x2 2 (mod 2 if 4 = x y 2, (mod 2 if ( = 1; 11 T3 (10, T3 (106, ( 3 T3 (586, 1 { ( 6 (x2 2 (mod 2 if 4 = x y 2, 0 (mod 2 if ( 19 = 1. { ( 78 (x2 2 (mod 2 if 4 = x y 2, 0 (mod 2 if ( 43 = 1. { ( 438 (x2 2 (mod 2 if ( = 1 & 4 = 67 x2 + 67y 2, 0 (mod 2 if ( = If 8893, then 1 T3 (71146, 1 Also, ( 3 { ( (x 2 2 (mod 2 if ( 163 x y 2, 0 (mod 2 if ( ( 3 T3 (2, 1 ( 1 ( 3 T3 (2, ( 3456 { ( 2 (x2 2 (mod 2 if 1 (mod 3 & 4 = x y 2, 0 (mod 2 if 2 (mod 3.

36 36 ZHI-WEI SUN And 1 2 ( 3 T3 (2, 1 (15 + 2( ( 3456 { 2( 2 (mod 2 if is a cubic residue mod, 0 (mod otherwise. Remar 4.6. The imaginary quadratic field Q( d has class number one for d = 7, 11, 19, 43, 67, 163. We observe that if > 3 is a rime m is an integer with m(3m (mod then 1 ( 3 T3 (m + 2, 1 (3m 3 ( m(3m ( 3 ( 6 3 ( 9m 24 3 (mod. Conjecture Let be an odd rime. (i When > 5 we have 1 ( T (38, ( 5 1 ( T (38, ( x 2 2 (mod 2 if ( 2 = ( = ( = 1, = 3 5 x2 + 30y 2, 12x 2 2 (mod 2 if ( = 1, ( 2 = ( = 1, = 3 5 3x2 + 10y 2, 2 8x 2 (mod 2 if ( 2 = 1, ( = ( = 1, = 3 5 2x2 + 15y 2, 20x 2 2 (mod 2 if ( 5 = 1, ( 2 = ( = 1, = 3 5x2 + 6y 2, δ,7 (mod 2 if ( 30 = 1. Also, 1 If 7, then 1 ( T 3 (38, 212 ( ( ( (mod 2. T 2 1 (4, 9 ( T (38, (mod ( 16 ( T 2 (4, 9 ( ( ( (mod 2.

37 CENTRAL BINOMIAL AND TRINOMIAL COEFFICIENTS 37 (ii When 7 we have 1 ( T (110, ( 14 1 ( T (110, ( 28 4x 2 2 (mod 2 if ( 2 = ( = ( = 1, = 3 7 x2 + 42y 2, 8x 2 2 (mod 2 if ( 2 = 1, ( = ( = 1, = 7 3 2x2 + 21y 2, 12x 2 2 (mod 2 if ( 2 = 1, ( = ( = 1, = 3 7 3x2 + 14y 2, 24x 2 2 (mod 2 if ( 2 = 1, ( 3 = ( = 1, = 7 6x2 + 7y 2, δ,19 (mod 2 if ( 42 = 1. Also, 1 ( T 3 (110, ( 5160 ( (mod 2. If > 3 11, 19, then 1 T 2 1 (5, 1 ( T (110, (mod ( T 2 (5, 1 ( ( (mod 2. Conjecture Let > 3 be a rime. When 5, we have 1 T 2 (9, x 2 2 (mod 2 if ( 1 = ( = ( = 1, = 3 11 x2 + 33y 2, 2x 2 2 (mod 2 if ( 1 = 1, ( = ( = 1, 2 = 3 11 x2 + 33y 2, 12x 2 2 (mod 2 if ( 1 = 1, ( = ( = 1, = x2 + 11y 2, 6x 2 2 (mod 2 if ( 1 = 1, ( 3 = ( = 1, 2 = 11 3x2 + 11y 2, 0 (mod 2 if ( 33 = 1.

38 38 ZHI-WEI SUN Provided 29, we have 1 T 2 (171, 171 ( x 2 2 (mod 2 if ( 1 = ( = ( = 1, = 7 19 x y 2, 2x 2 2 (mod 2 if ( 1 = 1, ( = ( = 1, 2 = 7 19 x y 2, 2 28x 2 (mod 2 if ( 1 = 1, ( = ( = 1, = x2 + 19y 2, 2 14x 2 (mod 2 if ( 1 = 1, ( = ( = 1, 2 = x2 + 19y 2, 0 (mod 2 if ( 133 = 1. Conjecture Let be an odd rime. When 23, we have 1 T 2 (7, x 2 2 (mod 2 if ( 2 = ( = ( = 1, = 5 7 x2 + 70y 2, 8x 2 2 (mod 2 if ( = 1, ( 2 = ( = 1, = 7 5 2x2 + 35y 2, 20x 2 2 (mod 2 if ( = 1, ( 2 = ( = 1, = 5 7 5x2 + 14y 2, 2 28x 2 (mod 2 if ( 2 = 1, ( = ( = 1, = 5 7 7x2 + 10y 2, 0 (mod 2 if ( 70 = 1. Provided 3, 7, 11, 17, 31, we have 1 T 2 (73, x 2 2 (mod 2 if ( 2 = ( = ( = 1, = 3 17 x y 2, 8x 2 2 (mod 2 if ( = 1, ( 2 = ( = 1, = x2 + 51y 2, 12x 2 2 (mod 2 if ( = 1, ( 2 = ( = 1, = x2 + 34y 2, 24x 2 2 (mod 2 if ( 2 = 1, ( = ( = 1, = x2 + 17y 2, 0 (mod 2 if ( 102 = 1.

39 CENTRAL BINOMIAL AND TRINOMIAL COEFFICIENTS 39 Conjecture Let > 3 be a rime. If 23, then 1 T 2 (23, x 2 2 (mod 2 if ( 2 = ( = ( = 1, = 3 13 x2 + 78y 2, 8x 2 2 (mod 2 if ( 2 = 1, ( = ( = 1, = x2 + 39y 2, 2 12x 2 (mod 2 if ( = 1, ( 2 = ( = 1, = x2 + 26y 2, 2 24x 2 (mod 2 if ( 3 = 1, ( 2 = ( = 1, = 13 6x2 + 13y 2, δ,7 (mod 2 if ( 78 = 1, where δ m,n taes 1 or 0 according as m = n or not. If 5, then 1 T 3 (1298, 6512 ( x 2 2 (mod 2 if ( 2 = ( = ( = 1, = 3 13 x2 + 78y 2, 2 8x 2 (mod 2 if ( 2 = 1, ( = ( = 1, = x2 + 39y 2, 12x 2 2 (mod 2 if ( = 1, ( 2 = ( = 1, = x2 + 26y 2, 2 24x 2 (mod 2 if ( 3 = 1, ( 2 = ( = 1, = 13 6x2 + 13y 2, (δ,7 + δ,31 (mod 2 if ( 78 = 1. Conjecture Let 2, 7, 11 be a rime. Then T 2 (7, 81 ( T 2 (20, x 2 2 (mod 2 if ( 11 = ( 2 = 1 & = x2 + 22y 2, 8x 2 2 (mod 2 if ( 11 = ( 2 = 1 & = 2x2 + 11y 2, 0 (mod 2 if ( 11 = ( 2. Conjecture Let 2, 7 be a rime. Then { T 2 (6, 2 4x 2 2 (mod 2 if ( = 1 & = 7 x2 + 7y 2, (mod 2 if ( = 1, ( T 2 (6, ( ( (mod 2.

40 40 ZHI-WEI SUN Conjecture Let > 3 be a rime. Then T 2 (7, 12 T 2 2 (3, x 2 2 (mod 2 if 1 (mod 12 & = x 2 + 9y 2, 4xy (mod 2 if 5 (mod 12 & = x 2 + y 2 (3 x y, 0 (mod 2 if 3 (mod 4. Conjecture Let be an odd rime. Then T 2 (19, 20 T 2 2 (9, x 2 2 (mod 2 if ( 1 = ( = 1 & = 5 x2 + y 2 (5 x, 4xy (mod 2 if ( 1 = ( = 1 & = 5 x2 + y 2 (5 x y, 0 (mod 2 if 3 (mod 4 & 11. We have many conjectures similar to Conjectures For examle, we find that 1 T 2 (b, c/m mod 2 is related to the reresentation = x 2 + dy 2 if (b, c, m; d is among (5, 4, 4; 10, (3, 4, 36; 13, (5, 4, 14 2 ; 30, (7, 1, 14 2 ; 30, (7, 28, 14 2 ; 21, (11, 49, 22 2 ; 42. Though we will not list many other conjectures similar to Conjectures , the above conjectures should convince the reader that our conjectural series for 1/π in the next section are indeed reasonable in view of the corresonding congruences. 5. Dualities new series for 1/π As mentioned in Section 1, for b > 0 c > 0 the main term of T n (b, c as n + is f n (b, c := (b + 2 c n+1/2 2 4 c. nπ Here we formulate a further refinement of this. Conjecture 5.1. For any ositive real numbers b c, we have ( T n (b, c = f n (b, c 1 + b 4 ( c 1 16n c + O n 2

41 CENTRAL BINOMIAL AND TRINOMIAL COEFFICIENTS 41 as n +. If c > 0 b = 4 c, then T n (b, c c n = T n (4, 1 = 3 ( 6n nπ 8n n + 21 ( n + O. 4 n 5 If c < 0 b R then lim n n Tn (b, c = b 2 4c. Let be an odd rime. Z.-H. Sun [S1] roved the congruence 1 16 x ( 1 16 (1 x (mod 2 (38 via Legendre olynomials; in fact this follows from the well-nown i- dentity P n ( x = ( 1 n P n (x with n = ( 1/2. In [Su8] the author managed to show the following congruences via the Zeilberger algorithm: 1 1 ( 4 2 x 64 ( 3 ( x 27 3 ( 3 ( x ( 2 1 ( ( (1 x (mod 2, (39 ( 3 (1 x (mod 2 ( 3, (40 27 ( 3 ( (1 x (mod 2 ( 3. (41 Our following result on dualities was motivated by (38-(41. Theorem 5.1. Let be an odd rime let b, c m 0 (mod be rational -adic integers. Then 1 1 ( 1 (16m T (b, c ( 4 ( 2 2 (64m T (b, c 1 1 (16m T (m b, c (mod 2, (42 ( 4 2 (64m T (m b, c (mod 2, (43

42 42 ZHI-WEI SUN Provided > 3 we also have ( ( (27m T (b, c 3 ( 3 ( 6 3 (432m T (b, c ( ( 3 (27m T (m b, c (mod 2, (44 ( 3 ( 6 3 (432m T (m b, c (mod 2. (45 Proof. Since the roofs of (42-(45 are very similar, we just show (43 in detail. For d = 0,..., 1, by taing differentiations of both sides (39 d times we get ( 4 ( 2 64 ( d x d In view of this, we have ( 4 2 (64m T (b, c 1 = 1 = j=0 1 = = j=0 ( 2 ( 2 ( 4 2 (64m ( j c j j m 2j ( j c j j m 2j 1 1 /2 j=0 ( ( 2 ( 4 2 (64m ( 4 2 ( 1 d 64 ( ( 2j b 2j c j 2j j ( 4 ( ( 2 b 64 2j m 1 ( /2 j=0 ( 2j 2j d ( ( 1 b 2j 2j m ( j (m b 2j c j j ( 4 2 (64m T (m b, c (mod 2. The roof of Theorem 5.1 is now comlete. (1 x d (mod 2. Examle 1. Let be an odd rime. By (42 we have ( 1 T (5, 4 T ( 4, 4 3 = ( 8 (mod 2.

43 CENTRAL BINOMIAL AND TRINOMIAL COEFFICIENTS 43 The author [Su4] conjectured that 3 { ( 8 4x 2 2 (mod 2 if = x 2 + y 2 (2 x, 0 (mod 2 if 3 (mod 4, this was recently confirmed by Z.-H. Sun [S2]. When > 3, by (44 we have ( 3 1 T (3, 1 ( 3 ( T ( 2, 1 = 27 ( 3 ( 27 (mod 2 ; the reader may consult [Su4, Conjecture 5.6] for 1 ( 3 /( 27 mod 2. Based on our investigations of congruences on sums of central binomial coefficients central trinomial coefficients, the author s hilosohy about series for 1/π stated in [Su7], we raise 61 conjectural series for 1/π of the following seven new tyes with a, b, c, d, m integers abcd(b 2 4cm nonzero. Tye I. (a + d( 2 2T (b, c/m. Tye II. (a + d( ( 2 3 T (b, c/m. Tye III. (a + d( 4 2 2( T (b, c/m. Tye IV. (a + d( 2 2T2 (b, c/m. Tye V. (a + d( ( 2 3 T3 (b, c/m. Tye VI. (a + dt (b, c 3 /m. Tye VII. (a + d( 2 T (b, c /m. Recall that a series a is said to converge at a geometric rate with ratio r if lim + a +1 /a = r (0, 1. All the series in Conjectures I-VII below converge at geometrical rates, they were found by the author in 2011 excet that (IV19-(IV21 were discovered in Conjecture I. We have the following identities: ( T ( 256 (1, 16 = 24 π, ( T ( 1024 (34, 1 = 12 π, (I1 (I2

44 44 ZHI-WEI SUN ( 2 T (194, 1 = 80 π, (I3 T (62, 1 = 16 3 π. (I4 Remar 5.1. (I1 was the first identity for 1/π involving generalized central trinomial coefficients; it was discovered by the author on Jan. 2, Different from classical Ramanujan-tye series for 1/π (cf. N. D. Baruah B. C. Berndt [BB], Berndt [Be, ] their nown generalizations (see, e.g., S. Cooer [C], the two coefficients in the linear art 30 1 of (I3 have different signs, also its corresonding -adic congruence (with > 3 a rime involves two Legendre symbols: 1 2 (30 1( T ( (194, ( 1 6 ( 3 (mod 2. Conjecture II. We have ( 3 ( 3 ( 3 ( ( 3 ( 3 ( ( 3 T (18, 6 = π, T (10, 1 = π, T (198, 1 = (II1 (II2 2π, (II3 T (970, 1 = 525 3, (II4 π T (730, 729 = 25 3 π, (II5 T (102, 1 = , (II6 2π

45 CENTRAL BINOMIAL AND TRINOMIAL COEFFICIENTS 45 ( ( ( ( ( ( T (198, 1 = , (II7 4π T (39202, 1 = , (II8 π ( ( 3 T (287298, 1 = , (II9 π ( ( 3 T (26, 729 = 8 3π ( , (II10 ( ( 3 T (70, 3645 = , (II11 6π ( ( 3 T (40, 1458 = 25 12π ( (II12 Remar In view of (44, we may view (II9 as the dual of (II7 since / = The series in (II7 converges raidly at a geometric rate with ratio 25/35937, but the series in (II9 converges very slow at a geometric rate with ratio 71825/ (II2, (II9 (II10 were motivated by the following congruences (with > 3 a rime: ( T (10, 1 { 4x 2 2 (mod 2 if 1, 3 (mod 8 & = x 2 + 2y 2, 0 (mod 2 if ( 2 = 1, i.e., 5, 7 (mod 8, 1 2 ( 3 T (10, 1 ( ( ( ( (mod 2 ( 5;

46 46 ZHI-WEI SUN 1 2 ( 3 T (287298, 1 ( ( ( ( (mod 2 ( 11; 1 ( ( T 24 3 (26, 729 ( ( ( Conjecture III. We have the following formulae: ( ( ( ( ( ( ( ( ( ( (mod 2. T (52, 1 = 33 33, (III1 π T (110, 1 = 3 6 π, (III2 T (98, 1 = π, (III3 T (257, 256 = π, (III4 ( ( 4 2 T (7, 4096 = , 2 8π (III5 ( ( 4 2 T (322, 1 = 126 7, 2 π (III6 ( ( 4 2 T (1442, 1 = π, (III7 ( ( 4 2 T (898, 1 = , 2 4π (III8 ( ( 4 2 T (12098, 1 = , 2 π (III9 ( ( 4 2 T (10402, 1 = , (III10 4π

47 CENTRAL BINOMIAL AND TRINOMIAL COEFFICIENTS ( 4 2 ( 4 2 ( ( 4 2 ( ( T ( , 1 = , π (III11 T (39202, 1 = , π (III12 T ( , 1 = π (III13 Remar 5.3. (III12 (III13 are dual in view of (43. Other dual airs include (III6 (III7, (III8 (III9, (III10 (III11. Below are the corresonding -adic congruences for (III1 (III13 (with > 3 a rime: 4 2 (85 + 2( T (52, ( ( ( (mod 2 ( 11, 1 ( 4 2 ( ( T ( , ( ( ( (mod 2 ( 19, 43. Conjecture IV. We have ( T 2(7, 1 = 48 ( π, (IV1 ( T 2(62, 1 = 120 ( π, (IV2 ( T 2(322, 1 = 4320 ( π, (IV3

48 48 ZHI-WEI SUN ( T 2(10, 1 = π, (IV4 ( T 2(38, 1 = π, (IV5 ( T 2(198, 1 = , (IV π ( T 2(18, 1 = 12 15, (IV π ( T 2(30, 1 = π, (IV8 ( T 2(110, 1 = π, (IV9 ( T 2(322, 1 = π, (IV10 ( T 2(198, 1 = π, (IV11 ( T 2(102, 1 = π, (IV12 ( T 2(1298, 1 = , (IV π ( T 2(1298, 1 = , (IV π ( T 2(4898, 1 = , (IV π ( T 2(5778, 1 = π, (IV16 ( T 2(5778, 1 = , (IV π ( T 2(54758, 1 = , (IV π

49 CENTRAL BINOMIAL AND TRINOMIAL COEFFICIENTS ( ( 2 T 2(10, 2 = 12 6 π, (IV19 ( 2 T 2(238, 14 = 288 2, π (IV20 ( 2 T 2(9918, 19 = π. (IV21 Remar 5.4. For (IV6, Mathematica indicates that if we set n ( s(n := T 2(198, then s(15 π < s(30 π < Here are corresonding -adic congruences of (IV9-(IV11 (IV18 with > 5 a rime: 1 2 T2 (110, 1 (56 + 3( ( ( (mod , 1 2 T2 (322, 1 (56 + 5( ( ( (mod , 1 2 T2 (198, 1 (10 + 1( ( ( ( 13 1 (mod ( 7, 1 2 T2 (54758, 1 ( ( ( ( (mod ( 13. For any rime > 3, the corresonding -adic congruence of (IV19 is 1 ( ( ( ( T 2(10, 2 6 (mod

50 50 ZHI-WEI SUN Conjecture V. We have the formula ( ( 3 T 3 (62, 1 = (V1 π Remar 5.5. (V1 was motivated by Conjecture 4.10; the series converges at a geometric rate with ratio 64/125. We conjecture that (IV1-(IV18 have exhausted all identities of the form 2 T2 (b, 1 (a + d( = C m π with a, d, m Z, b {1, 3, 4,...}, d > 0, C 2 rational. This comes from our following hyothesis motivated by (16 in the case h = 2 the author s hilosohy about series for 1/π stated in [Su7]. We have alied the hyothesis to see for series for 1/π of tye IV checed all those b = 1,..., 10 6 via comuter. Hyothesis 5.1 (i Suose that a + d m ( 2 T 2(b, 1 = C π with a, d, m Z, b Z + C 2 Q \ {0}. Then m is an integer dividing 16(b 2 4. Also, b = 7 or b 2 (mod 4. (ii Let ε {±1}, b, m Z + m 16(b 2 4. Then, there are a, d Z such that ( a + d 2 T 2(b, 1 = C (εm 2 π for some C 0 with C 2 rational, if only if m > 4(b T2 (b, 1 (εm 2 ( ε(b T2 (b, 1 (ε m 2 (mod 2 for all odd rimes b 2 4, where m = 16(b 2 4/m.

51 CENTRAL BINOMIAL AND TRINOMIAL COEFFICIENTS 51 Conjecture VI. We have the following formulae: ( T 3 (10, 11 2 = π, ( 80 T 3 3 (22, 21 2 = π, ( T 3 (62, 95 2 = π ( (VI1 (VI2 (VI3 Remar 5.6. The series (VI1-(VI3 converge at geometric rates with ratios 16 27, , resectively. The author would lie to offer $300 as the rize for the erson (not joint authors who can rovide first rigorous roofs of all the three identities (VI1-(VI3. (VI1 (VI3 were motivated by the author s following conjectural congruences for any rime > 3: 1 { T 3(10, 112 ( 2 (4x2 2 (mod 2 if = x 2 + 3y 2, ( (mod 2 if 2 (mod 3, ( T 3 (10, 11 2 ( ( 2 8 ( 6 (mod 2 ; 1 T 3 (62, 952 ( { ( 2 (4x2 2 (mod 2 if = x 2 + 7y 2, 0 (mod 2 if ( = 1, T ( (62, 95 2 ( ( ( 2 (mod 2.

52 52 ZHI-WEI SUN Conjecture VII. We have the following formulae: ( T (6, 2 = π, (VII1 ( T (4, 9 = 49 9π ( 3 + 6, (VII2 ( T (5, 1 = π, (VII3 ( T (7, 1 = , (VII4 2π ( T ( (3, 3 = 9 7 π, (VII ( ( T 2 (171, 171 = , 2π (VII6 ( T 2 (73, 576 = π. (VII7 Remar 5.7. The series (VII1-(VII7 converge at geometric rates with ratios , , , , 7 9, , resectively. The author found (VII2 (VII3 in light of Conjecture Similarly, (VII6-(VII7 were motivated by Conjectures Concerning the new identities in Conjectures I-VII, actually we first discovered congruences without linear arts related to binary quadratic forms (lie many congruences in Section 4, then found corresonding -adic congruences with linear arts, finally figured out the series for 1/π. Acnowledgments. The wor was suorted by the National Natural Science Foundation (grant of China, the initial version of this aer was osted to arxiv in Jan as a rerint with the ID arxiv: The rerint version of this aer available from arxiv has stimulated some others to wor on our conjectural series for 1/π of tyes I-V in Section 5.

53 CENTRAL BINOMIAL AND TRINOMIAL COEFFICIENTS 53 References [AKP] G. Almvist, C. Krattenthaler J. Petersson, Some new formulas for π, Exeriment. Math. 12 (2003, [BB] N. D. Baruah B. C. Berndt, Eisenstein series Ramanujan-tye series for 1/π, Ramanujan J. 23 (2010, [Be] B. C. Berndt, Ramanujan s Noteboos, Part IV, Sringer, New Yor, [BEW] B. C. Berndt, R. J. Evans K. S. Williams, Gauss Jacobi Sums, John Wiley & Sons, [CHV] J. S. Caughman, C. R. Haithcoc J. J. P. Veerman, A note on lattice chains Delannoy numbers, Discrete Math. 308 (2008, [CCL] H. H. Chan, S. H. Chan Z.-G. Liu, Domb s numbers Ramanujan- Sato tye series for 1/π, Adv. in Math. 186 (2004, [CC] H. H. Chan S. Cooer, Rational analogues of Ramanujan s series for [CDE] 1/π, Math. Proc. Cambridge Philos. Soc. 153 (2012, S. Chowla, B. Dwor R. J. Evans, On the mod 2 determination of ( ( 1/2 ( 1/4, J. Number Theory 24 (1986, [ChCh] D. V. Chudnovsy G. V. Chudnovsy, Aroximations comlex multilication according to Ramanujan, in: Ramanujan Revisited: Proc. of the Centenary Confer. (Urbana-Chamaign, ILL., 1987, (eds., G. E. Andrews, B. C. Berndt R. A. Ranin, Academic Press, Boston, MA, 1988, [C] S. Cooer, Soradic sequences, modular forms new series for 1/π, Ramanujan J. 29 (2012, [Co] D. A. Cox, Primes of the Form x 2 + ny 2, John Wiley & Sons, [HW] R. H. Hudson K. S. Williams, Binomial coefficients Jacobi sums, Trans. Amer. Math. Soc. 281 (1984, [I] T. Ishiawa, Suer congruence for the Aéry numbers, Nagoya Math. J. [IR] [L] [M1] [M2] [N] 118 (1990, K. Irel M. Rosen, A Classical Introduction to Modern Number Theory, 2nd Edition, Sringer, New Yor, L. Long, Hyergeometric evaluation identities suercongruences, Pacific J. Math. 249 (2011, E. Mortenson, Suercongruences between truncated 2 F 1 by geometric functions their Gaussian analogs, Trans. Amer. Math. Soc. 355 (2003, E. Mortenson, A -adic suercongruence conjecture of van Hamme, Proc. Amer. Math. Soc. 136 (2008, T. D. Noe, On the divisibility of generalized central trinomial coefficients, J. Integer Seq. 9 (2006, Article , 12. [PWZ] M. Petovše, H. S. Wilf D. Zeilberger, A = B, A K Peters, Wellesley, [R] S. Ramanujan, Modular equations aroximations to π, Quart. J. Math. (Oxford 45 (1914, [Ri] [RV] P. Ribenboim, The Boo of Prime Number Records, Second Edition, Sringer, New Yor, F. Rodriguez-Villegas, Hyergeometric families of Calabi-Yau manifolds, in: Calabi-Yau Varieties Mirror Symmetry (Toronto, ON, 2001,.