Citation: Chris R. Varney, Farida A. Selim. Color centers in YAG[J]. AIMS Materials Science, 2015, 2(4): 560-572. doi: 10.3934/matersci.2015.4.560
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In 1997, Van Hamme [19,(H.2)] proved the following supercongruence: for any prime
$ (p−1)/2∑k=0(12)3kk!3≡0(modp2), $ | (1.1) |
where
$ mp−1∑k=0(12)3kk!3≡0(modp2). $ | (1.2) |
The first purpose of this paper is to prove the following
Theorem 1.1. Let
$ mn−1∑k=0(1+q4k+1)(q2;q4)3k(1+q)(q4;q4)3kqk≡0(modΦn(q)2), $ | (1.3) |
$ [5pt]mn+(n−1)/2∑k=0(1+q4k+1)(q2;q4)3k(1+q)(q4;q4)3kqk≡0(modΦn(q)2). $ | (1.4) |
Here and in what follows, the
$ Φn(q)=∏1≤k≤ngcd(n,k)=1(q−ζk), $ |
where
The
In 2016, Swisher [18,(H.3) with
$ (p2−1)/2∑k=0(12)3kk!3≡p2(modp5), $ | (1.5) |
The second purpose of this paper is to prove the following
Theorem 1.2. Let
$ (n2−1)/2∑k=0(1+q4k+1)(q2;q4)3k(1+q)(q4;q4)3kqk≡[n2]q2(q3;q4)(n2−1)/2(q5;q4)(n2−1)/2q(1−n2)/2, $ | (1.6) |
$ [5pt]n2−1∑k=0(1+q4k+1)(q2;q4)3k(1+q)(q4;q4)3kqk≡[n2]q2(q3;q4)(n2−1)/2(q5;q4)(n2−1)/2q(1−n2)/2. $ | (1.7) |
Let
$ limq→1(q3;q4)(p2−1)/2(q5;q4)(p2−1)/2=(p2−1)/2∏k=14k−14k+1=(34)(p2−1)/2(54)(p2−1)/2. $ |
Therefore, we obtain the following conclusion.
Corollary 1. Let
$ (p2−1)/2∑k=0(12)3kk!3≡p2(34)(p2−1)/2(54)(p2−1)/2(modp4), $ | (1.8) |
$ [5pt]p2−1∑k=0(12)3kk!3≡p2(34)(p2−1)/2(54)(p2−1)/2(modp4). $ | (1.9) |
Comparing (1.5) and (1.8), we would like to propose the following conjecture, which was recently confirmed by Wang and Pan [20].
Conjecture 1. Let
$ (p2r−1)/2∏k=14k−14k+1≡1(modp2). $ | (1.10) |
Note that the
We need to use Watson's terminating
$ 8ϕ7[a,qa12,−qa12,b,c,d,e,q−na12,−a12,aq/b,aq/c,aq/d,aq/e,aqn+1;q,a2qn+2bcde]=(aq;q)n(aq/de;q)n(aq/d;q)n(aq/e;q)n4ϕ3[aq/bc, d, e, q−naq/b,aq/c,deq−n/a;q,q], $ | (2.1) |
where the basic hypergeometric
$ r+1ϕr[a1,a2,…,ar+1b1,…,br;q,z]:=∞∑k=0(a1;q)k(a2;q)k…(ar+1;q)k(q;q)k(b1;q)k⋯(br;q)kzk. $ |
The left-hand side of (1.4) with
$ 8ϕ7[q2,q5,−q5,q2,q,q2,q4+(4m+2)n,q2−(4m+2)nq,−q,q4,q5,q4,q2−(4m+2)n,q4+(4m+2)n;q4,q]. $ | (2.2) |
By Watson's transformation formula (2.1) with
$ (q6;q4)mn+(n−1)/2(q−(4m+2)n;q4)mn+(n−1)/2(q4;q4)mn+(n−1)/2(q2−(4m+2)n;q4)mn+(n−1)/2×4ϕ3[q3, q2,q4+(4m+2)n, q2−(4m+2)nq4,q5,q6;q4,q4]. $ | (2.3) |
It is not difficult to see that there are exactly
$ \frac{(q^3;q^4)_k (q^2;q^4)_k(q^{4+(4m+2)n};q^4)_k(q^{2-(4m+2)n};q^4)_k} {(q^4;q^4)_k^2(q^5;q^4)_k(q^6;q^4)_k}q^{4k} $ |
in the
It is easy to see that
The author and Zudilin [11,Theorem 1.1] proved that, for any positive odd integer
$ (n−1)/2∑k=0(1+q4k+1)(q2;q4)3k(1+q)(q4;q4)3kqk≡[n]q2(q3;q4)(n−1)/2(q5;q4)(n−1)/2q(1−n)/2(modΦn(q)2), $ | (3.1) |
which is also true when the sum on the left-hand side of (3.1) is over
It is easy to see that, for
$ \dfrac{[n^2]_{q^2}(q^3;q^4)_{(n^2-1)/2}} {(q^5;q^4)_{(n^2-1)/2}} q^{(1-n^2)/2}\equiv 0\pmod{\Phi_n(q)^2} $ |
because
Swisher's (H.3) conjecture also indicates that, for positive integer
$ (p2r−1)/2∑k=0(12)3kk!3≡p2r(modp2r+3). $ | (4.1) |
Motivated by (4.1), we shall give the following generalization of Theorem 1.2.
Theorem 4.1. Let
$ (n2r−1)/2∑k=0(1+q4k+1)(q2;q4)3k(1+q)(q4;q4)3kqk≡[n2r]q2(q3;q4)(n2r−1)/2(q5;q4)(n2r−1)/2q(1−n2r)/2, $ | (4.2) |
$ [5pt]n2r−1∑k=0(1+q4k+1)(q2;q4)3k(1+q)(q4;q4)3kqk≡[n2r]q2(q3;q4)(n2r−1)/2(q5;q4)(n2r−1)/2q(1−n2r)/2. $ | (4.3) |
Proof. Replacing
$ [n2r]q2(q3;q4)(n2r−1)/2(q5;q4)(n2r−1)/2q(1−n2r)/2≡0(modr∏j=1Φn2j−1(q)2). $ |
Further, by Theorem 1.1, we can easily deduce that the left-hand sides of (4.2) and (4.3) are also congruent to
Letting
Corollary 2. Let
$ (p2r−1)/2∑k=0(12)3kk!3≡p2r(34)(p2r−1)/2(54)(p2r−1)/2(modp2r+2), $ | (4.4) |
$ [5pt]p2r−1∑k=0(12)3kk!3≡p2r(34)(p2r−1)/2(54)(p2r−1)/2(modp2r+2). $ | (4.5) |
In light of (1.10), the supercongruence (4.4) implies that (4.1) holds modulo
It is known that
Conjecture 2 (Guo and Zudilin). Let
$ mn−1∑k=0(q;q2)2k(q2;q4)k(q2;q2)2k(q4;q4)kq2k≡0(modΦn(q)2),mn+(n−1)/2∑k=0(q;q2)2k(q2;q4)k(q2;q2)2k(q4;q4)kq2k≡0(modΦn(q)2). $ | (4.6) |
The author and Zudilin [10,Theorem 2] themselves have proved (4.6) for the
Conjecture 3. Let
$ (n2−1)/2∑k=0(q;q2)2k(q2;q4)k(q2;q2)2k(q4;q4)kq2k≡[n2](q3;q4)(n2−1)/2(q5;q4)(n2−1)/2,n2−1∑k=0(q;q2)2k(q2;q4)k(q2;q2)2k(q4;q4)kq2k≡[n2](q3;q4)(n2−1)/2(q5;q4)(n2−1)/2. $ |
There are similar such new
The author is grateful to the two anonymous referees for their careful readings of this paper.
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