Bilateral sums related to Ramanujan-like series

BILATERAL SUMS RELATED TO RAMANUJAN-LIKE SERIES

arXiv:1610.04839v1 [math.NT] 16 Oct 2016

´ GUILLERA JESUS

Abstract. Abstract. We define bilateral series related to Ramanujan-like series for 1/π 2 . Then, we conjecture a property of them and give some applications.

1. Introduction In [5] we constructed bilateral summations related to Ramanujan-type series for 1/π and applied them to prove a new kind of identities, which we called “the upside-down” counterpart. In this paper we consider bilateral sums related to Ramanujan-like series for 1/π 2 , conjecture a property of them and give some applications. We will need the following theorems Theorem 1. Suppose that f (x) is the sum over all integers n of g(n + x). Then, clearly f (x) is a periodic function of period x = 1. Theorem 2. Let f : C −→ C be a holomorphic function such that f (x) = O(ecπ|Im(x)| ), where c ≥ 0 is a constant, and suppose that f (x) admits a Fourier series: f (x) =

+∞ X

an e2πinx .

n=−∞

Then |2n| > c ⇒ an = 0.

Proof. It is known that the coefficients of a Fourier series are given by Z a+1+it an = f (x)e−2πinx dx. a+it

Then for n < 0 let t → +∞ and for n ≥ 0 let t → −∞.

2. Ramanujan-like series for 1/π 2 Let s0 = 1/2 , s3 = 1 − s1 , s4 = 1 − s2 . We recall that a Ramanujan-like series for 1/π 2 is a series of the form " 4 # ∞ X Y (si )n 1 (1) (a + bn + cn2 )z n = 2 , (1)n π n=0 i=0

where z, a, b and c are algebraic numbers and the possible couples (s1 , s2 ) are (1/2, 1/2), (1/2, 1/3), (1/2, 1/4), (1/2, 1/6), (1/3, 1/3), (1/3, 1/4), (1/3, 1/6), (1/4, 1/4), (1/4, 1/6), (1/6, 1/6), (1/5, 2/5), (1/8, 3/8), (1/10, 3/10), (1/12, 5/12). Up to date 11 convergent and 6 “divergent” formulas are known in this new family (see all in the Appendix). The value c (2) τ=√ , 1−z 1

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plays an important roll in the theory [2, eq. 3.47]. 3. Bilateral series related to Ramanujan-like series for 1/π 2 We introduce the function (3) f (x) =

4 Y cos πx − cos πsi X

1 − cos πsi

i=0

(−1)n

n∈Z

" 4 # Y (si )n+x i=0

(1)n+x

[a + b(n + x) + c(n + x)2](−z)n+x .

Applying Theorem 1, we see that it is a periodic function of period x = 1 because cos πx − cos πs0 = cos πx and (−1)n cos πx = cos π(n + x). Besides it is analytic due to the factor with the cosenus, which cancels the poles at x = −si and therefore all the other poles due to the periodicity of f (x). On the other hand we conjecture that f (x) = O(e5π|Im(x)| ). Hence by Theorem 2 we deduce that (4)

4 Y cos πx − cos πsi X i=0

(−1)

n

"

4 Y (si )n+x

#

[a + b(n + x) + c(n + x)2 ](−z)n+x

1 − cos πsi n∈Z (1)n+x i=0 1 = 2 u1 cos 2πx + u2 cos 4πx + (1 − u1 − u2 ) + v1 sin 2πx + v2 sin 4πx , π

where u1 , u2 , v1 , v2 do not depend on x, and we can determine their values by giving values to x. We conjecture that for the bilateral series corresponding to a Ramanujan-like series for 1/π 2 (that is, when z, a, b, c are algebraic numbers), the values of u1 , u2 , v1 and v2 are rational numbers. In addition, for the case of alternating series we conjecture that v1 = v2 = 0. These bilateral series are “divergent”, because the sum to the left or the sum to the right “diverges”. However we turn them convergent by analytic continuation. We can write (4) in the following way: (5)

∞ X

"

4 Y (si + x)n

#

[a + b(n + x) + c(n + x)2 ](−z)n+x (1 + x) n i=0 n=0 " 4 # ∞ Y (1 − x)n a + b(−n + x) + c(−n + x)2 X + x5 (−1)n (−z)−n+x = 5 (s − x) (−n + x) i n i=0 n=1 # " 4 1 Y (1)x u1 cos 2πx + u2 cos 4πx + (1 − u1 − u2 ) + v1 sin 2πx + v2 sin 4πx , π 2 i=0 (si )x csc2 πs1 csc2 πs2 cos πx(cos2 πx − cos2 πs1 )(cos2 πx − cos2 πs2 ) (−1)n

where we have splitted the summation in two sums and used the properties 1 (1)−n+x

= (−1)n+1

(1 − x)n x , (1)x n − x

(si )−n+x = (−1)n

(si )x . (1 − si − x)n

As the second sum is O(x5 ), formula (5) implies an expansion of the form # " 4 ∞ X Y (si + x)n 1 x2 x4 [a + b(n + x) + c(n + x)2 ](−z)n+x = 2 − k + jπ 2 + O(x5 ), (−1)n (1 + x)n π 2! 4! n=0 i=0


if z < 0 (alternating series), and an expansion of the form " 4 # ∞ X Y (si + x)n 1 x2 x4 [a + b(n + x) + c(n + x)2 ]z n+x = 2 − k + jπ 2 + O(x5 ), (1 + x)n π 2! 4! n=0 i=0

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if 0 < z < 1 (convergent series of positive terms). Hence, the conjecture that the coefficients of the Fourier expansion are rational is equivalent to the conjecture that k and j are rational stated in [2, 1]. The relation of τ with k and j is [2, eq. 3.48]:

k 2 5k j + + + 1 + (cot2 πs1 )(cot2 πs2 ) + (1 + k)(cot2 πs1 + cot2 πs2 ). 12 4 3 From (5) we deduce that # " 4 # " 4 ∞ X Y (si )n X Y (si )n (a + bn + cn2 )z n = (a + bn + cn2 )z n , (6) (1) (1) n n n=0 i=0 n∈Z i=0 τ2 =

and as the function (3) has period x = 1, we see that for all integer x, we have " 4 # Y (si )n+x X (7) (−1)n [a + b(n + x) + c(n + x)2 ](−z)n+x (1) n+x i=0 n∈Z " 4 # ∞ X Y (si )n = (−1)x (a + bn + cn2 )z n . (1) n n=0 i=0

If z > 0, we can write # " 4 Y (si )n+x X [a + b(n + x) + c(n + x)2 ](−z)n+x (8) (−1)n (1) n+x i=0 n∈Z " 4 # X Y (s ) i n+x = e−iπx [a + b(n + x) + c(n + x)2 ]z n+x . (1) n+x n∈Z i=0 Taking into account (7) and (8), we have written the following Maple procedure: bilater:=proc(s1,s2,z0,x,j) local p,s0,s3,s4,z1,h,hh,r,u,v; p:=(a,b)->pochhammer(a,b): s0:=1/2; s3:=1-s1; s4:=1-s2: h:=n->p(s0,n)*p(s1,n)*p(s2,n)*p(s3,n)*p(s4,n)/p(1,n)^5: if type(x,integer) then; hh:=n->h(n)*n^j: r:=evalf((-1)^x*subs(z=z0,sum(hh(n)*z^n,n=0..infinity))): return r; else; if z0(-1)^n*h(n+x)*(n+x)^j; else z1:=-z0; hh:=(j,n)->exp(-I*Pi*x)*h(n+x)*(n+x)^j; fi; u:=j->subs(z=z1,sum(hh(j,n)*(-z)^(n+x),n=0..infinity)): v:=j->subs(z=z1,sum(hh(j,-n)*(-z)^(-n+x),n=1..infinity)): r:=evalf(u(j)+v(j)): return r: fi; end:

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This procedure calculates the bilateral sum " 4 # Y (si )n+x X (−1)n (−z)n+x (n + x)j , (1) n+x i=0 n∈Z

and helps the reader to check the examples.

4. Examples of bilateral series We give some examples of bilateral series related to Ramanujan-like series for 1/π 2 . Example 1. Formula (25) in the Appendix is ∞

(9)

( 1 )5 20n2 + 8n + 1 1X 1 (−1)n 2 5n = 2. n 8 n=0 (1)n 4 π

Taking v1 = v2 = 0 and giving two values to x we get numerical approximations of u1 , u2 . We observe that these values look rational and do not change using other values of x. Hence, we conjecture that 1 5 2 1 − 21 cos 2πx + 12 cos 4πx 1X n ( 2 )n+x 20(n + x) + 8(n + x) + 1 (10) (−1) = , 8 n∈Z (1)5n+x 4n+x π 2 cos5 πx

after replacing u1 , u2 with their guessed rational values.

Example 2. Formula (32) in the Appendix is 1 2 1 5 6n ∞ 1 X 3 384 n 2 n 3 n 3 n 6 n 6 n . (1930n2 + 549n + 45) = (−1) 5 (1)n 4 π n=0 From it we conjecture the bilateral form 1 1 2 1 X n 2 n+x 3 n+x 3 n+x (−1) (11) 384 n∈Z (1)5n+x

1 6 n+x

× 1930(n + x)2 + 549(n + x) + 45 =

after identifying the coefficients.

5 6 n+x

6(n+x) 3 4

11 − 14 cos 2πx + 6 cos 4πx , π 2 cos πx (4 cos2 πx − 1)(4 cos2 πx − 3)

Example 3. From the following formula: 3 ∞ 1 3 1 2 X 32 2 n 4 n 4 n 120n + 34n + 3 (12) = , 5 n 2 (1) 16 π n n=0

which is (24) in the Appendix, we get the bilateral form n+x 1 3 1 3 1 X 2 n+x 4 n+x 4 n+x 1 (120(n + x)2 + 34(n + x) + 3) = 32 n∈Z (1)5n+x 16 eiπx

3 − 27 cos 2πx + 32 cos 4πx + 21 sin 2πx − 21 sin 4πx i , π 2 cos3 πx (2 cos2 πx − 1)


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after identifying the coefficients of the Fourier expansion from their numerical approximations. Observe that v1 and v2 are not null because the series in (12) is of positive terms. Observe also the factor eiπx , which comes from (−1)n /(−16)n+x Example 4. For the formula (29) in the Appendix we have the bilateral identity √ 1 1 3 5 7 7 X 2 n+x 8 n+x 8 n+x 8 n+x 8 n+x 1 (1920(n + x)2 + 304(n + x) + 15) 392 n∈Z (1)5n+x 74(n+x) 23 5 3 29 − 79 cos 2πx + cos 4πx + sin 2πx − sin 4πx i iπx 2 2 2 2 , =e π 3π π 2 2 2 2 2 2 2 π csc 8 csc 8 cos πx cos πx − cos 8 cos πx − cos 3π 8

after identifying the coefficients.

Example 5. Looking at the formula (31) in the Appendix which has z = (3/φ)3 , we had the intuition that another series with the other sign of the square root, namely z = −(3φ)3 , could exist and we were right because we discovered it by using the PSLQ algorithm. Here we recover it in a different way by writing the bilateral identity (13)

X n∈Z

1 3 2 n+x

1 3 n+x (1)5n+x

2 3 n+x

(−1)n (3φ)3(n+x) (c(n + x)2 + b(n + x) + a) = 3 u1 cos 2πx + u2 cos 4πx + (1 − u1 − u2 ) . 4π 2 cos3 πx cos2 πx − 14

Giving five values to x we get numerical approximations of a, b, c, u1 and u2 , which we could identify: (14)

u1 =

17 , 36

u2 =

3 , 16

c = 2408 +

216 , φ

b = 1800 +

162 , φ

a = 333 +

30 . φ

Replacing these values and taking x = 0, we arrive at the “divergent” formula (38), in the second list of the Appendix. Example 6. From the “divergent” series (36) in the Appendix, namely 2 ∞ 1 3 1 X 6 2 n 3 n 3 n (28n2 + 18n + 3)(−1)n 33n = 2 , 5 (1)n π n=0 we have the following bilateral sum evaluation: (15)

1 X ( 21 )3n+x ( 31 )n+x ( 23 )n+x 28(n + x)2 + 18(n + x) + 3 (−1)n 27n+x 5 6 n∈Z (1)n+x =

5 + 4 cos 2πx + 3 cos 4πx , 4π 2 cos3 πx(4 cos2 πx − 1)

after identifying the coefficients of the Fourier terminating expansion from their numerical approximations obtained giving values to x.

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5. Applications of the bilateral series If |z| > 1, then letting x → −1/2 or x → −s1 , etc in (5), we obtain the evaluation of some convergent series. In general, taking the limit of (5) as x → −sj , where j ∈ {0, 1, 2, 3, 4}, we deduce the following identity: " 4 # ∞ X Y (sj )n (16) a + b(−n − sj ) + c(−n − sj )2 z −n = (si + sj )n n=0 i=0 # " 4 sj Y (1)x u1 cos 2πx + u2 cos 4πx + (1 − u1 − u2 ) + v1 sin 2πx + v2 sin 4πx (−z) lim . 2 x→−s π (si )x csc2 πs1 csc2 πs2 cos πx(cos2 πx − cos2 πs1 )(cos2 πx − cos2 πs2 ) j i=0 Another application is explained in [5], but only proved for bilateral sums related to Ramanujan-type series for 1/π.

Example 7. From the formula (15), and taking sj = 1/2 in (16), we get n ∞ 1 5 X 3 28n2 + 10n + 1 −1 2 n (17) , = 3 1 5 6n + 1 27 π (1) n 6 n 6 n n=0

which is a new convergent series for 1/π. In the same way, from the ‘divergent” series (13-14), we get the convergent evaluation p 3n 216 54 3 ∞ 2 1 5 X (2408+ )n +(608 + )n + (35 + ) φ3 −1 3 φ φ φ 2 n (18) , = 5 6n + 1 3φ π (1)3n 61 6 n=0 n

n

where φ is the fifth power of the golden ratio.

Example 8. From (10), and letting sj → −1/2 in (16), we recover formula (34) in the Appendix. We observe that a formula with z implies another one with z −1 in the family s1 = s2 = 1/2, and we will refer to this property as “duality”. 6. The mirror map All the parameters of the bilateral sums related to Ramanujan-like series for 1/π 2 are algebraic, but the value of q (related to z by the mirror map) is not. Now we will define a function w(x) in which q is easily related to the coefficients of the Fourier expansion. First, we let u(x) and v(x) be the analytic functions " 4 # 4 Y (si )n+x Y cos πx − cos πsi X n (−1) (−z)n+x , u(x) = 1 − cos πsi n∈Z (1)n+x i=0 i=0

and

v(x) =

4 Y cos πx − cos πsi X i=0

1 − cos πsi

n∈Z

Then, we define the analytic function

(−1)

n

" 4 # Y (si )n+x i=0

(1)n+x

(n + x)(−z)n+x .

w(x) = v(0)u(x) − u(0)v(x).

We have the following Fourier expansion:

w(x) = a1 cos 2πx + a2 cos 4πx + (1 − a1 − a2 ) + b1 sin 2πx + b2 sin 4πx,


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The above identity for the function w(x) implies an expansion of the form (19)

" 4 # ∞ X Y (si )n n=0

i=0

(1)n

−

nz n

∞ X

(−1)n

n=0

" 4 # ∞ X Y (si )n n=0

i=0

(1)n

" 4 # Y (si )n+x

(−z)n+x

(1)n+x " 4 # ∞ Y (si )n+x X n n z (−1) (n + x)(−z)n+x (1)n+x i=0 n=0 i=0

= p1 x + p2 x2 + p3 x3 + p4 x4 + O(x5 )

if z < 0 (alternating series), and an expansion of the form " 4 # # " 4 ∞ ∞ X Y (si )n+x X Y (si )n nz n z n+x (20) (1) (1) n n+x n=0 i=0 n=0 i=0 " " 4 # # ∞ ∞ 4 X Y (si )n n X Y (si )n+x − z (n + x)z n+x (1) (1) n n+x n=0 i=0 n=0 i=0

= p1 x + p2 x2 + p3 x3 + p4 x4 + O(x5 )

if z > 0 and z 6= 1 (series of positive terms). From the differential equations for Calabi-Yau threefolds [8, 2, 1, 10], we deduce that −

1 p2 = t, π p1

q = −e−πt

or

q = e−πt

for (19) and (20) respectively, where z = z(q) is the mirror map, and 1 p4 5 2 2 − − cot πs1 − cot πs2 . k=2 π 2 p2 3 The method used in [1] is good for convergent Ramanujan series, but not when the series is “too divergent”. However the method based on bilateral series permits to calculate q with many digits in all the cases, when we know the value of z. Example 9. Taking s1 = 1/2, s2 = 1/3 and z = 27φ−3 (formula (31) in the Appendix), and using the formulas in [1], and q = e−πt , we get t1 = t(27φ−3) = 3.619403396730928522140860042453285904901. However for z = (−3φ)3 (formula (38) in the Appendix), we need to use the method based on bilateral sums. Then, we get that the value of t corresponding to (38), is with 40 correct digits (we can calculate many more) equal to t2 = t(−27φ3 ) = 1.233412165189594043723756118628903242841. √ √ We get the values of τ from (2), and they are τ1 = 4/3 · 10 and τ2 = 1/9 · 10. We thought that the values of t1 and t2 would be easily related but we have not succeeded in finding any relation.

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Example 10. For s1 = s2 = 1/2, the series with z and with z −1 satisfy duality. For the formulas (23) and (35) in the Appendix, we have t1 = t(−2−10 ) = 4.412809109031200738238212268698423552548, t2 = t(−210 ) = 1.252302434231184754606061614044396018505. √ √ We get τ1 = τ (−2−10 ) = 41 and τ2 = 1/16 · 41 from (2), and we observe that √ 41 + 3 t1 t2 = 2 √ . 41 − 3 Hence the values of t1 and t2 are nicely related. 7. Conclusion We have conjectured that the terminating Fourier expansion of the function (3) corresponding to a Ramanujan-like series for 1/π 2 has rational coefficients. A challenging problem is to prove it. That is, to obtain rigourosly the bilateral form corresponding to any Ramanujan-like series for 1/π 2 . This would suppose to give another step towards understanding the family of formulas for 1/π 2 . Below, we show how to solve the problem for the Ramanujan-type series for 1/π. Consider, for example the Ramanujan series 1 3 ∞ 1 X 3528 n 2 n 4 n 4 n 21460n + 1123 (−1) (21) = . 3 2n (1) 882 π n n=0

To solve the problem of finding the bilateral form we need to determine the value of u in 1 1 3 X 1 − u + u cos 2πx 2 n+x 4 n+x 4 n+x 21360(n + x) + 1123 (22) (−1)n = . 3 2(n+x) 2 πx − 1) (1) 3528 · 882 π cos πx(2 cos n+x n∈Z

Expanding the right side, comparing with [2, Expansion 1.1], and using [2, eq. 2.30-2.32], we rigorously obtain that u = 10, which is a rational number. 8. Appendix In the following two lists φ means the fifth power of the golden ratio. That is √ !5 1+ 5 φ= . 2 ?

The notation = means that the formula remains unproved, and the notation “ = ” means that we get the equality by analytic continuation. 8.1. List of convergent formulas. ∞ 1 5 n X 128 2 n (−1) (820n2 + 180n + 13) = 2 , (23) 5 10n (1)n 2 π n=0 (24)

∞ X n=0

1 3 2 n

1 4 n (1)5n

3 4 n

1 32 (120n2 + 34n + 3) = 2 , 4n 2 π


1 5 2 n (1)5n n=0

∞ X

(25)

(26)

∞ X

1 2 n

n=0

∞ X

(27)

1 4 n

1 2 n

n=0

(28)

∞ X

1 2 n

n=0

∞ X

(29)

n=0

1 2 n

(31)

n=0

(32)

∞ X n=0

2 3 n (1)5n

2 3 n (1)5n

1 4 n

3 8 n (1)5n

1 3 n (1)5n

5 6 n

5 8 n

√ (−1)n ? 256 3 2 (1640n + 278n + 15) = , 210n π2 3 4 n

1 6 n

1 3 2 n

5 6 n

(−1)n ? 48 (252n2 + 63n + 5) = 2 , n 48 π

√ (−1)n 128 5 ? (5418n2 + 693n + 29) = , 3n 2 80 π

7 8 n

√ 1 ? 56 7 2 (1920n + 304n + 15) = , 74n π2

3n 3 48 (74n2 + 27n + 3) = 2 , 4 π

2 3 n

2 3n 1 3 1 216 2 3 162 30 ? 2 n 3 n 3 n (32− )n +(18− )n+(3− ) = (1)5n φ φ φ φ 1 2 n

∞ X

(33)

n=0

∞ X

1 3 n

1 8 n

∞ X

(30)

1 6 n

1 3 n

(−1)n 8 (20n2 + 8n + 1) = 2 , 2n 2 π

3 4 n (1)5n

n=0

1 3 n

1 2 n

2 3 n (1)5n

1 6 n

1 3 n

2 3 n (1)5n

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5 6 n

1 6 n

3 , π2

6n 3 ? 384 (−1) (1930n2 + 549n + 45) = 2 , 4 π n

6n 3 ? 375 (532n2 + 126n + 9) = 2 . 5 4π

5 6 n

8.2. List of “divergent” formulas. ∞ 1 5 X 4 2 n (34) (10n2 + 6n + 1)(−1)n 4n “ = ” 2 , 5 (1)n π n=0 1 5 2 n (205n2 5 (1) n n=0

∞ X

(35)

∞ X

(36)

n=0

(37)

∞ X n=0

1 2 n

1 3 n

1 3 2 n

1 3 n (1)5n

2 3 n (1)5n

1 4 n

16 , π2

+ 160n + 32)(−1)n 210n “ = ”

2 3 n

3 4 n

?

(28n2 + 18n + 3)(−1)n 33n “ = ”

2

(172n + 75n + 9)(−1)

n

27 16

6 , π2

n

“=”

48 , π2

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(38)

∞ X n=0

(39)

2 1 3 1 216 2 162 30 2 n 3 n 3 n 3n (−3φ) (2408+ )n +(1800+ )n+(333+ ) (1)5n φ φ φ

∞ X n=0

1 2 n

1 5 n

2 5 n (1)5n

3 5 n

4 5 n

2

(483n + 245n + 30)(−1)

n

55 28

n

?

“=”

?

“=”

36 , π2

80 . π2

Formulas (23, 24, 25, 30, 34), were proved by the author using the WZ-method. The upside-down of (37) was proved in [7, Example 42], and we can extract from it the WZpair that we need to prove (37). All the other formulas are conjectured. The conjectured formula (33) is joint with Gert Almkvist. In [1] we recovered all the known convergent series for 1/π 2 and also two “divergent” ones: (34) and (37). However, the method used in [1] does not allow to discover “too divergent” series such as (35) or (38). We can check that the mosaic supercongruences pattern [3] hold for the formula (38). Inspired by these congruences, by the conjectured formulas [6, Conj. 1.1–1.6] and by [5], we observe that 3n ∞ X (1)5n −1 (2408 + 216φ−1)n2 − (1800 + 162φ−1)n + (333 + 30φ−1 ) 1 3 1 2 3φ n5 n=0 2 n 3 n 3 n ? 1125 √ (40) 5L5 (3) − 448ζ(3), = 4 seems true. As the convergence is fast, we can use it to get many digits of L5 (3). We discovered the formula (39) very recently. Related to it are the Zudilin-type supercongruences [9, 4]: 2 3 4 5 n p−1 1 1 X 5 ? 2 n 5 n 5 n 5 n 5 n 2 n (483n + 245n + 30)(−1) ≡ 30p3 (mod p5 ), (41) 5 8 (1)n 2 n=0 for primes p > 2, and the “upside-down” counterpart [5], namely 8 n ∞ X (1)5n 2 −483n2 + 245n − 30 ? n 1 2 3 4 (42) (−1) = 896 ζ(3), 1 5 5 n 5 n=1 2 n 5 n 5 n 5 n 5 n

which is a convergent series for ζ(3).

References [1] G. Almkvist and J. Guillera, Ramanujan-like series for 1/π 2 and String theory, Exp. Math., 21, (2012), 223-234. (eprint arXiv:1009.5202). [2] J. Guillera, A matrix form of Ramanujan-type series for 1/π; in Gems in Experimental Mathematics T. Amdeberhan, L.A. Medina, and V.H. Moll (eds.), Contemp. Math. 517 (2010), Amer. Math. Soc., 189–206; (eprint arXiv:0907.1547). [3] J. Guillera, Mosaic supercongruences of Ramanujan-type, Exp. Math. 21, (2012), 65-68. (e-print arXiv:1007.2290). [4] J. Guillera and W. Zudilin, “Divergent” Ramanujan-type supercongruences. Proc. Amer. Math. Soc. 140 (2012), 765–777. (e-print arXiv:1004.4337). [5] J. Guillera and M. Rogers, Ramanujan series upside-down. Journal of the Australian Math. Soc. 97, 78–106; (e-print arXiv:1206.3981). [6] Zhi-Wei Sun, List of conjectural formulas for powers of π and other constants. (arXiv:1102.5649). [7] W. Chu and W. Zhang Accelerating Dougall’s 5 F4 -sum and infinite series involving π, Journal: Math. Comp. 83 (2014), 475–512.


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[8] Y. Yang and W. Zudilin, On Sp4 modularity of Picard–Fuchs differential equations for Calabi–Yau threefolds, (with an appendix by V. Pasol); in Gems in Experimental Mathematics T. Amdeberhan, L.A. Medina, and V.H. Moll (eds.), Contemp. Math. 517 (2010), Amer. Math. Soc., 381–413; (e-print arXiv:0803.3322). [9] W. Zudilin, Ramanujan-type supercongruences. J. Number Theory 129:8 (2009), 1848–1857; (eprint arXiv:0805.2788). [10] W. Zudilin, Arithmetic hypergeometric series. Russian Math. Surveys 66:2 (2011), 369–420. Russian version in Uspekhi Mat. Nauk 66:2 (2011), 163–216; available at the author’s web site. Univ. de Zaragoza (Spain) E-mail address: [email protected]