Here's another batch of formulas for 1/pi. This is a pretty eclectic subject, so I'm splitting off a separate interest group for this sort of thing. I imagine it will simply be a list of a half dozen email addresses. Drop me a note if you want in. Rich rcs@xmission.com ----- Forwarded message from zwsun@nju.edu.cn ----- Date: Sun, 6 Feb 2011 13:44:41 -0600 From: Zhi-Wei SUN <zwsun@nju.edu.cn> Reply-To: Number Theory List <NMBRTHRY@listserv.nodak.edu>, Zhi-Wei SUN <zwsun@nju.edu.cn> Subject: A bunch of conjectural series for 1/pi To: NMBRTHRY@listserv.nodak.edu Dear number theorists, In 1914 Ramanujan recorded 17 Ramanujan-type series for 1/pi (which are all hyper-geometric). [Actually the identity Sum_{k=0,1,2,...}(4k+1)binom(-1/2,k)^3 = 2/pi on Ramanujan's list was first obtained by G. Bauer in 1859.] This is one of the main achievements of Ramanujan. Recall that for integers b and c we let T_n(b,c) denote the coefficient of x^n in the expansion of (x^2+bx+c)^n. In several messages sent to Number Theory List in 2011, I announced five conjectural series for 1/pi involving T_n(b,c) with bc and b^2-4c nonzero. As Ramanujan found nearly twenty series for 1/pi, I felt pressure in the past days, and now I'm glad that I can add 8 more new conjectural series for 1/pi so that I have a dozen of conjectural identities for 1/pi of the new type. Here is a list of the 8 new identities for 1/pi. CONJECTURE (Zhi-Wei Sun, Feb. 5-6, 2011). We have (1) Sum_{k=0,1,...}(340k+59)binom(2k,k)^2*T_{2k}(62,1)/(-480^2)^k = 120/pi, (2) Sum_{k=0,1,...}(13940k+1559)binom(2k,k)^2*T_{2k}(322,1)/(-5760^2)^k = 4320/pi, (3) Sum_{k=0,1,...}(10k+1)binom(2k,k)^2*T_{2k}(38,1)/240^{2k} = 15*sqrt(6)/(4*pi), (4) Sum_{k=0,1,...}(120k+13)binom(2k,k)^2*T_{2k}(18,1)/320^{2k} = 12*sqrt(15)/pi, (5) Sum_{k=0,1,...}(21k+2)binom(2k,k)^2*T_{2k}(30,1)/896^{2k} = 5*sqrt(7)/(2*pi), (6) Sum_{k=0,1,...}(56k+5)binom(2k,k)^2*T_{2k}(322,1)/48^{4k} = 72*sqrt(7)/(5*pi), (7) Sum_{k=0,1,...}(195k+14)binom(2k,k)^2*T_{2k}(102,1)/10400^{2k} = 85*sqrt(39)/(12*pi), (8) Sum_{k=0,1,...}(14280k+899)binom(2k,k)^2*T_{2k}(198,1)/39200^{2k} = 1155*sqrt(6)/pi. Note that all the series (1)-(8) converge at geometric rates with ratios -64/225, -81/1600, 4/9, 1/16, 1/49, 81/256, 1/625, 1/2401 respectively. In particular, the 8th series converges very fast. If we denote by r(n) the partial sum in (8) with k from 0 to n divided by the right-hand side of the identity, then via Mathematica we see that |r(15)-1|<10^{-50} and |r(30)-1|<10^{-100}. I include all my announced series for 1/pi and their related congruences in my preprint (arXiv:1101.0600) which will be polished later. Also, in my opinion, there should be some (but very few) series for 1/pi of the forms Sum_{k=0,1,...}(a+d*k)binom(6k,3k)binom(3k,k)T_k(b.c)/m^k and Sum_{k=0,1,...}(a+d*k)binom(2k,k)binom(3k,k)T_{3k}(b.c)/m^k, with a,b,c,d,m integers and m*bcd(b^2-4c) nonzero. I have not yet found such series for 1/pi. It seems that all known methods used to establish Ramanujan-type series do not work for my conjectural series which are of a more sophisticated type. In view of the 8+5 conjectural series for 1/pi, I hope that my conjectures could interest some mathematicians and stimulate the birth of a new tool or a new theory. For convenience, here I also collect the 5 announced identities for 1/pi conjectured by me. (9) Sum_{k=0,1,...}(30k+7)binom(2k,k)^2*T_k(1,16)/(-256)^k = 24/pi, (10) Sum_{k=0,1...}(15k+2)binom(2k,k)binom(3k,k)*T_k(18,6)/972^k = 45*sqrt(3)/(4*pi), (11) Sum_{k=0,1,...}(40k+3)binom(4k,2k)binom(2k,k)T_k(98,1)/112^{2k} = 70*sqrt(21)/(9*pi), (12) Sum_{k=0,1,...}(26k+5)binom(2k,k)^2*T_{2k}(7,1)/(-48^2)^k = 48/(5*pi), (13) Sum_{k=0,1,...}(8k+1)binom(2k,k)^2*T_{2k}(10,1)/96^{2k} = 10*sqrt(2)/(3*pi). Also, I'd like to mention that my following open conjectural identities announed in 2010 are still open. (14) Sum_{k>0}(10k-3)8^k/(k^3*binom(2k,k)^2*binom(3k,k)) = pi^2/2, (15) Sum_{k>0}(35k-8)81^k/(k^3*binom(2k,k)^2*binom(4k,2k)) = 12*pi^2, (16) Sum_{k>0}(28k^2-18k+3)(-64)^k/(k^5*binom(2k,k)^4*binom(3k,k)) = -14*zeta(3), (17) Sum_{k>0}(15k-4)(-27)^{k-1}/(k^3*binom(2k,k)^2*binom(3k,k)) = K, where K:=Sum_{k>0}(k/3)/k^2, (18) Sum_{k>0}(5k-1)(-144)^k/(k^3*binom(2k,k)^2*binom(4k,2k)) = -45K/2. Finally, I beg your pardon if you are not interested in my recent messages on series for powers of pi. Any comments are welcome! Zhi-Wei Sun http://math.nju.edu.cn/~zwsun ----- End forwarded message -----