Re: [math-fun] e^(pi rt 163) =
mrob> Your sequence appears in another series sum (but with non-alternating signs) in a 2003 paper (He and Jejjala, "Modular Matrix Models", arXiv:hep-th/0307293v2 , page 20 ) Possibly also related is the discussion at: mathoverflow.net/questions/4775/why-are-powers-of-exppisqrt163-almost-integers(near the end: David Speyer, Kevin Buzzard, and "FC") which appears to use the same power series. ----------------- Yow, I hope this merely reflects your deft googlemanship rather than your idea of fun reading. Weird! Their series is q + 744 + 196884/q + 167975456/q^2 + 180592706130/q^3 + 217940004309744/q^4 + 282054965806724344/q^5 + ... which they apparently get from 1/InverseSeries[1728*KleinInvariantJ[Log[q]/2/Pi/I]+O[q]^9] Plugging in q->-640320^3, both series seem to work!? Maybe just for this q? Maybe because they actually diverge? Maybe because this Mma kernel has (certifiably) lost its mind? Intriguingly, the paper gives a more direct formula as a modular transformation of a ratio of two 2F1s, but I can't get it to work. (Authors seem to underutilize the to-TeX features of their algebra systems.) Anyway, the inverse reciprocal J series is clearly what I was hoping for. My greedily calculated one was running out of convergence. Reciprocating now to q->E^(-Pi*Sqrt[163]), we have singular value #163 as K(k_163) = K((Theta2(q)/Theta3(q))^4) = K(1/4 (2 - 209 Sqrt[163]) + (50 (889 Sqrt[3] + 209 Sqrt[163]))/(1 + 557403 Sqrt[489])^(1/3) + (5 (889 Sqrt[3] - 209 Sqrt[163]) (1 + 557403 Sqrt[489])^(1/3))/5336) = Pi/2*EllipticTheta[3, 0, q]^2 So everything is made out of thetas, and the thetas are all made out of Eta[q] and Eta[q^2]. Recall the "Beegner" formula Eta[E^(-2*Sqrt[ 163]*Pi)]^8== Product[Gamma[k/ 163]^(2*JacobiSymbol[k, 163]),{k,166}]/(32*a* 163*Pi^2) where a is the real root of 3*a^2* 163*KleinInvariantJ[((Sqrt[163]*I + 1)/2)]^(1/3)*(-1)^(2/3) + 2*a^3== 163^3 i.e. a -> 163/ 2 (53360 + 1/3 (4102147072512054 - 30099762 Sqrt[489])^(1/3) + (2 (75965686528001 + 557403 Sqrt[489]))^(1/3)) (probably simplifiable) and Eta[E^(-Sqrt[163]*Pi)]/Eta[E^(-2*Sqrt[163]*Pi)] is algebraic: ((Eta[E^(-Sqrt[163]*Pi)]/Eta[E^(-2*Sqrt[163]*Pi)])^-8= 4270934400 - 334525400 Sqrt[163] + 72963522596556133540/(2492979348824909037130456032026 + 7295439987840023985 Sqrt[3] - 195265212657128312396161027135 Sqrt[163] - 571422960773571522 Sqrt[489])^( 1/3) - (5714944154168505080 Sqrt[ 163])/(2492979348824909037130456032026 + 7295439987840023985 Sqrt[3] - 195265212657128312396161027135 Sqrt[163] - 571422960773571522 Sqrt[489])^(1/3) + 1/2 (2492979348824909037130456032026 + 7295439987840023985 Sqrt[3] - 195265212657128312396161027135 Sqrt[163] - 571422960773571522 Sqrt[489])^(1/3), probably simplifiable.) Just for yucks, the "closed" form is EllipticEta[E^(-2 Sqrt[163]*Pi)]^4-> -(4096 2^(129/163) (-20 - (200 2^(2/3))/(40027 + 1881 Sqrt[489])^( 1/3) + (40027 + 1881 Sqrt[489])^(1/3)/2^(2/3)) Pi^17 G[1/ 326] G[9/326] G[21/326] G[25/326] G[33/326] G[41/326] G[49/ 326] G[53/326] G[57/326] G[61/326] G[65/326] G[69/326] G[77/ 326] G[81/326] G[43/163] G[46/163] G[47/163] G[51/163] G[54/ 163] G[55/163] G[56/163] G[58/163] G[60/163] G[167/326] G[179/ 326] G[199/326] G[227/326])/(3 G[-(1/163)] G[3/326] G[5/326] G[ 7/326] G[11/326] G[13/326] G[17/326] G[19/326] G[23/326] G[27/ 326] G[29/326] G[31/326] G[37/326] G[45/326] G[59/326] G[63/ 326] G[67/326] G[73/326] G[75/326] G[79/326] G[42/163] G[89/ 326] G[99/326] G[50/163] G[101/326] G[103/326] G[105/326] G[107/ 326] G[109/326] G[117/326] G[123/326] G[125/326] G[127/326] G[ 129/326] G[66/163] G[137/326] G[139/326] G[141/326] G[72/163] G[ 147/326] G[149/326] G[153/326] G[157/326] G[159/326] G[175/ 326] G[183/326] G[191/326] G[193/326] G[207/326] G[211/326] G[ 215/326] G[231/326] G[233/326] G[239/326] G[241/326] G[243/ 326] G[128/163] G[130/163] G[138/163] G[287/326] G[305/326] G[ 154/163] G[311/326])/.G->Gamma --rwg
Yes, merely search skills and nothing more -- and it isn't even a full match, they had 217940004309744 and you had 217940004309743. But I did find a real match too, a link to which is now in the OEIS entry. Alas, learning the math behind this would take me months, so I can't help much more... (-: On Friday, December 24, 2010, Bill Gosper <billgosper@gmail.com> wrote:
mrob>
----------------- Yow, I hope this merely reflects your deft googlemanship rather than your idea of fun reading. Weird! Their series is q + 744 + 196884/q + 167975456/q^2 + 180592706130/q^3 + 217940004309744/q^4 + 282054965806724344/q^5 + ...
which they apparently get from [...]
-- Robert Munafo -- mrob.com Follow me at: mrob27.wordpress.com - twitter.com/mrob_27 - youtube.com/user/mrob143 - rilybot.blogspot.com
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