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a(n) = (6*n+1) * (2*n)!^3 / n!^6.
0

%I #18 Oct 24 2024 05:00:38

%S 1,56,2808,152000,8575000,496093248,29188893888,1738242215424,

%T 104455598247000,6321316756040000,384702925005146176,

%U 23520160000755565056,1443504313932496274368,88879637239345064000000,5487711609457595160000000,339644002672064899081728000,21065385579274083203741943000

%N a(n) = (6*n+1) * (2*n)!^3 / n!^6.

%H S. Ramanujan, <a href="http://ramanujan.sirinudi.org/Volumes/published/ram06.pdf">Modular equations and approximations to Pi</a>, Quarterly Journal of Mathematics, XLV, 1914, p. 45.

%F a(n) = (6*n+1) * A002897(n).

%F a(n) ~ 3*2^(6*n+1)/sqrt(n*Pi^3). - _Stefano Spezia_, Oct 17 2024

%F D-finite with recurrence n^3*a(n) +8*(56*n^3-252*n^2+330*n-141)*a(n-1) -4096*(2*n-3)^3*a(n-2)=0. - _R. J. Mathar_, Oct 24 2024

%e G.f.: A(x) = 1 + 56*x + 2808*x^2 + 152000*x^3 + 8575000*x^4 + 496093248*x^5 + 29188893888*x^6 + 1738242215424*x^7 + ...

%e where

%e A(x) = 1 + 7*(1/2)^3*64*x + 13*((1*3)/(2*4))^3*64^2*x^2 + 19*((1*3*5)/(2*4*6))^3*64^3*x^3 + 25*((1*3*5*7)/(2*4*6*8))^3*64^4*x^4 + ... + (6*n+1)*(2*n)!^3/n!^6*x^n + ...

%e SPECIFIC VALUES.

%e At x = 1/256 we have the series

%e 4/Pi = 1 + 7*(1/2)^3/4 + 13*((1*3)/(2*4))^3/4^2 + 19*((1*3*5)/(2*4*6))^3/4^3 + 25*((1*3*5*7)/(2*4*6*8))^3/4^4 + ... = 1.273239544735162686...

%e see formula 28 in the Ramanujan link for details.

%t a[n_]:=(6*n+1) * (2*n)!^3 / n!^6; Array[a,17,0] (* _Stefano Spezia_, Oct 17 2024 *)

%Y Cf. A002897.

%K nonn

%O 0,2

%A _Paul D. Hanna_, Oct 17 2024