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Numerator of I(n) = sqrt(10)*(Integral_{x=0 to 1/3} 1/(1+x^2)^(n+1/2) dx).
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%I #12 Feb 06 2019 02:21:06

%S 1,29,1403,95115,8298105,885611805,111797745795,16298030927115,

%T 2694941727973425,498439798319375325,101970858789466224075,

%U 22865056868419298361675,5576927510911134523293225

%N Numerator of I(n) = sqrt(10)*(Integral_{x=0 to 1/3} 1/(1+x^2)^(n+1/2) dx).

%C The denominator is b(n) = 10^(n-1)*(2*n)!/(n!*2^n).

%H G. C. Greubel, <a href="/A091994/b091994.txt">Table of n, a(n) for n = 1..100</a>

%e The third term is 1403 since I(3) = 1403/1500.

%t Table[ Sqrt[10]*10^(n - 1)*(2*n)!/(n!*2^n)*Integrate[1/(1 + x^2)^(n + 1/2), {x, 0, 1/3}], {n, 14}] (* _Robert G. Wilson v_, Apr 23 2004 *)

%K nonn

%O 1,2

%A Al Hakanson (hawkuu(AT)excite.com), Mar 17 2004

%E Edited by _Robert G. Wilson v_, Apr 23 2004