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%I #21 Feb 07 2024 01:35:24
%S 1,28,421,26528,2148803,7878956,2765513941,74668877408,3808112752813,
%T 651187280816108,2511722368895123,173308843453994432,
%U 7798897955430811787,1895132203169713822916,54958833891921780540589
%N Numerators of partial sums of series for 3*arctanh(1/3) = (3/2)*log(2).
%C Denominators are given in A096950.
%C From the series of log((1+x)/(1-x)) for x = 1/3, i.e., for log(2) = 2*Sum_{k>=0} (1/3)^(2*k+1)/(2*k+1).
%F a(n) = numerator(A(n)) with the rational number A(n) := Sum_{k=0..n} (1/3)^(2*k)/(2*k+1) in lowest terms.
%F (3/2)*log(2) = a(n)/A096950(n) + 3*Integral_{x >= 3} 1/(x^(2*n+4) - x^(2*n+2)) dx. - _Peter Bala_, Feb 05 2024
%e n=3: 26528/A096950(3) = 26528/25515 = 1.0397021... approximates 3*arctanh(1/3) = 1.039720771...
%e n = 3: Sum_{k = 0..3} (1/3)^(2*k + 1)/(2*k + 1) = 1/3 + 1/81 + 1/1215 + 1/15309 = 26528/76545. Hence a(3) = 26528. - _Peter Bala_, Feb 06 2024
%Y Cf. A096950.
%K nonn,easy,frac
%O 0,2
%A _Wolfdieter Lang_, Jul 16 2004