This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A096952 #14 Apr 29 2023 23:01:35
%S A096952 1,7,11,463,4039,35839,320503,575267,25854247,232557151,298927153,
%T A096952 18830313487,6778577311,1525146340543,13726182847159,123535108753519,
%U A096952 1111813831298023,2001263178349523,90056808665990167,810511140554958031
%N A096952 Numerators of upper bounds for Lagrange remainder in Taylor's expansion of log((1+x)/(1-x)) for x=1/3, multiplied by 6/5.
%C A096952 An upper bound for the Lagrange remainder in the expansion of log((1+x)/(1-x)) for x=1/3, i.e., for log(2), is R(2*n) = (1/2^(2*n+1) + 1/3^(2*n+1))/(2*n+1), n >= 0.
%C A096952 The denominators are found in A096953.
%C A096952 log(2) = 2*Sum_{k>=1} ((1/3)^(2*k-1))/(2*k-1), from the Taylor series of log((1+x)/(1-x)) for x=1/3.
%D A096952 M. Barner and F. Flohr, Analysis I, de Gruyter, 5. Auflage, 2000; p. 293.
%H A096952 W. Lang, <a href="/A096952/a096952.txt">More comments</a>.
%F A096952 a(n) = numerator(A(n)), where A(n) = (6/5)*(1/2^(2*n+1) + 1/3^(2*n+1))/(2*n+1) = A096951(n)/((2*n+1)*6^(2*n)).
%e A096952 n=3: R(2*3)=(5/6)* a(3)/A096953(3) = (5/6)*463/326592 = 2315/1959552 = 0.001181..., therefore log(2) - 2*Sum_{k=1..3} ((1/3)^(2*k-1))/(2*k-1) < 0.001181... . In fact, the partial sum is 0.0001430654... .
%K A096952 nonn,easy,frac
%O A096952 0,2
%A A096952 _Wolfdieter Lang_, Jul 16 2004