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%I #25 Aug 02 2023 14:01:10
%S 1,79,14087,3990557,217474889,10326377909,19001942777579,
%T 3306285643032971,3279846716611480357,121354235196693865579,
%U 19902098013482397470501,1711580361934007500382731,9009759106282339175994464009,59689653955233943488755746919,3820137854975012405338172218301
%N Numerator of log(2) + (-1/4)^n*Integral_{x=0..1} (1 - x)^(4*n+2)/(1 + x^2) dx.
%C From _M. F. Hasler_, Jul 07 2023: (Start)
%C Equivalently, numerator of Sum c(n,k)/(k+1), where Sum c(n,k)*x^k = ((1 - x)^(4*n+2)/(-4)^n + 2*x)/(1 + x^2), a polynomial: The integrand (with factor (-1/4)^n) plus 2*x/(1 + x^2) is a polynomial that is easily integrated to yield the fraction a(n)/A363516(n), while Integral(-2*x/(1 + x^2)) = -log(1 + x^2) cancels the log(2).
%C Since the integrand/integral as a whole is less than 1/4^n in absolute value, it tends to zero and the fraction tends to log(2). (End)
%H Mathematica Stack Exchange, <a href="https://mathematica.stackexchange.com/questions/286529/how-to-prove-using-mathematica-that-the-sequence-converges-to-log2#286535">How to prove using Mathematica that the sequence converges to log(2)?</a>
%F Numerator of log(2) + HypergeometricPFQ([1/2, 1, 1], [2*(1 + n), 5/2 + 2*n], -1)/((3 + 4*n)*(-4)^n).
%F Limit_{n->oo} a(n)/A363516(n) = log(2).
%e n a(n)/A363516(n) approximate value
%e - ------------------- ------------------
%e 0 1 1
%e 1 79/120 0.6583333333...
%e 2 14087/20160 0.6987599206...
%e 3 3990557/5765760 0.6921129218...
%e 4 217474889/313657344 0.6933518158...
%e ...
%e From _M. F. Hasler_, Jul 07 2023: (Start)
%e Let f[n] = (-1/4)^n*(1 - x)^(4*n+2)/(1 + x^2), the rational fraction to be integrated from 0 to 1. We have:
%e f[0] = 1 - 2*x/(1 + x^2), with primitive F[0] = x/2 - log(1 + x^2), whence an integral equal to 1/2 - log(2).
%e f[1] = -x^4/4 + (3/2)*x^3 - (7/2)*x^2 + (7/2)*x - 1/4 - 2*x/(1 + x^2), and the term-wise integration of the polynomial part gives -1/20 + 3/8 - 7/6 + 7/4 - 1/4 = 79/120, whence a(1) = 79 and A363516(1) = 120.
%e f[2] = (1/16)*x^8 - (5/8)*x^7 + (11/4)*x^6 - (55/8)*x^5 + (83/8)*x^4 - (71/8)*x^3 + (11/4)*x^2 + (11/8)*x + 1/16 - 2*x/(1 + x^2), so the integration gives 1/144 - 5/64 + 11/28 - 55/48 + 83/40 - 71/32 + 11/12 + 11/16 + 1/16 - log(2) = 14087/20160 - log(2), whence a(2) = 14087 and A363516(2) = 20160, etc. (End)
%t Numerator[Simplify[Table[Log[2]+(-1)^n Integrate[(1-x)^(4n+2)/(1+x^2),{x,0,1}]/4^n,{n,0,14}]]]
%o (PARI) A363515(n) = numerator(subst(intformal(((1-x)^(4*n+2)/(-4)^n+2*x)/(1+x^2)),x,1)) \\ The argument of intformal is a polynomial that is trivially integrated over [0, 1]. - _M. F. Hasler_, Jul 07 2023
%Y Cf. A002162, A004767, A016825, A262710, A363516 (denominator).
%K nonn,frac,less
%O 0,2
%A _Alexander R. Povolotsky_ and _Stefano Spezia_, Jun 07 2023
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