A363515 Numerator of log(2) + (-1/4)^n*Integral_{x=0..1} (1 - x)^(4*n+2)/(1 + x^2) dx.
1, 79, 14087, 3990557, 217474889, 10326377909, 19001942777579, 3306285643032971, 3279846716611480357, 121354235196693865579, 19902098013482397470501, 1711580361934007500382731, 9009759106282339175994464009, 59689653955233943488755746919, 3820137854975012405338172218301
Offset: 0
Examples
n a(n)/A363516(n) approximate value - ------------------- ------------------ 0 1 1 1 79/120 0.6583333333... 2 14087/20160 0.6987599206... 3 3990557/5765760 0.6921129218... 4 217474889/313657344 0.6933518158... ... From _M. F. Hasler_, Jul 07 2023: (Start) 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: 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). 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. 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)
Links
- Mathematica Stack Exchange, How to prove using Mathematica that the sequence converges to log(2)?
Programs
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Mathematica
Numerator[Simplify[Table[Log[2]+(-1)^n Integrate[(1-x)^(4n+2)/(1+x^2),{x,0,1}]/4^n,{n,0,14}]]] -
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
Formula
Numerator of log(2) + HypergeometricPFQ([1/2, 1, 1], [2*(1 + n), 5/2 + 2*n], -1)/((3 + 4*n)*(-4)^n).
Limit_{n->oo} a(n)/A363516(n) = log(2).
Comments