A181049 - OEIS

%I #47 Jun 13 2024 17:32:56

%S 2,4,3,7,4,7,7,4,7,1,9,9,6,8,0,5,2,4,1,7,9,9,7,5,0,8,3,6,3,2,3,0,2,7,

%T 1,1,0,0,1,4,8,0,0,5,4,9,9,8,6,7,7,6,5,1,4,3,6,3,1,7,0,6,2,8,2,1,4,6,

%U 9,3,4,6,8,6,3,9,2,7,1,4,8,5,8,8,0,8,1,3,3,0,2,2,7,7,8,2,3,4,0,6,3,5,6,3,4

%N Decimal expansion of (Pi/2 - log(1+sqrt(2)))/(2*sqrt(2)) = Sum_{k>=0} (-1)^k/(4k+3).

%C Let N be a positive integer divisible by 4. We have the asymptotic expansion 2*((Pi/2 - log(1 + sqrt(2)))/(2*sqrt(2)) - Sum_{k = 0..N/4 - 1} (-1)^k/(4*k + 3)) ~ 1/N - 1/N^2 - 3/N^3 + 11/N^4 + 57/N^5 - - ..., where the sequence of coefficients [1, -1, -3, 11, 57, ...] is A212435. This follows from Borwein et al., Lemma 2 with f(x) = 1/x and then set x = N/4 and h = 3/4. An example is given below. Cf. A181048. - _Peter Bala_, Sep 23 2016

%H Gheorghe Coserea, <a href="/A181049/b181049.txt">Table of n, a(n) for n = 0..2015</a>

%H J. M. Borwein, P. B. Borwein, and K. Dilcher, <a href="http://www.jstor.org/stable/2324715">Pi, Euler numbers and asymptotic expansions</a>, Amer. Math. Monthly, 96 (1989), 681-687.

%H Eric W. Weisstein, <a href="https://mathworld.wolfram.com/EulersSeriesTransformation.html">Euler's Series Transformation</a>.

%F Equals Integral_{x=0..1} (x^2 dx)/(1+x^4).

%F Equals (1/2) * Integral_{x = 0..Pi/4} sqrt(tan(x)) dx. Cf. A247719. - _Peter Bala_, Sep 23 2016

%F Equals Sum_{n >= 0} 2^(n-1)*n!/(Product_{k = 0..n} 4*k + 3) = Sum_{n >= 0} 2^(n-1)*n!/A008545(n+1) (apply Euler's series transformation to Sum_{k >= 0} (-1)^k/(4*k+3)). - _Peter Bala_, Dec 01 2021

%F From _Peter Bala_, Mar 03 2024: (Start)

%F Continued fraction: 1/(3 + 3^2/(4 + 7^2/(4 + 11^2/(4 + 15^2/(4 + ... ))))) due to Euler.

%F Equals (1/3)*hypergeom([3/4, 1], [7/4], -1).

%F Gauss's continued fraction: 1/(3 + 3^2/(7 + 4^2/(11 + 7^2/(15 + 8^2/(19 + 11^2/(23 + 12^2/(27 + 15^2/(31 + 16^2/(35 + 19^2/(39 + ... )))))))))). (End)

%F Equals Integral_{x=1..oo, y=1..oo} 1/(x^4 + y^4) dx. - _Vaclav Kotesovec_, Jun 13 2024

%e 0.2437477471996805241799750836323027110...

%e From _Peter Bala_, Sep 23 2016: (Start)

%e At N = 100000 the truncated series Sum_{k = 0..N/4 - 1} (-1)^k/(4*k + 3) ) = 0.4874(8)5494(4)9936(4)048(24)99(444)67(625)6... to 32 digits The bracketed numbers show where this decimal expansion differs from that of 2*A181049. The numbers 1, -1, -3, 11, 57, -361 must be added to the bracketed numbers to give the correct decimal expansion to 32 digits: 2*( Pi/2 - log(1+sqrt(2)))/(2*sqrt(2) ) = 0.4874(9)5494(3)9936(1)048(35)99(501)67(264)6.... (End)

%t First@ RealDigits[N[(Pi/2 - Log[1 + Sqrt@ 2])/(2 Sqrt@ 2), 105]] (* _Michael De Vlieger_, Oct 07 2015 *)

%o (PARI)

%o default(realprecision, 106);

%o eval(vecextract(Vec(Str(sumalt(n=0, (-1)^(n)/(4*n+3)))), "3..-2")) \\ _Gheorghe Coserea_, Oct 06 2015

%o (PARI) (Pi/2 - log(1+sqrt(2)))/(2*sqrt(2)) \\ _G. C. Greubel_, Nov 28 2017

%o (Magma) C<i> := ComplexField(); [(Pi(C)/2 - Log(1+Sqrt(2)))/(2*Sqrt(2))]; // _G. C. Greubel_, Nov 28 2017

%Y Cf. A113476, A001586, A093954, A181048, A212435, A247719.

%K cons,nonn

%O 0,1

%A _Jonathan D. B. Hodgson_, Oct 01 2010, Oct 05 2010