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A196525 Decimal expansion of log(1+sqrt(2))/sqrt(2).

%I A196525 #77 Feb 16 2025 00:59:42
%S A196525 6,2,3,2,2,5,2,4,0,1,4,0,2,3,0,5,1,3,3,9,4,0,2,0,0,8,0,2,5,0,5,6,8,0,
%T A196525 0,2,6,5,0,6,9,5,3,1,2,3,4,6,5,6,7,2,5,2,8,9,8,7,1,4,7,7,6,0,9,6,1,7,
%U A196525 0,0,0,4,5,4,7,0,1,4,1,8,0,4,6,7,6,6,9,0,7,3,2,3,5,6,2,6,6
%N A196525 Decimal expansion of log(1+sqrt(2))/sqrt(2).
%H A196525 G. C. Greubel, <a href="/A196525/b196525.txt">Table of n, a(n) for n = 0..10000</a>
%H A196525 R. J. Mathar, <a href="http://arxiv.org/abs/1008.2547">Table of Dirichlet L-series and Prime Zeta Modulo Functions for Small Moduli</a>, arXiv:1008.2547 [math.NT], 2010-2015, Table section 2.2, L(m=8, r=2, s=1).
%H A196525 Paul J. Nahin, <a href="https://doi.org/10.1007/978-3-030-43788-6">Inside interesting integrals</a>, Undergrad. Lecture Notes in Physics, Springer (2020), (2.2.3)
%H A196525 <a href="/index/Tra#transcendental">Index entries for transcendental numbers</a>.
%F A196525 Equals Sum_{n>=1} A091337(n)/n = 1 - 1/3 - 1/5 + 1/7 + 1/9 - 1/11 - ...
%F A196525 Equals 2*Sum_{n>=1} (-1)^n/A001539(n). - _Michel Marcus_, Sep 27 2017
%F A196525 From _Fred Daniel Kline_, May 23 2019: (Start)
%F A196525 Equals arcsinh(1)/sqrt(2).
%F A196525 Equals Sum_{n>=1} 1/A118417(n-1) = Sum_{n>=1} 1/((2*n - 1)*2^n). (End)
%F A196525 From _Peter Bala_, Nov 01 2019: (Start)
%F A196525 Equals (1/sqrt(2))*arccoth(sqrt(2)).
%F A196525 Equals 1 - 8*Sum_{n >= 0} (-1)^(n+1)*n/(16*n^2 - 1).
%F A196525 Equals 1 - Integral_{x = 0..inf} exp(-2*x)*cosh(x)/cosh(2*x) dx.
%F A196525 Equals 2*Integral_{x = 0..inf} exp(x)*(exp(2*x) + 1)*(exp(4*x) - 1)/(exp(4*x) + 1)^2 dx - 1. (End)
%F A196525 From _Amiram Eldar_, Aug 16 2020: (Start)
%F A196525 Equals Sum_{k>=0} (-1)^k * (2*k)!!/(2*k+1)!!.
%F A196525 Equals Integral_{x=0..Pi/4} 1/(cos(x) + sin(x)) dx. (End)
%F A196525 From _Peter Bala_, Dec 01 2021: (Start)
%F A196525 Equals 2*Sum_{k >= 0} (-1)^k/((4*k + 1)*(4*k + 3)).
%F A196525 Let N be a positive integer divisible by 4. We have the asymptotic expansion (1/sqrt(2))*log(1 + sqrt(2)) - 2*Sum_{k = 0..N/4 - 1} (-1)^k/((4*k + 1)*(4*k + 3)) ~ 1/N^2 - 11/N^4 + 361/N^6 - 24611/N^8 + ..., where the sequence of unsigned coefficients [1, 11, 361, 24611, ...] is A000464. See A181048 and A181049. An example is given below. (End)
%F A196525 Equals 1/Product_{p prime} (1 - Kronecker(8,p)/p), where Kronecker(8,p) = 0 if p = 2, 1 if p == 1 or 7 (mod 8) or -1 if p == 3 or 5 (mod 8). - _Amiram Eldar_, Dec 17 2023
%F A196525 Equals integral_{x=0..Pi/2} sin^2(x)/(sin(x)+cos(x)) dx [Nahin]. - _R. J. Mathar_, May 16 2024
%e A196525 0.6232252401402305133940200802505680... = A091648/A002193.
%e A196525 From _Peter Bala_, Dec 01 2021: (Start)
%e A196525 With N = 10000, the truncated series Sum_{k = 0..N/4 - 1} (-1)^k/((4*k + 1)*(4*k+3)) = 0.6232252[3]014023[16]1339[3659]080... to 27 decimal places. The square bracketed numbers show where this decimal expansion differs from that of (1/sqrt(2))*log(1+sqrt(2)) = 0.6232252(4)014023(05) 1339(4020)080.... The numbers 1, -11, 361 must be added to the square bracketed numbers to give the correct decimal expansion to 27 decimal places. (End)
%t A196525 RealDigits[Log[1+Sqrt[2]]/Sqrt[2],10,120][[1]] (* _Harvey P. Dale_, Dec 27 2011 *)
%t A196525 RealDigits[Sum[1/((2 n - 1) 2^n), {n, 1, Infinity}], 10, 120][[1]] (* _Fred Daniel Kline_, May 23 2019 *)
%o A196525 (PARI) log(sqrt(2)+1)/sqrt(2) \\ _Michel Marcus_, Sep 27 2017
%o A196525 (Magma) SetDefaultRealField(RealField(100)); Log(Sqrt(2)+1)/Sqrt(2); // _G. C. Greubel_, Oct 05 2018
%Y A196525 Cf. A000464, A002193, A001539, A091648, A181048, A181049.
%K A196525 nonn,cons,easy
%O A196525 0,1
%A A196525 _R. J. Mathar_, Oct 03 2011