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A019675 Decimal expansion of Pi/8.

%I A019675 #83 Jun 25 2025 22:24:17
%S A019675 3,9,2,6,9,9,0,8,1,6,9,8,7,2,4,1,5,4,8,0,7,8,3,0,4,2,2,9,0,9,9,3,7,8,
%T A019675 6,0,5,2,4,6,4,6,1,7,4,9,2,1,8,8,8,2,2,7,6,2,1,8,6,8,0,7,4,0,3,8,4,7,
%U A019675 7,0,5,0,7,8,5,7,7,6,1,2,4,8,2,8,5,0,4,3,5,3,1,6,7,7,6,4,6,3,3
%N A019675 Decimal expansion of Pi/8.
%C A019675 Equals Integral_{x>=0} sin(4*x)/(4*x) dx. - _Jean-François Alcover_, Feb 28 2013
%C A019675 Consider 4 circles inscribed in a square. Inscribe a square in each circle. And finally, inscribe 4 circles inside each four small squares. Totally we get 16 small circles. Pi/8 is the ratio of the area of the 16 small circles to the area of initial square. See the link. - _Kirill Ustyantsev_, Apr 30 2020
%D A019675 Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 8.4, p. 492.
%H A019675 Ivan Panchenko, <a href="/A019675/b019675.txt">Table of n, a(n) for n = 0..1000</a>
%H A019675 Kirill Ustyantsev, <a href="/A019675/a019675.pdf">Geometric sense of Pi/8</a>.
%H A019675 <a href="/index/Tra#transcendental">Index entries for transcendental numbers</a>.
%F A019675 From _Peter Bala_, Nov 15 2016: (Start)
%F A019675 Pi/8 = Sum_{k >= 1} (-1)^k/((2*k - 3)*(2*k - 1)*(2*k + 1)).
%F A019675 More generally, for n >= 0 we have 1/(2*n)! * Pi/4 = Sum_{k >= 1} (-1)^(k+n-1) * 1/Product_{j = -n..n} (2*k + 2*j - 1): when n = 0 we get the Madhava-Gregory-Leibniz series for Pi/4.
%F A019675 For N divisible by 4, we have the asymptotic expansion Pi/8 - Sum_{k = 1..N/2} (-1)^k/((2*k - 3)*(2*k - 1)*(2*k + 1)) ~ -1/2*(1/N^3 - 2/N^5 + 31/N^7 - 692/N^9 + ...), where the sequence of unsigned coefficients [1, 2, 31, 692, ...] equals A024235. (End)
%F A019675 Equals Integral_{x = 0..1} x*sqrt(1 - x^4) dx. - _Peter Bala_, Oct 27 2019
%F A019675 Equals Integral_{x = 0..oo} sin(x)^6/x^4 dx = Sum_{n >= 1} sin(n)^6/n^4, by the Abel-Plana formula. - _Peter Bala_, Nov 04 2019
%F A019675 From _Amiram Eldar_, Jul 12 2020: (Start)
%F A019675 Equals arctan(sqrt(2) - 1).
%F A019675 Equals Sum_{k>=0} (-1)^k/(4*k+2).
%F A019675 Equals Sum_{k>=0} 1/((4*k+1)*(4*k+3)) = Sum_{k>=0} 1/A001539(k).
%F A019675 Equals Integral_{x=0..oo} dx/(x^2 + 16).
%F A019675 Equals Integral_{x=0..oo} dx/(x^4 + 4) = Integral_{x=0..oo} x/(x^4 + 4) dx.
%F A019675 Equals Integral_{x=0..oo} x/(x^4 + 1)^2 dx = Integral_{x=0..1} x/(x^4 + 1) dx.
%F A019675 Equals Integral_{x=0..1} x * arcsin(x) dx. (End)
%F A019675 From _Kritsada Moomuang_, Jun 18 2025: (Start)
%F A019675 Equals Integral_{x=0..oo} (x*log(x + 1))/((x^2 + 1)^2) dx.
%F A019675 Equals Integral_{x=0..oo} (x^3 - 3*x + 3*arctan(x))/(3*x^5) dx. (End)
%e A019675 Pi/8 = 0.392699081698724154807830422909937860524646174921888227621868... - _Vladimir Joseph Stephan Orlovsky_, Dec 02 2009
%t A019675 RealDigits[N[Pi/8,6! ]] (* _Vladimir Joseph Stephan Orlovsky_, Dec 02 2009 *)
%o A019675 (PARI)
%o A019675 default(realprecision, 1002);
%o A019675 eval(vecextract(Vec(Str(sumalt(n=0, (-1)^(n)/(4*n+2)))), "3..-2"))  \\ _Gheorghe Coserea_, Oct 06 2015
%o A019675 (Magma) pi:=Pi(RealField(110)); Reverse(Intseq(Floor(10^100*(pi)/8))); // _Vincenzo Librandi_, Oct 07 2015
%Y A019675 Cf. A195909, A195913, A195697. - _Mohammad K. Azarian_, Oct 11 2011
%Y A019675 Cf. A000796, A003881, A019670, A024235, A244978, A334422 (Pi/128).
%Y A019675 Cf. A001539, A061550.
%K A019675 nonn,cons
%O A019675 0,1
%A A019675 _N. J. A. Sloane_