A195697 -id:A195697 - cp's OEIS frontend

A019675 Decimal expansion of Pi/8.

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, 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, 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

Offset: 0

N. J. A. Sloane

Equals Integral_{x>=0} sin(4*x)/(4*x) dx. - Jean-François Alcover, Feb 28 2013

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

			Pi/8 = 0.392699081698724154807830422909937860524646174921888227621868... - _Vladimir Joseph Stephan Orlovsky_, Dec 02 2009

Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 8.4, p. 492.

Ivan Panchenko, Table of n, a(n) for n = 0..1000
Kirill Ustyantsev, Geometric sense of Pi/8.
Index entries for transcendental numbers.

Cf. A195909, A195913, A195697. - Mohammad K. Azarian, Oct 11 2011
Cf. A000796, A003881, A019670, A024235, A244978, A334422 (Pi/128).
Cf. A001539, A061550.

pi:=Pi(RealField(110)); Reverse(Intseq(Floor(10^100*(pi)/8))); // Vincenzo Librandi, Oct 07 2015

RealDigits[N[Pi/8,6! ]] (* Vladimir Joseph Stephan Orlovsky, Dec 02 2009 *)

default(realprecision, 1002);
eval(vecextract(Vec(Str(sumalt(n=0, (-1)^(n)/(4*n+2)))), "3..-2"))  \\ Gheorghe Coserea, Oct 06 2015

From Peter Bala, Nov 15 2016: (Start)
Pi/8 = Sum_{k >= 1} (-1)^k/((2*k - 3)*(2*k - 1)*(2*k + 1)).
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.
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)
Equals Integral_{x = 0..1} x*sqrt(1 - x^4) dx. - Peter Bala, Oct 27 2019
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
From Amiram Eldar, Jul 12 2020: (Start)
Equals arctan(sqrt(2) - 1).
Equals Sum_{k>=0} (-1)^k/(4*k+2).
Equals Sum_{k>=0} 1/((4*k+1)*(4*k+3)) = Sum_{k>=0} 1/A001539(k).
Equals Integral_{x=0..oo} dx/(x^2 + 16).
Equals Integral_{x=0..oo} dx/(x^4 + 4) = Integral_{x=0..oo} x/(x^4 + 4) dx.
Equals Integral_{x=0..oo} x/(x^4 + 1)^2 dx = Integral_{x=0..1} x/(x^4 + 1) dx.
Equals Integral_{x=0..1} x * arcsin(x) dx. (End)
From Kritsada Moomuang, Jun 18 2025: (Start)
Equals Integral_{x=0..oo} (x*log(x + 1))/((x^2 + 1)^2) dx.
Equals Integral_{x=0..oo} (x^3 - 3*x + 3*arctan(x))/(3*x^5) dx. (End)

A016655 Decimal expansion of log(32) = 5*log(2).

Original entry on oeis.org

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

Offset: 1

N. J. A. Sloane

The function exp(x) has its maximum curvature where x = -(1/10)*5*log(2) = -log(2)/2 = 0.34657... - Dimitri Papadopoulos, Oct 27 2022

This maximum curvature occurs at the point with coordinates (x_M = -log(2)/2 = -(this constant)/10; y_M = sqrt(2)/2 = A010503) and is equal to 2*sqrt(3)/9 = A212886. - Bernard Schott, Dec 23 2022

			3.465735902799726547086160607290882840377500671801276270603400047466968...

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 2.

Harry J. Smith, Table of n, a(n) for n = 1..20000
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
O. Espinosa, V. H. Moll, On some integrals involving the Hurwitz Zeta function: Part 1, Raman. J. 6 (2002) 159-188, eq. (5.7)
Steven R. Finch, Errata and Addenda to Mathematical Constants, arXiv:2001.00578 [math.HO], 2020-2021.
R. S. Melham and A. G. Shannon, Inverse Trigonometric Hyperbolic Summation Formulas Involving Generalized Fibonacci Numbers, The Fibonacci Quarterly, Vol. 33, No. 1 (1995), pp. 32-40.
Paul J. Nahin, Inside interesting integrals, Undergrad. Lecture Notes in Physics, Springer (2020), chap 7.1
H. M. Srivastava, M. L. Glasser, V. S. Adamchik, Some definite integrals associated with the Riemann Zeta Function, Z. Anal. Anw. 19 (3) (2000) 831-846
Index to sequences related to curves.
Index entries for transcendental numbers.

Cf. A000045, A000032, A060851, A195909, A195913, A195697, A016460 (continued fraction).

Cf. A010503, A212886.

[5*Log(2)]; // Vincenzo Librandi, Jan 02 2016

RealDigits[5 N [Log[2], 100]] [[1]] (* Vincenzo Librandi, Jan 02 2016 *)

log(32) \\ Charles R Greathouse IV, Jan 24 2012

log(2)/2 = (1 - 1/2 - 1/4) + (1/3 - 1/6 - 1/8) + (1/5 - 1/10 - 1/12) + ... [Jolley, Summation of Series, Dover (1961) eq (73)]
Equals 10*log(2)/2 = 5*log(2) = 5*A002162, so 10*(1/2 - 1/4 + 1/6 - 1/8 + 1/10 - ... + (-1)^(k+1)/(2*k) + ...) = log(32). - Eric Desbiaux, Nov 26 2008
-log(2)/2 = lim_{n->oo} ((Sum_{k=2..n} arctanh(1/k)) - log(n)). - Jean-François Alcover, Aug 07 2014, after Steven Finch
Equals log(sqrt(2)) with offset 0. - Michel Marcus, Feb 19 2017
Equals (5/4)*Sum_{k=1..4} (-1)^(k+1) gamma(0, k/4) where gamma(n,x) denotes the generalized Stieltjes constants. - Peter Luschny, May 16 2018
From Amiram Eldar, Jun 29 2020: (Start)
log(2)/2 = arctanh(1/3) = arcsinh(1/sqrt(8)).
log(2)/2 = Integral_{x=0..Pi/4} tan(x) dx.
log(2)/2 = Sum_{k>=0} (-1)^k/(2*k+2).
log(2)/2 = Sum_{k>=1} 1/A060851(k). (End)
log(2)/2 = Sum_{k>=1} (-1)^(k+1) * arctanh(Lucas(2*k+3)/Fibonacci(2*k+3)^2) (Melham and Shannon, 1995). - Amiram Eldar, Jan 15 2022
Equals 10 * Integral_{1..oo} dx/(x*(1+x^2)). [Nahin] - R. J. Mathar, May 22 2024
Equals -10*Integral_{q=0..1} q*log(sin(Pi*q))dq. [Espinosa] - R. J. Mathar, Aug 13 2024
log(2)/2 = Sum_{k>=2} (-1)^(k) * arccoth(k). - Antonio Graciá Llorente, Sep 16 2024
-0.34657359... = Sum_{k>=0} zeta(2k)/((2k+1)*2^(2k)), [Srivastava (2.20)] - R. J. Mathar, Feb 12 2020
Equals 10*Integral_{x=0..1} Ei((1 + sqrt(2))*log(x)) - li(x) dx, where Ei is the exponential integral and li is the logarithmic integral. - Kritsada Moomuang, Jun 06 2025

A195909 First numerator and then denominator in a fraction expansion of log(2) - Pi/8.

Original entry on oeis.org

1, 2, -1, 3, 1, 12, 1, 30, -1, 35, 1, 56, 1, 90, -1, 99, 1, 132, 1, 182, -1, 195, 1, 240, 1, 306, -1, 323, 1, 380, 1, 462, -1, 483, 1, 552, 1, 650, -1, 675, 1, 756, 1, 870, -1, 899, 1, 992, 1, 1122, -1, 1155, 1

Offset: 1

Mohammad K. Azarian, Sep 26 2011

			1/2 - 1/3 + 1/12 + 1/30 - 1/35 + 1/56 + 1/90 - 1/99 + 1/132 + 1/182 - 1/195 + 1/240 + ... = [(1 - 1/2) + (1/3 - 1/4) + (1/5 - 1/6) + (1/7 - 1/8) + (1/9 - 1/10) + (1/11 - 1/12) + ... ] - (1/2)*[(1 - 1/3) + (1/5 - 1/7) + (1/9 - 1/11) + (1/13 - 1/15) + ... ] = log(2) - Pi/8.

Mohammad K. Azarian, Problem 1218, Pi Mu Epsilon Journal, Vol. 13, No. 2, Spring 2010, p. 116. Solution published in Vol. 13, No. 3, Fall 2010, pp. 183-185.
Granino A. Korn and Theresa M. Korn, Mathematical Handbook for Scientists and Engineers, McGraw-Hill Book Company, New York (1968).

Cf. A195913, A195697, A195947, A164833, A118324, A098289, A075549, A016655, A019675, A161685, A144981, A168056, A004772.

log(2) - Pi/8 = Sum_{n>=1} (-1)^(n+1)*(1/n) + (-1/2)*Sum_{n>=0} (-1)^n*(1/(2*n+1)).

Empirical g.f.: x*(1+2*x-2*x^2+x^3+2*x^4+9*x^5-2*x^6+14*x^7+2*x^8+3*x^9-2*x^10+3*x^11+x^12) / ((1-x)^3*(1+x)^3*(1-x+x^2)^2*(1+x+x^2)^2). - Colin Barker, Dec 17 2015

A195913 The denominator in a fraction expansion of log(2)-Pi/8.

Original entry on oeis.org

2, 3, 12, 30, 35, 56, 90, 99, 132, 182, 195, 240, 306, 323, 380, 462, 483, 552, 650, 675, 756, 870, 899, 992, 1122, 1155, 1260, 1406, 1443, 1560, 1722, 1763, 1892, 2070, 2115, 2256, 2450, 2499, 2652, 2862, 2915

Offset: 1

Mohammad K. Azarian, Sep 25 2011

The minus sign in front of a fraction is considered the sign of the numerator and hence the sign of the fraction does not appear in this sequence.

			1/2 - 1/3 + 1/12 + 1/30 - 1/35 + 1/56 + 1/90 - 1/99 + 1/132 + 1/182 - 1/195 + 1/240 + ... = [(1 - 1/2) + (1/3 - 1/4) + (1/5 - 1/6) + (1/7 - 1/8) + (1/9 - 1/10) + (1/11 - 1/12) + ...] - (1/2)*[(1 - 1/3) + (1/5 - 1/7) + (1/9 - 1/11) + (1/13 - 1/15) + ... ] = log(2) - Pi/8.

Granino A. Korn and Theresa M. Korn, Mathematical Handbook for Scientists and Engineers, McGraw-Hill Book Company, New York (1968).

Mohammad K. Azarian, Problem 1218, Pi Mu Epsilon Journal, Vol. 13, No. 2, Spring 2010, p. 116. Solution published in Vol. 13, No. 3, Fall 2010, pp. 183-185.

Cf. A195909, A195697, A195947, A164833, A118324, A098289, A075549, A016655, A019675, A161685, A144981, A168056, A004772.

log(2) - Pi/8 = Sum_{n>=1} (-1)^(n+1)*(1/n) + (-1/2)*Sum_{n>=0} (-1)^n*(1/(2*n+1)).
Empirical g.f.: x*(2+x+9*x^2+14*x^3+3*x^4+3*x^5) / ((1-x)^3*(1+x+x^2)^2). - Colin Barker, Dec 17 2015
From Bernard Schott, Aug 11 2019: (Start)
k >= 1, a(3*k) = (4*k-1) * 4*k,
k >= 0, a(3*k+1) = (4*k+1) * (4*k+2),
k >= 0, a(3*k+2) = (4*k+1) * (4*k+3).
The even terms a(3*k) and a(3*k+1) come from log(2) and the odd terms a(3*k+2) come from - Pi/8. (End)

A161685 Continued fraction for (3*Pi)/8.

Original entry on oeis.org

1, 5, 1, 1, 1, 1, 2, 12, 12, 6, 1, 1, 1, 1, 11, 1, 4, 1, 3, 2, 1, 6, 1, 4, 4, 3, 3, 4, 10, 1, 4, 37, 1, 1, 20, 1, 1, 1, 4, 1, 1, 3, 1, 1, 1, 1, 2, 2, 15, 2, 1, 2, 2, 1, 1, 1, 3, 1, 4, 3, 83, 9, 19, 1, 2, 2, 4, 1, 1, 1, 4, 3, 4, 1, 1, 8, 2, 1, 2, 1, 7, 3, 2, 1, 1, 5, 1, 6, 1, 5, 1, 7, 17, 9, 6, 3, 1, 1, 1, 1, 2

Offset: 0

Harry J. Smith, Jun 18 2009

			1.178097245096172464423491268... = 1 + 1/(5 + 1/(1 + 1/(1 + 1/(1 + ...))))

Harry J. Smith, Table of n, a(n) for n = 0..20000

Cf. A195909, A195913, A195697. - Mohammad K. Azarian, Oct 11 2011

Cf. A093828 (decimal expansion).

SetDefaultRealField(RealField(105)); R:= RealField(); ContinuedFraction(3*Pi(R)/8); // G. C. Greubel, Aug 11 2019

convert(3*Pi/8, confrac, 105); # G. C. Greubel, Aug 11 2019

ContinuedFraction[(3*Pi)/8,120] (* Harvey P. Dale, Mar 13 2016 *)

{ allocatemem(932245000); default(realprecision, 21000); x=contfrac(3*Pi/8); for (n=0, 20000, write("b161685.txt", n, " ", x[n+1])); }

continued_fraction_list(3*pi/8, nterms=105) # G. C. Greubel, Aug 11 2019

A164833 Decimal expansion of Pi/8 - log(2)/2.

Original entry on oeis.org

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

Offset: 0

Jonathan Vos Post, Aug 27 2009

Digits and formula given at Waldschmidt, p. 4.

			0.0461254914187515000992143621808495764868961077417606...
1/(2*3*4) + 1/(6*7*8) + 1/(10*11*12) + 1/(14*15*16) + ... [_Bruno Berselli_, Mar 17 2014]

Mohammad K. Azarian, Problem 1218, Pi Mu Epsilon Journal, Vol. 13, No. 2, Spring 2010, p. 116. Solution published in Vol. 13, No. 3, Fall 2010, pp. 183-185.
L. B. W. Jolley, Summation of series, Dover Publications Inc. (New York), 1961, p. 46 (series n. 251).
A. J. Van Der Poorten, Effectively computable bounds for the solutions of certain Diophantine equations, Acta Arith., 33 (1977), pp. 195-207.

G. C. Greubel, Table of n, a(n) for n = 0..10000
Michel Waldschmidt, Perfect Powers: Pillai's works and their developments, Aug 27, 2009.

Cf. A001597, A019675, A016655.
Cf. A195909, A195913, A195697. - Mohammad K. Azarian, Oct 11 2011
Cf. A239362: Sum_{k>=1} 1/((3k-2)*(3k-1)*(3k)).

SetDefaultRealField(RealField(130)); R:= RealField(); (Pi(R)-4*Log(2))/8; // G. C. Greubel, Aug 11 2019

evalf[130]((Pi - 4*log(2))/8 ); # G. C. Greubel, Aug 11 2019

Join[{0},RealDigits[Pi/8-Log[2]/2,10,120][[1]]] (* Harvey P. Dale, Nov 13 2012 *)

default(realprecision, 130); (Pi - 4*log(2))/8 \\ G. C. Greubel, Aug 11 2019

numerical_approx((pi-4*log(2))/8, digits=130) # G. C. Greubel, Aug 11 2019

Equals Sum_{n>=0} Sum_{m>=0} 1/((4*n+3)^(2*m+1)).

Equals Sum_{k>=1} 1/( (4*k-2)*(4*k-1)*(4*k) ). - Bruno Berselli, Mar 17 2014

Normalized offset and leading zeros - R. J. Mathar, Sep 27 2009

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A019675 Decimal expansion of Pi/8.

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A016655 Decimal expansion of log(32) = 5*log(2).

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A195909 First numerator and then denominator in a fraction expansion of log(2) - Pi/8.

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A195913 The denominator in a fraction expansion of log(2)-Pi/8.

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A161685 Continued fraction for (3*Pi)/8.

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A164833 Decimal expansion of Pi/8 - log(2)/2.

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