A193887 -id:A193887 - cp's OEIS frontend

A093954 Decimal expansion of Pi/(2*sqrt(2)).

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

Offset: 1

Eric W. Weisstein, Apr 19 2004

The value is the length Pi*sqrt(2)/4 of the diagonal in the square with side length Pi/4 = Sum_{n>=0} (-1)^n/(2n+1) = A003881. The area of the circumcircle of this square is Pi*(Pi*sqrt(2)/8)^2 = Pi^3/32 = A153071. - Eric Desbiaux, Jan 18 2009
This is the value of the Dirichlet L-function of modulus m=8 at argument s=1 for the non-principal character (1,0,1,0,-1,0,-1,0). See arXiv:1008.2547. - R. J. Mathar, Mar 22 2011
Archimedes's-like scheme: set p(0) = sqrt(2), q(0) = 1; p(n+1) = 2*p(n)*q(n)/(p(n)+q(n)) (harmonic mean, i.e., 1/p(n+1) = (1/p(n) + 1/q(n))/2), q(n+1) = sqrt(p(n+1)*q(n)) (geometric mean, i.e., log(q(n+1)) = (log(p(n+1)) + log(q(n)))/2), for n >= 0. The error of p(n) and q(n) decreases by a factor of approximately 4 each iteration, i.e., approximately 2 bits are gained by each iteration. Set r(n) = (2*q(n) + p(n))/3, the error decreases by a factor of approximately 16 for each iteration, i.e., approximately 4 bits are gained by each iteration. For a similar scheme see also A244644. - A.H.M. Smeets, Jul 12 2018
The area of a circle circumscribing a unit-area regular octagon. - Amiram Eldar, Nov 05 2020

			1.11072073453959156175397...
From _Peter Bala_, Mar 03 2015: (Start)
Asymptotic expansion at n = 5000.
The truncated series Sum_{k = 0..5000 - 1} (-1)^floor(k/2)/(2*k + 1) = 1.110(6)207345(42)591561(18)3970(5238)1.... The bracketed digits show where this decimal expansion differs from that of Pi/(2*sqrt(2)). The numbers 1, -3, 57, -2763 must be added to the bracketed numbers to give the correct decimal expansion to 30 digits: Pi/(2*sqrt(2)) = 1.110(7)207345(39)591561(75)3970 (2475)1.... (End)
From _Peter Bala_, Nov 24 2016: (Start)
Case m = 1, n = 1:
Pi/(2*sqrt(2)) = 4*Sum_{k >= 0} (-1)^(1 + floor(k/2))/((2*k - 1)*(2*k + 1)*(2*k + 3)).
We appear to have the following asymptotic expansion for the tails of this series: for N divisible by 4, Sum_{k >= N/2} (-1)^floor(k/2)/((2*k - 1)*(2*k + 1)*(2*k + 3)) ~ 1/N^3 - 14/N^5 + 691/N^7 - 62684/N^9 - ..., where the coefficient sequence [1, 0, -14, 0, 691, 0, -62684, ...] appears to come from the e.g.f. (1/2!)*cosh(x)/cosh(2*x)*sinh(x)^2 = x^2/2! - 14*x^4/4! + 691*x^6/6! - 62684*x^8/8! + .... Cf. A019670.
For example, take N = 10^5. The truncated series Sum_{k = 0..N/2 -1} (-1)^(1+floor(k/2))/((2*k - 1)*(2*k + 1)*(2*k + 3)) = 0.27768018363489(8)89043849(11)61878(80026)6163(351171)58.... The bracketed digits show where this decimal expansion differs from that of (1/4)*Pi/(2*sqrt(2)). The numbers -1, 14, -691, 62684 must be added to the bracketed numbers to give the correct decimal expansion: (1/4)*Pi/(2*sqrt(2)) = 0.27768018363489(7) 89043849(25)61878(79335)6163(413855)58... (End)

J. M. Arnaudiès, P. Delezoide et H. Fraysse, Exercices résolus d'Analyse du cours de mathématiques - 2, Dunod, 1993, Exercice 5, p. 240.
George Boros and Victor H. Moll, Irresistible integrals, Cambridge University Press, 2006, p. 149.
Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 1.4.1, p. 20.
L. B. W. Jolley, Summation of Series, Dover, 1961, eq. 76, page 16.
Joel L. Schiff, The Laplace Transform: Theory and Applications, Springer-Verlag New York, Inc. (1999). See p. 149.
David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See p. 53.

Harry J. Smith, Table of n, a(n) for n = 1..20000
J. M. Borwein, P. B. Borwein, and K. Dilcher, Pi, Euler numbers and asymptotic expansions, Amer. Math. Monthly, 96 (1989), 681-687.
R. J. Mathar, Table of Dirichlet L-series and prime zeta modulo functions for small moduli, arXiv:1008.2547 [math.NT], 2010-2015, table 7 and section 2.2, value of L(m=8,r=4,s=1).
Michael Penn, Newton's sum (2023), YouTube video.
Michael I. Shamos, A catalog of the real numbers, (2007). See p. 149.
Eric Weisstein's World of Mathematics, Bifoliate.
Index entries for transcendental numbers.

Cf. A161684 (continued fraction).
Cf. A002388, A019670.
Cf. A000281, A063448, A247719, A193887, A244976, A181048, A181049.
Cf. A003881, A251809, A188510.
Cf. A352324, A248897, A019669.

simplify( sum((cos((1/2)*k*Pi)+sin((1/2)*k*Pi))/(2*k+1), k = 0 .. infinity) );  # Peter Bala, Mar 09 2015

RealDigits[Pi/Sqrt@8, 10, 111][[1]] (* Michael De Vlieger, Sep 23 2016 and slightly modified by Robert G. Wilson v, Jul 23 2018 *)

default(realprecision, 20080); x=Pi*sqrt(2)/4; for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b093954.txt", n, " ", d)); \\ Harry J. Smith, Jun 17 2009

Equals 1/A112628.
Equals Integral_{x=0..oo} 1/(x^4+1) dx. - Jean-François Alcover, Apr 29 2013
From Peter Bala, Feb 05 2015: (Start)
Pi/(2*sqrt(2)) = Sum_{k >= 0} binomial(2*k,k)*1/(2*k + 1)*(1/8)^k.
The integer sequences A(n) := 2^n*(2*n + 1)! and B(n) := A(n)*( Sum {k = 0..n} binomial(2*k,k)*1/(2*k + 1)*(1/8)^k ) both satisfy the second order recurrence equation u(n) = (12*n^2 + 1)*u(n-1) - 4*(n - 1)*(2*n - 1)^3*u(n-2). From this observation we can obtain the continued fraction expansion Pi/(2*sqrt(2)) = 1 + 1/(12 - 4*3^3/(49 - 4*2*5^3/(109 - 4*3*7^3/(193 - ... - 4*(n - 1)*(2*n - 1)^3/((12*n^2 + 1) - ... ))))). Cf. A002388 and A019670. (End)
From Peter Bala, Mar 03 2015: (Start)
Pi/(2*sqrt(2)) = Sum_{k >= 0} (-1)^floor(k/2)/(2*k + 1) = limit (n -> infinity) Sum_{k = -n .. n - 1} (-1)^k/(4*k + 1). See Wells.
We conjecture the asymptotic expansion Pi/(2*sqrt(2)) - Sum {k = 0..n - 1} (-1)^floor(k/2)/(2*k + 1) ~ 1/(2*n) - 3/(2*n)^3 + 57/(2*n)^5 - 2763/(2*n)^7 + ..., where n is a multiple of 4 and the sequence of unsigned coefficients [1, 3, 57, 2763, ...] is A000281. An example with n = 5000 is given below. (End)
From Peter Bala, Sep 21 2016: (Start)
c = 2 * Sum_{k >= 0} (-1)^k * (4*k + 2)/((4*k + 1)*(4*k + 3)) = A181048 + A181049. The asymptotic expansion conjectured above follows from the asymptotic expansions given in A181048 and A181049.
c = 1/2 * Integral_{x = 0..Pi/2} sqrt(tan(x)) dx. (End)
From Peter Bala, Nov 24 2016: (Start)
Let m be an odd integer and n a nonnegative integer. Then Pi/(2*sqrt(2)) = 2^n*m^(2*n)*(2*n)!*Sum_{k >= 0} (-1)^(n+floor(k/2)) * 1/Product_{j = -n..n} (2*k + 1 + 2*m*j). Cf. A003881.
In the particular case m = 1 the result has the equivalent form: for n a nonnegative integer, Pi/(2*sqrt(2)) = 2^n*(2*n)!*Sum_{k >= 0} (-1)^(n+k)*(8*k + 4)* 1/Product_{j = -n..n+1} (4*k + 2*j + 1). The case m = 1, n = 1 is considered in the Example section below.
Let m be an odd integer and n a nonnegative integer. Then Pi/(2*sqrt(2)) = 4^n*m^(2*n)*(2*n)!*Sum_{k >= 0} (-1)^(n+floor(k/2)) * 1/Product_{j = -n..n} (2*k + 1 + 4*m*j). (End)
Equals Integral_{x = 0..oo} cosh(x)/cosh(2*x) dx. - Peter Bala, Nov 01 2019
Equals Sum_{k>=1} A188510(k)/k = Sum_{k>=1} Kronecker(-8,k)/k = 1 + 1/3 - 1/5 - 1/7 + 1/9 + 1/11 - 1/13 - 1/15 + ... - Jianing Song, Nov 16 2019
From Amiram Eldar, Jul 16 2020: (Start)
Equals Product_{k>=1} (1 - (-1)^k/(2*k+1)).
Equals Integral_{x=0..oo} dx/(x^2 + 2).
Equals Integral_{x=0..Pi/2} dx/(sin(x)^2 + 1). (End)
Equals Integral_{x=0..oo} x^2/(x^4 + 1) dx (Arnaudiès). - Bernard Schott, May 19 2022
Equals Integral_{x = 0..1} 1/(2*x^2 + (1 - x)^2) dx. - Peter Bala, Jul 22 2022
Equals Integral_{x = 0..1} 1/(1 - x^4)^(1/4) dx. - Terry D. Grant, Mar 17 2023
Equals 1/Product_{p prime} (1 - Kronecker(-8,p)/p), where Kronecker(-8,p) = 0 if p = 2, 1 if p == 1 or 3 (mod 8) or -1 if p == 5 or 7 (mod 8). - Amiram Eldar, Dec 17 2023
Equals A068465*A068467. - R. J. Mathar, Jun 27 2024
From Stefano Spezia, Jun 05 2025: (Start)
Equals Sum_{k>=1} (-1)^(k+1)(1/(4*k - 3) + 1/(4*k - 1)).
Equals Product_{k=0..oo} (1 + (-1)^k/(2*k + 3)).
Equals Integral_{x=0..oo} 1/(2*x^2 + 1).
Equals Integral_{x=0..1} 1/((1 + x^2)*sqrt(1 - x^2)). (End)

A063448 Decimal expansion of Pi * sqrt(2).

Original entry on oeis.org

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

Offset: 1

Jason Earls, Jul 24 2001

Hypotenuse of the right triangle with legs Pi and Pi. - Zak Seidov, May 04 2005
Circumference of the circumcircle of the unit square. - Jonathan Sondow, Nov 23 2017
Half-perimeter of the closed curve with implicit Cartesian equation x^2 + y^2 = abs(x) + abs(y). - Stefano Spezia, Oct 20 2020

			4.4428829381583662470158809900606936986146216893756902230853...

Harry J. Smith, Table of n, a(n) for n = 1..20000
Peter Bala, Series for sqrt(2)*Pi
Eric Weisstein's World of Mathematics, Gauss Multiplication Formula.
Wikipedia, Bailey-Borwein-Plouffe formula.
Index entries for transcendental numbers

Cf. A063447 (continued fraction), A093954, A153799, A193887, A244976, A247719.

RealDigits[N[Pi*Sqrt[2], 200]][[1]] (* Vladimir Joseph Stephan Orlovsky, Mar 21 2011*)

PARI
```
\p 400; Pi * sqrt(2)
```

default(realprecision, 20080); x=Pi*sqrt(2); for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b063448.txt", n, " ", d)) \\ Harry J. Smith, Aug 21 2009

# Use some guard digits when computing.
# BBP formula (1/8) P(1, 64, 12, (32, 0, 8, 0, 8, 0, -4, 0, -1, 0, -1, 0))
from decimal import Decimal as dec, getcontext
def BBPpisqrt2(n: int) -> dec:
    getcontext().prec = n
    s = dec(0); f = dec(1); g = dec(64)
    for k in range(int(n * 0.5536546824812272) + 1):
        twk = dec(12 * k)
        s += f * ( dec(32) / (twk + 1) + dec(8)  / (twk + 3)
                 + dec(8)  / (twk + 5) - dec(4)  / (twk + 7)
                 - dec(1)  / (twk + 9) - dec(1)  / (twk + 11))
        f /= g
    return s / dec(8)
print(BBPpisqrt2(200))  # Peter Luschny, Nov 03 2023

Equals Gamma(1/4)*Gamma(3/4). - Jean-François Alcover, Nov 24 2014
From Amiram Eldar, Aug 06 2020: (Start)
Equals Integral_{x=0..oo} log(1 + 1/x^4) dx.
Equals Integral_{x=0..oo} log(1 + 2/x^2) dx.
Equals Integral_{x=-oo..oo} exp(x/4)/(exp(x) + 1) dx.
Equals Integral_{x=0..2*Pi} 1/(cos(x)^2 + 1) dx = Integral_{x=0..2*Pi} 1/(sin(x)^2 + 1) dx. (End)
From Andrea Pinos, Jul 03 2023: (Start)
Equals (Product_{k=1..4} Gamma(k/8)*Gamma(1 - k/8))^(1/4).
General result: 2*Pi/(4*y)^(1/(2*y)) = (Product_{k=1..y} Gamma(k/(2*y))*Gamma(1 - k/(2*y)) )^(1/y). (End)
From Peter Bala, Oct 22 2023: (Start)
sqrt(2)*Pi = 4 + 8*Sum_{n >= 0} (-1)^n/(16*n^2 + 32*n + 15). See A141759.
In the following the Eisenstein summation convention is assumed: that is,
Sum_{n = -oo..oo} means Limit_{N -> oo} Sum_{n = -N..N}:
sqrt(2)*Pi = 4*Sum_{n = -oo..oo} (-1)^n/(4*n + 1).
More generally, it appears that for k >= 0, k not of the form 4*m + 1,
sqrt(2)*Pi = -sign(cos(Pi*(k - 3)/4)) * 4*(2^floor(k/2))*k! * Sum_{n = -oo..oo} (-1)^n/((4*n + 1)*(4*n + 3)*...*(4*n + 2*k + 1)) (verified up to k = 50).
sqrt(2)*Pi = (2^4)*Sum_{n >= 0} (-1)^n * (2*n + 1)/((4*n + 1)*(4*n + 3)) = 512/105 - (2^6)*4!*Sum_{n >= 0} (-1)^n * (2*n + 3)/((4*n + 1)*(4*n + 3)*...*(4*n + 11)).
sqrt(2)*Pi = 4 + (2^3)*Sum_{n >= 0} (-1)^n * (4*n + 1)/((4*n + 1)*(4*n + 3)*(4*n + 5)) = 1408/315 - (2^5)*5!*Sum_{n >= 0} (-1)^n * (4*n + 1)/((4*n + 1)*(4*n + 3)*...*(4*n + 13)).
sqrt(2)*Pi = 16/3 - (2^4)*3!*Sum_{n >= 0} (-1)^n/((4*n + 1)*(4*n + 3)*(4*n + 5)*(4*n + 7)) = 14848/3465 + (2^6)*7!*Sum_{n >= 0} (-1)^n/((4*n + 1)*(4*n + 3)*...*(4*n + 15)). (End)
From Peter Bala, Nov 19 2023: (Start)
sqrt(2)*Pi = 512*Sum_{k >= 1} (-1)^(k+1) * k^2/((16*k^2 - 1)*(16*k^2 - 9)).
This is the case n = 1 of the more general result: for n >= 1,
sqrt(2)*Pi = (-1)^(n+1) * 2^(n+7) * (2*n)!/(2*n - 1) * Sum_{k >= 1} (-1)^(k+1) * k^2/( Product_{i = 0..n} (16*k^2 - (2*i+1)^2) ). Cf. A334422. (End)
Equals Integral_{x=-oo..oo} (x^2 + 1)/(x^4 + 1) dx. - Kritsada Moomuang, Jun 04 2025

Edited by N. J. A. Sloane, May 05 2007

Corrected by Neven Juric, Apr 24 2008

A247719 Decimal expansion of Integral_{t=0..Pi/2} sqrt(tan(t)) dt.

Original entry on oeis.org

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

Offset: 1

Jean-François Alcover, Sep 23 2014

			2.22144146907918312350794049503034684930731...

G. C. Greubel, Table of n, a(n) for n = 1..10000
D. H. Bailey and J. M. Borwein, Highly Parallel, High-Precision Numerical Integration p. 7. (2005) Lawrence Berkeley National Laboratory.
Philippe Flajolet and Robert Sedgewick, Analytic Combinatorics, Cambridge Univ. Press, 2009, pp. 235-236.

Cf. A063448, A093954, A193887, A244976, A181048, A181049, A186706.

SetDefaultRealField(RealField(100)); R:= RealField(); Pi(R)/Sqrt(2); // G. C. Greubel, Sep 07 2018

RealDigits[Pi/Sqrt[2], 10, 104] // First

default(realprecision, 100); Pi/sqrt(2) \\ G. C. Greubel, Sep 07 2018

Equals Pi/sqrt(2).
Equals A063448/2.
c = 2*( Sum_{k >= 0} (-1)^k/(4*k + 1) + Sum_{k >= 0} (-1)^k/(4*k + 3) ) = 2*(A181048 + A181049). - Peter Bala, Sep 21 2016
From Amiram Eldar, Aug 07 2020: (Start)
Equals Integral_{x=0..Pi} 1/(cos(x)^2 + 1) dx = Integral_{x=0..Pi} 1/(sin(x)^2 + 1) dx.
Equals Integral_{x=-oo..oo} 1/(x^4 + 1) dx.
Equals Integral_{x=-oo..oo} x^2/(x^4 + 1) dx.
Equals Integral_{x=0..oo} log(1 + 1/(2 * x^2)) dx. (End)
Equals Integral_{x=0..2*Pi} 1/(3 + sin(x)) dx; since for a>1: Integral_{x=0..2*Pi} 1/(a + sin(x)) dx = 2*Pi/sqrt(a^2-1). - Bernard Schott, Feb 19 2023
Equals 20/9 - 160*Sum_{n >= 1} 1/((64*n^2 - 1)*(64*n^2 - 4)*(64*n^2 - 9)). - Peter Bala, Nov 09 2023

cp's OEIS Frontend

A093954 Decimal expansion of Pi/(2*sqrt(2)).

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A063448 Decimal expansion of Pi * sqrt(2).

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A247719 Decimal expansion of Integral_{t=0..Pi/2} sqrt(tan(t)) dt.

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A244976 Decimal expansion of Pi/(8*sqrt(2)).

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A047554 Numbers that are congruent to {1, 2, 6, 7} mod 8.

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