OFFSET
1,1
COMMENTS
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
LINKS
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.
FORMULA
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)
EXAMPLE
4.4428829381583662470158809900606936986146216893756902230853...
MATHEMATICA
RealDigits[N[Pi*Sqrt[2], 200]][[1]] (* Vladimir Joseph Stephan Orlovsky, Mar 21 2011*)
PROG
(PARI) \p 400; Pi * sqrt(2)
(PARI) 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
(Python) # 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
CROSSREFS
KEYWORD
cons,nonn
AUTHOR
Jason Earls, Jul 24 2001
EXTENSIONS
Edited by N. J. A. Sloane, May 05 2007
Corrected by Neven Juric, Apr 24 2008
STATUS
approved