A063448 Decimal expansion of Pi * sqrt(2).
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
Examples
4.4428829381583662470158809900606936986146216893756902230853...
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.
- Index entries for transcendental numbers
Programs
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Mathematica
RealDigits[N[Pi*Sqrt[2], 200]][[1]] (* Vladimir Joseph Stephan Orlovsky, Mar 21 2011*)
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PARI
\p 400; Pi * sqrt(2)
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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
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)
Equals Integral_{x=-oo..oo} (x^2 + 1)/(x^4 + 1) dx. - Kritsada Moomuang, Jun 04 2025
Extensions
Edited by N. J. A. Sloane, May 05 2007
Corrected by Neven Juric, Apr 24 2008
Comments