cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A000796 Decimal expansion of Pi (or digits of Pi).

Original entry on oeis.org

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

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Keywords

Comments

Sometimes called Archimedes's constant.
Ratio of a circle's circumference to its diameter.
Also area of a circle with radius 1.
Also surface area of a sphere with diameter 1.
A useful mnemonic for remembering the first few terms: How I want a drink, alcoholic of course, after the heavy lectures involving quantum mechanics ...
Also ratio of surface area of sphere to one of the faces of the circumscribed cube. Also ratio of volume of a sphere to one of the six inscribed pyramids in the circumscribed cube. - Omar E. Pol, Aug 09 2012
Also surface area of a quarter of a sphere of radius 1. - Omar E. Pol, Oct 03 2013
Also the area under the peak-shaped even function f(x)=1/cosh(x). Proof: for the upper half of the integral, write f(x) = (2*exp(-x))/(1+exp(-2x)) = 2*Sum_{k>=0} (-1)^k*exp(-(2k+1)*x) and integrate term by term from zero to infinity. The result is twice the Gregory series for Pi/4. - Stanislav Sykora, Oct 31 2013
A curiosity: a 144 X 144 magic square of 7th powers was recently constructed by Toshihiro Shirakawa. The magic sum = 3141592653589793238462643383279502884197169399375105, which is the concatenation of the first 52 digits of Pi. See the MultiMagic Squares link for details. - Christian Boyer, Dec 13 2013 [Comment revised by N. J. A. Sloane, Aug 27 2014]
x*Pi is also the surface area of a sphere whose diameter equals the square root of x. - Omar E. Pol, Dec 25 2013
Also diameter of a sphere whose surface area equals the volume of the circumscribed cube. - Omar E. Pol, Jan 13 2014
From Daniel Forgues, Mar 20 2015: (Start)
An interesting anecdote about the base-10 representation of Pi, with 3 (integer part) as first (index 1) digit:
358 0
359 3
360 6
361 0
362 0
And the circle is customarily subdivided into 360 degrees (although Pi radians yields half the circle)...
(End)
Sometimes referred to as Archimedes's constant, because the Greek mathematician computed lower and upper bounds of Pi by drawing regular polygons inside and outside a circle. In Germany it was called the Ludolphian number until the early 20th century after the Dutch mathematician Ludolph van Ceulen (1540-1610), who calculated up to 35 digits of Pi in the late 16th century. - Martin Renner, Sep 07 2016
As of the beginning of 2019 more than 22 trillion decimal digits of Pi are known. See the Wikipedia article "Chronology of computation of Pi". - Harvey P. Dale, Jan 23 2019
On March 14, 2019, Emma Haruka Iwao announced the calculation of 31.4 trillion digits of Pi using Google Cloud's infrastructure. - David Radcliffe, Apr 10 2019
Also volume of three quarters of a sphere of radius 1. - Omar E. Pol, Aug 16 2019
On August 5, 2021, researchers from the University of Applied Sciences of the Grisons in Switzerland announced they had calculated 62.8 trillion digits. Guinness World Records has not verified this yet. - Alonso del Arte, Aug 23 2021
The Hermite-Lindemann (1882) theorem states, that if z is a nonzero algebraic number, then e^z is a transcendent number. The transcendence of Pi then results from Euler's relation: e^(i*Pi) = -1. - Peter Luschny, Jul 21 2023
The 10000 words of the book "Not A Wake" by Michael Keith, written in Pilish, match in length the first 10000 digits of Pi. - Paolo Xausa, Aug 07 2025

Examples

			3.1415926535897932384626433832795028841971693993751058209749445923078164062\
862089986280348253421170679821480865132823066470938446095505822317253594081\
284811174502841027019385211055596446229489549303819...
		

References

  • Mohammad K. Azarian, A Summary of Mathematical Works of Ghiyath ud-din Jamshid Kashani, Journal of Recreational Mathematics, Vol. 29(1), pp. 32-42, 1998.
  • J. Arndt & C. Haenel, Pi Unleashed, Springer NY 2001.
  • P. Beckmann, A History of Pi, Golem Press, Boulder, CO, 1977.
  • Florian Cajori, A History of Mathematical Notations, Dover edition (2012), par. 396.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 24, 237-239.
  • J.-P. Delahaye, Le fascinant nombre pi, Pour la Science, Paris 1997.
  • P. Eyard and J.-P. Lafon, The Number Pi, Amer. Math. Soc., 2004.
  • S. R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, Section 1.4.
  • Le Petit Archimede, Special Issue On Pi, Supplement to No. 64-5, May 1980 ADCS Amiens.
  • Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see p. 31.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 1, equations 1:7:1, 1:7:2 at pages 12-13.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See pp. 48-55.

Crossrefs

Cf. A001203 (continued fraction).
Pi in base b: A004601 (b=2), A004602 (b=3), A004603 (b=4), A004604 (b=5), A004605 (b=6), A004606 (b=7), A006941 (b=8), A004608 (b=9), this sequence (b=10), A068436 (b=11), A068437 (b=12), A068438 (b=13), A068439 (b=14), A068440 (b=15), A062964 (b=16), A224750 (b=26), A224751 (b=27), A060707 (b=60). - Jason Kimberley, Dec 06 2012
Decimal expansions of expressions involving Pi: A002388 (Pi^2), A003881 (Pi/4), A013661 (Pi^2/6), A019692 (2*Pi=tau), A019727 (sqrt(2*Pi)), A059956 (6/Pi^2), A060294 (2/Pi), A091925 (Pi^3), A092425 (Pi^4), A092731 (Pi^5), A092732 (Pi^6), A092735 (Pi^7), A092736 (Pi^8), A163973 (Pi/log(2)).
Cf. A001901 (Pi/2; Wallis), A002736 (Pi^2/18; Euler), A007514 (Pi), A048581 (Pi; BBP), A054387 (Pi; Newton), A092798 (Pi/2), A096954 (Pi/4; Machin), A097486 (Pi), A122214 (Pi/2), A133766 (Pi/4 - 1/2), A133767 (5/6 - Pi/4), A166107 (Pi; MGL).

Programs

  • Haskell
    -- see link: Literate Programs
    import Data.Char (digitToInt)
    a000796 n = a000796_list (n + 1) !! (n + 1)
    a000796_list len = map digitToInt $ show $ machin' `div` (10 ^ 10) where
       machin' = 4 * (4 * arccot 5 unity - arccot 239 unity)
       unity = 10 ^ (len + 10)
       arccot x unity = arccot' x unity 0 (unity `div` x) 1 1 where
         arccot' x unity summa xpow n sign
          | term == 0 = summa
          | otherwise = arccot'
            x unity (summa + sign * term) (xpow `div` x ^ 2) (n + 2) (- sign)
          where term = xpow `div` n
    -- Reinhard Zumkeller, Nov 24 2012
    
  • Haskell
    -- See Niemeijer link and also Gibbons link.
    a000796 n = a000796_list !! (n-1) :: Int
    a000796_list = map fromInteger $ piStream (1, 0, 1)
       [(n, a*d, d) | (n, d, a) <- map (\k -> (k, 2 * k + 1, 2)) [1..]] where
       piStream z xs'@(x:xs)
         | lb /= approx z 4 = piStream (mult z x) xs
         | otherwise = lb : piStream (mult (10, -10 * lb, 1) z) xs'
         where lb = approx z 3
               approx (a, b, c) n = div (a * n + b) c
               mult (a, b, c) (d, e, f) = (a * d, a * e + b * f, c * f)
    -- Reinhard Zumkeller, Jul 14 2013, Jun 12 2013
    
  • Macsyma
    py(x) := if equal(6,6+x^2) then 2*x else (py(x:x/3),3*%%-4*(%%-x)^3); py(3.); py(dfloat(%)); block([bfprecision:35], py(bfloat(%))) /* Bill Gosper, Sep 09 2002 */
    
  • Magma
    pi:=Pi(RealField(110)); Reverse(Intseq(Floor(10^105*pi))); // Bruno Berselli, Mar 12 2013
    
  • Maple
    Digits := 110: Pi*10^104:
    ListTools:-Reverse(convert(floor(%), base, 10)); # Peter Luschny, Oct 29 2019
  • Mathematica
    RealDigits[ N[ Pi, 105]] [[1]]
    Table[ResourceFunction["NthDigit"][Pi, n], {n, 1, 102}] (* Joan Ludevid, Jun 22 2022; easy to compute a(10000000)=7 with this function; requires Mathematica 12.0+ *)
  • PARI
    { default(realprecision, 20080); x=Pi; for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b000796.txt", n, " ", d)); } \\ Harry J. Smith, Apr 15 2009
    
  • PARI
    A796=[]; A000796(n)={if(n>#A796, localprec(n*6\5+29); A796=digits(Pi\.1^(precision(Pi)-3))); A796[n]} \\ NOTE: as the other programs, this returns the n-th term of the sequence, with n = 1, 2, 3, ... and not n = 1, 0, -1, -2, .... - M. F. Hasler, Jun 21 2022
    
  • PARI
    first(n)= default(realprecision, n+10); digits(floor(Pi*10^(n-1))) \\ David A. Corneth, Jun 21 2022
    
  • PARI
    lista(nn, p=20)= {my(u=10^(nn+p+1), f(x, u)=my(n=1, q=u\x, r=q, s=1, t); while(t=(q\=(x*x))\(n+=2), r+=(s=-s)*t); r*4); digits((4*f(5, u)-f(239, u))\10^(p+2)); } \\ Machin-like, with p > the maximal number of consecutive 9-digits to be expected (A048940) - Ruud H.G. van Tol, Dec 26 2024
    
  • Python
    from sympy import pi, N; print(N(pi, 1000)) # David Radcliffe, Apr 10 2019
    
  • Python
    from mpmath import mp
    def A000796(n):
        if n >= len(A000796.str): mp.dps = n*6//5+50; A000796.str = str(mp.pi-5/mp.mpf(10)**mp.dps)
        return int(A000796.str[n if n>1 else 0])
    A000796.str = '' # M. F. Hasler, Jun 21 2022
    
  • SageMath
    m=125
    x=numerical_approx(pi, digits=m+5)
    a=[ZZ(i) for i in x.str(skip_zeroes=True) if i.isdigit()]
    a[:m] # G. C. Greubel, Jul 18 2023

Formula

Pi = 4*Sum_{k>=0} (-1)^k/(2k+1) [Madhava-Gregory-Leibniz, 1450-1671]. - N. J. A. Sloane, Feb 27 2013
From Johannes W. Meijer, Mar 10 2013: (Start)
2/Pi = (sqrt(2)/2) * (sqrt(2 + sqrt(2))/2) * (sqrt(2 + sqrt(2 + sqrt(2)))/2) * ... [Viete, 1593]
2/Pi = Product_{k>=1} (4*k^2-1)/(4*k^2). [Wallis, 1655]
Pi = 3*sqrt(3)/4 + 24*(1/12 - Sum_{n>=2} (2*n-2)!/((n-1)!^2*(2*n-3)*(2*n+1)*2^(4*n-2))). [Newton, 1666]
Pi/4 = 4*arctan(1/5) - arctan(1/239). [Machin, 1706]
Pi^2/6 = 3*Sum_{n>=1} 1/(n^2*binomial(2*n,n)). [Euler, 1748]
1/Pi = (2*sqrt(2)/9801) * Sum_{n>=0} (4*n)!*(1103+26390*n)/((n!)^4*396^(4*n)). [Ramanujan, 1914]
1/Pi = 12*Sum_{n>=0} (-1)^n*(6*n)!*(13591409 + 545140134*n)/((3*n)!*(n!)^3*(640320^3)^(n+1/2)). [David and Gregory Chudnovsky, 1989]
Pi = Sum_{n>=0} (1/16^n) * (4/(8*n+1) - 2/(8*n+4) - 1/(8*n+5) - 1/(8*n+6)). [Bailey-Borwein-Plouffe, 1989] (End)
Pi = 4 * Sum_{k>=0} 1/(4*k+1) - 1/(4*k+3). - Alexander R. Povolotsky, Dec 25 2008
Pi = 4*sqrt(-1*(Sum_{n>=0} (i^(2*n+1))/(2*n+1))^2). - Alexander R. Povolotsky, Jan 25 2009
Pi = Integral_{x=-oo..oo} dx/(1+x^2). - Mats Granvik and Gary W. Adamson, Sep 23 2012
Pi - 2 = 1/1 + 1/3 - 1/6 - 1/10 + 1/15 + 1/21 - 1/28 - 1/36 + 1/45 + ... [Jonas Castillo Toloza, 2007], that is, Pi - 2 = Sum_{n>=1} (1/((-1)^floor((n-1)/2)*(n^2+n)/2)). - José de Jesús Camacho Medina, Jan 20 2014
Pi = 3 * Product_{t=img(r),r=(1/2+i*t) root of zeta function} (9+4*t^2)/(1+4*t^2) <=> RH is true. - Dimitris Valianatos, May 05 2016
From Ilya Gutkovskiy, Aug 07 2016: (Start)
Pi = Sum_{k>=1} (3^k - 1)*zeta(k+1)/4^k.
Pi = 2*Product_{k>=2} sec(Pi/2^k).
Pi = 2*Integral_{x>=0} sin(x)/x dx. (End)
Pi = 2^{k + 1}*arctan(sqrt(2 - a_{k - 1})/a_k) at k >= 2, where a_k = sqrt(2 + a_{k - 1}) and a_1 = sqrt(2). - Sanjar Abrarov, Feb 07 2017
Pi = Integral_{x = 0..2} sqrt(x/(2 - x)) dx. - Arkadiusz Wesolowski, Nov 20 2017
Pi = lim_{n->oo} 2/n * Sum_{m=1,n} ( sqrt( (n+1)^2 - m^2 ) - sqrt( n^2 - m^2 ) ). - Dimitri Papadopoulos, May 31 2019
From Peter Bala, Oct 29 2019: (Start)
Pi = Sum_{n >= 0} 2^(n+1)/( binomial(2*n,n)*(2*n + 1) ) - Euler.
More generally, Pi = (4^x)*x!/(2*x)! * Sum_{n >= 0} 2^(n+1)*(n+x)!*(n+2*x)!/(2*n+2*x+1)! = 2*4^x*x!^2/(2*x+1)! * hypergeom([2*x+1,1], [x+3/2], 1/2), valid for complex x not in {-1,-3/2,-2,-5/2,...}. Here, x! is shorthand notation for the function Gamma(x+1). This identity may be proved using Gauss's second summation theorem.
Setting x = 3/4 and x = -1/4 (resp. x = 1/4 and x = -3/4) in the above identity leads to series representations for the constant A085565 (resp. A076390). (End)
Pi = Im(log(-i^i)) = log(i^i)*(-2). - Peter Luschny, Oct 29 2019
From Amiram Eldar, Aug 15 2020: (Start)
Equals 2 + Integral_{x=0..1} arccos(x)^2 dx.
Equals Integral_{x=0..oo} log(1 + 1/x^2) dx.
Equals Integral_{x=0..oo} log(1 + x^2)/x^2 dx.
Equals Integral_{x=-oo..oo} exp(x/2)/(exp(x) + 1) dx. (End)
Equals 4*(1/2)!^2 = 4*Gamma(3/2)^2. - Gary W. Adamson, Aug 23 2021
From Peter Bala, Dec 08 2021: (Start)
Pi = 32*Sum_{n >= 1} (-1)^n*n^2/((4*n^2 - 1)*(4*n^2 - 9))= 384*Sum_{n >= 1} (-1)^(n+1)*n^2/((4*n^2 - 1)*(4*n^2 - 9)*(4*n^2 - 25)).
More generally, it appears that for k = 1,2,3,..., Pi = 16*(2*k)!*Sum_{n >= 1} (-1)^(n+k+1)*n^2/((4*n^2 - 1)* ... *(4*n^2 - (2*k+1)^2)).
Pi = 32*Sum_{n >= 1} (-1)^(n+1)*n^2/(4*n^2 - 1)^2 = 768*Sum_{n >= 1} (-1)^(n+1)*n^2/((4*n^2 - 1)^2*(4*n^2 - 9)^2).
More generally, it appears that for k = 0,1,2,..., Pi = 16*Catalan(k)*(2*k)!*(2*k+2)!*Sum_{n >= 1} (-1)^(n+1)*n^2/((4*n^2 - 1)^2* ... *(4*n^2 - (2*k+1)^2)^2).
Pi = (2^8)*Sum_{n >= 1} (-1)^(n+1)*n^2/(4*n^2 - 1)^4 = (2^17)*(3^5)*Sum_{n >= 2} (-1)^n*n^2*(n^2 - 1)/((4*n^2 - 1)^4*(4*n^2 - 9)^4) = (2^27)*(3^5)*(5^5)* Sum_{n >= 3} (-1)^(n+1)*n^2*(n^2 - 1)*(n^2 - 4)/((4*n^2 - 1)^4*(4*n^2 - 9)^4*(4*n^2 - 25)^4). (End)
For odd n, Pi = (2^(n-1)/A001818((n-1)/2))*gamma(n/2)^2. - Alan Michael Gómez Calderón, Mar 11 2022
Pi = 4/phi + Sum_{n >= 0} (1/phi^(12*n)) * ( 8/((12*n+3)*phi^3) + 4/((12*n+5)*phi^5) - 4/((12*n+7)*phi^7) - 8/((12*n+9)*phi^9) - 4/((12*n+11)*phi^11) + 4/((12*n+13)*phi^13) ) where phi = (1+sqrt(5))/2. - Chittaranjan Pardeshi, May 16 2022
Pi = sqrt(3)*(27*S - 36)/24, where S = A248682. - Peter Luschny, Jul 22 2022
Equals Integral_{x=0..1} 1/sqrt(x-x^2) dx. - Michal Paulovic, Sep 24 2023
From Peter Bala, Oct 28 2023: (Start)
Pi = 48*Sum_{n >= 0} (-1)^n/((6*n + 1)*(6*n + 3)*(6*n + 5)).
More generally, it appears that for k >= 0 we have Pi = A(k) + B(k)*Sum_{n >= 0} (-1)^n/((6*n + 1)*(6*n + 3)*...*(6*n + 6*k + 5)), where A(k) is a rational approximation to Pi and B(k) = (3 * 2^(3*k+3) * (3*k + 2)!) / (2^(3*k+1) - (-1)^k). The first few values of A(k) for k >= 0 are [0, 256/85, 65536/20955, 821559296/261636375, 6308233216/2008080987, 908209489444864/289093830828075, ...].
Pi = 16/5 - (288/5)*Sum_{n >= 0} (-1)^n * (6*n + 1)/((6*n + 1)*(6*n + 3)*...*(6*n + 9)).
More generally, it appears that for k >= 0 we have Pi = C(k) + D(k)*Sum_{n >= 0} (-1)^n* (6*n + 1)/((6*n + 1)*(6*n + 3)*...*(6*n + 6*k + 3)), where C(k) and D(k) are rational numbers. The case k = 0 is the Madhava-Gregory-Leibniz series for Pi.
Pi = 168/53 + (288/53)*Sum_{n >= 0} (-1)^n * (42*n^2 + 25*n)/((6*n + 1)*(6*n + 3)*(6*n + 5)*(6*n + 7)).
More generally, it appears that for k >= 1 we have Pi = E(k) + F(k)*Sum_{n >= 0} (-1)^n * (6*(6*k + 1)*n^2 + (24*k + 1)*n)/((6*n + 1)*(6*n + 3)*...*(6*n + 6*k + 1)), where E(k) and F(k) are rational numbers. (End)
From Peter Bala, Nov 10 2023: (Start)
The series representation Pi = 4 * Sum_{k >= 0} 1/(4*k + 1) - 1/(4*k + 3) given above by Alexander R. Povolotsky, Dec 25 2008, is the case n = 0 of the more general result (obtained by the WZ method): for n >= 0, there holds
Pi = Sum_{j = 0.. n-1} 2^(j+1)/((2*j + 1)*binomial(2*j,j)) + 8*(n+1)!*Sum_{k >= 0} 1/((4*k + 1)*(4*k + 3)*...*(4*k + 2*n + 3)).
Letting n -> oo gives the rapidly converging series Pi = Sum_{j >= 0} 2^(j+1)/((2*j + 1)*binomial(2*j,j)) due to Euler.
More generally, it appears that for n >= 1, Pi = 1/(2*n-1)!!^2 * Sum_{j >= 0} (Product_{i = 0..2*n-1} j - i) * 2^(j+1)/((2*j + 1)*binomial(2*j,j)).
For any integer n, Pi = (-1)^n * 4 * Sum_{k >= 0} 1/(4*k + 1 + 2*n) - 1/(4*k + 3 - 2*n). (End)
Pi = Product_{k>=1} ((k^3*(k + 2)*(2*k + 1)^2)/((k + 1)^4*(2*k - 1)^2))^k. - Antonio Graciá Llorente, Jun 13 2024
Equals Integral_{x=0..2} sqrt(8 - x^2) dx - 2 (see Ambrisi and Rizzi). - Stefano Spezia, Jul 21 2024
Equals 3 + 4*Sum_{k>0} (-1)^(k+1)/(4*k*(1+k)*(1+2*k)) (see Wells at p. 53). - Stefano Spezia, Aug 31 2024
Equals 4*Integral_{x=0..1} sqrt(1 - x^2) dx = lim_{n->oo} (4/n^2)*Sum_{k=0..n} sqrt(n^2 - k^2) (see Finch). - Stefano Spezia, Oct 19 2024
Equals Beta(1/2,1/2) (see Shamos). - Stefano Spezia, Jun 03 2025
From Kritsada Moomuang, Jun 18 2025: (Start)
Equals 2 + Integral_{x=0..1} 1/(sqrt(x)*(1 + sqrt(1 - x))) dx.
Equals 2 + Integral_{x=0..1} log(1 + sqrt(1 - x))/sqrt(x) dx. (End)
Pi = 2*arccos(1/phi) + arccos(1/phi^3) = 4*arcsin(1/phi) + 2*arcsin(1/phi^3) where phi = (1+sqrt(5))/2. - Chittaranjan Pardeshi, Jul 02 2025
Pi = Sum_{n >= 0} zeta(2*n)*(2^(2*n - 1) - 1)/2^(4*n - 3). - Andrea Pinos, Jul 29 2025

Extensions

Additional comments from William Rex Marshall, Apr 20 2001

A002388 Decimal expansion of Pi^2.

Original entry on oeis.org

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

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Author

Keywords

Comments

Also equals the volume of revolution of the sine or cosine curve for one full period, Integral_{x=0..2*Pi} sin(x)^2 dx. - Robert G. Wilson v, Dec 15 2005
Equals Sum_{n>0} 20/A026424(n)^2 where A026424 are the integers such that the number of prime divisors (counted with multiplicity) is odd. - Michel Lagneau, Oct 23 2015

Examples

			9.869604401089358618834490999876151135313699407240790626413349376220044...
		

References

  • W. E. Mansell, Tables of Natural and Common Logarithms. Royal Society Mathematical Tables, Vol. 8, Cambridge Univ. Press, 1964, p. XVIII.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See p. 76.

Crossrefs

Programs

  • Magma
    R:= RealField(100); Pi(R)^2; // G. C. Greubel, Mar 08 2018
    
  • Maple
    Digits:=100: evalf(Pi^2); # Wesley Ivan Hurt, Jul 13 2014
  • Mathematica
    RealDigits[Pi^2, 10, 111][[1]] (* Robert G. Wilson v, Dec 15 2005 *)
  • PARI
    default(realprecision, 20080); x=Pi^2; for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b002388.txt", n, " ", d)); \\ Harry J. Smith, May 31 2009
    
  • Python
    # Use some guard digits when computing.
    # BBP formula (9 / 8) P(2, 64, 6, (16, -24, -8, -6,  1, 0)).
    from decimal import Decimal as dec, getcontext
    def BBPpi2(n: int) -> dec:
        getcontext().prec = n
        s = dec(0); f = dec(1); g = dec(64)
        for k in range(int(n * 0.5536546824812272) + 1):
            sixk = dec(6 * k)
            s += f * ( dec(16) / (sixk + 1) ** 2 - dec(24) / (sixk + 2) ** 2
                     - dec(8)  / (sixk + 3) ** 2 - dec(6)  / (sixk + 4) ** 2
                     + dec(1)  / (sixk + 5) ** 2 )
            f /= g
        return (s * dec(9)) / dec(8)
    print(BBPpi2(200))  # Peter Luschny, Nov 03 2023

Formula

Pi^2 = 11/2 + 16 * Sum_{k>=2} (1+k-k^3)/(1-k^2)^3. - Alexander R. Povolotsky, May 04 2009
Pi^2 = 3*(Sum_{n>=1} ((2*n+1)^2/Sum_{k=1..n} k^3)/4 - 1). - Alexander R. Povolotsky, Jan 14 2011
Pi^2 = (3/2)*(Sum_{n>=1} ((7*n^2+2*n-2)/(2*n^2-1)/(n+1)^5) - zeta(3) - 3*zeta(5) + 22 - 7*polygamma(0,1-1/sqrt(2)) + 5*sqrt(2)*polygamma(0,1-1/sqrt(2)) - 7*polygamma(0,1+1/sqrt(2)) - 5*sqrt(2)*polygamma(0,1+1/sqrt(2)) - 14*EulerGamma). - Alexander R. Povolotsky, Aug 13 2011
Also equals 32*Integral_{x=0..1} arctan(x)/(1+x^2) dx. - Jean-François Alcover, Mar 25 2013
From Peter Bala, Feb 05 2015: (Start)
Pi^2 = 20 * Integral_{x = 0 .. log(phi)} x*coth(x) dx, where phi = (1/2)*(1 + sqrt(5)) is the golden ratio.
Pi^2 = 10 * Sum_{k >= 0} binomial(2*k,k)*(1/(2*k + 1)^2)*(-1/16)^k. Similar series expansions hold for Pi/3 (see A019670) and (7*/216)*Pi^3 (see A091925).
The integer sequences A(n) := 2^n*(2*n + 1)!^2/n! and B(n) := A(n)*( Sum_{k = 0..n} binomial(2*k,k)*1/(2*k + 1)^2*(-1/16)^k ) both satisfy the second order recurrence equation u(n) = (24*n^3 + 44*n^2 + 2*n + 1)*u(n-1) + 8*(n - 1)*(2*n - 1)^5*u(n-2). From this observation we can obtain the continued fraction expansion Pi^2/10 = 1 - 1/(72 + 8*3^5/(373 + 8*2*5^5/(1051 + ... + 8*(n - 1)*(2*n - 1)^5/((24*n^3 + 44*n^2 + 2*n + 1) + ... )))). Cf. A093954. (End)
Pi^2 = A304656 * A093602 = (gamma(0, 1/6) - gamma(0, 5/6))*(gamma(0, 2/6) - gamma(0, 4/6)), where gamma(n,x) are the generalized Stieltjes constants. This formula can also be expressed by the polygamma function. - Peter Luschny, May 16 2018
Equals 8 + Sum_{k>=1} 1/(k^2 - 1/4)^2 = -8 + Sum_{k>=0} 1/(k^2 - 1/4)^2. - Amiram Eldar, Aug 21 2020
From Peter Bala, Dec 10 2021: (Start)
Pi^2 = (2^6)*Sum_{n >= 1} n^2/(4*n^2 - 1)^2 = (2^11)*Sum_{n >= 1} n^2/ ((4*n^2 - 1)^2*(4*n^2 - 3^2)^2) = ((2^19)*(3^2)/7) * Sum_{n >= 1} n^2/((4*n^2 - 1)^2*(4*n^2 - 3^2)^2*(4*n^2 - 5^2)^2).
More generally, it appears that for k >= 0 we have Pi^2 = (2*k+1)*2^(4*k+6) * (2*k)!^4/(4*k)! * Sum_{n >= 1} n^2/((4*n^2 - 1)^2*...*(4*n^2 - (2*k+1)^2)^2).
It also appears that for k >= 0 we have Pi^2 = (-1)^k * 2^(6*k+8)*(2*k+1)^3/(6*k+1) * ((2*k)!^6 * (3*k)!)/(k!^3 * (6*k)!) * Sum_{n >= 1} n^2/((4*n^2 - 1)^3*...*(4*n^2 - (2*k+1)^2)^3). (End)
From Peter Bala, Oct 27 2023: (Start)
Pi^2 = 10 - Sum_{n >= 1} 1/(n*(n + 1))^3.
Pi^2 = 6217/630 + (648/35)*Sum_{n >= 1} 1/(n*(n + 1)*(n + 2)*(n + 3))^3.
The general result (verified using the WZ method - see Wilf) is : for n >= 0,
Pi^2 = A(n) + (-1)^(n+1) * B(n)*Sum_{k >= 1} 1/(k*(k + 1)*...*(k + 2*n + 1))^3, where A(n) = 10 - Sum_{i = 1..n} (-1)^(i+1) * (56*i^2 + 24*i + 3)*(2*i)!^3*(3*i)!/(2*i^2*(2*i + 1)*(6*i + 1)!*i!^3) and B(n) = (2*n + 1)!^6 * (3*n)! / ( (2*n + 1)*(6*n + 1)!*n!^3 ).
Letting n -> oo gives the fast converging alternating series
Pi^2 = 10 - Sum_{i >= 1} (-1)^(i+1) * (56*i^2 + 24*i + 3)*(2*i)!^3 * (3*i)!/(2*i^2*(2*i + 1)*(6*i + 1)!*i!^3). The i-th summand of the series is asymptotic to (14/3) * 1/(i^2 * 27^i) so taking 70 terms of the series gives a value for Pi^2 accurate to more than 100 decimal places.
The series representation Pi^2 = 3*Sum_{k >= 1} (2*k)/k^3 can be accelerated to give the faster converging series
Pi^2 = 99/10 - (8/5)*Sum_{k >= 1} (2*k + 2)/(k*(k + 1)*(k + 2))^3 and
Pi^2 = 54715/5544 + (41472/385)*Sum_{k >= 1} (2*k + 4)/(k*(k + 1)*(k + 2)*(k + 3)*(k + 4))^3.
The general result is: for n >= 1, Pi^2 = C(n) + (-1)^n * D(n)*Sum_{k >= 1} (2*k + 2*n)/(k*(k + 1)*...*(k + 2*n))^3, where C(n) = A(n) - 10*(-1)^n*(3*n)!*(2*n)!^3/((2*n + 1)*n!^3*(6*n + 1)!) and D(n) = (2*n)!^6 * (3*n)! / ( 2*n*(6*n - 1)!*n!^3 ). (End)
Equals 9 + 3*Sum_{n>=1} 1/((n^2*(n+1)^2)). - Davide Rotondo, May 29 2025

Extensions

More terms from Robert G. Wilson v, Dec 15 2005

A019692 Decimal expansion of 2*Pi.

Original entry on oeis.org

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

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Author

Keywords

Comments

Pi/5 or 2*Pi/10 is the expected surface area containing completely a Brownian curve (trajectory) on a plane. - Lekraj Beedassy, Jul 28 2005
Bob Palais considers this a more fundamental constant than Pi, see the Palais reference and link. - Jonathan Vos Post, Sep 10 2010
The Persian mathematician Jamshid al-Kashi seems to have been the first to use the circumference divided by the radius as the circle constant. In Treatise on the Circumference published 1424 he calculated the circumference of a unit circle to 9 sexagesimal places. - Peter Harremoës, John W. Nicholson, Aug 02 2012
"Proponents of a new mathematical constant tau (τ), equal to two times π, have argued that a constant based on the ratio of a circle's circumference to its radius rather than to its diameter would be more natural and would simplify many formulas" (from Wikipedia). - Jonathan Sondow, Aug 15 2012
The constant 2*Pi appears in the formula for the period T of a simple gravity pendulum. For small angles this period is given by Christiaan Huygens’s law, i.e., T = 2*Pi*sqrt(L/g), see for more information A223067. - Johannes W. Meijer, Mar 14 2013
There are seven consecutive nines at positions 762 to 768. - Roland Kneer, Jul 05 2013
Volume of a cylinder in which a sphere of radius 1 can be inscribed. - Omar E. Pol, Sep 25 2013
2*Pi is also the surface area of a sphere whose diameter equals the square root of 2. More generally, x*Pi is also the surface area of a sphere whose diameter equals the square root of x. - Omar E. Pol, Dec 18 2013
From Bernard Schott, Jan 31 2020: (Start)
Also, (2*Pi)*a^2 is the area of the deltoid (an hypocycloid with three cusps) whose Cartesian parametrization is:
x = a * (2*cos(t) + cos(2*t)),
y = a * (2*sin(t) - sin(2*t)).
The length of this deltoid is 16*a. See the curve at the Mathcurve link. (End)
Pi/5 = 0.1 * 2*Pi is the mean area of the plane triangles formed by 3 points independently and uniformly chosen at random on the surface of a unit-radius sphere. - Amiram Eldar, Aug 06 2020

Examples

			6.283185307179586476925286766559005768394338798750211641949889184615632...
		

References

  • Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 1.4, p. 17.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See p. 69.

Crossrefs

Cf. A058291 (continued fraction).
Cf. A093828 (astroid), A180434 (loop of strophoid), A197723 (cardioid).

Programs

  • Julia
    using Nemo
    RR = RealField(334)
    tau = const_pi(RR) + const_pi(RR)
    tau |> println # Peter Luschny, Mar 14 2018
    
  • Magma
    R:= RealField(100); 2*Pi(R); // G. C. Greubel, Mar 08 2018
    
  • Mathematica
    RealDigits[N[2 Pi, 6! ]] (* Vladimir Joseph Stephan Orlovsky, Dec 02 2009 *)
  • PARI
    default(realprecision, 20080); x=2*Pi; for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b019692.txt", n, " ", d)); \\ Harry J. Smith, May 31 2009
    
  • Python
    # Use some guard digits when computing.
    # BBP formula P(1, 16, 8, (0, 8, 4, 4, 0, 0, -1, 0)).
    from decimal import Decimal as dec, getcontext
    def BBPtau(n: int) -> dec:
        getcontext().prec = n
        s = dec(0); f = dec(1); g = dec(16)
        for k in range(n):
            ek = dec(8 * k)
            s += f * ( dec(8) / (ek + 2) + dec(4) / (ek + 3)
                     + dec(4) / (ek + 4) - dec(1) / (ek + 7))
            f /= g
        return s
    print(BBPtau(200))  # Peter Luschny, Nov 03 2023

Formula

e^(Zeta'(0)/Zeta(0)) = 2*Pi. - Peter Luschny, Jun 17 2018
From Peter Bala, Oct 30 2019: (Start)
2*Pi = Sum_{n >= 0} (-1)^n*( 1/(n + 1/6) + 1/(n + 5/6) ).
2*Pi = Sum_{n >= 0} (-1)^n*( 1/(n + 1/10) - 1/(n + 3/10) - 1/(n + 7/10) + 1/(n + 9/10) ). Cf. A091925 and A244979. (End)
From Amiram Eldar, Aug 06 2020: (Start)
Equals Gamma(1/6)*Gamma(5/6).
Equals Integral_{x=0..oo} log(1 + 1/x^6) dx.
Equals Integral_{x=0..oo} log(1 + 4/x^2) dx.
Equals Integral_{x=-oo..oo} exp(x/6)/(exp(x) + 1) dx.
Equals Sum_{k>=0} 1/((k + 1/4)*(k + 3/4)). (End)
Equals 4*Integral_{x=0..1} 1/sqrt(1 - x^2) dx (see Finch). - Stefano Spezia, Oct 19 2024

A019670 Decimal expansion of Pi/3.

Original entry on oeis.org

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

Views

Author

N. J. A. Sloane, Dec 11 1996

Keywords

Comments

With an offset of zero, also the decimal expansion of Pi/30 ~ 0.104719... which is the average arithmetic area of the 0-winding sectors enclosed by a closed Brownian planar path, of a given length t, according to Desbois, p. 1. - Jonathan Vos Post, Jan 23 2011
Polar angle (or apex angle) of the cone that subtends exactly one quarter of the full solid angle. See comments in A238238. - Stanislav Sykora, Jun 07 2014
60 degrees in radians. - M. F. Hasler, Jul 08 2016
Volume of a quarter sphere of radius 1. - Omar E. Pol, Aug 17 2019
Also smallest positive zero of Sum_{k>=1} cos(k*x)/k = -log(2*|sin(x/2)|). Proof of this identity: Sum_{k>=1} cos(k*x)/k = Re(Sum_{k>=1} exp(k*x*i)/k) = Re(-log(1-exp(x*i))) = -log(2*|sin(x/2)|), x != 2*m*Pi, where i = sqrt(-1). - Jianing Song, Nov 09 2019
The area of a circle circumscribing a unit-area regular dodecagon. - Amiram Eldar, Nov 05 2020

Examples

			Pi/3 = 1.04719755119659774615421446109316762806572313312503527365831486...
From _Peter Bala_, Nov 16 2016: (Start)
Case n = 1. Pi/3 = 18 * Sum_{k >= 0} (-1)^(k+1)( 1/((6*k - 5)*(6*k + 1)*(6*k + 7)) + 1/((6*k - 1)*(6*k + 5)*(6*k + 11)) ).
Using the methods of Borwein et al. we can find the following asymptotic expansion for the tails of this series: for N divisible by 6 there holds Sum_{k >= N/6} (-1)^(k+1)( 1/((6*k - 5)*(6*k + 1)*(6*k + 7)) + 1/((6*k - 1)*(6*k + 5)*(6*k + 11)) ) ~ 1/N^3 + 6/N^5 + 1671/N ^7 - 241604/N^9 + ..., where the sequence [1, 0, 6, 0, 1671, 0, -241604, 0, ...] is the sequence of coefficients in the expansion of ((1/18)*cosh(2*x)/cosh(3*x)) * sinh(3*x)^2 = x^2/2! + 6*x^4/4! + 1671*x^6/6! - 241604*x^8/8! + .... Cf. A024235, A278080 and A278195. (End)
		

References

  • Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 8.3, p. 489.

Crossrefs

Integral_{x=0..oo} 1/(1+x^m) dx: A013661 (m=2), A248897 (m=3), A093954 (m=4), A352324 (m=5), this sequence (m=6), A352125 (m=8), A094888 (m=10).

Programs

Formula

A third of A000796, a sixth of A019692, the square root of A100044.
Sum_{k >= 0} (-1)^k/(6k+1) + (-1)^k/(6k+5). - Charles R Greathouse IV, Sep 08 2011
Product_{k >= 1}(1-(6k)^(-2))^(-1). - Fred Daniel Kline, May 30 2013
From Peter Bala, Feb 05 2015: (Start)
Pi/3 = Sum {k >= 0} binomial(2*k,k)*1/(2*k + 1)*(1/16)^k = 2F1(1/2,1/2;3/2;1/4). Similar series expansions hold for Pi^2 (A002388), Pi^3 (A091925) and Pi/(2*sqrt(2)) (A093954.)
The integer sequences A(n) := 4^n*(2*n + 1)! and B(n) := A(n)*( Sum {k = 0..n} binomial(2*k,k)*1/(2*k + 1)*(1/16)^k ) both satisfy the second-order recurrence equation u(n) = (20*n^2 + 4*n + 1)*u(n-1) - 8*(n - 1)*(2*n - 1)^3*u(n-2). From this observation we can obtain the continued fraction expansion Pi/3 = 1 + 1/(24 - 8*3^3/(89 - 8*2*5^3/(193 - 8*3*7^3/(337 - ... - 8*(n - 1)*(2*n - 1)^3/((20*n^2 + 4*n + 1) - ... ))))). Cf. A002388 and A093954. (End)
Equals Sum_{k >= 1} arctan(sqrt(3)*L(2k)/L(4k)) where L=A000032. See also A005248 and A056854. - Michel Marcus, Mar 29 2016
Equals Product_{n >= 1} A016910(n) / A136017(n). - Fred Daniel Kline, Jun 09 2016
Equals Integral_{x=-oo..oo} sech(x)/3 dx. - Ilya Gutkovskiy, Jun 09 2016
From Peter Bala, Nov 16 2016: (Start)
Euler's series transformation applied to the series representation Pi/3 = Sum_{k >= 0} (-1)^k/(6*k + 1) + (-1)^k/(6*k + 5) given above by Greathouse produces the faster converging series Pi/3 = (1/2) * Sum_{n >= 0} 3^n*n!*( 1/(Product_{k = 0..n} (6*k + 1)) + 1/(Product_{k = 0..n} (6*k + 5)) ).
The series given above by Greathouse is the case n = 0 of the more general result Pi/3 = 9^n*(2*n)! * Sum_{k >= 0} (-1)^(k+n)*( 1/(Product_{j = -n..n} (6*k + 1 + 6*j)) + 1/(Product_{j = -n..n} (6*k + 5 + 6*j)) ) for n = 0,1,2,.... Cf. A003881. See the example section for notes on the case n = 1.(End)
Equals Product_{p>=5, p prime} p/sqrt(p^2-1). - Dimitris Valianatos, May 13 2017
Equals A019699/4 or A019693/2. - Omar E. Pol, Aug 17 2019
Equals Integral_{x >= 0} (sin(x)/x)^4 = 1/2 + Sum_{n >= 0} (sin(n)/n)^4, by the Abel-Plana formula. - Peter Bala, Nov 05 2019
Equals Integral_{x=0..oo} 1/(1 + x^6) dx. - Bernard Schott, Mar 12 2022
Pi/3 = -Sum_{n >= 1} i/(n*P(n, 1/sqrt(-3))*P(n-1, 1/sqrt(-3))), where i = sqrt(-1) and P(n, x) denotes the n-th Legendre polynomial. The first twenty terms of the series gives the approximation Pi/3 = 1.04719755(06...) correct to 8 decimal places. - Peter Bala, Mar 16 2024
Equals Integral_{x >= 0} (2*x^2 + 1)/((x^2 + 1)*(4*x^2 + 1)) dx. - Peter Bala, Feb 12 2025

A092425 Decimal expansion of Pi^4.

Original entry on oeis.org

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

Views

Author

Mohammad K. Azarian, Mar 22 2004

Keywords

Examples

			97.40909103400243723644033268870511124972758567268542169146785938997085...
		

Crossrefs

Cf. A000796 (Pi), A002388 (Pi^2), A091925 (Pi^3), A092731 (Pi^5), A092732 (Pi^6), A092735 (Pi^7), A092736 (Pi^8), A058286 (continued fraction), A013662.

Programs

  • Magma
    R:= RealField(150); (Pi(R))^4; // G. C. Greubel, Mar 09 2018
    
  • Magma
    R:=RealField(110); SetDefaultRealField(R); n:=Pi(R)^4; Reverse(Intseq(Floor(10^98*n))); // Bruno Berselli, Mar 12 2018
  • Mathematica
    RealDigits[Pi^4, 10, 100][[1]] (* G. C. Greubel, Mar 09 2018 *)
  • PARI
    default(realprecision, 20080); x=Pi^4/10; for (n=2, 20000, d=floor(x); x=(x-d)*10; write("b092425.txt", n, " ", d)); \\ Harry J. Smith, Jun 22 2009
    

Formula

Equals 120 * Sum_{j>=1} Sum_{i=1..j-1} 1/(i*j)^2. - Enrique Pérez Herrero, Jun 29 2012
Equals Sum_{k>=1} k*(k+1)*(k+2)*zeta(k+3)/2^(k-1). - Amiram Eldar, May 21 2021
From Peter Bala, Oct 21 2023: (Start)
Pi^4 = 90*Sum_{n >= 1} 1/n^4 (Euler).
The following faster converging series representations for the constant Pi^4 may be easily verified using partial fraction expansions of the summands of the series. Presumably, these are the first three cases of an infinite family of similar results.
Let P(n) = n*(n + 1)*(n + 2)/2!. Then Pi^4 = 1575/16 - 15*Sum_{n >= 1} d/dn(P(n))/P(n)^4.
Let Q(n) = n*(n + 1)*(n + 2)*(n + 3)*(n + 4)/4!. Then Pi^4 = 673165/6912 + Sum_{n >= 1} d/dn(Q(n))/Q(n)^4.
Let R(n) = n*(n + 1)*(n + 2)*(n + 3)*(n + 4)*(n + 5)*(n + 6)/6!. Then Pi^4 = 5610787/57600 - (3/56)*Sum_{n >= 1} d/dn(R(n))/R(n)^4.
Taking 10 terms of the last series gives the approximation Pi^4 = 97.4090910340
024372(50...), correct to 16 decimal places. (End)

A092731 Decimal expansion of Pi^5.

Original entry on oeis.org

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

Views

Author

Mohammad K. Azarian, Apr 12 2004

Keywords

Examples

			306.0196847852814532
		

Crossrefs

Programs

Formula

From Peter Bala, Oct 31 2019: (Start)
Pi^5 = (4!/(2*305)) * Sum_{n >= 0} (-1)^n*( 1/(n + 1/6)^5 + 1/(n + 5/6)^5 ), where 305 = ((3^5 + 1)/4)*A000364(2) = A002437(2).
Pi^5 = (4!/(2*3905)) * Sum_{n >= 0} (-1)^n*( 1/(n + 1/10)^5 - 1/(n + 3/10)^5 - 1/(n + 7/10)^5 + 1/(n + 9/10)^5 ), where 3905 = ((5^5 - 1)/4)*A000364(2).
Cf. A019692, A091925 and A092735. (End)

A092742 Decimal expansion of 1/Pi^2.

Original entry on oeis.org

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

Views

Author

Mohammad K. Azarian, Apr 12 2004

Keywords

Comments

The asymptotic density of squarefree numbers that are divisible by 5. - Amiram Eldar, Mar 25 2021

Examples

			0.101321183642337771443879463209727638904358774672246548845609...
		

References

  • Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 3.6.1, p. 220.

Crossrefs

Cf. A000796 (Pi), A002388 (Pi^2), A091925 (Pi^3), A092425 (Pi^4), A092731 (Pi^5), A092732 (Pi^6), A092735 (Pi^7), A092736 (Pi^8).
Cf. A049541 (1/Pi), A092743 (1/Pi^3), A092744 (1/Pi^4), A092745 (1/Pi^5), A092746 (1/Pi^6), A092747 (1/Pi^7), A092748 (1/Pi^8).

Programs

A092735 Decimal expansion of Pi^7.

Original entry on oeis.org

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

Views

Author

Mohammad K. Azarian, Apr 12 2004

Keywords

Comments

Wentworth (1903) shows how to compute the tangent of 15 degrees (A019913) to five decimal places by the laborious process of adding up the first few terms of Pi/12 + Pi^3/5184 + 2Pi^5/3732480 + 17Pi^7/11287019520 + ... - Alonso del Arte, Mar 13 2015

Examples

			3020.293227776792067514206493...
		

References

  • George Albert Wentworth, New Plane and Spherical Trigonometry, Surveying, and Navigation. Boston: The Atheneum Press (1903): 240.

Crossrefs

Programs

Formula

From Peter Bala, Oct 30 2019: (Start)
Pi^7 = (6!/(2*33367)) * Sum_{n >= 0} (-1)^n*( 1/(n + 1/6)^7 + 1/(n + 5/6)^7 ), where 33367 = ((3^7 + 1)/4)*A000364(3) = A002437(3).
Pi^7 = (6!/(2*1191391)) * Sum_{n >= 0} (-1)^n*( 1/(n + 1/10)^7 - 1/(n + 3/10)^7 - 1/(n + 7/10)^7 + 1/(n + 9/10)^7 ), where 1191391 = ((5^7 - 1)/4)*A000364(3). (End)

A164109 Decimal expansion of Pi^4/3.

Original entry on oeis.org

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

Views

Author

R. J. Mathar, Aug 10 2009

Keywords

Comments

Surface area of the 8-dimensional unit sphere.

Examples

			32.4696970113341457454801108962350370832425285...
		

Crossrefs

Programs

Formula

Equals 8*A164108 = A019670*A091925 = A092425/3.
Pi^4/30 = Sum_{k>=1} H(k)*H(k+1)/(k+1)^2, where H(k) = A001008(k)/A002805(k) is the k-th harmonic number. - Amiram Eldar, Aug 19 2025

A092039 Decimal expansion of cube root of Pi.

Original entry on oeis.org

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

Views

Author

Mohammad K. Azarian, Mar 27 2004

Keywords

Comments

This is the diameter of a sphere with volume = (Pi^2)/6 = zeta(2) = A013661. - Eric Desbiaux, Jan 21 2009
Edge of a cube with volume Pi. - Omar E. Pol, Aug 09 2012

Examples

			1.4645918875615232630...
		

Crossrefs

Cf. A000796 (Pi), A091925 (Pi^3), A093204 (Pi^(-1/3)), A002161 (sqrt(Pi)), A197111 (cont.frac.).

Programs

Formula

1/A093204. - M. F. Hasler, Oct 07 2014
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