A060294 -id:A060294 - cp's OEIS frontend

A277234 Numerators of partial sums of a Ramanujan series converging to 2/Pi = A060294.

1, 3, 435, 1855, 1678635, 8178093, 831557727, 4362807735, 26663516457435, 146862472576105, 13439367283090749, 76661183599555737, 54390019021528255975, 318658997759516188425, 27581665786275463543575, 165068987339858265879975, 7173478080571052213369487675

Offset: 0

Wolfdieter Lang, Nov 13 2016

The denominators seem to be A241756.

One of Ramanujan's series is 1 - 5*(1/2)^3 + 9*(1*3/(2*4))^3 - 13*(1*3*5/(2*4*6))^3 +- ... = Sum_{k>=0} (-1)^k*(1+4*k)*(risefac(1/2,k)/k!)^3 where risefac(x,k) = Product_{j=0..k-1} (x+j), and risefac(x,0) = 1. See the Hardy reference, p. 7, eq. (1.2) and p. 105, eq. (7.4.2) for s=1/2. The limit is Buffon's constant 2/Pi given in A060294.

			The rationals r(n) begin: 1, 3/8, 435/512, 1855/4096, 1678635/2097152, 8178093/16777216, 831557727/1073741824, 4362807735/8589934592, ...
The limit r(n), n -> oo, is 2/Pi = 0.6366197723... given in A060294.

G. H. Hardy, Ramanujan, AMS Chelsea Publ., Providence, RI, 2002, pp. 7, 105.

Cf. A060294, A241756, A277232.

a(n) = numerator(r(n)), with the rationals r(n) = Sum_{k=0..n} (-1)^k*(1+4*k)*(risefac(1/2,k)/k!)^3 = Sum_{k=0..n} (1+4*k)*(binomial(-1/2,k))^3 = Sum_{k=0..n} (-1)^k*(1+4*k)*((2*k-1)!!/(2*k)!!)^3. The rising factorial has been defined in a comment above. The double factorials are given in A001147 and A000165 with (-1)!! := 1.

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

N. J. A. Sloane

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

			3.1415926535897932384626433832795028841971693993751058209749445923078164062\
862089986280348253421170679821480865132823066470938446095505822317253594081\
284811174502841027019385211055596446229489549303819...

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Florian Cajori, A History of Mathematical Notations, Dover edition (2012), par. 396.
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N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
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David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See pp. 48-55.

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G. M. Phillips, Table of contents of "Pi: A source Book"
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Sizes, pi
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Wikipedia, Pilish.
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Alexander J. Yee & Shigeru Kondo, Round 2... 10 Trillion Digits of Pi
Index entries for sequences related to the number Pi
Index entries for "core" sequences
Index entries for transcendental numbers

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).
Cf. A001622, A048940, A248682.

-- 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

-- 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

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 */

pi:=Pi(RealField(110)); Reverse(Intseq(Floor(10^105*pi))); // Bruno Berselli, Mar 12 2013

Digits := 110: Pi*10^104:
ListTools:-Reverse(convert(floor(%), base, 10)); # Peter Luschny, Oct 29 2019

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+ *)

{ 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

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

first(n)= default(realprecision, n+10); digits(floor(Pi*10^(n-1))) \\ David A. Corneth, Jun 21 2022

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

from sympy import pi, N; print(N(pi, 1000)) # David Radcliffe, Apr 10 2019

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

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

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

Additional comments from William Rex Marshall, Apr 20 2001

A016742 Even squares: a(n) = (2*n)^2.

Original entry on oeis.org

0, 4, 16, 36, 64, 100, 144, 196, 256, 324, 400, 484, 576, 676, 784, 900, 1024, 1156, 1296, 1444, 1600, 1764, 1936, 2116, 2304, 2500, 2704, 2916, 3136, 3364, 3600, 3844, 4096, 4356, 4624, 4900, 5184, 5476, 5776, 6084, 6400, 6724, 7056, 7396, 7744, 8100, 8464

Offset: 0

N. J. A. Sloane

4 times the squares.
Number of edges in the complete bipartite graph of order 5n, K_{n,4n} - Roberto E. Martinez II, Jan 07 2002
It is conjectured (I think) that a regular Hadamard matrix of order n exists iff n is an even square (cf. Seberry and Yamada, Th. 10.11). A Hadamard matrix is regular if the sum of the entries in each row is the same. - N. J. A. Sloane, Nov 13 2008
Sequence arises from reading the line from 0, in the direction 0, 16, ... and the line from 4, in the direction 4, 36, ... in the square spiral whose vertices are the squares A000290. - Omar E. Pol, May 24 2008
The entries from a(1) on can be interpreted as pair sums of (2, 2), (8, 8), (18, 18), (32, 32) etc. that arise from a re-arrangement of the subshell orbitals in the periodic table of elements. 8 becomes the maximum number of electrons in the (2s,2p) or (3s,3p) orbitals, 18 the maximum number of electrons in (4s,3d,4p) or (5s,3d,5p) shells, for example. - Julio Antonio Gutiérrez Samanez, Jul 20 2008
The first two terms of the sequence (n=1, 2) give the numbers of chemical elements using only n types of atomic orbitals, i.e., there are a(1)=4 elements (H,He,Li,Be) where electrons reside only on s-orbitals, there are a(2)=16 elements (B,C,N,O,F,Ne,Na,Mg,Al,Si,P,S,Cl,Ar,K,Ca) where electrons reside only on s- and p-orbitals. However, after that, there is 37 (which is one more than a(3)=36) elements (from Sc, Scandium, atomic number 21 to La, Lanthanum, atomic number 57) where electrons reside only on s-, p- and d-orbitals. This is because Lanthanum (with the electron configuration [Xe]5d^1 6s^2) is an exception to the Aufbau principle, which would predict that its electron configuration is [Xe]4f^1 6s^2. - Antti Karttunen, Aug 14 2008.
Number of cycles of length 3 in the king's graph associated with an (n+1) X (n+1) chessboard. - Anton Voropaev (anton.n.voropaev(AT)gmail.com), Feb 01 2009
a(n+1) is the molecular topological index of the n-star graph S_n. - Eric W. Weisstein, Jul 11 2011
a(n) is the sum of two consecutives odd numbers 2*n^2-1 and 2*n^2+1 and the difference of two squares (n^2+1)^2 - (n^2-1)^2. - Pierre CAMI, Jan 02 2012
For n > 3, a(n) is the area of the irregular quadrilateral created by the points ((n-4)*(n-3)/2,(n-3)*(n-2)/2), ((n-2)*(n-1)/2,(n-1)*n/2), ((n+1)*(n+2)/2,n*(n+1)/2), and ((n+3)*(n+4)/2,(n+2)*(n+3)/2). - J. M. Bergot, May 27 2014
Number of terms less than 10^k: 1, 2, 5, 16, 50, 159, 500, 1582, 5000, 15812, 50000, 158114, 500000, ... - Muniru A Asiru, Jan 28 2018
Right-hand side of the binomial coefficient identity Sum_{k = 0..2*n} (-1)^(k+1)* binomial(2*n,k)*binomial(2*n + k,k)*(2*n - k) = a(n). - Peter Bala, Jan 12 2022

R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 2nd ed., 1994, p. 99.
Seberry, Jennifer and Yamada, Mieko; Hadamard matrices, sequences and block designs, in Dinitz and Stinson, eds., Contemporary design theory, pp. 431-560, Wiley-Intersci. Ser. Discrete Math. Optim., Wiley, New York, 1992.
W. D. Wallis, Anne Penfold Street and Jennifer Seberry Wallis, Combinatorics: Room squares, sum-free sets, Hadamard matrices, Lecture Notes in Mathematics, Vol. 292, Springer-Verlag, Berlin-New York, 1972. iv+508 pp.

Vincenzo Librandi, Table of n, a(n) for n = 0..900
R. P. Boas and N. J. A. Sloane, Correspondence, 1974.
Leo Tavares, Illustration: X Squares
Various, Electron Configuration (Discussion in Physics Forums).
Eric Weisstein's World of Mathematics, Graph Cycle.
Eric Weisstein's World of Mathematics, King Graph.
Eric Weisstein's World of Mathematics, Molecular Topological Index.
Wikipedia, Aufbau principle.
Index entries for sequences related to Hadamard matrices
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000290, A001105, A001539, A016754, A016802, A016814, A016826, A016838, A007742, A033991, A245058.
Sequences on the four axes of the square spiral: Starting at 0: A001107, A033991, A007742, A033954; starting at 1: A054552, A054556, A054567, A033951.
Sequences on the four diagonals of the square spiral: Starting at 0: A002939 = 2*A000384, A016742 = 4*A000290, A002943 = 2*A014105, A033996 = 8*A000217; starting at 1: A054554, A053755, A054569, A016754.
Sequences obtained by reading alternate terms on the X and Y axes and the two main diagonals of the square spiral: Starting at 0: A035608, A156859, A002378 = 2*A000217, A137932 = 4*A002620; starting at 1: A317186, A267682, A002061, A080335.
Cf. sequences listed in A254963.
Other n X n king graph cycle counts: A288918 (4-cycles), A288919 (5-cycles), A288920 (6-cycles).
Cf. A000384, A014105.
Cf. A016813.

List([0..100], n -> (2*n)^2); # Muniru A Asiru, Jan 28 2018

a016742 = (* 4) . (^ 2)
a016742_list = 0 : map (subtract 4) (zipWith (+) a016742_list [8, 16 ..])
-- Reinhard Zumkeller, Jun 28 2015, Apr 20 2015

[(2*n)^2: n in [0..50]]; // Vincenzo Librandi, Apr 26 2011

seq((2*n)^2, n=0..100); # Muniru A Asiru, Jan 28 2018

Table[(2n)^2, {n, 0, 46}] (* Alonso del Arte, Apr 26 2011 *)

makelist((2*n)^2,n,0,20); /* Martin Ettl, Jan 22 2013 */

a(n)=4*n^2 \\ Charles R Greathouse IV, Jul 28 2015

O.g.f.: 4*x*(1+x)/(1-x)^3. - R. J. Mathar, Jul 28 2008
a(n) = A000290(n)*4 = A001105(n)*2. - Omar E. Pol, May 21 2008
a(n) = A155955(n,2) for n > 1. - Reinhard Zumkeller, Jan 31 2009
Sum_{n>=1} 1/a(n) = (1/4)*Pi^2/6 = Pi^2/24. - Ant King, Nov 04 2009
a(n) = a(n-1) + 8*n - 4 (with a(0)=0). - Vincenzo Librandi, Nov 19 2010
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) with a(0) = 0, a(1) = 4, a(2) = 16. - Philippe Deléham, Mar 26 2013
a(n) = A118729(8n+3). - Philippe Deléham, Mar 26 2013
Pi = 2*Product_{n>=1} (1 + 1/(a(n)-1)). - Adriano Caroli, Aug 04 2013
Pi = Sum_{n>=0} 8/(a(2n+1)-1). - Adriano Caroli, Aug 06 2013
E.g.f.: exp(x)*(4x^2 + 4x). - Geoffrey Critzer, Oct 07 2013
a(n) = A000384(n) + A014105(n). - Bruce J. Nicholson, Nov 11 2017
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/48 (A245058). - Amiram Eldar, Oct 10 2020
From Amiram Eldar, Jan 25 2021: (Start)
Product_{n>=1} (1 + 1/a(n)) = sinh(Pi/2)/(Pi/2) (A308716).
Product_{n>=1} (1 - 1/a(n)) = sin(Pi/2)/(Pi/2) = 2/Pi (A060294). (End)
a(n) = A016754(n) - A016813(n). - Leo Tavares, Feb 24 2022

More terms from Sabir Abdus-Samee (sabdulsamee(AT)prepaidlegal.com), Mar 13 2006

A019669 Decimal expansion of Pi/2.

Original entry on oeis.org

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

Offset: 1

N. J. A. Sloane

With offset 2, decimal expansion of 5*Pi. - Omar E. Pol, Oct 03 2013
Decimal expansion of the number of radians in a quadrant. - John W. Nicholson, Oct 07 2013
Not the same as A085679. First differing term occurs at 10^-49, as list -49, or 51st counting term (a(-49)= 5 and A085679(-49) = 4). - John W. Nicholson, Oct 07 2013
5*Pi is also the surface area of a sphere whose diameter equals the square root of 5. More generally x*Pi is also the surface area of a sphere whose diameter equals the square root of x. - Omar E. Pol, Dec 22 2013
Pi/2 is also the radius of a sphere whose surface area equals the volume of the circumscribed cube. - Omar E. Pol, Dec 27 2013

			Pi/2 = 1.570796326794896619231321691639751442098584699...
5*Pi = 15.70796326794896619231321691639751442098584699...

Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Sections 1.4.1 and 1.4.2, pp. 20-21.

Harry J. Smith, Table of n, a(n) for n = 1..20000
David H. Bailey and Richard E. Crandall, On the Random Character of Fundamental Constant Expansions, Experimental Mathematics, Vol. 10 (2001), Issue 2, p. 185 (preprint draft).
Richard J. Mathar, Chebyshev approximation of x^m (-log x)^l in the interval 0 <= x <= 1, arXiv:2408.15212 [math.CA], 2024. See p. 2.
Michael Penn, A nice sum from the Harvard MIT math trust, YouTube video, 2022.
L. D. Servi, Nested Square Roots of 2, The American Mathematical Monthly 110:4 (Apr. 2003), pp. 326-330.
Johan Wästlund, An Elementary Proof of the Wallis Product Formula for pi, The American Mathematical Monthly 114:10 (Dec. 2007), pp. 914-917.
Eric W. Weisstein and Jonathan Sondow, Wallis Formula, MathWorld.
Wikipedia, Viète's formula.
Index entries for transcendental numbers.

Cf. A053300 (continued fraction), A060294 (2/Pi).

Cf. A000796, A019692, A122952, A019694 (Pi through 4*Pi), A106854.

Digits:=100: evalf(Pi/2); # Wesley Ivan Hurt, Oct 26 2016

RealDigits[N[Pi/2, 200]] (* Vladimir Joseph Stephan Orlovsky, Dec 02 2009 *)

default(realprecision, 20080); x=Pi/2; for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b019669.txt", n, " ", d)); \\ Harry J. Smith, May 31 2009

Pi/2 = log(i)/i, where i = sqrt(-1). - Eric Desbiaux, Jun 27 2009
Pi/2 = Product_{n>=1} (n/(n+1))^((-1)^n) = 2 * 2/3 * 4/3 * 4/5 * 6/5 * 6/7 * 8/7 * 8/9 * 10/9 * ... (Wallis formula). - William Keith and Alonso del Arte, Jun 24 2012
Equals Sum_{k>1} 2^k/binomial(2*k,k). - Bruno Berselli, Sep 11 2015
The previous result is the particular case n = 1 of the more general identity: Pi/2 = 4^(n-1) * n!/(2*n)! * Sum_{k >= 2} 2^(k+1)*(k + n - 1)!*(k + 2*n - 2)!/(2*k + 2*n - 2)! valid for n = 0,1,2,... . - Peter Bala, Oct 26 2016
Pi/2 = Product_{n>=1} (4*n^2)/(4*n^2-1). - Fred Daniel Kline, Oct 29 2016
Pi/2 = lim_{n->oo} F(2^(n+3))/2, with one half of the area of a regular 2^(n+3)-gon, for n >= 0, inscribed in the unit circle, written as iterated square roots of 2 as F(2^(n+3))/2 = 2^n*sqrt(2 + sq2(n)), with sq2(n) = sqrt(2 + sq2(n-1)), n >= 1, with input sq2(0) = 0 (2 appears n times in sq2(n)). Viète's infinite product formula works with the partial product F(2^(n+2))/2 = Product_{j=1..n} (2/sq2(j)), n >= 1, which corresponds to the above given formula. - Wolfdieter Lang, Jul 06 2018
Pi/2 = Integral_{x = 0..oo} sin(x)^2/x^2 dx = 1/2 + Sum_{n >= 1} sin(n)^2/n^2, by the Abel-Plana formula. - Peter Bala, Nov 05 2019
From Amiram Eldar, Aug 15 2020: (Start)
Equals Sum_{k>=0} k!/(2*k + 1)!!.
Equals Sum_{k>=0} (-1)^k/(k + 1/2).
Equals Integral_{x=0..oo} 1/(x^2 + 1) dx.
Equals Integral_{x=0..oo} sin(x)/x dx.
Equals Integral_{x=0..oo} exp(x/2)/(exp(x) + 1) dx.
Equals Product_{p prime > 2} p/(p + (-1)^((p-1)/2)). (End)
Pi/2 = Integral_{x = 0..oo} 1/(1 - x^2 + x^4) dx = (1 + 2/3 + 1/5) - (1/7 + 2/9 + 1/11) + (1/13 + 2/15 + 1/17) - .... - Peter Bala, Jul 22 2022
Equals arcsin(9/10) + sqrt(19)*Sum_{k >= 1} A106854(k-1)/(k*10^k) (see Bailey and Crandall, 2001). - Paolo Xausa, Jul 15 2024
Equals 2F1(1/2,1/2 ; 3/2; 1). - R. J. Mathar, Aug 20 2024
Pi/2 = [1;1,1/2,1/3,...,1/n,...] by Wallis's approximation. - Thomas Ordowski, Oct 19 2024
From Stefano Spezia, Oct 21 2024: (Start)
Equals Sum_{k>=0} 2^k/((2*k + 1)*binomial(2*k,k)) (see Finch).
Equals Limit_{n->oo} 2^(4*n)/((2*n + 1)*binomial(2*n,n)^2) (see Finch). (End)
Equals Integral_{x=-oo..oo} sech((2*x^3 + x^2 - 5*x)/(x^2 - 1)) dx. - Kritsada Moomuang, May 29 2025

A088538 Decimal expansion of 4/Pi.

Original entry on oeis.org

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

Offset: 1

Benoit Cloitre, Nov 16 2003

Average length of chord formed from two randomly chosen points on the circumference of a unit circle (see Weisstein/MathWorld link). - Rick L. Shepherd, Jun 19 2006
Suppose u(0) = 1 + i where i^2 = -1 and u(n+1) = (1/2)*(u(n) + |u(n)|). Conjecture: lim_{n -> oo} Real(u(n)) = 4/Pi. - Yalcin Aktar, Jul 18 2007
Ratio of the arc length of the cycloid for one period to the circumference of the corresponding circle rolling on a line. Thus, for any integral number n of revolutions of a circle of radius r, a point on the circle travels (4/Pi)*2*Pi*r*n = 8*r*n (while the center of the circle moves only 2*Pi*r*n). This ratio varies for partial revolutions and depends upon the initial position of the point with points nearest the line moving the slowest (see Dudeney, who explains how the tops of bicycle wheels move faster than the parts nearest the ground). - Rick L. Shepherd, May 05 2014
Average distance traveled in two steps of length 1 for a random walk in the plane starting at the origin. - Jean-François Alcover, Aug 04 2014
Ratio of the circle area to the area of a square having equal perimeters. - Iaroslav V. Blagouchine, May 06 2016
This is also the value of a special case (n=1) of an n-family of series considered by Hardy (see A278145): 1 + (1/2)*(1/2)^2 + (1/3)*(1*3/(2*4))^2 + (1/4)*((1*3*5) / (2*4*6))^2 + ... = Sum_{k>=0} (1/(k+1))*((2*k-1)!!/(2*k)!!)^2. - Wolfdieter Lang, Nov 14 2016
Minimum ratio of the area of a rectangle to one of its inscribed ellipses, or only existing ratio if the rectangle is a square and then the only inscribed ellipse is a circle. This ellipse has its semiaxes parallel to the sides of the rectangle. If a rectangle has sides of length 2a and 2b, its area is 4*a*b, while the ellipse inscribed has a and b as semiaxes, therefore its area is a*b*Pi. Thus the ratio is (4*a*b)/(a*b*Pi) = 4/Pi. - Giovanni Zedda, Jun 20 2019
The diameter of the conventional spherical earth is (4/Pi)*10000 km = 12732.395... km. - Jean-François Alcover, Oct 30 2021
From Jianing Song, Aug 06 2022: (Start)
sign(sin(x)) = (4/Pi) * Sum_{n>=0} sin((2*n+1)*x)/(2*n+1), for all x in R;
sign(cos(x)) = (4/Pi) * Sum_{n>=0} (-1)^n*cos((2*n+1)*x)/(2*n+1), for all x in R. (End)

			4/Pi = 1.2732395.... = 1/0.78539816...

Florian Cajori, A History of Mathematical Notations, Dover edition (2012), par. 195.
H. E. Dudeney, 536 Puzzles & Curious Problems, Charles Scribner's Sons, New York, 1967, pp. 99, 300-301, #294.
Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, p. 86.
G. H. Hardy, Ramanujan, AMS Chelsea Publ., Providence, RI, 2002, p. 105, eq. (7.5.1) for n=1.
L. B. W. Jolley, Summation of Series, Dover (1961).

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
J.-P. Allouche, On a formula of T. Rivoal, arXiv:1307.3906 [math.NT], 2013.
Friedrich L. Bauer, Historische Notizen / Wallis-artige Kettenprodukte, Informatik Spektrum 31,4 (2008) 348-352.
J. M. Borwein, A. Straub, J. Wan, and W. Zudilin, Densities of short uniform random walks, arXiv:1103.2995 [math.CA], 2011.
Manuel Benito, Luis M. Navas and Juan Luis Varona, Möbius inversion from the point of view of arithmetical semigroup flows, Proceedings of the “Segundas Jornadas de Teoría de Números” (Madrid, 2007), 63-81
R. J. Mathar, Chebyshev Series Expansion of Inverse Polynomials, arXiv:0403344 [math.CA], 2004-2005.
Michael I. Shamos, A catalog of the real numbers, (2007). See p. 309.
Eric Weisstein's World of Mathematics, Circle Line Picking.
Eric Weisstein's World of Mathematics, Cycloid.
Index entries for transcendental numbers.

Cf. A079097 for terms of a generalized continued fraction for 4/Pi. Inverse of A003881. A060294, A278145, A049541 (1/Pi).

RealDigits[N[4/Pi, 6!]][[1]] (* Vladimir Joseph Stephan Orlovsky, Jun 18 2009 *)

4/Pi \\ Charles R Greathouse IV, Jun 21 2013

4/Pi = Product_(1-(-1)^((p-1)/2)/p) where p runs through the odd primes.
Arcsin x = (4/Pi) Sum_{n = 1, 3, 5, 7, ...} T_n(x)/n^2 (Chebyshev series of arcsin; App C of math.CA/0403344). - R. J. Mathar, Jun 26 2006
Equals 1 + Sum_{n >= 1} ((2n-3)!!/(2n)!!)^2. [Jolley eq 274]. - R. J. Mathar, Nov 03 2011
Equals binomial(1,1/2). - Bruno Berselli, May 17 2016
2*A060294 (twice Buffon's constant) = 1/Gamma(3/2)^2. - Wolfdieter Lang, Nov 14 2016
Equals 1 + Sum_{n>=0} (Catalan(n)/2^(2*n+1))^2, with Catalan(n) = A000108(n). This is the rewritten Jolley (274) series. See the above R. J. Mathar entry with (-1)!! := 1. - Ralf Steiner, Sep 18 2018
4/Pi = 1 + (1/4)*hypergeometric([1, 1/2, 1/2], [2, 2], 1) = hypergeometric([-1/2, -1/2], [1], 1). From the g.f. of Catalan^2 given in A001246. - Wolfdieter Lang, Sep 18 2018
Equals Product_{k>=1} (1 + 1/(4*k*(k+1))). - Amiram Eldar, Aug 05 2020
From Stefano Spezia, Oct 26 2024: (Start)
4/Pi = 1 + K_{n>=1} n^2/(2*n + 1), where K is the Gauss notation for an infinite continued fraction. In the expanded form, 4/Pi = 1 + 1^2/(3 + 2^2/(5 + 3^2/(7 + 4^2/(9 + 5^2/(11 + ...))))) (see Finch at p. 23).
4/Pi = Sum_{n>=0} tan(Pi/2^(n+2))/2^n (see Shamos). (End)
4/Pi = Sum_{n >= 0} (-1)^n * mu(2*n+1)/(2*n + 1), where mu(n) is the Möbius function A008683 (see, for example, Benito et al., p. 77). - Peter Bala, Jan 07 2025
Equals Integral_{x=0..1} (2*EllipticK(x))/Pi dx. - Kritsada Moomuang, Jun 04 2025

A086201 Decimal expansion of 1/(2*Pi).

Original entry on oeis.org

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

Offset: 0

Eric W. Weisstein, Jul 12 2003

If a single hump of cycloid, with arc length 8*radius (generating circle), is inside a rectangle with width=2*radius and length=2*Pi*radius, then the radius must be 1/(2*Pi) (this sequence) to have (2/Pi), A060294, as semi arc of cycloid (arc = 4/Pi = A088538) and the rectangle... length = 1, width = 1/Pi. I suppose that in 3D geometry, gliding along a cycloid, in all directions around, from a point A at the height of 1/Pi, gives Pi*point B. - Eric Desbiaux, Dec 21 2008
Radius of circle having circumference 1. - Clark Kimberling, Jan 06 2014
The number of primitive Pythagorean triangles with hypotenuse less than N is approximately N/(2*Pi), found by Lehmer, cf. Knott link. - Frank Ellermann, Mar 27 2020

			0.15915494309189533576888376337251...

Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 8.4, p. 493.

Ron Knott, 9.6 Pythagorean Triples and Pi, 2019.
Eric Weisstein's World of Mathematics, Plouffe's Constants.
Eric Weisstein's World of Mathematics, Pythagorean Triple.
Index entries for transcendental numbers.

Cf. A000796 (Pi), A019692 (2*Pi).

RealDigits[N[1/(2 Pi), 100]][[1]] (* Vladimir Joseph Stephan Orlovsky, Jun 18 2009 *)

PARI
```
1/(2*Pi) \\ Michel Marcus, Mar 28 2020
```

Link corrected by Fred Daniel Kline, Jul 29 2015

A089491 Decimal expansion of Buffon's constant 3/Pi.

Original entry on oeis.org

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

Offset: 0

Robert G. Wilson v, Nov 04 2003

Whereas 2/Pi (A060294) is the probability that a needle will land on one of many parallel lines, this is the probability that a needle will land on one of many lines making up a grid.
The probability that the boundary of an equilateral triangle will intersect one of the parallel lines if the triangle edge length l (almost) equals the distance d between each pair of lines. This follows directly from the Weisstein/MathWorld Buffon's Needle Problem link's statement P=p/(Pi*d), where P is the probability of intersection with any convex polygon's boundary if the generalized diameter of that polygon is less than d and p is the perimeter of the polygon. (Take d=l, then p=3d.) - Rick L. Shepherd, Jan 11 2006
Related grid problems are discussed in the Weisstein/MathWorld Buffon-Laplace Needle Problem link. - Rick L. Shepherd, Jan 11 2006
The area of a regular dodecagon circumscribed in a unit-area circle. - Amiram Eldar, Nov 05 2020
From Bernard Schott, Apr 19 2022: (Start)
For any non-obtuse triangle ABC (see Mitrinović and Oppenheim links):
(a/A + b/B + c/C)/(a+b+c) >= 3/Pi,
(a^2/A + b^2/B + c^2/C)/(a^2+b^2+c^2) <= 3/Pi,
where (A,B,C) are the angles (measured in radians) and (a,b,c) the side lengths of this triangle.
Equality stands iff triangle ABC is equilateral. (End)

			3/Pi = 0.95492965855137201461330258023508617220675787444273869248600...

Joe Portney, Portney's Ponderables, Litton Systems, Inc., Appendix 2, 'Buffon's Needle' by Lawrence R. Weill, 200, pp. 135-138.

Harry Khamis, Buffon's Needle Problem.
D. S. Mitrinović, J. E. Pečarić, and V. Volenec, Recent Advances In Geometric Inequalities, Kluwer Academic Publishers, 1989, Inequalities 4.11, p. 170.
A. Oppenheim, Problem E 2649, American Mathematical Monthly, 84 (1977), p. 294.
Kevin Peterson, A Problem in Geometric Probability: Buffon's Needle Problem.
George Reese, Buffon's Needle, An Analysis and Simulation.
Shodor Education Foundation, Inc., Buffon's needle.
Washington and Lee University, Problem 18: Buffon's Needle Again. [Broken link]
Eric Weisstein's World of Mathematics, Buffon's needle problem.
Eric Weisstein's World of Mathematics, Buffon-Laplace needle problem.
Eric Weisstein's World of Mathematics, Generalized Diameter.
Index entries for transcendental numbers

Cf. A000796 (Pi), A060294 (2/Pi).

Mathematica
```
RealDigits[ N[ 3/Pi, 111]][[1]]
```
PARI
```
3/Pi \\ Michel Marcus, Nov 05 2020
```

Equals sinc(Pi/6). - Peter Luschny, Oct 04 2019
From Amiram Eldar, Aug 20 2020: (Start)
Equals Product{k>=1} cos(Pi/(6*2^k)).
Equals Product{k>=1} (1 - 1/(6*k)^2). (End)

A137245 Decimal expansion of Sum_{p prime} 1/(p * log p).

Original entry on oeis.org

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

Offset: 1

R. J. Mathar, Mar 09 2008

Sum_{p prime} 1/(p^s * log p) equals this value here if s=1, equals A221711 if s=2, 0.22120334039... if s=3. See arXiv:0811.4739.
Erdős (1935) proved that for any sequence where no term divides another, the sum of 1/(x log x) is at most some constant C. He conjectures (1989) that C can be taken to be this constant 1.636..., that is, the primes maximize this sum. - Charles R Greathouse IV, Mar 26 2012 [The conjecture has been proved by Lichtman 2022. - Pontus von Brömssen, Jun 23 2022]
Note that sum 1/(p * log p) is almost (a tiny bit less than) 1 + 2/Pi = 1+A060294 = 1.63661977236758... (Why is it so close?) - Daniel Forgues, Mar 26 2012
Sum 1/(p * log p) is quite close to sum 1/n^2 = Pi^2/6 = 1.644934066... (Cf. David C. Ullrich, "Re: What is Sum(1/p log p)?" for why this is so; mentions A115563.) - Daniel Forgues, Aug 13 2012

			1.63661632335...

Henri Cohen, Number Theory, Volume II: Analytic and Modern Tools, GTM Vol. 240, Springer, 2007; see pp. 208-209.
Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 2.27.2, p. 186.

Karim Belabas and Henri Cohen, Computation of sum_{p prime} 1/(p^s log(p)), PARI/GP script, 2020.
Thomas Bloom, Problem 164, Erdős Problems.
Henri Cohen, High-precision computation of Hardy-Littlewood constants, (1998).
Henri Cohen, High-precision computation of Hardy-Littlewood constants. [pdf copy, with permission]
Paul Erdős, Note on sequences of integers no one of which is divisible by any other, J. London Math. Soc. 10 (1935), pp. 126-128, [DOI].
Paul Erdős, Some problems and results on combinatorial number theory, Graph theory and its applications: East and West (Jinan, 1986), Ann. New York Acad. Sci., 576, pp. 132-145, New York Acad. Sci., New York, 1989.
Brady Haran and Jared Duker Lichtman, Primes and Primitive Sets, Numberphile video (2022).
Jared Duker Lichtman, Almost primes and the Banks-Martin conjecture, arXiv:1909.00804 [math.NT], 2019.
Jared Duker Lichtman, A proof of the Erdős primitive set conjecture, arXiv:2202.02384 [math.NT], 2022.
Jared Duker Lichtman, Proving the Erdős Primitive Set Conjecture, Oxford Mathematics YouTube video, 2022.
Jared Duker Lichtman, A proof of the Erdős primitive set conjecture, YouTube video from Combinatorial and additive number theory conference (CANT), 2022.
Richard J. Mathar, Twenty digits of some integrals of the prime zeta function, arXiv:0811.4739 [math.NT], 2008-2009, table in Section 2.4.
Terence Tao, Erdős problem database, see no. 164.
David C. Ullrich, Re: What is Sum(1/p log p)?, posting in newsgroup sci.math.research, 11 Feb 2006.

Cf. A000040, A060294, A221711 (p squared), A115563, A319231 (log squared), A319232 (p and log squared), A354952.

digits = 105;
precision = digits + 10;
tmax = 500; (* integrand considered negligible beyond tmax *)
kmax = 500; (* f(k) considered negligible beyond kmax *)
InLogZeta[k_] := NIntegrate[Log[Zeta[t]], {t, k, tmax}, WorkingPrecision -> precision, MaxRecursion -> 20];
f[k_] := With[{mu = MoebiusMu[k]}, If[mu == 0, 0, (mu/k^2)*InLogZeta[k]]];
s = 0;
Do[s = s + f[k]; Print[k, " ", s], {k, 1, kmax}];
RealDigits[s][[1]][[1 ;; digits]] (* Jean-François Alcover, Feb 06 2021, updated Jun 22 2022 *)

\\ See Belabas, Cohen link. Run as SumEulerlog(1) after setting the required precision.

default(realprecision, 200); s=0; for(k=1, 500, s = s + moebius(k)/k^2 * intnum(x=k,[[1], 1], log(zeta(x))); print(s)); \\ Vaclav Kotesovec, Jun 12 2022

Equals Sum_{n>=1} 1/(A000040(n)*log A000040(n)).

More terms from Hugo Pfoertner, Feb 01 2020

More terms from Vaclav Kotesovec, Jun 12 2022

A070750 0 if n-th prime is even, 1 if n-th prime is == 1 (mod 4), and -1 if n-th prime is == 3 (mod 4).

Original entry on oeis.org

0, -1, 1, -1, -1, 1, 1, -1, -1, 1, -1, 1, 1, -1, -1, 1, -1, 1, -1, -1, 1, -1, -1, 1, 1, 1, -1, -1, 1, 1, -1, -1, 1, -1, 1, -1, 1, -1, -1, 1, -1, 1, -1, 1, 1, -1, -1, -1, -1, 1, 1, -1, 1, -1, 1, -1, 1, -1, 1, 1, -1, 1, -1, -1, 1, 1, -1, 1, -1, 1, 1, -1

Offset: 1

Reinhard Zumkeller, May 04 2002

Also, sin(prime(n)*Pi/2), where prime(n) = A000040(n), Pi=3.1415... (original definition).
Also imaginary part of primes mapped as defined in A076340, A076341: a(n) = A076341(A000040(n)), real part = A076342.
Legendre symbol (-1/prime(n)) for n > 1. - T. D. Noe, Nov 05 2003
For n > 1, let p = prime(n) and m = (p-1)/2. Then c(m) - a(n) == 0 (mod p), where c(m) = (2*m)!/(m!)^2 = A000984(m) is the central binomial coefficient. [Proof: By definition, c(m)*(m!)^2 - (p-1)! = 0 and therefore c(m)*(m!)^2*(-1)^(m+1) - (p-1)!*(-1)^(m+1) = 0. Now apply Wilson's theorem, (p-1)! == 1 (mod p), and its corollary, (m!)^2 == (-1)^(m+1) (mod p), and finally use the formula by T. D. Noe listed below to replace (-1)^m with a(n).] Similarly, C_m - 2*a(n) == 0 (mod p), with C_m = A000108(m) being the m-th Catalan number. [Proof: By definition, C_m*(p+1)*(m!)^2 - 2*(p-1)! = 0. The result follows proceeding as in the first proof.] - Stanislav Sykora, Aug 11 2014

			p = 4*k+1 (see A002144): a(p) = sin((4*k+1)*Pi/2) = sin(2*k*Pi + Pi/2) = sin(Pi/2) = 1.
p = 4*k+3 (see A002145): a(p) = sin((4*k+3)*Pi/2) = sin(2*k*Pi + 3*Pi/2) = sin(3*Pi/2) = -1.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Legendre Symbol.
Wikipedia, Wilson's theorem.

Cf. A000040, A039702, A070748, A070749, A002144, A002145, A000108, A000984, A134323, A257834, A076340, A076341.

Cf. A060294, A088538.

a070750 = (2 -) . (`mod` 4) . a000040  -- Reinhard Zumkeller, Feb 28 2012

a[n_] := JacobiSymbol[-1, Prime[n]]; a[1] = 0; Table[a[n], {n, 1, 72}] (* Jean-François Alcover, Oct 05 2012, after T. D. Noe *)
Table[Which[EvenQ[p],0,Mod[p,4]==1,1,True,-1],{p,Prime[Range[80]]}] (* Harvey P. Dale, Mar 16 2020 *)

apply(n->2-n%4,primes(100)) \\ Charles R Greathouse IV, Aug 21 2011

a(n) = 2 - prime(n) mod 4 = 2 - A039702(n).
a(n) = (-1)^((prime(n)-1)/2) for n > 1. - T. D. Noe, Nov 05 2003
From Amiram Eldar, Dec 24 2022: (Start)
Product_{n>=1} (1 - a(n)/prime(n)) = 4/Pi (A088538).
Product_{n>=1} (1 + a(n)/prime(n)) = 2/Pi (A060294). (End)

Wording of definition changed by N. J. A. Sloane, Jun 21 2015

A132696 Decimal expansion of 6/Pi.

Original entry on oeis.org

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

Offset: 1

Omar E. Pol, Aug 26 2007, Nov 02 2007

6/Pi = Volume of the cuboid (If L1>L2>L3) / Volume of the inscribed ellipsoid.
6/Pi = Volume of the cuboid (If L1>(L2=L3)) / Volume of the inscribed spheroid.
6/Pi = Volume of the regular hexahedron (or cube) / Volume of the inscribed Sphere.
6/Pi = 1 / Arc of 30 degrees.
6/Pi = Volume of the cuboid (If L1<(L2=L3)) / Volume of the inscribed spheroid.
6/Pi = Surface area of the regular hexahedron (or cube) / surface area of the inscribed sphere.

			1.90985931710274402922660516047... .

Index entries for transcendental numbers

Cf. A049541, A060294, A089491, A088538, A086201, A059956.

RealDigits[N[6/Pi, 200]] (* Erich Friedman, Mar 22 2008 *)

6/Pi \\ Charles R Greathouse IV, Dec 31 2011

Equals Product_{k>=1} (2k+1)^3 / ( (2k)^2*(2k+3) ). - Federico Provvedi, Nov 09 2024

More terms from Erich Friedman, Mar 22 2008

cp's OEIS Frontend

A277234 Numerators of partial sums of a Ramanujan series converging to 2/Pi = A060294.

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

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A016742 Even squares: a(n) = (2*n)^2.

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A019669 Decimal expansion of Pi/2.

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A088538 Decimal expansion of 4/Pi.

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