A019693 -id:A019693 - cp's OEIS frontend

A019692 Decimal expansion of 2*Pi.

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

N. J. A. Sloane

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

			6.283185307179586476925286766559005768394338798750211641949889184615632...

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.

Harry J. Smith, Table of n, a(n) for n = 1..20000
Robert Ferréol, Deltoid, Mathcurve.
Christophe Garban and José A. Trujillo Ferreras, The expected area of the filled planar Brownian loop is pi/5, Communications in mathematical physics, Vol. 264, No. 3 (2006), pp. 797-810, preprint, arXiv:math/0504496 [math.PR], 2005.
Peter Harremoës, Al-Kashi’s constant
Michael Hartl, The Tau Manifesto.
Melissa Larson, Verifying and discovering BBP-type formulas, 2008.
Bob Palais, Web page about "Pi is wrong!".
Bob Palais, Pi is wrong!, The Mathematical Intelligencer Volume 23, Number 3, 2001, pp. 7-8.
Grant Sanderson, How pi was almost 6.283185..., 3Blue1Brown video (2018).
Michael I. Shamos, A catalog of the real numbers, (2007). See p. 318.
Wikipedia, Tau proposals.
Wikipedia, Bailey-Borwein-Plouffe formula.
Index entries for transcendental numbers.

Cf. A058291 (continued fraction).
Cf. A000796, A019693, A019699, A091925, A244979.
Cf. A093828 (astroid), A180434 (loop of strophoid), A197723 (cardioid).

using Nemo
RR = RealField(334)
tau = const_pi(RR) + const_pi(RR)
tau |> println # Peter Luschny, Mar 14 2018

R:= RealField(100); 2*Pi(R); // G. C. Greubel, Mar 08 2018

RealDigits[N[2 Pi, 6! ]] (* Vladimir Joseph Stephan Orlovsky, Dec 02 2009 *)

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

# 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

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

N. J. A. Sloane, Dec 11 1996

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

			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)

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

Ivan Panchenko, Table of n, a(n) for n = 1..1000
Kunle Adegoke, Infinite arctangent sums involving Fibonacci and Lucas numbers, arXiv:1603.08097 [math.NT], 2016.
J. M. Borwein, P. B. Borwein, and K. Dilcher, Pi, Euler numbers and asymptotic expansions, Amer. Math. Monthly, 96 (1989), 681-687.
Jean Desbois and Stéphane Ouvry, Algebraic and arithmetic area for m planar Brownian paths, arXiv:1101.4135 [math-ph], Jan 21, 2011.
Index entries for transcendental numbers.

Cf. A000796 (Pi), A019669 (Pi/2), A019692 (2*Pi), A122952 (3*Pi), A003881(Pi/4), A137914, A238238, A002388, A091925, A093954, A016910, A019694, A136017, A352324.

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).

RealDigits[N[Pi/3,6! ]] (* Vladimir Joseph Stephan Orlovsky, Dec 02 2009 *)

Pi/3 \\ Charles R Greathouse IV, Sep 08 2011

N=150; default(realprecision, 3+N); digits(Pi*10^N\3) \\ M. F. Hasler, Jul 08 2016

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

A019699 Decimal expansion of 2Pi/15 = (4Pi/3)/10.

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane

With offset 1, decimal expansion of 4*Pi/3, the volume of a sphere of radius 1. - Omar E. Pol, Aug 27 2007, Sep 25 2013
2*Pi/15 is the common value of the base angles of the isosceles triangle formed at the common vertex of the figure obtained by gluing a hexagon and a pentagon, both regular, along a common side, as shown in the CNRS link. - Michel Marcus, Mar 06 2015
This is also the surface area (in some cubic length unit (l.u.)) of a sphere with a central cylinder symmetrical hole of length 2 l.u. Thanks to Sven Heinemeyer for reminding me of this classical astonishing result. See e.g., Bild der Wissenschaft, Januar 1964, p. 75, or the Gardner reference, Problem 7 on p. 51. In two dimensions things are different. See A258146. - Wolfdieter Lang, May 31 2015

			2*Pi/15 = 0.418879020478639098461685784437267...
4*Pi/3 = 4.18879020478639098461685784437267... - _Omar E. Pol_, Sep 25 2013

Bild der Wissenschaft, Januar 1964.
Martin Gardner, Mathematische Rätsel und Probleme, 3. Auflage, Friedr. Vieweg + Sohn, Braunschweig, 1975, p. 51 (in German). In English: Mathematical Puzzles and Diversions from "Scientific American", Simon and Schuster, N. Y. 1959/1961.

Ivan Panchenko, Table of n, a(n) for n = 0..1000
Ana Rechtman, Février 2015, 4ème défi, Images des Mathématiques, CNRS, 2015.
Index entries for transcendental numbers

Cf. A000796, A019692, A019693.

RealDigits[(2 Pi)/15,10,120][[1]] (* Harvey P. Dale, Mar 22 2015 *)

2*Pi/15 \\ Charles R Greathouse IV, Jan 09 2018

(1/10)*volume of the unit sphere in R^3 = (1/10)*Pi^(3/2)/gamma(1+3/2). - Benoit Cloitre, Jun 19 2003

A175887 Numbers that are congruent to {1, 14} mod 15.

Original entry on oeis.org

1, 14, 16, 29, 31, 44, 46, 59, 61, 74, 76, 89, 91, 104, 106, 119, 121, 134, 136, 149, 151, 164, 166, 179, 181, 194, 196, 209, 211, 224, 226, 239, 241, 254, 256, 269, 271, 284, 286, 299, 301, 314, 316, 329, 331, 344, 346, 359, 361, 374, 376, 389, 391, 404

Offset: 1

Bruno Berselli, Oct 08 2010 - Nov 17 2010

Cf. property described by Gary Detlefs in A113801: more generally, these numbers are of the form (2*h*n+(h-4)*(-1)^n-h)/4 (h, n natural numbers), therefore ((2*h*n+(h-4)*(-1)^n-h)/4)^2-1==0 (mod h); in this case, a(n)^2-1==0 (mod 15).

Bruno Berselli, Table of n, a(n) for n = 1..10000.
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A000217, A019693, A019976, A113801, A175886, A091998, A175885, A090771, A056020, A047522, A047336, A007310, A047209, A005408, A001651.

Cf. A132240 (primes).

a175887 n = a175887_list !! (n-1)
a175887_list = 1 : 14 : map (+ 15) a175887_list
-- Reinhard Zumkeller, Jan 07 2012

[n: n in [1..450] | n mod 15 in [1,14]];

[(30*n+11*(-1)^n-15)/4: n in [1..55]]; // Vincenzo Librandi, Aug 19 2013

Select[Range[1, 450], MemberQ[{1,14}, Mod[#, 15]]&]
CoefficientList[Series[(1 + 13 x + x^2) / ((1 + x) (1 - x)^2), {x, 0, 55}], x] (* Vincenzo Librandi, Aug 19 2013 *)

a(n)=(30*n+11*(-1)^n-15)/4 \\ Charles R Greathouse IV, Sep 28 2015

G.f.: x*(1+13*x+x^2)/((1+x)*(1-x)^2).
a(n) = (30*n+11*(-1)^n-15)/4.
a(n) = -a(-n+1) = a(n-1)+a(n-2)-a(n-3).
a(n) = 15*A000217(n-1) -2*sum(a(i), i=1..n-1) +1 for n>1.
a(n) = A047209(A047225(n+1)).
Sum_{n>=1} (-1)^(n+1)/a(n) = (Pi/15)*cot(Pi/15) = A019693 * A019976 / 10. - Amiram Eldar, Dec 04 2021
E.g.f.: 1 + ((30*x - 15)*exp(x) + 11*exp(-x))/4. - David Lovler, Sep 05 2022
From Amiram Eldar, Nov 25 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = (Pi/15)*cosec(Pi/15).
Product_{n>=2} (1 + (-1)^n/a(n)) = 2*cos(Pi/15). (End)

A122857 Expansion of (phi(q)^2 + phi(q^3)^2) / 2 in powers of q where phi() is a Ramanujan theta function.

Original entry on oeis.org

1, 2, 2, 2, 2, 4, 2, 0, 2, 2, 4, 0, 2, 4, 0, 4, 2, 4, 2, 0, 4, 0, 0, 0, 2, 6, 4, 2, 0, 4, 4, 0, 2, 0, 4, 0, 2, 4, 0, 4, 4, 4, 0, 0, 0, 4, 0, 0, 2, 2, 6, 4, 4, 4, 2, 0, 0, 0, 4, 0, 4, 4, 0, 0, 2, 8, 0, 0, 4, 0, 0, 0, 2, 4, 4, 6, 0, 0, 4, 0, 4, 2, 4, 0, 0, 8, 0, 4, 0, 4, 4, 0, 0, 0, 0, 0, 2, 4, 2, 0, 6, 4, 4, 0, 4

Offset: 0

Michael Somos, Sep 14 2006

Ramanujan theta functions: f(q) (see A121373), phi(q) (A000122), psi(q) (A010054), chi(q) (A000700).

			G.f. = 1 + 2*q + 2*q^2 + 2*q^3 + 2*q^4 + 4*q^5 + 2*q^6 + 2*q^8 + 2*q^9 + 4*q^10 + ...

B. C. Berndt, Ramanujan's Notebooks Part IV, Springer-Verlag, see p. 197, Entry 44.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Michael Somos, Introduction to Ramanujan theta functions, 2019.
Eric Weisstein's World of Mathematics, Ramanujan Theta Functions.

Cf. A002654, A008441, A019693, A035154, A113446, A125079, A125061, A132003.

Cf. A000122, A000700, A010054, A121373.

a[ n_] := If[ n < 1, Boole[n == 0], 2 DivisorSum[ n, KroneckerSymbol[ -36, #] &]]; (* Michael Somos, Jul 09 2013 *)
a[ n_] := SeriesCoefficient[ (EllipticTheta[ 3, 0, q]^2 + EllipticTheta[ 3, 0, q^3]^2) / 2, {q, 0, n}]; (* Michael Somos, Jul 09 2013 *)

{a(n) = if( n<1, n==0, 2 * sumdiv( n, d, kronecker( -36, d)))};

{a(n) = my(A, p, e); if( n<1, n==0, A = factor(n); 2 * prod( k=1, matsize(A)[1], [p, e] = A[k, ]; if( p<5, 1, p%12<6, e+1, !(e%2) )))};

{a(n) = my(A, p, e); if( n<1, n==0, A = factor(n); prod( k=1, matsize(A)[1],
     [p, e] = A[k, ]; if( p==2, 1, p==3, 1+e%2*2, p%4==1, e+1, !(e%2) )))};

Expansion of eta(q^2)^3 * eta(q^3)^2 * eta(q^6) / (eta(q)^2 * eta(q^4)* eta(q^12)) in powers of q.
Expansion of 2 * psi(q) * psi(q^2) * psi(q^3) / psi(q^6) - phi(q^3)^2 in powers of q. - Michael Somos, Jul 09 2013
Euler transform of period 12 sequence [ 2, -1, 0, 0, 2, -4, 2, 0, 0, -1, 2, -2, ...].
Moebius transform is period 12 sequence [ 2, 0, 0, 0, 2, 0, -2, 0, 0, 0, -2, 0, ...].
a(12*n + 7) = a(12*n + 11) = 0.
a(n) = 2 * b(n) where b(n) is multiplicative and b(2^e) = b(3^e) = 1, b(p^e) = e+1 if p == 1, 5 (mod 12), a(p^e) == (1-(-1)^e)/2 if p == 7, 11 (mod 12).
G.f. is a period 1 Fourier series which satisfies f(-1 / (12 t)) = 4 (t/i) g(t) where q = exp(2 Pi i t) and g() is the g.f. for A125061.
A035154(n) = a(n) / 2 if n > 0. A008441(n) = a(4*n + 1) / 2. A125079(n) = a(2*n + 1) / 2. A113446(3*n + 1) = A002654(3*n + 1) = a(3*n + 1) / 2.
a(n) = (-1)^n * A132003(n). Expansion of (phi(-q^3) / phi(-q)) * phi(-q^2) * phi(-q^6) in powers of q where phi() is a Ramanujan theta function. - Michael Somos, Mar 05 2023
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 2*Pi/3 = 2.0943951... (A019693). - Amiram Eldar, Nov 21 2023

A145438 Decimal expansion of sum_{n=1..inf} 1/(n^3*binomial(2n,n)).

Original entry on oeis.org

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

Offset: 0

R. J. Mathar, Feb 08 2009

Alexander Apelblat, Tables of Integrals and Series, Harri Deutsch, (1996), 4.1.47 gives Pi*sqrt(3)*(psi(2/3)-psi(1/3))/72-Zeta(3)/3 which is negative and therefore not correct.

Comment from Mikhail Kalmykov (kalmykov.mikhail(AT)googlemail.com), Jun 01 2009: Analytical results for this sum were also given in Eq. (8) of the Kalmykov and Veretin paper. These results confirm the last comment from Alois P. Heinz.

			0.522946...

J. M. Borwein, R. Girgensohn, Evaluation of Binomial Series, CECM-02-188 (2002).
A. I. Davydychev, M. Yu. Kalmykov, Massive Feynman diagrams and inverse binomial sums, Nucl. Phys. B 699 (2004), 3-64.
M. Yu. Kalmykov and O. Veretin, Single-scale diagrams and multiple binomial sums, Phys. Lett. B 483 (2000) 315-323.
R. J. Mathar, Corrigenda to "Interesting Series involving..", arXiv:0905.0215 [math.CA]

RealDigits[ N[1/18*(Sqrt[3]* Pi*(-PolyGamma[1, 2/3] + PolyGamma[1, 4/3] + 9) - 24*Zeta[3]), 105]][[1]] (* Jean-François Alcover, Nov 08 2012, after R. J. Mathar *)

Comment from Alois P. Heinz, Feb 08 2009: Maple's answer to this is: a:= sum(1/(n^3*binomial(2*n,n)), n=1..infinity); a:= 1/2 hypergeom([1, 1, 1, 1], [2, 2, 3/2], 1/4); evalf (a, 140); .522946192133335108491185183527303540163044591743977841465941014...

Equals A019693*A143298-4*A002117/3 =2*Pi*Cl_2(Pi/3)/3-4*zeta(3)/3. [From R. J. Mathar, Feb 09 2009]

A035181 a(n) = Sum_{d|n} Kronecker(-9, d).

Original entry on oeis.org

1, 2, 1, 3, 2, 2, 0, 4, 1, 4, 0, 3, 2, 0, 2, 5, 2, 2, 0, 6, 0, 0, 0, 4, 3, 4, 1, 0, 2, 4, 0, 6, 0, 4, 0, 3, 2, 0, 2, 8, 2, 0, 0, 0, 2, 0, 0, 5, 1, 6, 2, 6, 2, 2, 0, 0, 0, 4, 0, 6, 2, 0, 0, 7, 4, 0, 0, 6, 0, 0, 0, 4, 2, 4, 3, 0, 0, 4, 0, 10, 1, 4, 0, 0, 4, 0, 2, 0, 2, 4, 0, 0, 0, 0, 0, 6, 2, 2, 0, 9, 2, 4, 0, 8, 0

Offset: 1

N. J. A. Sloane

			x + 2*x^2 + x^3 + 3*x^4 + 2*x^5 + 2*x^6 + 4*x^8 + x^9 + 4*x^10 + 3*x^12 + ...

G. C. Greubel, Table of n, a(n) for n = 1..5000

Cf. A008441, A019693, A125079.

Sum_{d|n} Kronecker(k, d): A035143..A035181 (k=-47..-9, skipping numbers that are not cubefree), A035182 (k=-7), A192013 (k=-6), A035183 (k=-5), A002654 (k=-4), A002324 (k=-3), A002325 (k=-2), A035184 (k=-1), A000012 (k=0), A000005 (k=1), A035185 (k=2), A035186 (k=3), A001227 (k=4), A035187..A035229 (k=5..47, skipping numbers that are not cubefree).

a[ n_] := If[ n < 1, 0, Sum[ KroneckerSymbol[ -9, d], { d, Divisors[ n]}]] (* Michael Somos, Jun 24 2011 *)

{a(n) = if( n<1, 0, sumdiv( n, d, kronecker( -9, d)))} \\ Michael Somos, Jun 24 2011

{a(n) = if( n<1, 0, direuler( p=2, n, 1 / ((1 - X) * (1 - kronecker( -9, p) * X))) [n])} \\ Michael Somos, Jun 24 2011

{a(n) = local(A, p, e); if( n<0, 0, A = factor(n); prod(k=1, matsize(A)[1], if(p = A[k, 1], e = A[k, 2]; if( p==2, e+1, if( p==3, 1, if( p%4==1, e+1, (1 + (-1)^e)/2))))))} \\ Michael Somos, Jun 24 2011

A035181(n)=sumdivmult(n,d,kronecker(-9,d)) \\ M. F. Hasler, May 08 2018

From Michael Somos, Jun 24 2011: (Start)
a(n) is multiplicative with a(2^e) = e + 1, a(3^e) = 1, a(p^e) = e + 1 if p == 1 (mod 4), a(p^e) = (1 + (-1)^e) / 2 if p == 3 (mod 4) and p > 3.
Dirichlet g.f.: zeta(s) * L(chi,s) where chi(n) = Kronecker(-9, n). Sum_{n>0} a(n) / n^s = Product_{p prime} 1 / ((1 - p^-s) * (1 - Kronecker(-9, p) * p^-s)). (End)
a(3*n) = a(n). a(2*n + 1) = A125079(n). a(4*n + 1) = A008441(n).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 2*Pi/3 = 2.094395... (A019693). - Amiram Eldar, Oct 17 2022

A352453 Decimal expansion of the area of intersection of 4 unit-radius circles that have the vertices of a unit-side square as centers.

Original entry on oeis.org

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

Offset: 0

Amiram Eldar, Mar 16 2022

The solution to a problem in Jones (1932): "At each corner of a garden, surrounded by a wall n yards square, a goat is tied with a rope n yards long. Find the area of the part of the garden common to the four goats." (When the square is taken to be of unit size, the common area is this constant.)
The perimeter of the shape formed by the intersection is 2*Pi/3 (A019693).
The solution to the three-dimensional version of this problem is A352454.

			0.31514674362772045262676811958729526112291787931465...

Donald L. Chambers, Problem 3684, School Science and Mathematics, Vol. 77, No. 5 (1977), p. 443; Solution by J. Philip Smith, ibid., Vol. 78, No. 4 (1978), pp. 354-355.
Amiram Eldar, Illustration.
Samuel Isaac Jones, Mathematical Nuts: For Lovers of Mathematics, 1932, Problems 9 and 10, pp. 86, 301-302.
Missouri State University, Problem #8, Finding the Area (resp. Volume) of Overlapping Circles (resp. Spheres), Advanced Problem Archive,; Solution to Problem #8, by Raymond Roan.
Bruce Shawyer, Problem 6, APICS 1999 Mathematics Competition, The Academy Corner, Crux Mathematicorum, Vol. 25, No. 8, 1999, p. 453; Solutions by Richard Tod and Catherine Shevlin, Vol. 26, No. 4, 2000, pp. 193-194.
Charles W. Trigg, Problem 686, Crux Mathematicorum, Vol. 7, No. 9, 1981, p. 275; Solution by Jordan Dou, Vol. 8, No. 9, 1982, p. 294.

Cf. A002194, A019670, A019693.

Cf. A075838, A133731, A192930, A352454.

RealDigits[1 + Pi/3 - Sqrt[3], 10, 100][[1]]

Equals 1 + Pi/3 - sqrt(3) = 1 + A019670 - A002194.

A282195 a(n) is the numerator of Sum_{m=0..n}(Sum_{k=0..m} ((k+1)/(m-k+1)^2) * (Catalan(k)/(2^(2k)))^2)(Sum_{k=0..n-m} ((k+1)/(n-m-k+1)^2) * (Catalan(k)/(2^(2*k)))^2).

Original entry on oeis.org

1, 3, 299, 1691, 4451729, 13446833, 16372396819, 208298035171, 1669160962863, 446401251163753, 6516008708737202119, 44233149340111747277, 5029067414956952883994601, 5810809342741928035310687, 46442062699559407155897191, 1018306138326248284055588777, 369103117042133718901423551221401

Offset: 0

Ralf Steiner, Feb 08 2017

The series a(n)/A282196(n) is absolutely convergent to (2/3 Pi)^2.

Paolo P. Lava, Table of n, a(n) for n = 0..100

Cf. A281070, A280723, A282196 (denominators).

Cf. A000108 (Catalan), A019693 (2 Pi/3).

b[n_]=(Sum[((k+1)/(n-k+1)^2)((CatalanNumber[k])/(2^(2k)))^2, {k, 0, n}]); a[n_] = Sum[(b[k]*b[n - k]), {k, 0, n}]; Numerator /@a/@ Range[0, 10]

C(n) = binomial(2*n,n)/(n+1);
b(n) = sum(k=0, n, ((k+1)/(n-k+1)^2) * (C(k)/(2^(2*k)))^2);
a(n) = numerator(sum(k=0, n, b(k)*b(n-k))); \\ Michel Marcus, Feb 11 2017

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