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User: William H. Richardson

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William H. Richardson has authored 3 sequences.

A244644 Consider the method used by Archimedes to determine the value of Pi (A000796). This sequence denotes the number of iterations of his algorithm which would result in a difference of less than 1/10^n from that of Pi.

Original entry on oeis.org

0, 1, 3, 5, 6, 8, 10, 11, 13, 15, 16, 18, 20, 21, 23, 25, 26, 28, 29, 31, 33, 34, 36, 38, 39, 41, 43, 44, 46, 48, 49, 51, 53, 54, 56, 58, 59, 61, 63, 64, 66, 68, 69, 71, 73, 74, 76, 78, 79, 81, 83, 84, 86, 88, 89, 91, 93, 94, 96, 98, 99, 101, 103, 104, 106, 108, 109, 111, 113, 114
Offset: 0

Views

Author

William H. Richardson and Robert G. Wilson v, Jul 03 2014

Keywords

Comments

It takes on average 5/3 iterations to yield another digit in the decimal expansion of Pi.
The side of a 96-gon inscribed in a unit circle is equal to sqrt(2-sqrt(2+sqrt(2+sqrt(2+sqrt(3))))). This is the size of one of the two polygons that Archimedes used to derive that 3 + 10/70 < Pi < 3 + 10/71.
In the Mathematica program, we started with an inscribed triangle and a circumscribed triangle of a unit circle and used decimal precision to just over a 1000 places.
The perimeter of the circumscribed 3*2^n-polygon exceeds Pi by more than the deficit of the perimeter of the inscribed 3*2^n-polygon. If we were to give twice the weight of the inscribed 3*2^n-polygon to that of the circumscribed 3*2^n-polygon, then the convergence would be twice as fast!
From A.H.M. Smeets, Jul 12 2018: (Start)
Archimedes's scheme: set upp(0) = 2*sqrt(3), low(0) = 3 (hexagons); upp(n+1) = 2*upp(n)*low(n)/(upp(n)+low(n)) (harmonic mean, i.e., 1/upp(n+1) = (1/upp(n) + 1/low(n))/2), low(n+1) = sqrt(upp(n+1)*low(n)) (geometric mean, i.e., log(low(n+1)) = (log(upp(n+1)) + log(low(n)))/2), for n >= 0. Invariant: low(n) < Pi < upp(n); variant function: upp(n)-low(n) tends to zero for n -> inf. The error of low(n) and upp(n) decreases by a factor of approximately 4 each iteration, i.e., approximately 2 bits are gained by each iteration.
From Archimedes's scheme, set r(n) = (2*low(n) + upp(n))/3, r(n) > Pi and the error decreases by a factor of approximately 16 for each iteration, i.e., approximately 4 bits are gained by each iteration. This is often called "Snellius acceleration".
For similar schemes see also A014549 (in this case with quadratically convergence), A093954, A129187, A129200, A188615, A195621, A202541.
Note that replacing "5/3" by "log(10)/log(4)" would be better in the first comment. (End)

Examples

			Just averaging the initial two triangles (3.89711) would yield Pi to one place of accuracy, i.e., the single digit '3'. Therefore a(0) = 0.
The first iteration yields, as the perimeters of the two hexagons, 4*sqrt(3) and 6. Their average is ~ 3.2320508 which is within 1/10 of the true value of Pi. Therefore a(1) = 1.
a(3) = 5 since it takes 5 iterations of Archimedes's algorithm to drive the averaged value of the circumscribed 96-gon and the inscribed 96-gon to yield a value within 0.001 of the correct value of Pi.
a(4) = 6 since it takes 6 iterations of Archimedes's algorithm to drive the averaged value of the circumscribed 3*2^6-gon and the inscribed 3*2^6-gon to yield a value within 0.0001 of the correct value of Pi.

References

  • Petr Beckmann, A History of Pi, 5th Ed. Boulder, Colorado: The Golem Press (1982).
  • Jonathan Borwein and David Bailey, Mathematics by Experiment, Second Edition, A. K. Peters Ltd., Wellesley, Massachusetts 2008.
  • Jonathan Borwein & Keith Devlin, The Computer As Crucible, An Introduction To Experimental Mathematics, A. K. Peters, Ltd., Wellesley, MA, Chapter 7, 'Calculating [Pi]' pp. 71-79, 2009.
  • Eli Maor, The Pythagorean Theorem, Princeton Science Library, Table 4.1, page 55.
  • Daniel Zwillinger, Editor-in-Chief, CRC Standard Mathematical Tables and Formulae, 31st Edition, Chapman & Hall/CRC, Boca Raton, London, New York & Washington, D.C., 2003, §4.5 Polygons, page 324.

Links

Crossrefs

Cf. A000796.

Programs

  • Mathematica
    a[n_] := a[n] = N[2 a[n - 1] b[n - 1]/(a[n - 1] + b[n - 1]), 2^10]; b[n_] := b[n] = N[ Sqrt[ b[n - 1] a[n]], 2^10]; a[-1] = 2Sqrt[27]; b[-1] = a[-1]/2; f[n_] := Block[{k = 0}, While[ 10^n*((a[k] + b[k])/4 -Pi) > 1, k++]; k]; Array[f, 70]

Formula

Conjecture: There exists a c such that a(n) = floor(n*log(10)/log(4)+c); where c is in the range [0.08554,0.10264]. Critical values to narrow the range are believed to be at a(74), a(133), a(192), a(251), a(310), a(366), a(425), a(484). - A.H.M. Smeets, Jul 23 2018

A211174 Johannes Kepler's polyhedron circumscribing constant.

Original entry on oeis.org

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

Views

Author

William H. Richardson and Robert G. Wilson v, Feb 01 2013

Keywords

Comments

The finite solid analogy to the plane polygon circumscribing constant (A051762).
The five Platonic solids are the tetrahedron, the hexahedron or cube, the octahedron, the dodecahedron and the icosahedron.
The geometric interpretation is as follows. Begin with a unit sphere. Circumscribe a tetrahedron and then circumscribe a sphere. Circumscribe a cube and then circumscribe a sphere. Circumscribe an octahedron and then circumscribe a sphere. Circumscribe a dodecahedron and then a sphere. Circumscribe an icosahedron and then a sphere. The constant is the radius of this last sphere. In actuality, it makes no difference the order of the five solids.

Examples

			14.25232921501135639390462188851108328620660858097761088937154874783...

Links

Crossrefs

Cf. A051762.

Programs

  • Mathematica
    RealDigits[ 9(15 - 6 * Sqrt[5]), 10, 111][[1]]

Formula

= 9*(15 - 6*sqrt(5)).

Extensions

Offset corrected by Rick L. Shepherd, Dec 31 2013

A194940 The Square Peg in the Round Hole constant.

Original entry on oeis.org

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

Views

Author

William H. Richardson and Robert G. Wilson v, Sep 05 2011

Keywords

Comments

Given a unit circle and a square of equal area, what is the amount of the square peg shavings (or filings) which would allow the peg to be inserted into the circle? It turns out to be not quite two sevenths.

Examples

			0.28445855040980187815920101812693174533005283078946269804587750...

References

  • Daniel Zwillinger, Editor, CRC Standard Mathematical Tables and Formulae, 31st Edition, Chapman & Hall/CRC, Boca Raton, Section 4.6.6 Circles, page 334 & figure 4.18, 2003.

Links

Crossrefs

Cf. A000796, A019704, A127454.

Programs

  • Mathematica
    RealDigits[ 4*ArcCos[ Sqrt[Pi]/2] - Sqrt[ Pi(4 - Pi)], 10, 111][[1]]
RealDigits[Pi + Sqrt[ 2Pi(2 - Sqrt[Pi (4 - Pi)])] - 4 ArcSin[ Sqrt[Pi/4]], 10, 111][[1]] (* Robert G. Wilson v, Sep 20 2011 *)
  • PARI
    4*acos(sqrt(Pi)/2) - sqrt(Pi*(4-Pi)) \\ G. C. Greubel, Mar 28 2017

Formula

Area = 4*arccos(sqrt(Pi)/2) - sqrt(Pi*(4-Pi)).
Area = Pi + sqrt(2*Pi(2 - sqrt(Pi*(4 - Pi)))) - 4*arcsin(sqrt(Pi/4)). - Robert G. Wilson v, Mar 19 2014

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