Displaying 51-60 of 85 results found.
Decimal expansion of (5*sqrt(3) + sqrt(15))/(6*Pi).
+10
4
6, 6, 4, 9, 0, 8, 8, 9, 4, 2, 0, 5, 3, 2, 6, 6, 4, 3, 1, 1, 4, 4, 2, 8, 4, 4, 6, 7, 0, 8, 6, 3, 3, 7, 1, 6, 1, 6, 4, 8, 7, 6, 5, 8, 0, 5, 5, 5, 6, 9, 1, 9, 3, 8, 1, 0, 5, 7, 5, 9, 2, 6, 0, 5, 7, 2, 2, 9, 6, 4, 7, 1, 8, 1, 8, 7, 7, 3, 2, 5, 9, 7, 4, 9, 7, 0, 8, 9, 0, 0, 2, 6, 9, 2, 0, 9, 2, 5, 9, 8, 9, 8, 2, 8, 0
COMMENTS
The ratio of the volume of a regular dodecahedron to the volume of the circumscribed sphere (which has circumradius a*(sqrt(3) + sqrt(15))/4 = a*( A002194 + A010472)/4, where a is the dodecahedron's edge length; see MathWorld link). For similar ratios for other Platonic solids, see A165922, A049541, A165952, and A165954. A063723 shows the order of these by size.
FORMULA
Equals (5 + sqrt(5))/(2*Pi*sqrt(3)).
EXAMPLE
0.6649088942053266431144284467086337161648765805556919381057592605722964718...
MATHEMATICA
RealDigits[(5*Sqrt[3]+Sqrt[15])/(6*Pi), 10, 120][[1]] (* Harvey P. Dale, Feb 16 2018 *)
PROG
(PARI) (5*sqrt(3)+sqrt(15))/(6*Pi)
CROSSREFS
Cf. A000796, A002194, A010472, A165922, A049541, A165952, A165954, A063723, A002163, A020760, A010527.
Decimal expansion of the surface area of pentagonal rotunda with edge length 1.
+10
4
2, 2, 3, 4, 7, 2, 0, 0, 2, 6, 5, 3, 9, 4, 1, 2, 8, 2, 7, 6, 7, 9, 8, 4, 1, 4, 1, 5, 8, 1, 8, 8, 6, 1, 3, 0, 7, 3, 8, 1, 8, 0, 1, 3, 5, 1, 3, 4, 3, 1, 6, 2, 2, 6, 1, 2, 9, 7, 9, 9, 7, 6, 3, 1, 6, 7, 1, 0, 2, 0, 4, 7, 1, 6, 7, 6, 3, 5, 2, 4, 7, 7, 6, 8, 3, 3, 9, 9, 7, 2, 1, 9, 3, 8, 6, 4, 1, 1, 4, 7, 0, 3, 3, 2, 0
COMMENTS
Pentagonal rotunda: 20 vertices, 35 edges, and 17 faces.
FORMULA
Digits of sqrt(5*(145+58*sqrt(5)+2*sqrt(30*(65+29*sqrt(5)))))/2.
EXAMPLE
22.3472002653941282767984141581886130738180135134316226129799763167102...
MATHEMATICA
RealDigits[N[Sqrt[5*(145+58*Sqrt[5]+2*Sqrt[30*(65+29*Sqrt[5])])]/2, 200]]
Decimal expansion of sqrt(7)/2.
+10
4
1, 3, 2, 2, 8, 7, 5, 6, 5, 5, 5, 3, 2, 2, 9, 5, 2, 9, 5, 2, 5, 0, 8, 0, 7, 8, 7, 6, 8, 1, 9, 6, 3, 0, 2, 1, 2, 8, 5, 5, 1, 2, 9, 5, 9, 1, 5, 4, 1, 2, 2, 5, 0, 9, 0, 1, 8, 4, 1, 6, 7, 2, 2, 9, 6, 0, 0, 5, 3, 4, 4, 1, 1, 6, 1, 5, 1, 4, 1, 8, 1, 3, 8, 8, 0, 1, 9, 6, 4, 4, 3, 2, 3, 7, 2, 7
COMMENTS
Absolute value of the imaginary part of any of the nontrivial divisors of 2 in O_Q(sqrt(-7)).
The incircle of a triangle with sides of lengths 4, 5, 6 units respectively has a radius of sqrt(7)/2.
With a different offset, decimal expansion of 5 * sqrt(7).
In a regular hexagon inscribed in a circle with a radius of 1 unit the three distinct distances between any vertex and the middle of the sides are 1/2, sqrt(7)/2 and sqrt(13)/2.
The continued fraction expansion of sqrt(7)/2 is 1, repeat(3, 10, 3, 2). The convergents are given in A294972/ A294973. (End)
FORMULA
(1/2 - sqrt(-7)/2)(1/2 + sqrt(-7)/2) = 2.
EXAMPLE
1.32287565553229529525...
MATHEMATICA
RealDigits[Sqrt[7]/2, 10, 100][[1]]
PROG
(PARI) { default(realprecision, 20080); x=sqrt(7)/2; for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b242703.txt", n, " ", d)); } \\ Iain Fox, Nov 18 2017
Decimal expansion of (1 + sqrt(3))/2, unique positive root of x^2 - x - 1/2.
+10
4
1, 3, 6, 6, 0, 2, 5, 4, 0, 3, 7, 8, 4, 4, 3, 8, 6, 4, 6, 7, 6, 3, 7, 2, 3, 1, 7, 0, 7, 5, 2, 9, 3, 6, 1, 8, 3, 4, 7, 1, 4, 0, 2, 6, 2, 6, 9, 0, 5, 1, 9, 0, 3, 1, 4, 0, 2, 7, 9, 0, 3, 4, 8, 9, 7, 2, 5, 9, 6, 6, 5, 0, 8, 4, 5, 4, 4, 0, 0, 0, 1, 8, 5, 4, 0, 5, 7, 3, 0, 9, 3, 3, 7, 8, 6, 2, 4, 2, 8, 7, 8, 3, 7, 8
COMMENTS
Also, max {a, b} where {a,b} is the unique solution of a + b = 1 and a^2 + b^2 = 2 (implying also ab = -1/2 and a^3 + b^3 = 5/2 without solving for a, b). See A332122 for a generalization to 3 variables {a, b, c}.
This is a non-integer element of the quadratic number field Q(sqrt(3)) with the given monic minimal polynomial. The other negative root is -(-1 + sqrt(3))/2 = - A152422. - Wolfdieter Lang, Aug 30 2022
FORMULA
Equals 1/2 + Sum_{n>=0} ((-1)^(n + 1)*binomial(2*n, n))/(2^(3*n + 1/2)*(2*n - 1)). - Antonio GraciĆ” Llorente, Nov 11 2024
EXAMPLE
1.3660254037844386467637231707529361834714026269051903140279...
MATHEMATICA
RealDigits[(1 + Sqrt[3])/2, 10, 120][[1]] (* Amiram Eldar, Jun 21 2023 *)
PROG
(PARI) localprec(111); digits(solve(a=0, 2, a^2-a-1/2)\.1^99)
Decimal expansion of the maximum possible volume of a polyhedron with 8 vertices inscribed in the unit sphere.
+10
4
1, 8, 1, 5, 7, 1, 6, 1, 0, 4, 2, 2, 4, 4, 2, 0, 3, 9, 7, 5, 0, 8, 4, 9, 4, 9, 3, 0, 6, 3, 3, 1, 7, 7, 7, 8, 9, 0, 1, 3, 1, 0, 0, 9, 5, 5, 2, 7, 5, 4, 3, 9, 8, 3, 7, 6, 6, 6, 3, 7, 2, 9, 1, 6, 9, 1, 8, 4, 8, 9, 9, 3, 7, 0, 0, 0, 2, 8, 9, 3, 8, 6, 5, 2, 7, 0, 3
COMMENTS
Berman and Hanes (see link, page 81) proved in 1970 that an arrangement of 8 points on the surface of a sphere with 4 points with node degree 4 and 4 points with node degree 5 is the one with a maximum volume of their convex hull.
FORMULA
Equals sqrt((475 + 29*sqrt(145))/250).
EXAMPLE
1.8157161042244203975084949306331777890131009552754398376663729...
MATHEMATICA
RealDigits[Sqrt[(475 + 29*Sqrt[145])/250], 10, 120][[1]] (* Amiram Eldar, Jun 01 2023 *)
PROG
(PARI) sqrt((475+29*sqrt(145))/250)
Decimal expansion of the conjecturally maximum possible volume of a polyhedron with 9 vertices inscribed in the unit sphere.
+10
4
2, 0, 4, 3, 7, 5, 0, 1, 1, 5, 8, 9, 9, 6, 3, 9, 8, 4, 1, 1, 6, 6, 3, 6, 5, 4, 6, 4, 2, 2, 6, 9, 8, 5, 3, 3, 3, 8, 6, 3, 2, 6, 0, 6, 1, 5, 2, 9, 4, 7, 5, 1, 8, 1, 8, 7, 1, 8, 2, 1, 5, 7, 9, 5, 6, 8, 7, 1, 0, 4, 2, 6, 4, 0, 9, 2, 7, 7, 1, 4, 0, 6, 1, 7, 8, 5, 9
FORMULA
Equals 3*sqrt(2*sqrt(3) - 3).
EXAMPLE
2.0437501158996398411663654642269853338632606152947518187182157956871...
MATHEMATICA
RealDigits[3*Sqrt[2*Sqrt[3] - 3], 10, 120][[1]] (* Amiram Eldar, Jun 28 2023 *)
PROG
(PARI) 3*sqrt(2*sqrt(3) - 3)
Decimal expansion of the conjecturally maximum possible volume of a polyhedron with 10 vertices inscribed in the unit sphere.
+10
4
2, 2, 1, 8, 7, 1, 1, 1, 3, 1, 5, 4, 5, 3, 9, 9, 4, 0, 3, 2, 4, 7, 2, 8, 2, 7, 5, 1, 1, 2, 8, 4, 1, 7, 0, 1, 3, 8, 1, 0, 7, 2, 5, 3, 7, 4, 6, 6, 3, 3, 4, 4, 3, 8, 1, 7, 5, 0, 0, 4, 9, 0, 8, 4, 2, 0, 1, 0, 0, 8, 1, 2, 7, 9, 9, 0, 9, 1, 8, 1, 4, 8, 8, 4, 6, 3, 3
COMMENTS
The polyhedron (see linked illustration) has vertices at the poles and two square rings of vertices rotated by Pi/4 against each other, with a polar angle of approx. +-62.89908285 degrees against the poles. The polyhedron is completely described by this angle and its order 16 symmetry. It would be desirable to know a closed formula representation of this angle and the volume.
EXAMPLE
2.218711131545399403247282751128417013810725374663344381750049084201...
Continued fraction expansion of sqrt(3)/2.
+10
3
0, 1, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6
COMMENTS
Also continued fraction expansion of sin(Pi/3).
FORMULA
a(n) = 2*(2+(-1)^n) for n>1. a(n) = a(n-2) for n>3. G.f.: -x*(x^2+6*x+1) / ((x-1)*(x+1)). - Colin Barker, Jan 10 2014
MATHEMATICA
ContinuedFraction[Sqrt[3]/2, 120] (* or *) PadRight[{0, 1}, 120, {6, 2}] (* Harvey P. Dale, May 14 2016 *)
Decimal expansion of (sqrt(3)-1)/2.
+10
3
3, 6, 6, 0, 2, 5, 4, 0, 3, 7, 8, 4, 4, 3, 8, 6, 4, 6, 7, 6, 3, 7, 2, 3, 1, 7, 0, 7, 5, 2, 9, 3, 6, 1, 8, 3, 4, 7, 1, 4, 0, 2, 6, 2, 6, 9, 0, 5, 1, 9, 0, 3, 1, 4, 0, 2, 7, 9, 0, 3, 4, 8, 9, 7, 2, 5, 9, 6, 6, 5, 0, 8, 4, 5, 4, 4, 0, 0, 0, 1, 8, 5, 4, 0, 5, 7, 3, 0, 9, 3, 3, 7, 8, 6, 2, 4, 2, 8, 7
COMMENTS
The number has continued fraction [0, 2, 1, 2, 1, 2, 1, ...].
The iterated function z^2 - 1/2 gives a good rational approximation of this number times -1 after sixty steps. - Alonso del Arte, Apr 10 2016
FORMULA
Equals Product_{k>=1} (1 + (-1)^k * 2/(6*k-3)). - Amiram Eldar, Aug 10 2020
EXAMPLE
0.36602540378443864676372317...
Decimal expansion of the conjecturally maximum possible volume of a polyhedron with 11 vertices inscribed in the unit sphere.
+10
3
2, 3, 5, 4, 6, 3, 4, 4, 9, 5, 0, 6, 8, 6, 1, 5, 2, 0, 3, 2, 3, 6, 8, 8, 0, 5, 9, 2, 6, 3, 8, 9, 2, 6, 5, 4, 1, 6, 0, 3, 4, 4, 8, 6, 4, 2, 6, 9, 3, 4, 2, 1, 6, 8, 5, 9, 9, 6, 0, 7, 5, 6, 6, 0, 7, 9, 8, 5, 4, 5, 8, 3, 1, 4, 8, 1, 5, 5, 5, 3, 1, 5, 0, 1, 9, 4, 5
COMMENTS
The polyhedron (see linked illustration) with a symmetry group of order 4 has a vertex in the north pole on its axis of symmetry. The remaining 10 vertices are diametrically opposite in pairs relative to this axis of symmetry. The polar vertex has vertex degree 6. 8 vertices have vertex degree 5. 2 vertices have vertex degree 4.
This allocation seems to be the best possible approximation of a medial distribution of the vertex degrees, which is a known necessary condition for maximum volume. Of the 25 possible triangulations with vertex degree >= 4, all the others have more than 2 vertices with vertex degree 4, which leads to more pointed corners and therefore smaller volumes.
EXAMPLE
2.35463449506861520323688059263892654160344864269342168599607566...
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