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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A376870 Reduced numerators of Newton's iteration for 1/sqrt(3), starting with 1/3.

Original entry on oeis.org

1, 4, 130, 2739685, 21055737501685791580, 9337539302589041654242365815942422114384262970589593842110
Offset: 0

Views

Author

Steven Finch, Oct 07 2024

Keywords

Comments

An explicit formula for a(n) is not known, although it arises from a recurrence and the corresponding denominators are simply 3^((3^n + 1)/2) = 3*A134799(n).
Next term is too large to include.

Examples

			a(1) = 4 because b(1) = (3/2)*(1/3)*(1 - 1/9) = 4/9.
1/3, 4/9, 130/243, 2739685/4782969, ... = A376870(n)/(3*A134799(n)).
		

Crossrefs

Programs

  • Mathematica
    Module[{n = 0}, NestList[#*(3^(3^n++ + 1) - #^2)/2 &, 1, 6]] (* Paolo Xausa, Oct 17 2024 *)
  • Python
    from itertools import count, islice
    def A376870_gen(): # generator of terms
        p = 1
        for k in count(0):
            yield p
            p = p*(3**(3**k+1)-p**2)>>1
    A376870_list = list(islice(A376870_gen(),6)) # Chai Wah Wu, Oct 11 2024

Formula

a(n) is the reduced numerator of b(n) = (3/2)*b(n-1)*(1 - b(n-1)^2); b(0) = 1/3.
Limit_{n -> oo} a(n)/(3*A134799(n)) = 1/sqrt(3) = A020760.
a(n+1) = a(n)*(3^(3^n+1)-a(n)^2)/2. - Chai Wah Wu, Oct 11 2024

A134799 a(n) = 3^((3^n - 1)/2).

Original entry on oeis.org

1, 3, 81, 1594323, 12157665459056928801, 5391030899743293631239539488528815119194426882613553319203
Offset: 0

Views

Author

Yasutoshi Kohmoto, Jan 09 2008

Keywords

Comments

Number of partitions into "bus routes" of the graph G_{n+1} defined below.
These seem to be one-third the reduced denominators of Newton's iteration for 1/sqrt(3), starting with 1/3. - Steven Finch, Oct 08 2024

Examples

			.........|..................G_1
****
.......__|__................G_2
.........|
****
.__|_____|_____|__..........G_3
...|.....|.....|
.........|
.......__|__
.........|
****.
..._|_........._|_..........G_4
_|__|_____|_____|__|_
.|._|_....|...._|_.|
....|.....|.....|
......_|__|__|_
.......|._|_.|
..........|
****
G_1 = o---. = rooted tree with one edge and one leaf node. For n > 0, G_{n+1} is obtained from G_n by splitting each leaf node into three.
		

Crossrefs

Programs

  • Mathematica
    3^((3^Range[0, 6] - 1)/2) (* Paolo Xausa, Oct 17 2024 *)

Formula

a(n) is conjectured to be one-third the reduced denominator of b(n) = (3/2)*b(n-1)*(1 - b(n-1)^2); b(0) = 1/3. - Steven Finch, Oct 08 2024
Limit_{n -> oo} A376870(n)/(3*a(n)) = 1/sqrt(3) = A020760. - Steven Finch, Oct 08 2024

Extensions

Edited by N. J. A. Sloane, Jan 29 2008
a(5) from Andrew Howroyd, Oct 07 2024

A156309 Decimal expansion of the absolute value of the larger solution of (n^2+n)/2 = -1/12. (Real root q of 6n^2 + 6n + 1, the other root being p = -1-q.)

Original entry on oeis.org

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

Views

Author

Daniele P. Morelli, Feb 07 2009

Keywords

Comments

The formula returning the n-th triangular number (A000217) is (n^2+n)/2. On the other hand, Ramanujan's identity claims that the value of the infinite sum 1+2+3+.... is -1/12. This irrational number is the solution of the equation (n^2+n)/2 = -1/12, that is, the "limit" triangular number.
Equals the Knuth's random generators constant, that is, the ratio c/m in congruence random number generators of the type X_(n+1) = (aX_n +c) mod (m) which minimizes the correlation between successive values. - Stanislav Sykora, Nov 13 2013
It is also the fraction of the full solid angle cut out by a cone having the magic angle (A195696) as its polar angle. - Stanislav Sykora, Nov 13 2013

Examples

			The two roots of 6n^2 + 6n + 1 = 0 are -0.21132... and -0.78867513... (Cf. A020769.)
		

References

  • B. Candelpergher, Ramanujan summation of divergent series. Lectures notes in mathematics 2185, Springer 2017.
  • D. E. Knuth, The Art of Computer Programming, Vol. 2, Addison-Wesley, 1969, Chapter 3.3.3.

Crossrefs

Programs

  • Mathematica
    First[RealDigits[(3 - Sqrt[3])/6, 10, 100]] (* Paolo Xausa, Jun 25 2024 *)
  • PARI
    abs(solve(n=-1/2, 0, 6*n^2+6*n+1)) \\ Michel Marcus, Oct 05 2013

Formula

(1 - 1/sqrt(3))/2 = (1 - A020760)/2 = 1/2 - A020769. - R. J. Mathar, Feb 10 2009
Equals - HurwitzZeta(-1, (9 - sqrt(3))/6). - Peter Luschny, Jul 05 2020
Equals (3 - sqrt(3))/6. - Michel Marcus, Jun 10 2021
Equals 1/A165663 = A334843/3. - Hugo Pfoertner, Jun 25 2024

Extensions

Flipped sign of definition, corrected offset, simplified formula R. J. Mathar, Feb 10 2009

A212886 Decimal expansion of 2/(3*sqrt(3)) = 2*sqrt(3)/9.

Original entry on oeis.org

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

Views

Author

Rick L. Shepherd, May 29 2012

Keywords

Comments

Consider any cubic polynomial f(x) = a(x - r)(x - (r + s))(x -(r + 2s)), where a, r, and s are real numbers with s > 0 and nonzero a; i.e., any cubic polynomial with three distinct real roots, of which the middle root, r + s, is equidistant (with distance s) from the other two. Then the absolute value of f's local extrema is |a|*s^3*(2*sqrt(3)/9). They occur at x = r + s +- s*(sqrt(3)/3), with the local maximum, M, at r + s - s*sqrt(3)/3 when a is positive and at r + s + s*sqrt(3)/3 when a is negative (and the local minimum, m, vice versa). Of course m = -M < 0.
A quadratic number with denominator 9 and minimal polynomial 27x^2 - 4. - Charles R Greathouse IV, Apr 21 2016
This constant is also the maximum curvature of the exponential curve, occurring at the point M of coordinates [x_M = -log(2)/2 = (-1/10)*A016655; y_M = sqrt(2)/2 = A010503]. The corresponding minimum radius of curvature is (3*sqrt(3))/2 = A104956 (see the reference Eric Billault and the link MathStackExchange). - Bernard Schott, Feb 02 2020

Examples

			0.384900179459750509672765853667971637098401167513417917345734...
		

References

  • Eric Billault, Walter Damin, Robert Ferréol et al., MPSI - Classes Prépas, Khôlles de Maths, Ellipses, 2012, exercice 17.07 pages 386, 389-390.

Crossrefs

Programs

  • Mathematica
    RealDigits[2/(3*Sqrt[3]), 10, 105] (* T. D. Noe, May 31 2012 *)
  • PARI
    default(realprecision, 1000); 2*sqrt(3)/9

Formula

(2/9)*sqrt(3) = (2/9)*A002194.

A343057 Decimal expansion of tan(Pi/32).

Original entry on oeis.org

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

Views

Author

Seiichi Manyama, Apr 04 2021

Keywords

Examples

			0.098491403357164253077197...
		

Crossrefs

Cf. A343055 (sin(Pi/32)), A343056 (cos(Pi/32)).
tan(Pi/m): A002194 (m=3), A019934 (m=5), A020760 (m=6), A343058 (m=7), A188582 (m=8), A019918 (m=9), A019916 (m=10), A019913 (m=12), A343059 (m=14), A019910 (m=15), A343060 (m=16), A343061 (m=17), A019908 (m=18), A019907 (m=20), A343062 (m=24), A019904 (m=30), A343057 (m=32), A019903 (m=36).

Programs

  • Magma
    R:= RealField(125); Tan(Pi(R)/32); // G. C. Greubel, Sep 30 2022
    
  • Mathematica
    RealDigits[Tan[Pi/32], 10, 120, -1][[1]] (* Amiram Eldar, Apr 27 2021 *)
  • PARI
    tan(Pi/32)
    
  • PARI
    sqrt((2-sqrt(2+sqrt(2+sqrt(2))))/(2+sqrt(2+sqrt(2+sqrt(2)))))
    
  • SageMath
    numerical_approx(tan(pi/32), digits=125) # G. C. Greubel, Sep 30 2022

Formula

Equals sqrt( (2-sqrt(2+sqrt(2+sqrt(2))))/(2+sqrt(2+sqrt(2+sqrt(2)))) ).
Equals A232738/(1+A232737). - R. J. Mathar, Sep 06 2025

A359837 Decimal expansion of the unsigned ratio of similitude between an equilateral reference triangle and its first Morley triangle.

Original entry on oeis.org

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

Views

Author

Frank M Jackson, Jan 14 2023

Keywords

Comments

The first Morley triangle of any reference triangle is always equilateral. Therefore a reference equilateral triangle and its first Morley triangle will be in a homothetic relationship. This sequence is the real terms of a constant that is the ratio of similitude of the homothety. The constant is negative.
If an equilateral triangle has a side a, a circumradius R and a first Morley triangle with side a', then a = R*sqrt(3) and a' = 8*R*(sin(Pi/9))^3, so the ratio of similitude between the two triangles is a'/a = (8/sqrt(3)) * (sin(Pi/9))^3. - Bernard Schott, Feb 06 2023

Examples

			0.1847925309040953727013520475722037558709135597926517172356...
		

Crossrefs

Programs

  • Mathematica
    RealDigits[Sin[Pi/18]/Cos[Pi/9], 10, 100][[1]]
    N[Solve[x^3 + 3*x^2 - 6*x + 1 == 0, {x}][[2]], 90]
  • PARI
    sin(Pi/18)/cos(Pi/9) \\ Michel Marcus, Jan 15 2023

Formula

Equals sin(Pi/18)/cos(Pi/9).
A root of x^3+3*x^2-6*x+1.
Equals A019819/A019879. - Michel Marcus, Jan 15 2023
Equals 8 * A020760 * A019829^3. - Bernard Schott, Feb 06 2023

A184977 a(n) = Sum_{k=1..n} floor(k*gamma) where gamma is Euler's constant (A001620).

Original entry on oeis.org

0, 1, 2, 4, 6, 9, 13, 17, 22, 27, 33, 39, 46, 54, 62, 71, 80, 90, 100, 111, 123, 135, 148, 161, 175, 190, 205, 221, 237, 254, 271, 289, 308, 327, 347, 367, 388, 409, 431, 454, 477, 501, 525, 550, 575, 601, 628, 655, 683, 711, 740, 770, 800, 831, 862, 894, 926, 959, 993, 1027, 1062
Offset: 1

Views

Author

Michel Lagneau, Mar 27 2011

Keywords

Comments

a(n) = A183143(n) for n = 1..96 where A183143(n) is the sequence floor(1/r) + floor(2/r) + ... + floor(n/r) and r = sqrt(3). It is interesting to note that a(n)/n^2 converges to gamma/2.
gamma = 0.57721566490153286060651209... (A002852)
1/sqrt(3) = 0.577350269189625764509148... (A020760)
Starts to differ from A183143 at a(97). - R. J. Mathar, Aug 28 2025

Crossrefs

Programs

  • Magma
    R:=RealField(100); [(&+[Floor(k*EulerGamma(R)): k in [1..n]]): n in [1..50]]; // G. C. Greubel, Aug 27 2018
  • Maple
    with(numtheory):Digits:=500:s:=0:c:=evalf(gamma(0)):for n from 1 to 100 do:
      s:=s+floor(n*c):printf(`%d, `,s):od:
  • Mathematica
    Table[Sum[Floor[k*EulerGamma], {k, 1, n}], {n, 50}] (* G. C. Greubel, Jun 02 2017 *)
  • PARI
    a(n) = sum(k=1, n, floor(k*Euler)); \\ Michel Marcus, Apr 02 2017
    

Formula

Partial sums of A038128.

Extensions

Name edited by Jon E. Schoenfield, Apr 02 2017

A263574 Beatty sequence for 1/sqrt(3) - log(phi)/3575 where phi is the golden ratio, A001622.

Original entry on oeis.org

0, 0, 1, 1, 2, 2, 3, 4, 4, 5, 5, 6, 6, 7, 8, 8, 9, 9, 10, 10, 11, 12, 12, 13, 13, 14, 15, 15, 16, 16, 17, 17, 18, 19, 19, 20, 20, 21, 21, 22, 23, 23, 24, 24, 25, 25, 26, 27, 27, 28, 28, 29, 30, 30, 31, 31, 32, 32, 33, 34, 34, 35, 35, 36, 36, 37, 38, 38, 39, 39, 40, 40, 41, 42, 42, 43
Offset: 0

Views

Author

Karl V. Keller, Jr., Oct 21 2015

Keywords

Comments

The number 1/sqrt(3) - log(phi)/3575 (=0.577215664483...) is an approximation to Euler's constant (A001620) (=0.577215664901...).
M. Hudson found a similar Euler-Mascheroni constant approximation (see link), 1/sqrt(3)-1/7429 (=0.57721566157...).

Examples

			For n=9, floor(9*(0.577215664483)) = floor(5.194940980347) = 5.
		

Crossrefs

Cf. A001620, A020760 (1/sqrt(3)), A038128 (Beatty sequence for Euler's constant), A097337 (Beatty sequence for 1/sqrt(3)).

Programs

  • Magma
    phi:= (1+Sqrt(5))/2; [Floor(n*(1/Sqrt(3) - Log(phi)/3575)): n in [0..100]]; // G. C. Greubel, Sep 05 2018
  • Mathematica
    Table[Floor[n (1/Sqrt@ 3 - Log[GoldenRatio]/3575)], {n, 0, 75}] (* Michael De Vlieger, Nov 12 2015 *)
  • PARI
    {phi = (1+sqrt(5))/2}; vector(100, n, n--; floor(n*(1/sqrt(3) - log(phi)/3575))) \\ G. C. Greubel, Sep 05 2018
    
  • Python
    from sympy import floor, log, sqrt
    for n in range(0,101):print(floor(n*(1/sqrt(3)-log(1/2+sqrt(5)/2)/3575)),end=', ')
    

Formula

a(n) = floor(n*(1/sqrt(3) - log(phi)/3575)).
a(n) = A038128(n) for n < 58628.

A270230 Decimal expansion of 3/(4*Pi).

Original entry on oeis.org

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

Views

Author

Stanislav Sykora, Mar 13 2016

Keywords

Comments

Consider generic prisms with triangular bases (tp), enclosed by a sphere, and let f(tp) be the fraction of the sphere volume occupied by any of them (i.e., the ratio of the prism volume to the sphere volume). Then this constant is the supremum of f(tp). It is attained by prisms which have as their base equilateral triangles with edge lengths r*sqrt(2), and rectangular side faces that are r*sqrt(2) wide and r*2/sqrt(3) high, where r is the radius of the enclosing, circumscribed sphere.
An intriguing fact is that the volume of such a best-fitting prism is exactly r^3. Hence, 1/a is the volume of a sphere with radius 1.
Examples of similar constants obtained for other shapes enclosed by spheres are: A020760 for cylinders and A165952 for cuboids.

Examples

			0.238732414637843003653325645058771543051689468610684673121501016...
		

Crossrefs

Cf. A002193, A019699 (one tenth of 1/a), A020760, A020832 (one tenth of 2/sqrt(3)), A165952.

Programs

  • Mathematica
    First@ RealDigits[N[3/4/Pi, 120]] (* Michael De Vlieger, Mar 15 2016 *)
  • PARI
    3/4/Pi

A270395 Denominators of r-Egyptian fraction expansion for sqrt(1/3), where r(k) = 1/Fibonacci(k+1).

Original entry on oeis.org

2, 7, 57, 2713, 4918440, 22223269372702, 355194969748884199331083933, 896996605353313749663062291034129550464167047150212163, 710754225314643793883316602476833806192189702887005360976366457324682443530843343068467237316280025378530303
Offset: 1

Views

Author

Clark Kimberling, Mar 22 2016

Keywords

Comments

Suppose that r is a sequence of rational numbers r(k) <= 1 for k >= 1, and that x is an irrational number in (0,1). Let f(0) = x, n(k) = floor(r(k)/f(k-1)), and f(k) = f(k-1) - r(k)/n(k). Then x = r(1)/n(1) + r(2)/n(2) + r(3)/n(3) + ..., the r-Egyptian fraction for x.
See A269993 for a guide to related sequences.

Examples

			sqrt(1/3) = 1/2 + 1/(2*7) + 1/(3*57) + 1/(5*2713) + ...
		

Crossrefs

Programs

  • Mathematica
    r[k_] := 1/Fibonacci[k+1]; f[x_, 0] = x; z = 10;
    n[x_, k_] := n[x, k] = Ceiling[r[k]/f[x, k - 1]]
    f[x_, k_] := f[x, k] = f[x, k - 1] - r[k]/n[x, k]
    x = Sqrt[1/3]; Table[n[x, k], {k, 1, z}]
  • PARI
    r(k) = 1/fibonacci(k+1);
    f(k,x) = if (k==0, x, f(k-1, x) - r(k)/a(k, x););
    a(k, x=sqrt(1/3)) = ceil(r(k)/f(k-1, x)); \\ Michel Marcus, Mar 22 2016
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