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.

Showing 1-10 of 12 results. Next

A002194 Decimal expansion of sqrt(3).

Original entry on oeis.org

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

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Author

Keywords

Comments

"The square root of 3, the 2nd number, after root 2, to be proved irrational, by Theodorus."
Length of a diagonal between any vertex of the unit cube and the one corresponding (opposite) vertex not part of the three faces meeting at the original vertex. (Diagonal is hypotenuse of a triangle with sides 1 and sqrt(2)). Hence the diameter of the sphere circumscribed around the unit cube; the ratio of the diameter of any sphere to the edge length of its inscribed cube. - Rick L. Shepherd, Jun 09 2005
The square root of 3 is the length of the minimal Y-shaped (symmetrical) network linking three points unit distance apart. - Lekraj Beedassy, Apr 12 2006
Continued fraction expansion is 1 followed by {1, 2} repeated. - Harry J. Smith, Jun 01 2009
Also, tan(Pi/3) = 2 sin(Pi/3). - M. F. Hasler, Oct 27 2011
Surface of regular tetrahedron with unit edge. - Stanislav Sykora, May 31 2012
This is the case n=6 of Gamma(1/n)*Gamma((n-1)/n)/(Gamma(2/n)*Gamma((n-2)/n)) = 2*cos(Pi/n), therefore sqrt(3) = A175379*A203145/(A073005*A073006). - Bruno Berselli, Dec 13 2012
Ratio of base length to leg length in the isosceles "vampire" triangle, that is, the only isosceles triangle without reflection triangle. The product of cosines of the internal angles of a triangle with sides 1, 1 and sqrt(3) and all similar triangles is -3/8. Hence its reflection triangle is degenerate. See the link below. - Martin Janecke, May 09 2013
Half of the surface of regular octahedron with unit edge (A010469), and one fifth that of a regular icosahedron with unit edge (i.e., 2*A010527). - Stanislav Sykora, Nov 30 2013
Diameter of a sphere whose surface area equals 3*Pi. More generally, the square root of x is also the diameter of a sphere whose surface area equals x*Pi. - Omar E. Pol, Nov 11 2018
Sometimes called Theodorus's constant, after the ancient Greek mathematician Theodorus of Cyrene (5th century BC). - Amiram Eldar, Apr 02 2022
For any triangle ABC, cotan(A) + cotan(B) + cotan(C) >= sqrt(3); equality is obtained only when the triangle is equilateral (see the Kiran S. Kedlaya link). - Bernard Schott, Sep 13 2022

Examples

			1.73205080756887729352744634150587236694280525381038062805580697945193...
		

References

  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 24, 184.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §3.4 Irrational Numbers and §12.4 Theorems and Formulas (Solid Geometry), pp. 84, 450.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers, Revised Edition, Penguin Books, London, England, 1997, page 23.

Crossrefs

Cf. A040001 (continued fraction), A220335.
Cf. A010469 (double), A010527 (half), A131595 (surface of regular dodecahedron).

Programs

  • Magma
    SetDefaultRealField(RealField(100)); Sqrt(3); // G. C. Greubel, Aug 21 2018
  • Maple
    evalf(sqrt(3), 100); # Michal Paulovic, Feb 24 2023
  • Mathematica
    RealDigits[Sqrt[3], 10, 100][[1]]
  • PARI
    default(realprecision, 20080); x=(sqrt(3)); for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b002194.txt", n, " ", d));  \\ Harry J. Smith, Jun 01 2009
    

Formula

Equals Sum_{k>=0} binomial(2*k,k)/6^k = Sum_{k>=0} binomial(2*k,k) * k/6^k. - Amiram Eldar, Aug 03 2020
sqrt(3) = 1 + 1/2 + 1/(2*3) + 1/(2*3*4) + 1/(2*3*4*2) + 1/(2*3*4*2*8) + 1/(2*3*4*2*8*14) + 1/(2*3*4*2*8*14*2) + 1/(2*3*4*2*8*14*2*98) + 1/(2*3*4*2*8*14*2*98*194) + .... (Define F(n) = (n-1)*sqrt(n^2 - 1) - (n^2 - n - 1). Show F(n) = 1/2 + 1/(2*(n+1)) + 1/(2*(n+1)*(2*n)) + 1/(2*(n+1)*(2*n))*F(2*n^2 - 1) for n >= 0; then iterate this identity at n = 2. See A220335.) - Peter Bala, Mar 18 2022
Equals i^(1/3) + i^(-1/3). - Gary W. Adamson, Jul 06 2022
Equals Product_{n>=1} 3^(1/3^n). - Michal Paulovic, Feb 24 2023
Equals Product_{n>=0} ((6*n + 2)*(6*n + 4))/((6*n + 1)*(6*n + 5)). - Antonio Graciá Llorente, Feb 22 2024
Equals tan(Pi/3) = A010527/(1/2). - R. J. Mathar, Aug 31 2025

Extensions

More terms from Robert G. Wilson v, Dec 07 2000

A220337 A modified Engel expansion for 3*sqrt(15) - 11.

Original entry on oeis.org

2, 5, 8, 2, 32, 62, 2, 1922, 3842, 2, 7380482, 14760962, 2, 108942999582722, 217885999165442, 2, 23737154316161495960243527682, 47474308632322991920487055362, 2, 1126904990058528673830897031906808442930637286502826475522
Offset: 1

Views

Author

Peter Bala, Dec 12 2012

Keywords

Comments

For a brief description of the modified Engel expansion of a real number see A220335.
Let p >= 2 be an integer and set Q(p) = (p - 1)*sqrt(p^2 - 1) - (p^2 - p - 1), so Q(4) = 3*sqrt(15) - 11. Iterating the identity Q(p) = 1/2 + 1/(2*(p+1)) + 1/(2*(p+1)*(2*p)) + 1/(2*(p+1)*(2*p))*Q(2*p^2-1) leads to a representation for Q(p) as an infinite series of unit fractions. The sequence of denominators of these unit fractions can be used to find the modified Engel expansion of Q(p). For further details see the Bala link. The present sequence is the case p = 4. For other cases see A220335 (p = 2), A220336 (p = 3) and A220338 (p = 5).

Crossrefs

Cf. A005828, A220335 (p = 2), A220336 (p = 3), A220338 (p = 5).

Formula

Define the harmonic sawtooth map h(x) := floor(1/x)*(x*ceiling(1/x) - 1). Let x = 3*sqrt(15) - 11. Then a(1) = ceiling(1/x) and for n >= 2, a(n) = floor(1/h^(n-2)(x))*ceiling(1/h^(n-1)(x)), where h^(n)(x) denotes the n-th iterate of the map h(x), with the convention h^(0)(x) = x.
a(3*n+2) = 1/2*{2 + (4 + sqrt(15))^(2^n) + (4 - sqrt(15))^(2^n)} and
a(3*n+3) = (4 + sqrt(15))^(2^n) + (4 - sqrt(15))^(2^n), both for n >= 0.
For n >= 0, a(3*n+1) = 2. For n >= 1, a(3*n+2) = 2*(A005828(n-1))^2 and a(3*n+3) = 4*(A005828(n-1))^2 - 2.
Recurrence equations:
For n >= 1, a(3*n+2) = 2*{a(3*n-1)^2 - 2*a(3*n-1) + 1} and
a(3*n+3) = 2*a(3*n+2) - 2.
Put P(n) = Product_{k=1..n} a(k). Then we have the infinite Egyptian fraction representation 3*sqrt(15) - 11 = Sum_{n>=1} 1/P(n) = 1/2 + 1/(2*5) + 1/(2*5*8) + 1/(2*5*8*2) + 1/(2*5*8*2*32) + ....

A220336 A modified Engel expansion for 4*sqrt(2) - 5.

Original entry on oeis.org

2, 4, 6, 2, 18, 34, 2, 578, 1154, 2, 665858, 1331714, 2, 886731088898, 1773462177794, 2, 1572584048032918633353218, 3145168096065837266706434, 2, 4946041176255201878775086487573351061418968498178, 9892082352510403757550172975146702122837936996354
Offset: 1

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Author

Peter Bala, Dec 12 2012

Keywords

Comments

For a brief description of the modified Engel expansion of a real number see A220335.
Let p >= 2 be an integer and set Q(p) = (p - 1)*sqrt(p^2 - 1) - (p^2 - p - 1), so Q(3) = 4*sqrt(2) - 5. Iterating the identity Q(p) = 1/2 + 1/(2*(p+1)) + 1/(2*(p+1)*(2*p)) + 1/(2*(p+1)*(2*p))*Q(2*p^2-1) leads to a representation for Q(p) as an infinite series of unit fractions. The sequence of denominators of these unit fractions can be used to find the modified Engel expansion of Q(p). For further details see the Bala link. The present sequence is the case p = 3. For other cases see A220335 (p = 2), A220337 (p = 4) and A220338 (p = 5).

Crossrefs

Cf. A001601, A028257, A220335 (p = 2), A220337 (p = 4), A220338 (p = 5).

Formula

Define the map h(x) := floor(1/x)*(x*ceiling(1/x) - 1). Let x = 4*sqrt(2) - 5. Then a(1) = ceiling(1/x) and for n >= 2, a(n) = floor(1/h^(n-2)(x))*ceiling(1/h^(n-1)(x)), where h^(n)(x) denotes the n-th iterate of the map h(x), with the convention h^(0)(x) = x.
a(3*n+2) = 1/2*{2 + (1+sqrt(2))^(2^(n+1)) + (1-sqrt(2))^(2^(n+1))},
a(3*n+3) = {(1 + sqrt(2))^(2^(n+1)) + (1 - sqrt(2))^(2^(n+1))}, both
for n >= 0.
For n >= 0, a(3*n+1) = 2. For n >= 1, a(3*n+2) = 2*A001601(n)^2 and a(3*n+3) = 4*A001601(n)^2 - 2.
Recurrence equations:
For n >= 1, a(3*n+2) = 2*{a(3*n-1)^2 - 2*a(3*n-1) + 1} and
a(3*n+3) = 2*a(3*n+2) - 2.
Put P(n) = Product_{k=1..n} a(k). Then we have the infinite Egyptian fraction representation 4*sqrt(2) - 5 = Sum_{n>=1} 1/P(n) = 1/2 + 1/(2*4) + 1/(2*4*6) + 1/(2*4*6*2) + 1/(2*4*6*2*18) + ....

A220338 A modified Engel expansion for 8*sqrt(6) - 19.

Original entry on oeis.org

2, 6, 10, 2, 50, 98, 2, 4802, 9602, 2, 46099202, 92198402, 2, 4250272665676802, 8500545331353602, 2, 36129635465198759610694779187202, 72259270930397519221389558374402, 2, 2610701117696295981568349760414651575095962187244375364404428802
Offset: 1

Views

Author

Peter Bala, Dec 12 2012

Keywords

Comments

For a brief description of the modified Engel expansion of a real number see A220335.
Let p >= 2 be an integer and set Q(p) = (p - 1)*sqrt(p^2 - 1) - (p^2 - p - 1), so Q(5) = 8*sqrt(6) - 19. Iterating the identity Q(p) = 1/2 + 1/(2*(p+1)) + 1/(2*(p+1)*(2*p)) + 1/(2*(p+1)*(2*p))*Q(2*p^2-1) leads to a representation for Q(p) as an infinite series of unit fractions. The sequence of denominators of these unit fractions can be used to find the modified Engel expansion of Q(p). For further details see the Bala link. The present sequence is the case p = 5. For other cases see A220335 (p = 2), A220336 (p = 3) and A220337 (p = 4).

Crossrefs

Cf. A084765, A220335 (p = 2), A220336 (p = 3), A220337 (p = 4).

Formula

Define the harmonic sawtooth map h(x) := floor(1/x)*(x*ceiling(1/x) - 1). Let x = 8*sqrt(6) - 19. Then a(1) = ceiling(1/x) and for n >= 2, a(n) = floor(1/h^(n-2)(x))*ceiling(1/h^(n-1)(x)), where h^(n)(x) denotes the n-th iterate of the map h(x), with the convention h^(0)(x) = x.
a(3*n+2) = 1/2*{2 + (5 + 2*sqrt(6))^(2^n) + (5 - 2*sqrt(6))^(2^n)} and
a(3*n+3) = {(5 + 2*sqrt(6))^(2^n) + (5 - 2*sqrt(6))^(2^n)} both for n >= 0.
For n >= 0, a(3*n+1) = 2. For n >= 1, a(3*n+2) = 2*A084765(n)^2 and a(3*n+3) = 4*A085765(n)^2 - 2.
Recurrence equations:
For n >= 1, a(3*n+2) = 2*{a(3*n-1)^2 - 2*a(3*n-1) + 1} and
a(3*n+3) = 2*a(3*n+2) - 2.
Put P(n) = Product_{k=1..n} a(k). Then we have the infinite Egyptian fraction representation 8*sqrt(6) - 19 = Sum_{n>=1} 1/P(n) = 1/2 + 1/(2*6) + 1/(2*6*10) + 1/(2*6*10*2) + 1/(2*6*10*2*50) + ....

A028257 Engel expansion of sqrt(3).

Original entry on oeis.org

1, 2, 3, 3, 6, 17, 23, 25, 27, 73, 84, 201, 750, 24981, 46882, 119318, 121154, 242807, 276226, 3009377, 3383197, 37871208, 45930966, 261728403, 281868388, 3021299588, 3230725884, 13646315477, 30951797814, 80602040381, 1016719946612, 49448385811777
Offset: 1

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Author

Naoki Sato (naoki(AT)math.toronto.edu)

Keywords

Crossrefs

Cf. A006784 (for definition of Engel expansion), A220335.

Programs

  • Mathematica
    EngelExp[A_,n_]:=Join[Array[1&,Floor[A]],First@Transpose@NestList[{Ceiling[1/Expand[ #[[1]]#[[2]]-1]],Expand[ #[[1]]#[[2]]-1]}&,{Ceiling[1/(A-Floor[A])],A-Floor[A]},n-1]]; EngelExp[N[3^(1/2),7! ],50] (* Vladimir Joseph Stephan Orlovsky, Jun 08 2009 *)

Formula

For a number x (here sqrt(3)), define a(1) <= a(2) <= a(3) <= ... so that x = 1/a(1) + 1/(a(1)*a(2)) + 1/(a(1)*a(2)*a(3)) + ... by x(1) = x, a(n) = ceiling(1/x(n)), x(n+1) = x(n)*a(n) - 1.

Extensions

Better name and more terms from Simon Plouffe
More terms from Sean A. Irvine, Dec 16 2019

A220393 A modified Engel expansion of Pi.

Original entry on oeis.org

1, 1, 1, 8, 14, 2, 2, 3, 4, 5, 96, 115, 8, 2, 2, 2, 81, 160, 2, 6, 355, 140, 4, 12, 6, 2, 2, 3, 4, 3, 46, 66, 4, 2, 9, 16, 3, 4, 3, 4, 2, 2, 4, 9, 4, 2, 4, 33, 20, 2, 3, 4, 2, 2
Offset: 1

Views

Author

Peter Bala, Dec 13 2012

Keywords

Comments

The Engel expansion of a positive real number x is the unique nondecreasing sequence {e(1), e(2), e(3), ...} of positive integers such that x = 1/e(1) + 1/(e(1)*e(2)) + 1/(e(1)*e(2)*e(3)) + .... The terms in the Engel expansion of x are obtained from the iterates of the mapping g(x) := x*(1 + floor(1/x)) - 1 by means of the formula e(n) = 1 + floor(1/g^(n)(x)).
Let h(x) = floor(1/x)*g(x). This is called the harmonic sawtooth map by Crowley. Then the modified Engel expansion of a real number 0 < x <= 1 is a sequence {a(1), a(2), a(3), ...} of positive integers such that x = 1/a(1) + 1/(a(1)*a(2)) + 1/(a(1)*a(2)*a(3)) + ... whose terms are obtained from the iterates of the harmonic sawtooth map h(x) by the formulas a(1) = 1 + floor(1/x) and, for n >= 2, a(n) = floor(1/h^(n-2)(x))*{1 + floor(1/h^(n-1)(x))}. Here h^(n)(x) = h(h^(n-1)(x)) denotes the n-th iterate of the map h(x), with the convention that h^(0)(x) = x. For further details see the Bala link.
When the real number x > 1 with, say, floor(x) = m, the modified Engel expansion of x is found by first calculating the modified Engel expansion of x - m and then prepending a sequence of m 1's to this.

Crossrefs

Formula

Let x = Pi - 3. Then a(1) = a(2) = a(3) = 1, a(4) = ceiling(1/x) and, for n >= 1, a(n+4) = floor(1/h^(n-1)(x))*ceiling(1/h^(n)(x)).
Put P(n) = Product_{k = 1..n} a(k). Then we have the Egyptian fraction series expansion Pi = Sum_{n>=1} 1/P(n) = 1/1 + 1/1 + 1/1 + 1/8 + 1/(8*14) + 1/(8*14*2) + 1/(8*14*2*2) + .... For n >= 4, the error made in truncating this series to n terms is less than the n-th term.

A220398 A modified Engel expansion of the golden ratio (1/2)*(1 + sqrt(5)) (A001622).

Original entry on oeis.org

1, 2, 5, 8, 3, 4, 4, 6, 2, 162, 322, 2, 51842, 103682, 2, 5374978562, 10749957122, 2, 57780789062419261442, 115561578124838522882, 2, 6677239169351578707225356193679818792962, 13354478338703157414450712387359637585922, 2
Offset: 1

Views

Author

Peter Bala, Dec 13 2012

Keywords

Comments

See A220393 for a definition of the modified Engel expansion of a positive real number. For further details see the Bala link.

Crossrefs

Formula

Let h(x) = x*(floor(1/x) + (floor(1/x))^2) - floor(1/x). Let x = 1/2*(1 + sqrt(5)) - 1. Then a(1) = 1, a(2) = ceiling(1/x) and, for n >= 1, a(n+2) = floor(1/h^(n-1)(x))*(1 + floor(1/h^(n)(x))).
Recurrence equations: For n >= 3, a(3*n) = 2. For n >= 4 we have a(3*n+2) = 2*a(3*n+1) - 2 and a(3*n+1) = 2*(a(3*n-2) - 1)^2.
Put P(n) = Product_{k = 1..n} a(k). Then we have the Egyptian fraction series expansion sqrt(2) = Sum_{n>=1} 1/P(n) = 1 + 1/2 + 1/(2*5) + 1/(2*5*8) + 1/(2*5*8*3) + 1/(2*5*8*3*4) + .... For n >= 2, the error made in truncating this series to n terms is less than the n-th term.

A220394 A modified Engel expansion of exp(1).

Original entry on oeis.org

1, 1, 2, 3, 4, 5, 8, 2, 10, 99, 20, 2, 2, 2, 2, 2, 2, 3, 6, 4, 8, 14, 2, 2, 4, 6, 10, 252, 81, 30, 28, 31, 60, 4, 6, 3, 4, 2, 2, 2, 2, 19, 54, 8, 6, 22, 63, 4, 2, 4, 6, 2, 2, 5, 12, 4, 2, 2, 2, 2, 6, 15, 10, 348, 172, 2, 2, 4, 6, 4, 30, 207, 220
Offset: 1

Views

Author

Peter Bala, Dec 13 2012

Keywords

Comments

See A220393 for a description of the modified Engel expansion of a positive real number. For further details see the Bala link.
The Engel expansion for exp(1) is the sequence of positive integers A000027.

Crossrefs

Formula

Let h(x) = x*(floor(1/x) + (floor(1/x))^2) - floor(1/x) where x = exp(1) - 2, then a(1) = a(2) = 1, a(3) = ceiling(1/x) and, for n >= 1, a(n+3) = floor(1/h^(n-1)(x))*(1 + floor(1/h^(n)(x))).
Put P(n) = Product_{k = 1..n} a(k). Then we have the Egyptian fraction series expansion exp(1) = Sum_{n>=1} 1/P(n) = 1/1 + 1/1 + 1/2 + 1/(2*3) + 1/(2*3*4) + 1/(2*3*4*5) + 1/(2*3*4*5*8) + .... For n >= 3, the error made in truncating this series to n terms is less than the n-th term.

A220395 A modified Engel expansion of log(2).

Original entry on oeis.org

2, 3, 8, 6, 2, 4, 93, 60, 2, 2, 2, 2, 3, 12, 10, 2, 2, 14, 52, 6, 5, 8, 2, 2, 5, 8, 2, 2, 3, 4, 14, 273, 40, 2, 3, 4, 4, 12, 27, 16, 14, 26, 4, 6, 4, 6, 2, 3, 12, 10, 4, 6, 14, 65, 12, 8, 6, 2, 7, 90, 294, 40, 2, 2, 32, 155, 8, 7, 12, 2, 2, 2, 2, 4, 6, 3, 10
Offset: 1

Views

Author

Peter Bala, Dec 13 2012

Keywords

Comments

See A220393 for the definition of the modified Engel expansion of a positive real number. For further details see the Bala link.

Crossrefs

Formula

Let h(x) = x*(floor(1/x) + (floor(1/x))^2) - floor(1/x). Let x = log(2). Then a(1) = 1 + floor(1/x) and, for n >= 1, a(n+1) = floor(1/h^(n-1)(x))*(1 + floor(1/h^(n)(x))).
Put P(n) = Product_{k = 1..n} a(k). Then we have the Egyptian fraction series expansion log(2) = Sum_{n>=1} 1/P(n) = 1/2 + 1/(2*3) + 1/(2*3*8) + 1/(2*3*8*6) + 1/(2*3*8*6*2) + .... The error made in truncating this series to n terms is less than the n-th term.

A220396 A modified Engel expansion of the Euler-Mascheroni constant gamma.

Original entry on oeis.org

2, 7, 18, 4, 2, 2, 3, 1466, 1464, 9, 24, 4, 2, 9, 104, 60, 8, 2, 3, 6, 4, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 32, 30, 2, 13, 36, 6, 4, 3, 6, 6, 4, 4, 6, 2, 4, 6, 2, 4, 6, 9, 24, 4, 5, 8, 2, 2, 2, 2, 2, 3, 20
Offset: 1

Views

Author

Peter Bala, Dec 13 2012

Keywords

Comments

See A220393 for the definition of the modified Engel expansion of a positive real number. For further details see the Bala link.

Crossrefs

Formula

Let h(x) = x*(floor(1/x) + (floor(1/x))^2) - floor(1/x). Let x = gamma (see A001620). Then a(1) = 1 + floor(1/x) and, for n >= 1, a(n+1) = floor(1/h^(n-1)(x))*(1 + floor(1/h^(n)(x))).
Put P(n) = Product_{k = 1..n} a(k). Then we have the Egyptian fraction series expansion sqrt(2) = Sum_{n>=1} 1/P(n) = 1/2 + 1/(2*7) + 1/(2*7*18) + 1/(2*7*18*4) + 1/(2*7*18*4*2) + .... The error made in truncating this series to n terms is less than the n-th term.
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