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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A384625 Decimal expansion of the surface area of a pentagonal orthobicupola with unit edge.

Original entry on oeis.org

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

Views

Author

Paolo Xausa, Jun 05 2025

Keywords

Comments

The pentagonal orthobicupola is Johnson solid J_30.
Also the surface area of a pentagonal gyrobicupola (Johnson solid J_31) with unit edge.

Examples

			17.771081820100127079336639808541900116171761474546...
		

Crossrefs

Cf. A384624 (volume).

Programs

  • Mathematica
    First[RealDigits[10 + Sqrt[5*(10 + Sqrt[5] + Sqrt[75 + 30*Sqrt[5]])/2], 10, 100]] (* or *)
    First[RealDigits[PolyhedronData["J30", "SurfaceArea"], 10, 100]]

Formula

Equals 10 + sqrt(5*(10 + sqrt(5) + sqrt(75 + 30*sqrt(5)))/2) = 10 + sqrt(5*(10 + A002163 + sqrt(75 + 30*A002163))/2).
Equals the largest root of x^8 - 80*x^7 + 2700*x^6 - 50000*x^5 + 552750*x^4 - 3710000*x^3 + 14628125*x^2 - 30562500*x + 25328125.

A108404 Expansion of (1-4x)/(1-8x+11x^2).

Original entry on oeis.org

1, 4, 21, 124, 761, 4724, 29421, 183404, 1143601, 7131364, 44471301, 277325404, 1729418921, 10784771924, 67254567261, 419404046924, 2615432135521, 16310012568004, 101710347053301, 634272638178364, 3955367287840601
Offset: 0

Views

Author

Philippe Deléham, Jul 04 2005

Keywords

Comments

Binomial transform of A098648. Second binomial transform of A001077. Third binomial transform of A084057. 4th binomial transform of (1, 0, 5, 0, 25, 0, 125, 0, 625, 0, 3125, ...).

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[(1-4x)/(1-8x+11x^2),{x,0,30}],x] (* or *) LinearRecurrence[{8,-11},{1,4},30] (* Harvey P. Dale, Jan 03 2012 *)

Formula

E.g.f.: exp(4x)cosh(sqrt(5)x).
a(n) = 8a(n-1) - 11a(n-2), a(0) = 1, a(1) = 4.
a(n) = ((4+sqrt(5))^n + (4-sqrt(5))^n)/2.
a(n+1)/a(n) converges to 4 + sqrt(5) = 6.2360679774997896964... = 4+A002163.
a(n) = A091870(n+1)-4*A091870(n). - R. J. Mathar, Nov 10 2013

A378975 Decimal expansion of the inradius of a triakis icosahedron with unit shorter edge length.

Original entry on oeis.org

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

Views

Author

Paolo Xausa, Dec 14 2024

Keywords

Comments

The triakis icosahedron is the dual polyhedron of the truncated dodecahedron.

Examples

			1.37451744701047164727510000639742367448102733307...
		

Crossrefs

Cf. A378973 (surface area), A378974 (volume), A378976 (midradius), A378977 (dihedral angle).
Cf. A002163.

Programs

  • Mathematica
    First[RealDigits[Sqrt[(477 + 199*Sqrt[5])/488], 10, 100]] (* or *)
    First[RealDigits[PolyhedronData["TriakisIcosahedron", "Inradius"], 10, 100]]

Formula

Equals sqrt((477 + 199*sqrt(5))/488) = sqrt((477 + 199*A002163)/488).

A384141 Decimal expansion of the surface area of an elongated pentagonal bipyramid with unit edge.

Original entry on oeis.org

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

Views

Author

Paolo Xausa, May 20 2025

Keywords

Comments

The elongated pentagonal bipyramid is Johnson solid J_16.

Examples

			9.3301270189221932338186158537646809173570131345...
		

Crossrefs

Cf. A384140 (volume).
Cf. A002163.
Essentially the same as A120011.

Programs

  • Mathematica
    First[RealDigits[5*(2 + Sqrt[3])/2, 10, 100]] (* or *)
    First[RealDigits[PolyhedronData["J16", "SurfaceArea"], 10, 100]]

Formula

Equals 5*(2 + sqrt(3))/2 = 5*(2 + A002194)/2.
Equals the largest root of 4*x^2 - 40*x + 25.
Equals 10*A019884^2. - R. J. Mathar, Sep 05 2025

A384213 Decimal expansion of the volume of an elongated pentagonal rotunda with unit edge.

Original entry on oeis.org

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

Views

Author

Paolo Xausa, May 23 2025

Keywords

Comments

The elongated pentagonal rotunda is Johnson solid J_21.

Examples

			14.611971811062835576338722470794915893557631368294...
		

Crossrefs

Cf. A179637 (surface area - 10).

Programs

  • Mathematica
    First[RealDigits[(45 + 17*Sqrt[5] + 30*Sqrt[5 + Sqrt[20]])/12, 10, 100]] (* or *)
    First[RealDigits[PolyhedronData["J21", "Volume"], 10, 100]]

Formula

Equals (45 + 17*sqrt(5) + 30*sqrt(5 + 2*sqrt(5)))/12 = (45 + 17*A002163 + 30*sqrt(5 + A010476))/12.
Equals the largest real root of 1296*x^4 - 19440*x^3 + 2340*x^2 + 70200*x + 43525.

A145433 Decimal expansion of Sum_{n>=1} (-1)^(n-1)*n/binomial(2n,n).

Original entry on oeis.org

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

Views

Author

R. J. Mathar, Feb 08 2009

Keywords

Examples

			0.274432715277120323...
		

References

  • Alexander Apelblat, Tables of Integrals and Series, Harri Deutsch, (1996), 4.1.39

Crossrefs

Programs

  • Maple
    6/25+4/125*5^(1/2)*ln(1/2+1/2*5^(1/2)) ;
  • Mathematica
    RealDigits[6/25 + 4*Sqrt[5]*Log[GoldenRatio]/125, 10, 93] // First (* Jean-François Alcover, Oct 27 2014 *)
    RealDigits[Hypergeometric2F1[2, 2, 3/2, -1/4]/2, 10, 93] // First (* Vaclav Kotesovec, Oct 27 2014 *)

Formula

Equals 2*(15+2*A002163*A002390)/125.

A176910 Decimal expansion of sqrt(145).

Original entry on oeis.org

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

Views

Author

Klaus Brockhaus, Apr 28 2010

Keywords

Comments

Continued fraction expansion of sqrt(145) is 12 followed by A010863.

Examples

			sqrt(145) = 12.04159457879229548012...
		

Crossrefs

Cf. A002163 (decimal expansion of sqrt(5)), Cf. A010484 (decimal expansion of sqrt(29)), A176907 (decimal expansion of (9+sqrt(145))/16), A176908 (decimal expansion of (7+sqrt(145))/16), A010863 (all 24's sequence).

Programs

  • Mathematica
    RealDigits[Sqrt[145],10,120][[1]] (* Harvey P. Dale, Oct 15 2023 *)

A305310 Numbers k(n) used for Cassels's Markoff forms MF(n) corresponding to the conjectured unique Markoff triples MT(n) with maximal entry m(n) = A002559(n), for n >= 1.

Original entry on oeis.org

0, 1, 2, 5, 12, 13, 34, 70, 75, 89, 179, 233, 408, 507, 610, 1120, 1597, 2378, 2673, 2923, 3468, 4181, 6089, 10946, 13860, 15571, 16725, 19760, 23763, 28657, 39916, 51709, 80782, 75025, 113922, 162867, 206855, 196418, 249755, 353702
Offset: 1

Views

Author

Wolfdieter Lang, Jun 26 2018

Keywords

Comments

For these Markoff forms see Cassels, p. 31. A link to the two original Markoff references is given in A305308.
MF(n) = f_{m(n)}(x, y) = m(n)*F_{m(n)}(x, y) = m(n)*x^2 + (3*m(n) - 2*k(n))*x*y + (l(n) - 3*k(n))*y^2, with the Markoff number m = m(n) = A002559(n) and l(n) = (k(n)^2 + 1)/m(n), for n >= 1.
Every m(n) is proved to appear as largest member of a Markoff triple MT(n) = (m_1(n), m_2(n), m(n)), with positive integers m_1(n) < m_2(n) < m(n) for n >= 3 (MT(1) = (1, 1, 1) and MT(2) = (1, 1, 2)) satisfying the Markoff equation m_1(n)^2 + m_2(n)^2 + m(n)^2 = 3*m_1(n)*m_2(n)*m(n). The famous Markoff uniqueness conjecture is that m(n) as largest member determines exactly one ordered triple MT(n). See, e.g., the Aigner reference, pp. 38-39, and Corollary 3.5, p. 48. [In numerating the sequence with n related to A002559(n) this conjecture is assumed to be true. - Wolfdieter Lang, Jul 29 2018]
The nonnegative integers k(n) are defined for the Markoff forms given by Cassels by k(n) = min{k1(n), k2(n)}, where m_1(n)*k1(n) - m_2(n) == 0 (mod m(n)), with 0 <= k1(n) < m(n), and m_2(n)*k2(n) - m_1(n) == 0 (mod m(n)), with 0 <= k2(n) < m(n). The k1 and k2 sequences are k1 = [0, 1, 2, 5, 17, 13, 34, 99, 119, 89, 179, 233, 577, 818, 610, 1777, 1597, 3363, 2673, 2923, 5609, 4181, 6089, 10946, 19601, 22095, 26536, 31881, 38447, 28657, 39916, 51709, 114243, 75025, 113922, 263522, 206855, 196418, 396263, 572063, ...], and k2 = [0, 1, 3, 8, 12, 21, 55, 70, 75, 144, 254, 377, 408, 507, 987, 1120, 2584, 2378, 3793, 4638, 3468, 6765, 8612, 17711, 13860, 15571, 16725, 19760, 23763, 46368, 56641, 83428, 80782, 121393, 180763, 162867, 292538, 317811, 249755, 353702, ...].
The discriminant of the form MF(n) = f_{m(n)}(x, y) is D(n) = 9*m(n)^2 - 4. D(n) = A305312(n), for n >= 1. Because D(n) > 0 (not a square) this is an indefinite binary quadratic form, for n >= 1. See Cassels Fig. 2 on p. 32 for the Markoff tree with these forms.
The quadratic irrational xi, determined by the solution with positive square root of f_{m(n)}(x, 1) = 0, is xi(n) = ((2*k - 3*m) + sqrt(D))/(2*m) (the argument n has been dropped). The regular continued fraction is eventually periodic, but not purely periodic. One can find equivalent Markoff forms determining purely periodic quadratic irrationals. The corresponding k sequence is given in A305311.
For the approximation of xi(n) with infinitely many rationals (in lowest terms) Perron's unimodular invariant M(xi) enters. For quadratic irrationals M(xi) < 3, and the values coincide with the discrete Lagrange spectrum < 3: M(xi(n)) = Lagrange(n) = sqrt{D(n)}/m(n), n >= 1. For n=1..4 see A002163, A010466, A200991 and A305308.

Examples

			n = 5: a(5) = k(5) = 12 because m(5) = A002559(5) = 29 with the triple MT(5) = (2, 5, 29). Whence 2*k1(5) - 5 == 0 (mod 29) for k1(5) = 17 < 29, and 5*k2(5) - 2 == 0 (mod 29) leads to k2(5) = 12. The smaller value is k2(5) = k(5) = 12. This leads to the form coefficients MF(5) = [29, 63, -31].
The forms MF(n) = [m(n), 3*m(n) - k(n), l(n) - 3*k(n)] with l(n) := (k(n)^2 + 1)/m(n) begin: [1, 3, 1], [2, 4, -2], [5, 11, -5], [13, 29, -13], [29, 63, -31], [34, 76, -34], [89, 199, -89], [169, 367, -181], [194, 432, -196], [233, 521, -233], [433, 941, -463], [610, 1364, -610], [985, 2139, -1055], [1325, 2961, -1327], [1597, 3571, -1597], [2897, 6451, -2927], [4181, 9349, -4181], [5741, 12467, -6149], [6466, 14052, -6914], [7561, 16837, -7639] ... .
The quadratic irrationals xi(n) = ((2*k(n) - 3*m(n)) + sqrt(D(n)))/(2*m(n)) begin: (-3 + sqrt(5))/2, -1 + sqrt(2), (-11 + sqrt(221))/10, (-29 + sqrt(1517))/26, (-63 + sqrt(7565))/58, (-19 + 5*sqrt(26))/17, (-199 + sqrt(71285))/178, (-367 + sqrt(257045))/338, (-108 + sqrt(21170))/97, (-521 + sqrt(488597))/466, (-941 + sqrt(1687397))/866, (-341 + sqrt(209306))/305, (-2139 + sqrt(8732021))/1970, (-2961 + sqrt(15800621))/2650, (-3571 + sqrt(22953677))/3194, (-6451 + sqrt(75533477))/5794, (-9349 + sqrt(157326845))/8362, (-12467 + 5*sqrt(11865269))/11482, (-3513 + 5*sqrt(940706))/3233, (-16837 + sqrt(514518485))/15122, ... .
The invariant M(xi(n)) = Lagrange(n) numbers begin with n >=1: sqrt(5), 2*sqrt(2), (1/5)*sqrt(221), (1/13)*sqrt(1517), (1/29)*sqrt(7565), (10/17)*sqrt(26), (1/89)*sqrt(71285), (1/169)*sqrt(257045), (2/97)*sqrt(21170), (1/233)*sqrt(488597), (1/433)*sqrt(1687397), (2/305)*sqrt(209306), (1/985)*sqrt(8732021), (1/1325)*sqrt(15800621), (1/1597)*sqrt(22953677), (1/2897)*sqrt(75533477), (1/4181)*sqrt(157326845), (5/5741)*sqrt(11865269), (10/3233)*sqrt(940706), (1/7561)*sqrt(514518485), ... .
		

References

  • Martin Aigner, Markov's Theorem and 100 Years of the Uniqueness Conjecture, Springer, 2013.
  • J. W. S. Cassels, An Introduction to Diophantine Approximation, Cambridge University Press, 1957, Chapter II, The Markoff Chain, pp. 18-44.
  • Julian Havil, The Irrationals, Princeton University Press, Princeton and Oxford, 2012, pp. 172-180 and 222-224.
  • Oskar Perron, Über die Approximation irrationaler Zahlen durch rationale, Sitzungsber. Heidelberger Akademie der Wiss., 1921, 4. Abhandlung, pp. 1-17 , and part II., 8. Abhandlung, pp.1-12, Carl Winters Universitätsbuchhandlung.

Crossrefs

Formula

a(n) = k(n) has been defined in terms of the (conjectured unique) ordered Markoff triple MT(n) = (m_1(n), m_2(n), m(n)) with m(n) = A002559(n) in the comment above as k(n) = min{k1(n), k2(n)}, where m_1(n)*k1(n) - m_2(n) == 0 (mod m(n)), with 0 <= k1(n) < m(n), and m_2(n)*k2(n) - m_1(n) == 0 (mod m(n)), with 0 <= k2(n) < m(n).

A377797 Decimal expansion of the volume of a truncated icosidodecahedron (great rhombicosidodecahedron) with unit edge length.

Original entry on oeis.org

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

Views

Author

Paolo Xausa, Nov 08 2024

Keywords

Examples

			206.80339887498948482045868343656381177203091798...
		

Crossrefs

Cf. A377796 (surface area), A377798 (circumradius), A377799 (midradius).

Programs

  • Mathematica
    First[RealDigits[95 + 50*Sqrt[5], 10, 100]] (* or *)
    First[RealDigits[PolyhedronData["TruncatedIcosidodecahedron", "Volume"], 10, 100]]

Formula

Equals 95 + 50*sqrt(5) = 95 + 50*A002163.
Equals 45 + 100*A001622.

A379134 Decimal expansion of the inradius of a pentakis dodecahedron with unit shorter edge length.

Original entry on oeis.org

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

Views

Author

Paolo Xausa, Dec 17 2024

Keywords

Comments

The pentakis dodecahedron is the dual polyhedron of the truncated icosahedron.

Examples

			1.4453319256522148283158512491020811977238711778430...
		

Crossrefs

Cf. A379132 (surface area), A379133 (volume), A379135 (midradius), A379136 (dihedral angle).
Cf. A002163.

Programs

  • Mathematica
    First[RealDigits[Sqrt[477/436 + 97*Sqrt[5]/218], 10, 100]] (* or *)
    First[RealDigits[PolyhedronData["PentakisDodecahedron", "Inradius"], 10, 100]]
  • PARI
    sqrt(477/436 + 97*sqrt(5)/218) \\ Charles R Greathouse IV, Feb 05 2025

Formula

Equals sqrt(477/436 + 97*sqrt(5)/218) = sqrt(477/436 + 97*A002163/218).
Equals the largest root of 1744*x^4 - 3816*x^2 + 361.
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