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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A093766 Decimal expansion of Pi/(2*sqrt(3)).

Original entry on oeis.org

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

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Author

Eric W. Weisstein, Apr 15 2004

Keywords

Comments

Density of densest packing of equal circles in two dimensions (achieved for example by the A2 lattice).
The number gives the areal coverage (90.68... percent) of the close hexagonal (densest) packing of circles in the plane. The hexagonal unit cell is a rhombus of side length 1 and height sqrt(3)/2; the area of the unit cell is sqrt(3)/2 and the four parts of circles add to an area of one circle of radius 1/2, which is Pi/4. - R. J. Mathar, Nov 22 2011
Ratio of surface area of a sphere to the regular octahedron whose edge equals the diameter of the sphere. - Omar E. Pol, Dec 09 2013

Examples

			0.906899682117108925297039128821077866142033124046370287784942...

References

  • J. H. Conway and N. J. A. Sloane, "Sphere Packings, Lattices and Groups", Springer, 3rd. ed., 1998. See p. xix.
  • Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 8.7, p. 506.
  • L. B. W. Jolley, Summation of Series, Dover (1961), Eq. (84) on page 16.
  • Joel L. Schiff, The Laplace Transform: Theory and Applications, Springer-Verlag New York, Inc. (1999). See p. 149.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987, p. 30.

Links

Crossrefs

Related constants: A020769, A020789, A093825, A222066, A222067, A222068, A222069, A222070, A222071, A222072, A260646, (1/2)*A093602, A346585, A346584, A346583.
Cf. A072691, A381671.

Programs

Formula

Equals (5/6)*(7/6)*(11/12)*(13/12)*(17/18)*(19/18)*(23/24)*(29/30)*(31/30)*..., where the numerators are primes > 3 and the denominators are the nearest multiples of 6.
Equals Sum_{n>=1} 1/A134667(n). [Jolley]
Equals Sum_{n>=0} (-1)^n/A124647(n). [Jolley eq. 273]
Equals A000796 / A010469. - Omar E. Pol, Dec 09 2013
Continued fraction expansion: 1 - 2/(18 + 12*3^2/(24 + 12*5^2/(32 + ... + 12*(2*n - 1)^2/((8*n + 8) + ... )))). See A254381 for a sketch proof. - Peter Bala, Feb 04 2015
From Peter Bala, Feb 16 2015: (Start)
Equals 4*Sum_{n >= 0} 1/((6*n + 1)*(6*n + 5)).
Continued fraction: 1/(1 + 1^2/(4 + 5^2/(2 + 7^2/(4 + 11^2/(2 + ... + (6*n + 1)^2/(4 + (6*n + 5)^2/(2 + ... ))))))). (End)
The inverse is (2*sqrt(3))/Pi = Product_{n >= 1} 1 + (1 - 1/(4*n))/(4*n*(9*n^2 - 9*n + 2)) = (35/32) * (1287/1280) * (8075/8064) * (5635/5632) * (72819/72800) * ... = 1.102657790843585... - Dimitris Valianatos, Aug 31 2019
From Amiram Eldar, Aug 15 2020: (Start)
Equals Integral_{x=0..oo} 1/(x^2 + 3) dx.
Equals Integral_{x=0..oo} 1/(3*x^2 + 1) dx. (End)
Equals 1 + Sum_{k>=1} ( 1/(6*k+1) - 1/(6*k-1) ). - Sean A. Irvine, Jul 24 2021
For positive integer k, Pi/(2*sqrt(3)) = Sum_{n >= 0} (6*k + 4)/((6*n + 1)*(6*n + 6*k + 5)) - Sum_{n = 0..k-1} 1/(6*n + 5). - Peter Bala, Jul 10 2024
From Stefano Spezia, Jun 05 2025: (Start)
Equals Sum_{k>=0} (-1)^k/((k + 1)*(3*k + 1)).
Equals Integral_{x=0..oo} 1/(x^4 + x^2 + 1) dx.
Equals Integral_{x=0..oo} x^2/(x^4 + x^2 + 1) dx. (End)
Equals sqrt(A072691) = 3*A381671. - Hugo Pfoertner, Jun 05 2025

Extensions

Entry revised by N. J. A. Sloane, Feb 10 2013

A254619 a(n) = 4^n*(2*n + 1)!/n!.

Original entry on oeis.org

1, 24, 960, 53760, 3870720, 340623360, 35424829440, 4250979532800, 578133216460800, 87876248902041600, 14763209815542988800, 2716430606059909939200, 543286121211981987840000, 117349802181788109373440000
Offset: 0

Views

Author

Peter Bala, Feb 03 2015

Keywords

Crossrefs

Cf. A002391, A034910, A073000, A254381, A254620.

Programs

  • Maple
    seq(4^n*(2*n + 1)!/n!, n = 0..13);
  • Mathematica
    Table[4^n (2n+1)!/n!,{n,0,20}] (* Harvey P. Dale, Oct 02 2021 *)

Formula

E.g.f.: 1/(1 - 16*x)^(3/2) = 1 + 24*x + 960*x^2/2! + 53760*x^3/3! + ....
Recurrence equation: a(n) = 8*(2*n + 1)*a(n-1) with a(0) = 1.
2nd order recurrence equation: a(n) = (20*n + 6)*a(n-1) - 16*(2*n - 1)^2*a(n-2) with a(0) = 1, a(1) = 24.
Define a sequence b(n) := a(n)*sum {k = 0..n} 1/((2*k + 1)*4^k) beginning [1, 26, 1052, 59032, 4251984, 374204832, 38917967808, ...]. It is not difficult to check that b(n) also satisfies the previous 2nd order recurrence equation (and so is an integer sequence). From this observation we can obtain the continued fraction expansion
log(3) = Sum {k >= 0} 1/((2*k + 1)*4^k) = 1 + 2/(24 - 16*3^2/(46 - 16*5^2/(66 - ... - 16*(2*n - 1)^2/((20*n + 6) - ... )))).
Alternative 2nd order recurrence equation: a(n) = (12*n + 10)*a(n-1) + 16*(2*n - 1)^2*a(n-2) with a(0) = 1, a(1) = 24.
Define now a sequence c(n) := a(n)*sum {k = 0..n} (-1)^k/((2*k + 1)*4^k) beginning [1, 22, 892, 49832, 3589584, 315853152, 32849393088, ...], which, along with a(n), satisfies the alternative 2nd order recurrence equation. From this observation we find the continued fraction expansion 2*arctan(1/2) = Sum {k >= 0} (-1)^k/((2*k + 1)*4^k) = 1 - 2/(24 + 16*3^2/(34 + 16*5^2/(46 + ... + 16*(2*n - 1)^2/((12*n + 10) + ... )))). Cf. A254381 and A254620.

A254620 a(n) = 9^n*(2*n + 1)!/n!.

Original entry on oeis.org

1, 54, 4860, 612360, 99202320, 19642059360, 4596241890240, 1240985310364800, 379741504971628800, 129871594700297049600, 49091462796712284748800, 20323865597838885886003200, 9145739519027498648701440000, 4444829406247364343268899840000
Offset: 0

Views

Author

Peter Bala, Feb 03 2015

Keywords

Crossrefs

Cf. A002162, A002391, A034910, A073000, A105531, A254381, A254619.

Programs

  • Maple
    seq(9^n*(2*n + 1)!/n!, n = 0..14);
  • Mathematica
    Table[9^n (2n+1)!/n!,{n,0,20}] (* Harvey P. Dale, Aug 13 2019 *)

Formula

E.g.f.: 1/(1 - 36*x)^(3/2) = 1 + 54*x + 4860*x^2/2! + 612360*x^3/3! + ....
Recurrence equation: a(n) = 18*(2*n + 1)*a(n-1) with a(0) = 1.
2nd order recurrence equation: a(n) = (40*n + 16)*a(n-1) - 36*(2*n - 1)^2*a(n-2) with a(0) = 1, a(1) = 54.
Define a sequence b(n) := a(n)*sum {k = 0..n} 1/((2*k + 1)*9^k) beginning [1, 56, 5052, 636672, 103142544, 20422253952, 4778808090048, ...]. It is not difficult to check that b(n) also satisfies the previous 2nd order recurrence equation (and so is an integer sequence). Using this observation we obtain the continued fraction expansion log(2) = 2/3*Sum {k >= 0} 1/((2*k + 1)*9^k) = 2/3*(1 + 2/(54 - 36*3^2/(96 - 36*5^2/(136 - ... - 36*(2*n - 1)^2/((40*n + 16) - ... ))))).
Alternative 2nd order recurrence equation: a(n) = (32*n + 20)*a(n-1) + 36*(2*n - 1)^2*a(n-2) with a(0) = 1, a(1) = 54.
Define now a sequence c(n) := a(n)*sum {k = 0..n} (-1)^k/((2*k + 1)*9^k) beginning [1, 52, 4692, 591072, 95755344, 18959527872, 4436530187328, ...], which, along with a(n), satisfies the alternative 2nd order recurrence equation. From this observation we find the continued fraction expansion arctan(1/3) = 1/3*Sum {k >= 0} (-1)^k/((2*k + 1)*9^k) = 1/3*(1 - 2/(54 + 36*3^2/(84 + 36*5^2/(116 + ... + 36*(2*n - 1)^2/((32*n + 20) + ... ))))). Cf. A254381 and A254619.
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