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 1-3 of 3 results.

A006268 A continued cotangent.

Original entry on oeis.org

3, 36, 46764, 102266868132036, 1069559300034650646049671039050649693658764
Offset: 0

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Keywords

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Continued cotangents: A006267, A006266, A006269, A145180, A145181, A145182, A145183, A145184, A145185, A145186, A145187, A145188, A145189 (k = 1 to 15 with k=4 being A006267(n+1)).

Programs

  • Mathematica
    a = {}; k = 3; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a (* Artur Jasinski, Oct 03 2008 *)
    Table[Round[N[(3/2 + Sqrt[13]/2)^(3^(n - 1)), 1000]], {n, 1, 8}] (* Artur Jasinski, Oct 03 2008 *)
  • PARI
    a(n) = if (n==0, 3, a(n-1)^3 + 3*a(n-1)); \\ Michel Marcus, Aug 28 2020

Formula

From Artur Jasinski, Oct 03 2008: (Start)
a(n+1) = a(n)^3 + 3*a(n) and a(0)=3.
a(n) = round((3/2 + sqrt(13)/2)^(3^(n - 1))). (End)
From Peter Bala, Jan 19 2022: (Start)
a(n) = (3/2 + sqrt(13)/2)^(3^(n-1)) + (3/2 - sqrt(13)/2)^(3^(n-1))
a(n) divides a(n+1) and b(n) = a(n+1)/a(n) satisfies the recurrence b(n+1) = b(n)^3 - 3*b(n-1)^2 + 3. For remarks about this recurrence see A002813.
1 + a(n)^2 = A006273(n+1). (End)

A006271 Numerators of a continued fraction for 1 + sqrt(2).

Original entry on oeis.org

2, 5, 197, 7761797, 467613464999866416197, 102249460387306384473056172738577521087843948916391508591105797
Offset: 0

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Comments

With b(n) = floor((1+sqrt(2))^n) (cf. A080039) the terms appear to be b(2*3^n). - Joerg Arndt, Apr 29 2013
Note that 1 + sqrt(2) = (c + sqrt(c^2+4))/2 and has regular continued fraction [c, c, ...] with c = 2. With b(n) = A006266(n), it can be expanded into an irregular continued fraction f(1) = b(1) and f(n) = (b[n-1]^2+1)/(b[n]-b[n-1]), and numerator(f(n)) = a(n) (cf. Shallit). - Michel Marcus, Apr 29 2013

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

For denominators see A006272. Cf. A002814, A006266, A006273, A006275, A006276.

Programs

  • Maple
    a := proc (n) option remember; if n = 1 then 5 else a(n-1)^3 + 3*a(n-1)^2 - 3 end if; end proc:
    seq(a(n), n = 1 .. 5); # Peter Bala, Jan 19 2022

Formula

From Peter Bala, Jan 18 2022: (Start)
a(n) = (3 + 2*sqrt(2))^3^(n-1) + (3 - 2*sqrt(2))^3^(n-1) - 1 for n >= 1.
a(n) = A006266(n)^2 + 1 for n >= 1.
a(1) = 5 and a(n) = a(n-1)^3 + 3*a(n-1)^2 - 3 for n >= 2.
a(1) = 5 and a(n) = 8*(Product_{k = 1..n-1} a(k))^2 - 3 for n >= 2.
2 - Product_{n = 1..N} (1 + 2/a(n))^2 = 8/(a(N+1) + 3). Therefore
sqrt(2) = (1 + 2/5) * (1 + 2/197) * (1 + 2/7761797) * (1 + 2/ 467613464999866416197) * ... - see Bauer.
The convergence is cubic - see Fine. The first six factors of the product give sqrt(2) correct to more than 500 decimal places. (End)

Extensions

Previous values for a(3) and a(4) were 776 and 1797. They have been merged into 7761797 to reflect the 2nd continued fraction on page 6 of Shallit paper by Michel Marcus, Apr 29 2013

A006274 First differences of A006268.

Original entry on oeis.org

33, 46728, 102266868085272, 1069559300034650646049671038948382825526728
Offset: 0

Views

Author

Keywords

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Extensions

Definition corrected by N. J. A. Sloane, May 23 2023 using a formula suggested by R. J. Mathar, Apr 26 2007.
Showing 1-3 of 3 results.