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 1-2 of 2 results.

A246645 Expansion of 1/(1 - 22*x + 81*x^2), used in A246643.

Original entry on oeis.org

1, 22, 403, 7084, 123205, 2136706, 37027927, 641541208, 11114644489, 192557340910, 3335975296411, 57794311907332, 1001260862952013, 17346399720450394, 300518663950795615, 5206352229561021616, 90197737270328030737, 1562635689352773925318, 27071968446864455867299
Offset: 0

Views

Author

Wolfdieter Lang, Sep 30 2014

Keywords

Comments

This sequence is used in the formula for the curvature in a touching circle problem considered in A247512 and A246643.

Links

Crossrefs

Cf. A247512, A246643, A246646.

Programs

  • Magma
    I:=[1, 22]; [n le 2 select I[n] else 22*Self(n-1) - 81*Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 20 2017
  • Mathematica
    CoefficientList[Series[1/(1 - 22*x + 81*x^2), {x, 0, 50}], x] (* or *) LinearRecurrence[{22,-81}, {1,22}, 50] (* G. C. Greubel, Dec 20 2017 *)
  • PARI
    Vec(1/(1 - 22*x + 81*x^2) + O(x^40)) \\ Michel Marcus, Sep 30 2014

Formula

O.g.f.: 1/(1 - 22*x + 81*x^2).
a(n) = 9^n*S(n, 22/9) with Chebyshev's S-polynomials (see A049310).
a(n) = 22*a(n-1) - 81*a(n-2), n >= 1, a(-1) = 0 and a(0) = 1.
a(n) = 9^n*(ap^(n+1) - am^(n+1))/(ap - am), n >= 1, with ap := (11 + 2*sqrt(10))/9 and am = 1/ap = (11 - 2*sqrt(10))/9 (Binet - de Moivre formula). a(0) = 1 (via L'Hopital's rule).
a(n) = 9^(n+1)*sinh(2*(n + 1)*arccsch(3))/(2*sqrt(10)). - Federico Provvedi, Feb 02 2021

A247512 The curvature (rounded down) of touching circles inscribed in a special way in the smaller segment of circle of radius 10/9 divided by a chord of length 4/3.

Original entry on oeis.org

9, 10, 13, 20, 35, 64, 119, 224, 428, 821, 1576, 3030, 5828, 11215, 21584, 41545, 79968, 153931, 296306, 570371, 1097933, 2113463, 4068308, 7831289, 15074840, 29018319, 55858826, 107525476, 206981225, 398428629, 766955420, 1476351286, 2841903278, 5470523390
Offset: 0

Views

Author

Kival Ngaokrajang, Sep 18 2014

Keywords

Comments

Refer to comment of A240926. This is the companion of A247335. After the first two terms, the curvatures seem to be non-integer.
The actual rational curvatures can be computed. See part II of the W. Lang link for the proofs of the statements given in the formula section.

Examples

			The first curvatures r(n) are 9, 10, 121/9, 1690/81, 25921/729, 420250/6561, 7027801/59049, 119508490/531441,... - _Wolfdieter Lang_, Sep 30 2014

Links

Crossrefs

Cf. A240926, A247335.
Cf. A246643, A049310, A246645. - Wolfdieter Lang, Sep 30 2014

Programs

  • Mathematica
    r[0] := 9; r[n_] := r[n] = (11*r[n - 1] - 9 + 20*Sqrt[(r[n - 1] - 9)*r[n - 1]/10])/9; Table[Floor[r[n]], {n, 0, 30}] (* G. C. Greubel, Dec 20 2017 *)
  • PARI
    {
r=0.1;print1(floor(9/(10*r)),", ");r1=r;
for (n=1,50,
if (n<=1,ab=2-r,ab=sqrt(ac^2+r^2));
ac=sqrt(ab^2-r^2);
if (n<=1,z=0,z=(Pi/2)-atan(ac/r)+asin((r1-r)/(r1+r));r1=r);
b=acos(r/ab)-z;
r=r*(1-cos(b))/(1+cos(b));
print1(floor(9/(10*r)),", ")
)
}

Formula

From Wolfdieter Lang, Sep 30 2014 (Start)
a(n) = floor(r(n)) with the rational curvatures r(n) satisfying the one step nonlinear recurrence relation r(n) = (11*r(n-1) - 9 + 20*sqrt((r(n-1) - 9)*r(n-1)/10))/9 with input r(0) = 9. (In the link r(n) is called b'(n).)
r(n) = A246643(n)/9^(n-1) = (9/2)*(1 + S(n, 22/9) - (11/9)*S(n-1, 22/9)), n >= 0, with Chebyshev/s S-polynomials (see A049310). 9^n*S(n, 22/9) = A246645(n). See A246643 for more details. (End)

Extensions

Edited: Keyword easy and Chebyshev index link added. Wolfdieter Lang, Sep 30 2014
Showing 1-2 of 2 results.

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