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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A239073 Given a circle of radius R into which small circles of radius R/2^n are packed in a "hexagonal pattern" (see Comments), a(n) is the maximum number of small circles.

Original entry on oeis.org

1, 2, 10, 48, 208, 874, 3600, 14622, 58976, 236824, 949094, 3800126, 15207812, 60846154, 243414326, 973716670, 3894985588, 15580180122, 62321195992, 249285735518, 997144844044, 3988583179554, 15954340324098, 63817376508852, 255269536476262
Offset: 0

Views

Author

Kival Ngaokrajang, Mar 10 2014

Keywords

Comments

The construction rule is: (1) Start with a unit circle (big circle). (2) Pack circles at radius 1/2^n (small circles) on the diameter line of the big circle. (3) Maintain hexagonal packing pattern of small circles on the rows above and below that of the previous step. The number of small circles in any row is limited so that the circumference of the last small circle does not cross the circumference of the big circle (but is allowed to contact it). (4) Repeat process to the top and bottom rows.
See illustration in links.
From Jon E. Schoenfield, Mar 23 2014: (Start)
There can never be exactly one small circle on the top row iMax (and, symmetrically, exactly one on the bottom row) that contacts the large circle, since the upper edge of that small circle would be at a radius of (iMax*sqrt(3)+1)/2^n from the center of the unit circle (which cannot equal 1 unless iMax=0, i.e., there are no rows above or below the center).
The number of small circles on the top row seems to be on the order of 2^(n/2). It seems nearly certain that, for n>0, there is never exactly one small circle on the top row. (End)
After Wolfdieter Lang's suggestions of an alternative definition, I found later that a(n) = A053417(2^n-1), n >= 1.

Crossrefs

Programs

  • Magma
    for n in [0..25] do
       x:=(2^n - 1)^2;
       c:=2^n; // for row i=0
       for i in [1..Isqrt(x div 3)] do
          t:=x-3*i*i;
          if IsOdd(i) then
             c:=c+2*(2*Floor(Sqrt(t)/2)+1);
          else
             c:=c+2*(2*Floor(Sqrt(t)/2+1/2));
          end if;
       end for;
       n, c;
    end for; // Jon E. Schoenfield, Mar 17 2014

Formula

a(n) = 2^n + 2 * Sum_{i odd, <= iMax} (2*floor(sqrt(x - 3 * i^2) / 2) + 1) + 2 * Sum_{i even, 0 < i <= iMax} 2*floor(sqrt(x - 3 * i^2) / 2 + 1/2) where x = (2^n-1)^2 and iMax = floor(sqrt(x/3)). - Jon E. Schoenfield, Mar 17, Mar 23 2014

Extensions

a(20) - a(24) and corrections to a(15), a(17), and a(18) from Jon E. Schoenfield, Mar 17 2014

A239074 Given a circle of radius R into which small circles of radius R/2^n are packed in a "hexagonal pattern" (see Comments), a(n) is the number of points at which a small circle is tangent to the big circle.

Original entry on oeis.org

2, 2, 6, 2, 6, 6, 6, 2, 18, 6, 2, 18, 6, 18, 54, 2, 6, 54, 6, 6, 90, 2, 2, 54, 54, 18, 54, 18, 6, 162, 6, 2, 18, 6, 18, 1458, 18, 18, 162, 18, 2, 810, 6, 18, 1458, 2, 6, 486, 18, 162, 486, 54, 6, 486, 18, 54, 162, 18, 2, 4374, 6, 18, 2430, 6, 54, 162, 18, 18, 18, 54
Offset: 1

Views

Author

Kival Ngaokrajang, Apr 06 2014

Keywords

Comments

The construction rule is: (1) Start with a unit circle (big R-circle). (2) Pack circles at radius 1/2^n (small r-circles) on the diameter line of the big circle. (3) Pack small circles in the rows above and below the row packed in the previous step, maintaining a hexagonal packing pattern. The number of small circles in any row is limited so that the circumference of the last small circle does not cross (but is allowed to contact) the circumference of the big circle. (4) Repeat process to the top and bottom rows.
The contact points are the points where the circumference of a small circle contacts the circumference of the big circle, i.e., they are mutually tangent.
See illustration in links.
Also, the number of integer solutions to the equation (2^n-1)^2 = 3*x^2 + y^2. - Andrew Howroyd, May 27 2018

Examples

			n=9 (see the link): In the first quadrant, shown there, there are 4 touching points with the large circle for rows x > 0, namely for the rows 52, 132, 280 and 292. With the trivial 2 touching points with the large circle for the row x=0 this adds to the total number 2 + 4*4 = 18 = a(9). - _Wolfdieter Lang_, Apr 06 2014
		

Crossrefs

Programs

  • Mathematica
    a[1] = 2; a[n_] := Module[{f = FactorInteger[2^n - 1]}, 2*Product[If[Mod[ f[[i, 1]], 3] == 1, 2*f[[i, 2]] + 1, 1] , {i, 1, Length[f]}]];
    Array[a, 70] (* Jean-François Alcover, Jul 03 2018, after Andrew Howroyd *)
  • PARI
    a(n) = {my(f=factor(2^n-1)); 2*prod(i=1, #f~, if(f[i, 1]%3==1, 2*f[i, 2]+1, 1))} \\ Andrew Howroyd, May 27 2018

Formula

a(n) = 2 + 4*A(n), n >= 1, with A(n) the number of integer solutions for x(n,j) = sqrt((2^n-1)^2 + 3*j^2), for j = 1, 2, ..., floor((2^n-1)/sqrt(3)). R = 2^n and r=1 (small radius) was used here. - Wolfdieter Lang, Apr 07 2014
a(n) = A033716((2^n - 1)^2). - Andrew Howroyd, May 27 2018

Extensions

Corrected and extended by Wolfdieter Lang, Apr 06 2014
a(26)-a(70) from Andrew Howroyd, May 27 2018
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