%I #47 Jul 03 2018 08:12:25

%S 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,

%T 162,6,2,18,6,18,1458,18,18,162,18,2,810,6,18,1458,2,6,486,18,162,486,

%U 54,6,486,18,54,162,18,2,4374,6,18,2430,6,54,162,18,18,18,54

%N 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.

%C 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.

%C 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.

%C See illustration in links.

%C Also, the number of integer solutions to the equation (2^n-1)^2 = 3*x^2 + y^2. - _Andrew Howroyd_, May 27 2018

%H Kival Ngaokrajang, <a href="/A239074/a239074_2.pdf">Illustration of initial terms</a>

%H Kival Ngaokrajang, <a href="/A239074/a239074_3.pdf">Illustration for n = 9</a>

%F 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

%F a(n) = A033716((2^n - 1)^2). - _Andrew Howroyd_, May 27 2018

%e 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

%t 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]}]];

%t Array[a, 70] (* _Jean-François Alcover_, Jul 03 2018, after _Andrew Howroyd_ *)

%o (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

%Y Cf. A033716, A239073, A239206.

%K nonn

%O 1,1

%A _Kival Ngaokrajang_, Apr 06 2014

%E Corrected and extended by _Wolfdieter Lang_, Apr 06 2014

%E a(26)-a(70) from _Andrew Howroyd_, May 27 2018