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a(n) is the number of geometrically distinct edge-unfoldings of the regular n-gonal cupola.
0

%I #31 May 13 2021 09:33:52

%S 308,3030,29757,294327,2911142,28814940,285214743,2823311133,

%T 27947663768,276653115090,2738581182417,27109156615827,

%U 268352962161482,2656420444277880,26295851254778283,260302091898387033,2576725065493516028,25506948561006315150

%N a(n) is the number of geometrically distinct edge-unfoldings of the regular n-gonal cupola.

%C The term "regular" applies only to the regular n-gon and 2n-gon (the "top and bottom" of the cupola), the other faces (the "sides") being n isosceles triangles and n sufficiently long rectangles. For n=3,4,5, regular triangles and squares can be used for the sides. That applies to n=6 if a two-sided (flat) polyhedron is allowed.

%C The first 25 terms of the auxiliary sequence c(n) in the Formula and Mathematica program match the 25 terms listed for sequence A085376.

%H Zsolt Lengvárszky and Rick Mabry, <a href="http://dx.doi.org/10.14232/actasm-016-789-y">Enumerating nets of prism-like polyhedra</a>, Acta Sci. Math. (Szeged) 83 (2017), no. 3-4, 377-392.

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Cupola_(geometry)">Cupola</a>

%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (11,-1,-109,109,1,-11,1).

%F Recursively define the sequence c(m) as follows: Let c(1) = 1, c(2) = 3, c(3) = 11, c(4) = 30, and for m > 4, let c(m) = 10*c(m-2) - c(m-4). Then for all n >= 3, the sequence a(n) can be given by a(n) = (c(2*n+1) + 5*c(2*n) - c(2*n-1) - c(2*n-2) - 5)/8 + (3 + (-1)^n)*c(n)/4.

%F a(n) = (c(2*n+1) + 5*c(2*n) - c(2*n-1) - c(2*n-2) - 5)/8 + (3 + (-1)^n)*c(n)/4 for n >= 3 where c(m) = 10*c(m-2) - c(m-4) for m > 4 and c(1) = 1, c(2) = 3, c(3) = 11, c(4) = 30.

%F G.f.: x^3*(308 - 358*x - 3265*x^2 + 3602*x^3 - 360*x^5 + 33 x^6)/(1 - 11*x + x^2 + 109*x^3 - 109*x^4 - x^5 + 11*x^6 - x^7). - _Stefano Spezia_, Apr 10 2021

%t a[n_]:=Sum[c[k],{k,1,2n-1}]+(1/2)c[2n]+If[OddQ[n],(1/2)c[n],c[n]];

%t c[1] = 1; c[2] = 3; c[3] = 11; c[4] = 30;

%t c[m_] := c[m] = 10 c[m - 2] - c[m - 4];

%Y Cf. A085376; see the sequence c(n) in the Formula and Mathematica program, but note that A085376 has only been conjectured to be the same as c(n).

%K nonn,easy

%O 3,1

%A _Rick Mabry_, Apr 10 2021