OFFSET
1,1
COMMENTS
These are figurate numbers produced in the construction of approximate pyritohedrons (dodecahedrons with irregular pentagonal faces) using points in a face-centered cubic lattice. The dihedral angle along the long edges for such a pyritohedron is 2*arctan(2) = A197376.
The n-th term is expressed as a summation which takes the (8*n-6)-th thickened cube number (A050492) and adds 6 stacks - each layer the number of points. The stacks increase by 4 layers with unit increment in n, totaling 4*n-3 layers. This corresponds to constructing a pyritohedron by taking a cube and placing a wedge on each of its 6 faces.
So, a(n) = A050492(8*n-6) + 6*(6*n - 4 + Sum_{i=2..n} [(6*n+2*i-8)*(8*i-14) + (6*n+2*i-9)*(8*i-13)] + [(6*n+2*i-7)*(8*i-12)+(6*n+2*i-8)*(8*i-11)] + [(6*n+2*i-8)*(8*i-10)+(6*n+2*i-7)*(8*i-9)] + [(6*n+2*i-7)*(8*i-8)+(6*n+2*i-6)*(8*i-7)]).
LINKS
Derek Delk, Table of n, a(n) for n = 1..10000
Derek Delk, Visualization of Six Terms.
Wikipedia, Cubic crystal system.
Wikipedia, Dodecahedron.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
FORMULA
G.f.: 2*x*(927*x^3 + 6583*x^2 + 2845*x + 13)/(1 - x)^4.
E.g.f.: 2*(927 + exp(x)*(1728*x^3 + 972*x^2 + 940*x - 927)). - Stefano Spezia, Dec 02 2025
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
a(n) = a(n-1) + 10368*n^2 - 27216*n + 18728.
a(n) = A050492(8*n-6) + 6*[2 + Sum_{i=2..n} (704*i^2 - 1848*i + 1272)], by summing wedge gnomons.
MATHEMATICA
a[n_]:=3456*n^3 - 8424*n^2 + 6848*n - 1854; Array[a, 30] (* James C. McMahon, Dec 09 2025 *)
(* Alternative: *)
LinearRecurrence[{ 4, -6, 4, -1}, {26, 5794, 36186, 111938}, 30] (* James C. McMahon, Dec 09 2025 *)
(* Alternative: *)
Rest[CoefficientList[Series[2*x*(927*x^3 + 6583*x^2 + 2845*x + 13)/(1 - x)^4, {x, 0, 30}], x]] (* James C. McMahon, Dec 09 2025 *)
CROSSREFS
KEYWORD
nonn,easy,changed
AUTHOR
Derek Delk, Dec 01 2025
STATUS
approved