OFFSET
1,2
COMMENTS
From Derek Delk, Oct 15 2025: (Start)
These are figurate numbers produced in the construction of approximate rhombic dodecahedrons using points in a face-centered cubic lattice. Each term is expressed as two summations.
One summation takes the (4n-3)-th octahedral number (A005900) and adds 8 stacks of triangular numbers (A000217), each consisting of n-1 layers. This corresponds to constructing a rhombic dodecahedron by starting with an octahedron and placing a triangular pyramid on each of its 8 faces.
Another summation takes the (2n-1)-th thickened cube number (A050492) and adds 6 stacks consisting of alternating layers of centered square numbers (A001844) and 4 times triangular numbers (A046092), totaling 2n-2 layers per stack. This corresponds to constructing a rhombic dodecahedron by starting with a cube and placing a square pyramid on each of its 6 faces.
REFERENCES
John H. Conway and Richard K. Guy, 'The Book of Numbers', Copernicus Pub. Co., 1996.
Peter Pearce and Susan Pearce, 'Polyhedra primer', Van Nostrand Reinhold, 1978.
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Derek Delk, Visualization of Six Terms
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
FORMULA
G.f.: x*(1 + 89*x + 251*x^2 + 43*x^3)/(1-x)^4. - R. J. Mathar, Nov 10 2011
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Vincenzo Librandi, Jul 01 2012
E.g.f.: 43 + (-43 + 44*x + 24*x^2 + 64*x^3)*exp(x). - G. C. Greubel, Nov 14 2019
MAPLE
seq(((8*n-7)^3 +(8*n-1))/8, n=1..40); # G. C. Greubel, Nov 14 2019
MATHEMATICA
CoefficientList[Series[(1+89*x+251*x^2+43*x^3)/(x-1)^4, {x, 0, 40}], x] (* Vincenzo Librandi, Jul 01 2012 *)
LinearRecurrence[{4, -6, 4, -1}, {1, 93, 617, 1957}, 30] (* Harvey P. Dale, Mar 16 2013 *)
Table[((8*n-7)^3 +(8*n-1))/8, {n, 40}] (* G. C. Greubel, Nov 14 2019 *)
PROG
(Magma) I:=[1, 93, 617, 1957]; [n le 4 select I[n] else 4*Self(n-1) - 6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..40]]; // Vincenzo Librandi, Jul 01 2012
(PARI) a(n)=64*n^3-168*n^2+148*n-43 \\ Charles R Greathouse IV, Apr 25 2016
(SageMath) [((8*n-7)^3 +(8*n-1))/8 for n in (1..40)] # G. C. Greubel, Nov 14 2019
(GAP) List([1..40], n-> ((8*n-7)^3 +(8*n-1))/8); # G. C. Greubel, Nov 14 2019
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Chris G. Spies-Rusk (chaosorder4(AT)gmail.com), May 05 2009, May 11 2009, May 19 2009
EXTENSIONS
Edited by G. C. Greubel, Nov 14 2019
STATUS
approved
