OFFSET
0,2
COMMENTS
Note that it is not possible to fill a 2 X 3 X (2*n-1) hole using 1 X 2 X 2 bricks.
a(n+1) of the Jacobsthal sequence A001045 gives the number of ways of filling a 2 X 2 X n hole with 1 X 2 X 2 bricks.
Will the pattern of rightmost digits (1,5,7,7) be continued? - Bill McEachen, May 20 2009
The answer to the question in a previous comment is: the linear recurrence proves that the pattern 1, 5, 7, 7 of the least significant digits will continue. - R. J. Mathar, Jun 20 2010
a(n) is the number of compositions of n when there are 5 types of 1 and 2 types of other natural numbers. - Milan Janjic, Aug 13 2010
LINKS
Martin Griffiths, Filling cuboidal holes with bricks, Mathematical Spectrum (Applied Probability Trust) 42(2) (2010), 91-92.
Index entries for linear recurrences with constant coefficients, signature (6,-3).
FORMULA
a(0)=1, a(1)=5 and a(n) = 6*a(n-1) - 3*a(n-2) for n > 1.
a(n) = (3^n) * 2F1[-((n + 1)/2), -(n/2); 1/2; 2/3], using Gauss' hypergeometric function.
From Martin Griffiths, Apr 02 2009: (Start)
G.f.: A(x) = (1-x)/(1-6x+3x^2).
a(n) = (1/6)*((3+sqrt(6))^(n+1) + (3-sqrt(6))^(n+1)). (End)
From R. J. Mathar, Mar 29 2009: (Start)
G.f.: -(-1+x)/(1-6*x+3*x^2).
G.f.: G(0)/(6*x) -1/(3*x), where G(k) = 1 + 1/(1 - x*(2*k-3)/(x*(2*k-1) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 13 2013
MATHEMATICA
Simplify[Table[ 1/6 * ((3 + Sqrt[6])^(n + 1) + (3 - Sqrt[6])^(n + 1)), {n, 0, 19}]]
Table[3^n * Hypergeometric2F1[ -((n + 1)/2), -(n/2), 1/2, 2/3], {n, 0, 19}]
LinearRecurrence[{6, -3}, {1, 5}, 30] (* Harvey P. Dale, May 28 2015 *)
PROG
(Sage)
def A158869(n): return 3^n*lucas_number2(n+1, 2, 1/3)/2
[A158869(n) for n in (0..19)] # Peter Luschny, May 06 2013
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Martin Griffiths, Mar 28 2009
EXTENSIONS
Edited by Charles R Greathouse IV, Mar 08 2011
STATUS
approved