A158869 Number of ways of filling a 2 X 3 X 2n hole with 1 X 2 X 2 bricks.

1, 5, 27, 147, 801, 4365, 23787, 129627, 706401, 3849525, 20977947, 114319107, 622980801, 3394927485, 18500622507, 100818952587, 549411848001, 2994014230245, 16315849837467, 88913056334067

Offset: 0

Author

Martin Griffiths, Mar 28 2009

Keywords

• easy
• nonn
• changed

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.
• Dimana Miroslavova Pramatarova, Investigating the Periodicity of Weighted Catalan Numbers and Generalizing Them to Higher Dimensions, MIT Res. Sci. Instit. (2025). See p. 12.
• Index entries for linear recurrences with constant coefficients, signature (6,-3).

Programs

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 *)

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

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).

a(n) = A138395(n+1) - A138395(n). (End)

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

Extensions

Edited by Charles R Greathouse IV, Mar 08 2011