0,2

Note that it is not possible to fill a 2 by 3 by (2*n-1) hole using 1 by 2 by 2 bricks.

a(n+1) of the Jacobsthal sequence A001045 gives the number of ways of filling a 2 by 2 by n hole with 1 by 2 by 2 bricks.

Will the pattern of rightmost digits) (1,5,7,7) be continue? [From Bill R McEachen (bmceache(AT)centralsan.org), May 20 2009]

M. Griffiths, Filling cuboidal holes with bricks, Mathematical Spectrum (Applied Probability Trust), (to appear).

a(0)=1, a(1)=5 and a(n)=6*a(n-1)-3*a(n-2) for n>1.

Using Mathematica notation, a(n) can be expressed in terms of Gauss' hypergeometric function as a(n)=(3^n)*Hypergeometric2F1[ -((n + 1)/2),-(n/2),1/2,2/3].

Contribution from Martin Griffiths (griffm(AT)essex.ac.uk), 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)

G.f.: -(-1+x)/(1-6*x+3*x^2). a(n)=A154234(n+1)-A154234(n). [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Mar 29 2009]

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}]

Sequence in context: A015535 A026292 A100193 this_sequence A162557 A134425 A083326

Adjacent sequences: A158866 A158867 A158868 this_sequence A158870 A158871 A158872

easy,nonn

Martin Griffiths (griffm(AT)essex.ac.uk), Mar 28 2009