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%I #39 Oct 22 2019 04:21:28
%S 1,5,27,147,801,4365,23787,129627,706401,3849525,20977947,114319107,
%T 622980801,3394927485,18500622507,100818952587,549411848001,
%U 2994014230245,16315849837467,88913056334067
%N Number of ways of filling a 2 X 3 X 2n hole with 1 X 2 X 2 bricks.
%C Note that it is not possible to fill a 2 X 3 X (2*n-1) hole using 1 X 2 X 2 bricks.
%C 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.
%C Will the pattern of rightmost digits (1,5,7,7) be continued? - _Bill McEachen_, May 20 2009
%C 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
%C 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
%H Martin Griffiths, <a href="http://www.appliedprobability.org/data/files/MS%20issues/Vol42_No2.pdf">Filling cuboidal holes with bricks</a>, Mathematical Spectrum (Applied Probability Trust) 42(2) (2010), 91-92.
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (6,-3).
%F a(0)=1, a(1)=5 and a(n) = 6*a(n-1) - 3*a(n-2) for n > 1.
%F a(n) = (3^n) * 2F1[-((n + 1)/2), -(n/2); 1/2; 2/3], using Gauss' hypergeometric function.
%F From _Martin Griffiths_, Apr 02 2009: (Start)
%F G.f.: A(x) = (1-x)/(1-6x+3x^2).
%F a(n) = (1/6)*((3+sqrt(6))^(n+1) + (3-sqrt(6))^(n+1)). (End)
%F From _R. J. Mathar_, Mar 29 2009: (Start)
%F G.f.: -(-1+x)/(1-6*x+3*x^2).
%F a(n) = A138395(n+1) - A138395(n). (End)
%F 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
%t Simplify[Table[ 1/6 * ((3 + Sqrt[6])^(n + 1) + (3 - Sqrt[6])^(n + 1)), {n, 0, 19}]]
%t Table[3^n * Hypergeometric2F1[ -((n + 1)/2), -(n/2), 1/2, 2/3], {n, 0, 19}]
%t LinearRecurrence[{6,-3},{1,5},30] (* _Harvey P. Dale_, May 28 2015 *)
%o (Sage)
%o def A158869(n): return 3^n*lucas_number2(n+1, 2, 1/3)/2
%o [A158869(n) for n in (0..19)] # _Peter Luschny_, May 06 2013
%K easy,nonn
%O 0,2
%A _Martin Griffiths_, Mar 28 2009
%E Edited by _Charles R Greathouse IV_, Mar 08 2011
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