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A158869
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Number of ways of filling a 2 X 3 X 2n hole with 1 X 2 X 2 bricks.
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1
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1, 5, 27, 147, 801, 4365, 23787, 129627, 706401, 3849525, 20977947, 114319107, 622980801, 3394927485, 18500622507, 100818952587, 549411848001, 2994014230245, 16315849837467, 88913056334067
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OFFSET
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0,2
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COMMENTS
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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
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LINKS
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FORMULA
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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.
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).
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
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MATHEMATICA
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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 *)
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PROG
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(Sage)
def A158869(n): return 3^n*lucas_number2(n+1, 2, 1/3)/2
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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