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A038754
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a(2n) = 3^n, a(2n+1) = 2*3^n.
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91
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1, 2, 3, 6, 9, 18, 27, 54, 81, 162, 243, 486, 729, 1458, 2187, 4374, 6561, 13122, 19683, 39366, 59049, 118098, 177147, 354294, 531441, 1062882, 1594323, 3188646, 4782969, 9565938, 14348907, 28697814, 43046721, 86093442, 129140163, 258280326, 387420489
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OFFSET
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0,2
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COMMENTS
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In general, for the recurrence a(n) = a(n-1)*a(n-2)/a(n-3), all terms are integers iff a(0) divides a(2) and first three terms are positive integers, since a(2n+k) = a(k)*(a(2)/a(0))^n for all nonnegative integers n and k.
Equals eigensequence of triangle A070909; (1, 1, 2, 3, 6, 9, 18, ...) shifts to the left with multiplication by triangle A070909. - Gary W. Adamson, May 15 2010
The a(n) represent all paths of length (n+1), n >= 0, starting at the initial node on the path graph P_5, see the second Maple program. - Johannes W. Meijer, May 29 2010
a(n) is the difference between numbers of multiple of 3 evil (A001969) and odious (A000069) numbers in interval [0, 2^(n+1)). - Vladimir Shevelev, May 16 2012
A "half-geometric progression": to obtain a term (beginning with the third one) we multiply the before previous one by 3. - Vladimir Shevelev, May 21 2012
Pisano period lengths: 1, 2, 1, 4, 8, 2, 12, 4, 1, 8, 10, 4, 6, 12, 8, 8, 32, 2, 36, 8, ... . - R. J. Mathar, Aug 10 2012
Numbers n such that the n-th cyclotomic polynomial has a root mod 3. - Eric M. Schmidt, Jul 31 2013
Range of row n of the circular Pascal array of order 6. - Shaun V. Ault, Jun 05 2014
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LINKS
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FORMULA
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a(n) = a(n-1)*a(n-2)/a(n-3) with a(0)=1, a(1)=2, a(2)=3.
a(2*n) = (3/2)*a(2*n-1) = 3^n, a(2*n+1) = 2*a(2*n) = 2*3^n.
a(1)=1, a(n)= 2*a(n-1) if a(n-1) is odd, or a(n)= (3/2)*a(n-1) if a(n-1) is even.
a(n) = (1/6)*(5-(-1)^n)*3^floor(n/2).
a(2*n) = a(2*n-1) + a(2*n-2) + a(2*n-3).
a(2*n+1) = a(2*n) + a(2*n-1). (End)
G.f.: (1+2*x)/(1-3*x^2). - Paul Barry, Aug 25 2003
a(n) = (1 + n mod 2) * 3^floor(n/2).
a(n) = sqrt(3)*(2+sqrt(3))*(sqrt(3))^n/6 - sqrt(3)*(2-sqrt(3))*(-sqrt(3))^n/6. - Paul Barry, Sep 16 2003
a(n+1) = a(n) + a(n - n mod 2). (End)
If p(i) = Fibonacci(i-3) and if A is the Hessenberg matrix of order n defined by A(i,j) = p(j-i+1), (i<=j), A(i,j)=-1, (i=j+1), and A(i,j)=0 otherwise. Then, for n>=1, a(n-1) = (-1)^n det A. - Milan Janjic, May 08 2010
a(n) = 2^((1-(-1)^n)/2)*3^((2*n-1+(-1)^n)/4). - Luce ETIENNE, Aug 11 2014
For n > 0: a(n+1) = a(n) + if a(n) odd then min{a(n), a(n-1)} else max{a(n), a(n-1)}, see also A128588. (End)
E.g.f.: (7*cosh(sqrt(3)*x) + 4*sqrt(3)*sinh(sqrt(3)*x) - 4)/3. - Stefano Spezia, Feb 17 2022
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EXAMPLE
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In the interval [0,2^5) we have 11 multiples of 3 numbers, from which 10 are evil and only one (21) is odious. Thus a(4) = 10 - 1 = 9. - Vladimir Shevelev, May 16 2012
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MAPLE
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a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=3*a[n-2]+2 od: seq(a[n]+1, n=0..34); # Zerinvary Lajos, Mar 20 2008
with(GraphTheory): P:=5: G:=PathGraph(P): A:= AdjacencyMatrix(G): nmax:=35; for n from 1 to nmax do B(n):=A^n; a(n):=add(B(n)[1, k], k=1..P) od: seq(a(n), n=1..nmax); # Johannes W. Meijer, May 29 2010
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MATHEMATICA
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LinearRecurrence[{0, 3}, {1, 2}, 40] (* Harvey P. Dale, Jan 26 2014 *)
CoefficientList[Series[(1+2x)/(1-3x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Aug 18 2016 *)
Module[{nn=20, c}, c=3^Range[0, nn]; Riffle[c, 2c]] (* Harvey P. Dale, Aug 21 2021 *)
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PROG
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(PARI) a(n)=(1/6)*(5-(-1)^n)*3^floor(n/2)
(PARI) a(n)=3^(n>>1)<<bittest(n, 0)
(Haskell)
import Data.List (transpose)
a038754 n = a038754_list !! n
a038754_list = concat $ transpose [a000244_list, a008776_list]
(Magma) [n le 2 select n else 3*Self(n-2): n in [1..40]]; // Vincenzo Librandi, Aug 18 2016
(SageMath) [2^(n%2)*3^((n-(n%2))/2) for n in range(61)] # G. C. Greubel, Oct 10 2022
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CROSSREFS
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KEYWORD
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easy,nice,nonn
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AUTHOR
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STATUS
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approved
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