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A038754
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a(2n) = 3^n, a(2n+1) = 2*3^n.
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61
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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
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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.
a(n) = A140740(n+2,2). - Reinhard Zumkeller, May 26 2008
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
sum_(n=0...infinity) 1/a(n) = 9/4. - Alexander R. Povolotsky, Aug 24 2012
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LINKS
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T. D. Noe, Table of n, a(n) for n = 0..400
V. Shevelev, On monotonic strengthening of Newman-like phenomenon on (2m+1)-multiples in base 2m
Index entries for sequences related to linear recurrences with constant coefficients, signature (0,3).
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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(2n) = (3/2)*a(2n-1)=3^n, a(2n+1)=2*a(2n)=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. - Benoit Cloitre, Apr 27 2003
a(n) = (1/6)*(5-(-1)^n)*3^floor(n/2); a(2n)=a(2n-1)+a(2n-2)+a(2n-3); a(2n+1)=a(2n)+a(2n-1) - Benoit Cloitre, Apr 27 2003
G.f.: (1+2x)/(1-3x^2) - Paul Barry, Aug 25 2003
a(n) = (1 + n mod 2) * 3^floor(n/2). a(n) = A087503(n) - A087503(n-1). - Reinhard Zumkeller, Sep 11 2003
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). - Reinhard Zumkeller, May 26 2008
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) = A182751(n) for n >= 2. - Jaroslav Krizek, Nov 27 2010
a(n) = sum {i=0,...,2^(n+1), i==0 mod 3}(-1)^A000120(i). - Vladimir Shevelev, May 16 2012
a(0)=1, a(1)=2, for n>=3, a(n)=3*a(n-2). - Vladimir Shevelev, May 21 2012
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EXAMPLE
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In interval [0,2^5) we have 11 multiple of 3 numbers, from which 10 evil and only one (21) 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 (zerinvarylajos(AT)yahoo.com), 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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lst={a=b=1}; Do[AppendTo[lst, b=2*a]; AppendTo[lst, a=b+a], {n, 0, 20}]; lst (* Vladimir Joseph Stephan Orlovsky, Apr 13 2011 *)
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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)
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CROSSREFS
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Cf. Somos sequences A006720, A006721, A006722, A006723.
a(n) = A094718(5, n).
Cf. A000079, A133464, A140730, A037124, A070909, A048328, A068911, A124302, A000045, A038754, A028495, A030436, A061551, A178381, A182751-A182757.
Sequence in context: A035522 A018311 A018481 * A182522 A165647 A191398
Adjacent sequences: A038751 A038752 A038753 * A038755 A038756 A038757
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KEYWORD
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easy,nice,nonn
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AUTHOR
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Henry Bottomley, May 03 2000
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STATUS
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approved
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