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A068156
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G.f.: (x+2)*(x+1)/((x-1)*(x-2)) = Sum(n=0,inf,a(n)*(x/2)^n).
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7
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1, 3, 9, 21, 45, 93, 189, 381, 765, 1533, 3069, 6141, 12285, 24573, 49149, 98301, 196605, 393213, 786429, 1572861, 3145725, 6291453, 12582909, 25165821, 50331645, 100663293, 201326589, 402653181, 805306365, 1610612733
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,2
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COMMENTS
| Number of moves to solve Hard Pagoda puzzle.
Partial sums of A058295. Binomial transform of (1,2,4,2,4,2,4 ....) - Paul Barry, Feb 28 2003
a(n)=a(n-1)+ 3*2^(n-1); a(1)=3. - Ctibor O. Zizka, Apr 17 2008
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REFERENCES
| Richard I. Hess, Compendium of Over 7000 Wire Puzzles, privately printed, 1991.
Richard I. Hess, Analysis of Ring Puzzles, booklet distributed at 13-th International Puzzle Party, Amsterdam, Aug 20 1993.
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LINKS
| Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index to sequences with linear recurrences with constant coefficients, signature (3,-2).
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FORMULA
| a(0)=1, a(n) = A060482(2n+1). For n >0, a(n+1) = 2*a(n)+3.
G.f.: (1+2*x^2)/((1-2*x)*(1-x)). - Paul Barry, Feb 28 2003
a(n) = 3*2^n+0^n-3 - Paul Barry, Sep 04 2003
a(n) = A099257(A033484(n)+1) = 2*A033484(n) + 1. - Reinhard Zumkeller, Oct 09 2004
a(n) = 3*a(n-1)-2*a(n-2), n>1. - Vincenzo Librandi, Nov 11 2011
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MATHEMATICA
| a=0; lst={1}; k=3; Do[a+=k; AppendTo[lst, a]; k+=k, {n, 0, 4!}]; lst [From Vladimir Orlovsky, Dec 15 2008]
a=3; lst={1, a}; k=6; Do[a+=k; AppendTo[lst, a]; k+=k, {n, 0, 5!}]; lst [From Vladimir Orlovsky, Dec 16 2008]
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PROG
| (MAGMA) [3*2^n+0^n-3 : n in [0..30]]; // VIncenzo Librandi, Nov 11 2011
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CROSSREFS
| Sequence in context: A192970 A110964 A107351 * A166452 A052101 A063830
Adjacent sequences: A068153 A068154 A068155 * A068157 A068158 A068159
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KEYWORD
| nonn,easy
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AUTHOR
| Benoit Cloitre (benoit7848c(AT)orange.fr), Mar 12 2002
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