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A026644
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a(n) = a(n-1) + 2*a(n-2) + 2, for n>=3, where a(0)= 1, a(1)= 2, a(2)= 4.
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15
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1, 2, 4, 10, 20, 42, 84, 170, 340, 682, 1364, 2730, 5460, 10922, 21844, 43690, 87380, 174762, 349524, 699050, 1398100, 2796202, 5592404, 11184810, 22369620, 44739242, 89478484, 178956970, 357913940, 715827882, 1431655764, 2863311530, 5726623060
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OFFSET
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0,2
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COMMENTS
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Number of moves to solve Chinese rings puzzle.
a(n-1) (with a(0):=0) enumerates all sequences of length m=1,2,...,floor(n/2) with nonzero integer entries n_i satisfying sum |n_i| <= n-m. Rephrasing K. A. Meissner's example p. 6. Example n=4: from length m=1: [1], [2], [3], each in 2 signed versions; from m=2: [1,1] in 2^2 = 4 signed versions. Hence a(3) = a(4-1) = 3*2 + 1*4 = 10.
Also the number of different 3-colorings (out of 4 colors) for the vertices of all triangulated planar polygons on a base with n+1 vertices if the colors of the two base vertices are fixed. - Patrick Labarque, Mar 23 2010
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REFERENCES
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Richard I. Hess, Compendium of Over 7000 Wire Puzzles, privately printed, 1991.
Richard I. Hess, Analysis of Ring Puzzles, booklet distributed at 13th International Puzzle Party, Amsterdam, Aug 20 1993.
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LINKS
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Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Nicolas Gastineau, O. Togni, On S-packing edge-colorings of cubic graphs, arXiv preprint arXiv:1711.10906 [cs.DM], 2017.
Lee Hae-hwang, Illustration of initial terms in terms of rosemary plants
Krzysztof A. Meissner, Black hole entropy in Loop Quantum Gravity, arXiv:gr-qc/0407052, 2004.
Index entries for linear recurrences with constant coefficients, signature (2,1,-2).
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FORMULA
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a(2k) = 2*a(2k-1), a(2k+1) = 2*a(2k) + 2. - Peter Shor, Apr 11 2002
For n>0: a(n+1) = a(n) + 2*b(n+1) + 4*b(n), where b(k) = A001045(k). - N. J. A. Sloane, May 16 2003
For n>0: if n mod 2 = 0 then (2^(n+2)-4)/3 else (2^(n+2)-2)/3. - Richard Hess
a(2n) = 2n-1 + sum_{k=0..2n-1} a(k), n>0; a(2n+1) = 2n+1 + sum_{k=0..n} a(k). - Lee Hae-hwang, Sep 17 2002; corrected by R. J. Mathar, Oct 21 2008
a(n) = 2n + 2*sum_{k=1..n-2} a(k), n>0. - Lee Hae-hwang, Sep 19 2002; corrected by R. J. Mathar, Oct 21 2008
G.f.: (1 - x^2 + 2x^3)/((1 - x)(1 - x - 2x^2)); a(n) = J(n+2) - 1 + 0^n, where J(n)=A001045(n); a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3); a(n) = 0^n + sum{k=0..n} (2 - 2*0^(n-k))J(k+1). - Paul Barry, Oct 24 2007
a(n) = -1 + (1/3)*(-1)^(n-1) + (8/3)*2^(n-1) + (C(2*n,n) mod 2). - Paolo P. Lava, Oct 03 2008
a(n) = A052953(n+1) - 2, n>0. [Moved from A020988, R. J. Mathar, Oct 21 2008]
a(n) = floor(A097074(n+1)/2), n>0. - Gary Detlefs Dec 19 2010
a(n) = A169969(2n-1) - 1, n>=2; a(n) = 3*2^(n-1) - 1 - A169969(2n-7), n>=5 . - Yosu Yurramendi, Jul 05 2016
a(n+3) = 3*2^(n+2) - 2 - a(n), n>=1, a(1)=2, a(2)=4, a(3)=10 . - Yosu Yurramendi, Jul 05 2016
a(n) + A084170(n) = 3*2^n - 2, n>=1 . - Yosu Yurramendi, Jul 05 2016
E.g.f: (3 - 4*cosh(x) + 4*cosh(2*x) - 2*sinh(x) + 4*sinh(2*x))/3. - Ilya Gutkovskiy, Jul 05 2016
a(n+3) = 9*2^n + A084170(n), n>=0 . - Yosu Yurramendi, Jul 07 2016
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MAPLE
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f:=n-> if n mod 2 = 0 then (2^(n+2)-4)/3 else (2^(n+2)-2)/3; fi;
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MATHEMATICA
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Join[{1}, Floor[(2^Range[3, 40] - 2)/3]] (* or *) LinearRecurrence[{2, 1, -2}, {1, 2, 4, 10}, 40] (* Vladimir Joseph Stephan Orlovsky, Jan 29 2012 *)
CoefficientList[Series[(1-x^2+2x^3)/((1-x)(1-x-2x^2)), {x, 0, 1001}], x] (* Vincenzo Librandi, Apr 04 2012 *)
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PROG
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(PARI) Vec((1-x^2+2*x^3)/(1-x)/(1-x-2*x^2)+O(x^99)) \\ Charles R Greathouse IV, Apr 04 2012
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CROSSREFS
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a(n) = T(n, 0) + T(n, 1) + ... + T(n, n), T given by A026637.
For n >= 1, equals twice A000975, also A001045 - 1.
A167030 is an essentially identical sequence.
Sequence in context: A318975 A255386 A167030 * A167193 A026666 A325508
Adjacent sequences: A026641 A026642 A026643 * A026645 A026646 A026647
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KEYWORD
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nonn,easy
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AUTHOR
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Clark Kimberling
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EXTENSIONS
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Recurrence in definition line found by Lee Hae-hwang, Apr 03 2002
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STATUS
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approved
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