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 A026644 a(n) = a(n-1) + 2*a(n-2) + 2, for n>=3, where a(0)= 1, a(1)= 2, a(2)= 4. 15
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS 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: , , , 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 REFERENCES 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. LINKS Vincenzo Librandi, Table of n, a(n) for n = 0..1000 Thomas Baruchel, Properties of the cumulated deficient binary digit sum, arXiv:1908.02250 [math.NT], 2019. 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). FORMULA 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 MAPLE f:=n-> if n mod 2 = 0 then (2^(n+2)-4)/3 else (2^(n+2)-2)/3; fi; MATHEMATICA 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 *) PROG (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 CROSSREFS 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 KEYWORD nonn,easy AUTHOR EXTENSIONS Recurrence in definition line found by Lee Hae-hwang, Apr 03 2002 STATUS approved

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Last modified May 29 07:22 EDT 2020. Contains 334697 sequences. (Running on oeis4.)