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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: [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

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

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 A238439

Adjacent sequences:  A026641 A026642 A026643 * A026645 A026646 A026647

KEYWORD

nonn,easy

AUTHOR

Clark Kimberling

EXTENSIONS

Recurrence in definition line found by Lee Hae-hwang, Apr 03 2002

STATUS

approved

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Last modified November 19 05:29 EST 2018. Contains 317333 sequences. (Running on oeis4.)