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Number of moves needed to solve an (n+1)-ring baguenaudier if two simultaneous moves of the two end rings are counted as one.
20

%I #133 Apr 28 2023 08:15:48

%S 1,1,4,7,16,31,64,127,256,511,1024,2047,4096,8191,16384,32767,65536,

%T 131071,262144,524287,1048576,2097151,4194304,8388607,16777216,

%U 33554431,67108864,134217727,268435456,536870911,1073741824

%N Number of moves needed to solve an (n+1)-ring baguenaudier if two simultaneous moves of the two end rings are counted as one.

%C Might be called the "Purkiss sequence", after Henry John Purkiss who in 1865 found that this is the number of moves for the accelerated Chinese Rings puzzle (baguenaudier). [Email from _Andreas M. Hinz_, Feb 15 2017, who also pointed out that there was an error in the definition in this entry]. - _N. J. A. Sloane_, Feb 18 2017

%C The row sums of triangle A166692. - _Paul Curtz_, Oct 20 2009

%C The inverse binomial transform equals (-1)^n*A062510(n) with an extra leading term 1. - _Paul Curtz_, Oct 20 2009

%C This is the sequence A(1,1;1,2;1) of the family of sequences [a,b:c,d:k] considered by G. Detlefs, and treated as A(a,b;c,d;k) in the W. Lang link given below. - _Wolfdieter Lang_, Oct 18 2010

%C Also, the decimal representation of the x-axis, from the origin to the right edge, of the n-th stage of growth of the two-dimensional cellular automaton defined by Rules 261, 269, 277, 285, 293, 301, 309, 317, 325, 333, 341, 349, 357, 365, 37, and 381, based on the 5-celled von Neumann neighborhood. - _Robert Price_, Jan 02 2017

%H Vincenzo Librandi, <a href="/A051049/b051049.txt">Table of n, a(n) for n = 0..1000</a>

%H Paul Barry, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL19/Barry/barry321.html">Jacobsthal Decompositions of Pascal's Triangle, Ternary Trees, and Alternating Sign Matrices</a>, Journal of Integer Sequences, 19, 2016, #16.3.5.

%H Andreas M. Hinz, <a href="https://www.fq.math.ca/Papers1/55-1/Hinz09152016.pdf">The Lichtenberg sequence</a>, Fib. Quart., 55 (2017), 2-12.

%H A. M. Hinz, S. Klavžar, U. Milutinović, and C. Petr, <a href="http://dx.doi.org/10.1007/978-3-0348-0237-6">The Tower of Hanoi - Myths and Maths</a>, Birkhäuser 2013. See page 56. <a href="http://tohbook.info">Book's website</a>

%H Wolfdieter Lang, <a href="/A051049/a051049.pdf">Notes on certain inhomogeneous three term recurrences.</a> [From _Wolfdieter Lang_, Oct 18 2010]

%H N. J. A. Sloane, <a href="http://arxiv.org/abs/1503.01168">On the Number of ON Cells in Cellular Automata</a>, arXiv:1503.01168 [math.CO], 2015.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Baguenaudier.html">Baguenaudier</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/ElementaryCellularAutomaton.html">Elementary Cellular Automaton</a>

%H S. Wolfram, <a href="http://wolframscience.com/">A New Kind of Science</a>

%H Wolfram Research, <a href="http://atlas.wolfram.com/">Wolfram Atlas of Simple Programs</a>

%H <a href="/index/Ce#cell">Index entries for sequences related to cellular automata</a>

%H <a href="https://oeis.org/wiki/Index_to_2D_5-Neighbor_Cellular_Automata">Index to 2D 5-Neighbor Cellular Automata</a>

%H <a href="https://oeis.org/wiki/Index_to_Elementary_Cellular_Automata">Index to Elementary Cellular Automata</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (2,1,-2).

%F a(n) = (2^(n+1) - (1 + (-1)^(n+1)))/2. - _Paul Barry_, Apr 24 2003

%F a(n+2) = a(n+1) + 2*a(n) + 1, a(0)=a(1)=1. - _Paul Barry_, May 01 2003

%F From _Paul Barry_, Sep 19 2003: (Start)

%F G.f.: (1 - x + x^2)/((1 - x^2)*(1 - 2*x));

%F e.g.f.: exp(2*x) - sinh(x). (End)

%F a(n) = ((Sum_{k=0..n} 2^k) + (-1)^n)/2 = (A000225(n+1) + (-1)^n)/2. - _Paul Barry_, May 27 2003

%F (a(n+1) - a(n))/3 = A001045(n). - _Paul Barry_, May 27 2003

%F a(n) = Sum_{k=0..floor(n/2)} binomial(n+1, 2*k). - _Paul Barry_, May 27 2003

%F a(n) = (Sum_{k=0..n} binomial(n,k) + (-1)^(n-k)) - 1. - _Paul Barry_, Jul 21 2003

%F a(n) = Sum_{k=0..n} Sum_{j=0..n-k, (j-k) mod 2 = 0} binomial(n-k, j). - _Paul Barry_, Jan 25 2005

%F Row sums of triangle A135221. - _Gary W. Adamson_, Nov 23 2007

%F a(n) = A001045(n+1) + A000975(n+1) - A000079(n). - _Paul Curtz_, Oct 20 2009

%F a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3), a(0) = a(1) = 1, a(2) = 4. Observed by G. Detlefs. See the W. Lang link. - _Wolfdieter Lang_, Oct 18 2010

%F a(n) = 3*a(n-1) - 2*a(n-2) + 3*(-1)^n. - _Gary Detlefs_, Dec 21 2010

%F a(n) = 3* A000975(n-1) + 1, n > 0. - _Gary Detlefs_, Dec 21 2010

%F a(n+2) = A001969(2^n+1) + A000069(2^n); evil + odious. - _Johannes W. Meijer_, Jun 24 2011, Jun 26 2011

%F E.g.f.: exp(2x) - sinh(x) = Q(0); Q(k) = 1 - k!*x^(k+1)/((2*k + 1)!*2^k - 2*(((2*k + 1)!*2^k)^2)/((2*k + 1)!*2^(k+1) - x^k*(k + 1)!/Q(k+1))); (continued fraction). - _Sergei N. Gladkovskii_, Nov 16 2011

%F a(n) = Sum_{k=0..n} Sum_{i=0..n} C(k-1,i). - _Wesley Ivan Hurt_, Sep 21 2017

%F a(n) = A000975(n+1) - A001045(n). - _Yuchun Ji_, Jul 08 2018

%F a(n) = A026147(2^(n-1)) for n > 0. - _Chunqing Liu_, Dec 18 2022

%p A051049:= proc(n): 2^n -(1-(-1)^n)/2 end: seq(A051049(n), n=0..40); # _Johannes W. Meijer_, Jun 24 2011

%t a[n_?EvenQ]:= 2^(n-1) -1; a[n_?OddQ]:= 2^(n-1); Table[a[n], {n,50}]

%t LinearRecurrence[{2,1,-2},{1,1,4},40] (* _Jean-François Alcover_, Jan 08 2019 *)

%o (Magma) [2^n -(1-(-1)^n)/2: n in [0..40]]; // _Vincenzo Librandi_, Aug 14 2011

%o (PARI) a(n)=2^(n-1)-(n%2==0) \\ _Charles R Greathouse IV_, Mar 22 2013

%o (SageMath) [2^n -(n%2) for n in range(41)] # _G. C. Greubel_, Apr 23 2023

%Y Cf. A000069, A000079, A000225, A000975, A001045.

%Y Cf. A001969, A026147, A062510, A135221.

%Y Row sums of A131086.

%Y Row sums of A166692.

%K nonn,easy

%O 0,3

%A _Eric W. Weisstein_

%E Edited and information added by _Johannes W. Meijer_, Jun 24 2011