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A051049
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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.
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20
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1, 1, 4, 7, 16, 31, 64, 127, 256, 511, 1024, 2047, 4096, 8191, 16384, 32767, 65536, 131071, 262144, 524287, 1048576, 2097151, 4194304, 8388607, 16777216, 33554431, 67108864, 134217727, 268435456, 536870911, 1073741824
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OFFSET
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0,3
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COMMENTS
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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
The inverse binomial transform equals (-1)^n*A062510(n) with an extra leading term 1. - Paul Curtz, Oct 20 2009
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
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
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LINKS
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FORMULA
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a(n) = (2^(n+1) - (1 + (-1)^(n+1)))/2. - Paul Barry, Apr 24 2003
a(n+2) = a(n+1) + 2*a(n) + 1, a(0)=a(1)=1. - Paul Barry, May 01 2003
G.f.: (1 - x + x^2)/((1 - x^2)*(1 - 2*x));
e.g.f.: exp(2*x) - sinh(x). (End)
a(n) = ((Sum_{k=0..n} 2^k) + (-1)^n)/2 = (A000225(n+1) + (-1)^n)/2. - Paul Barry, May 27 2003
a(n) = Sum_{k=0..floor(n/2)} binomial(n+1, 2*k). - Paul Barry, May 27 2003
a(n) = (Sum_{k=0..n} binomial(n,k) + (-1)^(n-k)) - 1. - Paul Barry, Jul 21 2003
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
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
a(n) = 3*a(n-1) - 2*a(n-2) + 3*(-1)^n. - Gary Detlefs, Dec 21 2010
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
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MAPLE
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MATHEMATICA
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a[n_?EvenQ]:= 2^(n-1) -1; a[n_?OddQ]:= 2^(n-1); Table[a[n], {n, 50}]
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PROG
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(SageMath) [2^n -(n%2) for n in range(41)] # G. C. Greubel, Apr 23 2023
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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