A051049 Number of moves needed to solve an (n+1)-ring baguenaudier if two simultaneous moves of the two end rings are counted as one.
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
Offset: 0
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..1000
- Paul Barry, Jacobsthal Decompositions of Pascal's Triangle, Ternary Trees, and Alternating Sign Matrices, Journal of Integer Sequences, 19, 2016, #16.3.5.
- Andreas M. Hinz, The Lichtenberg sequence, Fib. Quart., 55 (2017), 2-12.
- A. M. Hinz, S. Klavžar, U. Milutinović, and C. Petr, The Tower of Hanoi - Myths and Maths, Birkhäuser 2013. See page 56. Book's website
- Wolfdieter Lang, Notes on certain inhomogeneous three term recurrences. [From _Wolfdieter Lang_, Oct 18 2010]
- N. J. A. Sloane, On the Number of ON Cells in Cellular Automata, arXiv:1503.01168 [math.CO], 2015.
- Eric Weisstein's World of Mathematics, Baguenaudier
- Eric Weisstein's World of Mathematics, Elementary Cellular Automaton
- S. Wolfram, A New Kind of Science
- Wolfram Research, Wolfram Atlas of Simple Programs
- Index entries for sequences related to cellular automata
- Index to 2D 5-Neighbor Cellular Automata
- Index to Elementary Cellular Automata
- Index entries for linear recurrences with constant coefficients, signature (2,1,-2).
Crossrefs
Programs
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Magma
[2^n -(1-(-1)^n)/2: n in [0..40]]; // Vincenzo Librandi, Aug 14 2011
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Maple
A051049:= proc(n): 2^n -(1-(-1)^n)/2 end: seq(A051049(n), n=0..40); # Johannes W. Meijer, Jun 24 2011
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Mathematica
a[n_?EvenQ]:= 2^(n-1) -1; a[n_?OddQ]:= 2^(n-1); Table[a[n], {n,50}] LinearRecurrence[{2,1,-2},{1,1,4},40] (* Jean-François Alcover, Jan 08 2019 *) -
PARI
a(n)=2^(n-1)-(n%2==0) \\ Charles R Greathouse IV, Mar 22 2013
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SageMath
[2^n -(n%2) for n in range(41)] # G. C. Greubel, Apr 23 2023
Formula
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
From Paul Barry, Sep 19 2003: (Start)
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+1) - a(n))/3 = A001045(n). - 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
Row sums of triangle A135221. - Gary W. Adamson, Nov 23 2007
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
a(n) = 3* A000975(n-1) + 1, n > 0. - Gary Detlefs, Dec 21 2010
a(n+2) = A001969(2^n+1) + A000069(2^n); evil + odious. - Johannes W. Meijer, Jun 24 2011, Jun 26 2011
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
a(n) = Sum_{k=0..n} Sum_{i=0..n} C(k-1,i). - Wesley Ivan Hurt, Sep 21 2017
a(n) = A026147(2^(n-1)) for n > 0. - Chunqing Liu, Dec 18 2022
Extensions
Edited and information added by Johannes W. Meijer, Jun 24 2011
Comments