A294619 a(0) = 0, a(1) = 1, a(2) = 2 and a(n) = 1 for n > 2.
0, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
Offset: 0
Examples
For n = 2, the shadow of the Hopf link yields 2 two-component state diagrams (see example in A300453). Thus a(2) = 2.
References
- V. I. Arnold, Topological Invariants of Plane Curves and Caustics, American Math. Soc., 1994.
- L. H. Kauffman, Knots and Physics, World Scientific Publishers, 1991.
- V. Manturov, Knot Theory, CRC Press, 2004.
Links
- I. Altintas, An oriented state model for the Jones polynomial and its applications to alternating links, Appl. Math. Comput. 194 (2007) 168-178.
- J. A. Baldwin and A. S. Levine, A combinatorial spanning tree model for knot Floer homology, Advances in Mathematics, Vol. 231 (2012), 1886-1939.
- A. Banerjee, Knot theory [Foil knot family].
- D. Denton and P. Doyle, Shadow movies not arising from knots, arXiv preprint, arXiv:1106.3545 [math.GT], 2011.
- L. H. Kauffman, State models and the Jones polynomial, Topology, Vol. 26 (1987), 395-407.
- Franck Ramaharo, A generating polynomial for the two-bridge knot with Conway's notation C(n,r), arXiv:1902.08989 [math.CO], 2019.
- Index entries for linear recurrences with constant coefficients, signature (1).
Crossrefs
Programs
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Mathematica
CoefficientList[Series[(x + x^2 - x^3)/(1 - x), {x, 0, 100}], x] (* Wesley Ivan Hurt, Nov 05 2017 *) f[n_] := If[n > 2, 1, n]; Array[f, 105, 0] (* Robert G. Wilson v, Dec 27 2017 *) PadRight[{0,1,2},120,{1}] (* Harvey P. Dale, Feb 20 2023 *) -
Maxima
makelist((1 + (-1)^((n + 1)!))/2 + kron_delta(n, 2), n, 0, 100);
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PARI
a(n) = if(n>2, 1, n);
Formula
a(n) = ((-1)^2^(n^2 + 3*n + 2) + (-1)^2^(n^2 - n) - (-1)^2^(n^2 - 3*n + 2) + 1)/2.
a(n) = (1 + (-1)^((n + 1)!))/2 + Kronecker(n, 2).
a(n) = min(n, 3) - 2*(max(n - 2, 0) - max(n - 3, 0)).
a(n) = floor(F(n+1)/F(n)) for n > 0, with a(0) = 0, where F(n) = A000045(n) is the n-th Fibonacci number.
a(n) = a(n-1) for n > 3, with a(0) = 0, a(1) = 1, a(2) = 2 and a(3) = 1.
a(n+1) = 2^A185012(n+1), with a(0) = 0.
G.f.: (x + x^2 - x^3)/(1 - x).
E.g.f.: (2*exp(x) - 2 + x^2)/2.
Comments