A332750 The number of flips to go from Hamiltonian cycle alpha_n to beta_n in the Cameron graph of size n using Thomason's algorithm.
11, 65, 265, 1005, 3749, 13927, 51683, 191735, 711243, 2638305, 9786545, 36302213, 134659381, 499505271, 1852863915, 6873009871, 25494729643, 94570101217, 350798151929, 1301249991357, 4826854219941, 17904723777319, 66415748007763, 246362448161159, 913856392265003
Offset: 1
Links
- Colin Barker, Table of n, a(n) for n = 1..1000
- K. Cameron, Thomason's algorithm for finding a second hamiltonian circuit through a given edge in a cubic graph is exponential on Krawczyk's graphs, Discrete Mathematics (235), 2001, pp. 69-77.
- Donald Knuth, The Art of Computer Programming, Pre-fascicle 8a, Hamiltonian paths and cycles, exercise 77, pp. 16, 26-27 (retrieved Feb 22 2020).
- Index entries for linear recurrences with constant coefficients, signature (4,-1,0,-1,0,-1).
Crossrefs
Cf. A332751 (number of flips from beta_n to gamma_n, same growth rate).
Programs
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Mathematica
LinearRecurrence[{4, -1, 0, -1, 0, -1}, {11, 65, 265, 1005, 3749, 13927}, 20] (* Jinyuan Wang, Feb 22 2020 *) -
PARI
Vec(z*(1+z)*(11+10*z+6*z^2+4*z^3+z^4)/((1-z)*(1-3*z-2*z^2-2*z^3-z^4-z^5)) + O(z^30)) \\ Jinyuan Wang, Feb 22 2020
Formula
G.f.: z(1+z)(11+10z+6z^2+4z^3+z^4)/((1-z)(1-3z-2z^2-2z^3-z^4-z^5)).
a(n) = 4*a(n-1) - a(n-2) - a(n-4) - a(n-6) for n>6. - Colin Barker, Feb 22 2020
Extensions
More terms from Jinyuan Wang, Feb 22 2020