A194952 Number of Hamiltonian cycles in C_3 X C_n.
48, 126, 390, 1014, 2982, 8094, 23646, 66726, 196086, 568302, 1682382, 4954998, 14750310, 43833150, 130942398, 390959430, 1170256854, 3502513038, 10495480494, 31450265622, 94296270918, 282731526366
Offset: 3
Links
- Vincenzo Librandi, Table of n, a(n) for n = 3..2000
- Artem M. Karavaev, Hamilton Cycles page
- Eric Weisstein's World of Mathematics, Hamiltonian Cycle
- Eric Weisstein's World of Mathematics, Torus Grid Graph
- Index entries for linear recurrences with constant coefficients, signature (5,-1,-25,26,20,-24).
Crossrefs
Row 3 of A270273.
Programs
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Magma
[3^n + 3/4*n*2^n + (2^n-(-2)^n)/2 + (-1)^n - 4: n in [3..40]]; // Vincenzo Librandi, Sep 19 2011
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Maple
C3xCn := n->3^n+3/4*n*2^n+(2^n-(-2)^n)/2+(-1)^n-4:seq(C3xCn(n),n=3..16);
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PARI
a(n)=([0,1,0,0,0,0; 0,0,1,0,0,0; 0,0,0,1,0,0; 0,0,0,0,1,0; 0,0,0,0,0,1; -24,20,26,-25,-1,5]^(n-3)*[48;126;390;1014;2982;8094])[1,1] \\ Charles R Greathouse IV, Jul 08 2024
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Python
# Using graphillion from graphillion import GraphSet def make_CnXCk(n, k): grids = [] for i in range(1, k + 1): for j in range(1, n): grids.append((i + (j - 1) * k, i + j * k)) grids.append((i + (n - 1) * k, i)) for i in range(1, k * n, k): for j in range(1, k): grids.append((i + j - 1, i + j)) grids.append((i + k - 1, i)) return grids def A194952(n): universe = make_CnXCk(n, 3) GraphSet.set_universe(universe) cycles = GraphSet.cycles(is_hamilton=True) return cycles.len() print([A194952(n) for n in range(3, 30)]) # Seiichi Manyama, Nov 22 2020
Formula
a(n) = 3^n + 3/4*n*2^n + (2^n-(-2)^n)/2 + (-1)^n - 4, n>=3.
a(n) = 5*a(n-1)-a(n-2)-25*a(n-3)+26*a(n-4)+20*a(n-5)-24*a(n-6), for n>=9, with a(3)=48, a(4)=126, a(5)=390, a(6)=1014, a(7)=2982, a(8)=8094.
G.f.: -6*x^3*(-8+19*x+32*x^2-65*x^3-34*x^4+48*x^5) / ( (x-1)*(3*x-1)*(2*x+1)*(1+x)*(-1+2*x)^2 ). - R. J. Mathar, Sep 18 2011
Extensions
More terms from Alexander R. Povolotsky, Sep 07 2011
Comments