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From Colin Barker, Aug 21 2019: (Start)
G.f.: x^6*(1 - x) / (1 - 6*x + 11*x^2 - 9*x^3 + 4*x^4 + 4*x^5 - x^6).
a(n) = 6*a(n-1) - 11*a(n-2) + 9*a(n-3) - 4*a(n-4) - 4*a(n-5) + a(n-6) for n>7.
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(PARI) concat([0, 0, 0, 0, 0], Vec(x^6*(1 - x) / (1 - 6*x + 11*x^2 - 9*x^3 + 4*x^4 + 4*x^5 - x^6) + O(x^30))) \\ Colin Barker, Aug 21 2019
Colin Barker, <a href="/A092492/b092492.txt">Table of n, a(n) for n = 1..1000</a>
<a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (6,-11,9,-4,-4,1).
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Z. Stankova and J. West, Explicit enumeration of 321, hexagon-avoiding permutations, Discrete Math., 280 (2004), 165-189.
Z. Stankova and J. West, <a href="https://doi.org/10.1016/j.disc.2003.06.003">Explicit enumeration of 321, hexagon-avoiding permutations</a>, Discrete Math., 280 (2004), 165-189.
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a(n): = 2*A058094(n-3) - 5*A058094(n-4) + A058094(n-5) for n >= 6. - Emeric Deutsch, Jun 08 2004
b[1]:=1: b[2]:=2: b[3]:=5: b[4]:=14: b[5]:=42: b[6]:=132: for n from 6 to 45 do b[n+1]:=6*b[n]-11*b[n-1]+9*b[n-2]-4*b[n-3]-4*b[n-4]+b[n-5] od: a[1]:=0: a[2]:=0: a[3]:=0: a[4]:=0: a[5]:=0: for n from 6 to 40 do a[n]:=2*b[n-3]-5*b[n-4]+b[n-5] od: seq(a[n], n=1..40); (# _Emeric Deutsch)_, Jun 08 2004
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