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a(n) = A058094(n) - 3*A058094(n-1) + A058094(n-2) for n >=4.
5

%I #17 Aug 21 2019 06:52:11

%S 0,0,0,1,5,20,75,271,957,3337,11559,39896,137423,472808,1625632,

%T 5587228,19198971,65963978,226623902,778551761,2674604282,9188106871,

%U 31563807424,108430368827,372487292867,1279591674070,4395730089428

%N a(n) = A058094(n) - 3*A058094(n-1) + A058094(n-2) for n >=4.

%C A recurrence relation follows in a straightforward manner from the above formula and the recurrence relation for A058094.

%H Colin Barker, <a href="/A092490/b092490.txt">Table of n, a(n) for n = 1..1000</a>

%H 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.

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (6,-11,9,-4,-4,1).

%F G.f.: x^4*(1 - x + x^2 + x^3) / (1 - 6*x + 11*x^2 - 9*x^3 + 4*x^4 + 4*x^5 - x^6). - _R. J. Mathar_, Dec 02 2007

%F 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. - _Colin Barker_, Aug 21 2019

%p b[1]:=1:b[2]:=2:b[3]:=5:b[4]:=14:b[5]:=42:b[6]:=132: for n from 6 to 32 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:for n from 4 to 32 do a[n]:=b[n]-3*b[n-1]+b[n-2] od: seq(a[n],n=1..32); # _Emeric Deutsch_, Apr 12 2005

%o (PARI) concat([0,0,0], Vec(x^4*(1 - x + x^2 + x^3) / (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

%Y Cf. A058094, A092489, A092491, A092492.

%Y Cf. A058094.

%K nonn,easy

%O 1,5

%A _N. J. A. Sloane_, Apr 04 2004

%E Edited by _Emeric Deutsch_, Apr 12 2005