Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).
%I #30 Aug 21 2019 03:08:46
%S 1,1,2,5,14,42,132,429,1426,4806,16329,55740,190787,654044,2244153,
%T 7704047,26455216,90860572,312090478,1072034764,3682565575,
%U 12650266243,43456340025,149282561256,512821712570,1761669869321,6051779569463,20789398928496,71416886375493
%N Number of 321-hexagon-avoiding permutations in S_n, i.e., permutations of 1..n with no submatrix equivalent to 321, 56781234, 46781235, 56718234 or 46718235.
%C If y is 321-hexagon avoiding, there are simple explicit formulas for all the Kazhdan-Lusztig polynomials P_{x,y} and the Kazhdan-Lusztig basis element C_y is the product of C_{s_i}'s corresponding to any reduced word for y.
%H Alois P. Heinz, <a href="/A058094/b058094.txt">Table of n, a(n) for n = 0..1000</a>
%H S. C. Billey and G. S. Warrington, <a href="https://arxiv.org/abs/math/0005052">Kazhdan-Lusztig Polynomials for 321-hexagon-avoiding permutations</a>, arXiv:math/0005052 [math.CO], 2000; J. Alg. Combinatorics, Vol. 13, No. 2 (March 2001), 111-136, DOI:<a href="https://doi.org/10.1023/A:1011279130416">10.1023/A:1011279130416</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 a(n+1) = 6a(n) - 11a(n-1) + 9a(n-2) - 4a(n-3) - 4a(n-4) + a(n-5) for n >= 5.
%F O.g.f.: 1 -x*(1-4*x+4*x^2-3*x^3-x^4+x^5)/(-1+6*x-11*x^2+9*x^3-4*x^4 -4*x^5 +x^6). - _R. J. Mathar_, Dec 02 2007
%e Since the Catalan numbers count 321-avoiding permutations in S_n, a(8) = 1430 - 4 = 1426 subtracting the four forbidden hexagon patterns.
%p a[0]:=1: a[1]:=1: a[2]:=2: a[3]:=5: a[4]:=14: a[5]:=42: for n from 5 to 35 do a[n+1]:=6*a[n]-11*a[n-1]+9*a[n-2]-4*a[n-3]-4*a[n-4]+a[n-5] od: seq(a[n], n=0..35);
%t LinearRecurrence[{6,-11,9,-4,-4,1},{1,2,5,14,42,132},40] (* _Harvey P. Dale_, Nov 09 2012 *)
%Y Cf. A000108, A092489, A092490, A092491, A092492, A092493.
%K nice,nonn,easy
%O 0,3
%A _Sara Billey_, Dec 03 2000
%E More terms from _Emeric Deutsch_, May 04 2004
%E a(0) prepended by _Alois P. Heinz_, Sep 21 2014