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%I #33 Sep 08 2022 08:46:25
%S 1,2,6,15,37,93,238,616,1604,4189,10955,28667,75036,196430,514242,
%T 1346283,3524593,9227481,24157834,63246004,165580160,433494457,
%U 1134903191,2971215095,7778742072,20365011098,53316291198,139583862471,365435296189,956722026069
%N Number of permutations w in S_n that form Boolean intervals [s, w] in the Bruhat order for every simple reflection s in the support of w.
%H Colin Barker, <a href="/A331347/b331347.txt">Table of n, a(n) for n = 1..1000</a>
%H B. E. Tenner, <a href="https://arxiv.org/abs/2001.05011">Interval structures in the Bruhat and weak orders</a>, arXiv:2001.05011 [math.CO], 2020.
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (5,-8,5,-1).
%F a(n) = Fibonacci(2n-1) + n - 2 = A001519(n) + n - 2.
%F From _Colin Barker_, Jan 14 2020: (Start)
%F G.f.: x*(1 - 3*x + 4*x^2 - 4*x^3 + x^4) / ((1 - x)^2*(1 - 3*x + x^2)).
%F a(n) = 5*a(n-1) - 8*a(n-2) + 5*a(n-3) - a(n-4) for n>5.
%F (End)
%F E.g.f.: 1 + exp((1/2)*(3-sqrt(5))*x)*(3 + sqrt(5) + 2*exp(sqrt(5)*x))/(5 + sqrt(5)) + exp(x)*(x - 2). - _Stefano Spezia_, Jan 15 2020
%e a(4) = 15 because the permutations with this property in S_4 are all permutations of length < 4.
%t Join[{1},Table[Fibonacci[2n-1]+n-2,{n,2,30}]] (* or *) LinearRecurrence[ {5,-8,5,-1},{1,2,6,15,37},30] (* _Harvey P. Dale_, Feb 21 2020 *)
%o (PARI) Vec(x*(1 - 3*x + 4*x^2 - 4*x^3 + x^4) / ((1 - x)^2*(1 - 3*x + x^2)) + O(x^30)) \\ _Colin Barker_, Jan 14 2020
%o (Magma) R<x>:=PowerSeriesRing(Integers(), 30); Coefficients(R!( x*(1-3*x+4*x^2-4*x^3+x^4)/((1-x)^2*(1-3*x+x^2)))); // _Marius A. Burtea_, Jan 15 2020
%Y Cf. A001519.
%K easy,nonn
%O 1,2
%A _Bridget Tenner_, Jan 14 2020