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A368881 a(n) = binomial(n+3, 4) + binomial(n+1, 3) + 1.

%I A368881 #31 Jan 21 2025 18:52:14
%S A368881 1,2,7,20,46,91,162,267,415,616,881,1222,1652,2185,2836,3621,4557,
%T A368881 5662,6955,8456,10186,12167,14422,16975,19851,23076,26677,30682,35120,
%U A368881 40021,45416,51337,57817,64890,72591,80956,90022,99827,110410,121811,134071
%N A368881 a(n) = binomial(n+3, 4) + binomial(n+1, 3) + 1.
%C A368881 The number of bigrassmannian permutations in the type B hyperoctahedral group of order 2^n*n!, i.e., those with a unique left and right type B descent or the identity. This can be characterized by avoiding 18 signed permutation patterns.
%H A368881 Harvey P. Dale, <a href="/A368881/b368881.txt">Table of n, a(n) for n = 0..1000</a>
%H A368881 Joshua Swanson, <a href="/A368881/a368881.pdf">Bigrassmannians and pattern avoidance notes</a>.
%H A368881 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).
%F A368881 a(n) = (1/24)*(n^4 + 10*n^3 + 11*n^2 + 2*n + 24).
%F A368881 G.f.: (x^4 - 5x^3 + 7x^2 - 3x + 1)/(1-x)^5.
%F A368881 E.g.f.: exp(x)*(24 + 24*x + 48*x^2 + 16*x^3 + x^4)/24. - _Stefano Spezia_, Jan 09 2024
%e A368881 For n=2, all eight 2 X 2 signed permutation matrices are bigrassmannian except the negative of the identity matrix, or equivalently the one with window notation [-1 -2], so a(2) = 7.
%t A368881 Table[Binomial[n + 3, 4] + Binomial[n + 1, 3] + 1, {n, 0, 20}]
%t A368881 LinearRecurrence[{5,-10,10,-5,1},{1,2,7,20,46},50] (* _Harvey P. Dale_, Jan 21 2025 *)
%o A368881 (Python)
%o A368881 def A368881(n): return 1+(n*(n*(n*(n + 10) + 11) + 2))//24 # _Chai Wah Wu_, Jan 27 2024
%Y A368881 Cf. A050407.
%Y A368881 It appears that this is equal to {A005712}+1, also ({A212039}+2)/3 .
%K A368881 nonn,easy
%O A368881 0,2
%A A368881 _Joshua Swanson_, Jan 08 2024