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A362388 a(n) = a(n-1) + a(n-3) + a(n-4), where a(0) = 1, a(1) = 2, a(2) = 5, a(3) = 7.

%I A362388 #45 Jun 22 2023 13:32:07
%S A362388 1,2,5,7,10,17,29,46,73,119,194,313,505,818,1325,2143,3466,5609,9077,
%T A362388 14686,23761,38447,62210,100657,162865,263522,426389,689911,1116298,
%U A362388 1806209,2922509,4728718,7651225,12379943,20031170,32411113,52442281,84853394
%N A362388 a(n) = a(n-1) + a(n-3) + a(n-4), where a(0) = 1, a(1) = 2, a(2) = 5, a(3) = 7.
%C A362388 For n >= 3, a(n) is also the number of ways to tile this "central staircase" figure of length n with squares and dominoes; this is the picture for length n=10:
%C A362388              _
%C A362388    _________|_|_______
%C A362388   |_|_|_|_|_|_|_|_|_|_|
%C A362388           |_|
%H A362388 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,1,1).
%F A362388 a(n) = a(n-1) + a(n-3) + a(n-4), where a(0) = 1, a(1) = 2, a(2) = 5, a(3) = 7.
%F A362388 G.f.: (1 + x + 3*x^2 + x^3)/((1 +x^2)*(1-x-x^2)).
%F A362388 a(2*n) = F(n+2)^2 + F(n)^2, a(2*n+1) = F(n+2)^2 + F(n+1)*L(n+1) for F(n) and L(n) the Fibonacci and Lucas numbers.
%F A362388 Sum_{k=0..n} a(k) = A295681(n+5) - 3.
%F A362388 5*a(n) = 3*A000032(n+2) -2*A000034(n+1)*(-1)^floor(n/2). - _R. J. Mathar_, Jun 22 2023
%F A362388 a(n)+a(n+2) = 3*A000045(n+3). - _R. J. Mathar_, Jun 22 2023
%e A362388 Here is one of the a(10)=194 tilings for length n=10:
%e A362388              _
%e A362388    _________|_|_______
%e A362388   |___|___| |___|_|___|
%e A362388           |_|
%t A362388 LinearRecurrence[{1, 0, 1, 1}, { 1, 2, 5, 7}, 50]
%Y A362388 Cf. A000032, A000045, A195971, A295681, A347902.
%K A362388 nonn,easy
%O A362388 0,2
%A A362388 _Greg Dresden_ and Jiaqi Wang, Jun 18 2023