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a(0) = a(1) = a(2) = a(3) = a(4) = 1; for n > 4, a(n) = ( a(n-1)+a(n-3)+1 )*( a(n-2)+a(n-4)+1 ) / a(n-5).
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%I #33 Dec 14 2021 16:54:01

%S 1,1,1,1,1,9,33,385,13825,5474305,8430415841,1398605982547209,

%T 30625582893143965429313,3098236789946633955987434183345281,

%U 17332850039068891068793031113694107707268123637761

%N a(0) = a(1) = a(2) = a(3) = a(4) = 1; for n > 4, a(n) = ( a(n-1)+a(n-3)+1 )*( a(n-2)+a(n-4)+1 ) / a(n-5).

%H Seiichi Manyama, <a href="/A276284/b276284.txt">Table of n, a(n) for n = 0..21</a>

%F a(n) = (8-4*(-1)^n)*a(n-1)*a(n-3) - a(n-2) - a(n-4) - 1 for n>3.

%t RecurrenceTable[{a[n] == (a[n - 1] + a[n - 3] + 1) (a[n - 2] + a[n - 4] + 1)/a[n - 5], a[0] == a[1] == a[2] == a[3] == a[4] == 1}, a, {n, 0, 14}] (* _Michael De Vlieger_, Aug 27 2016 *)

%t nxt[{a_,b_,c_,d_,e_}]:={b,c,d,e,(e+c+1) (d+b+1)/a}; NestList[nxt,{1,1,1,1,1},15][[All,1]] (* _Harvey P. Dale_, Dec 14 2021 *)

%o (Ruby)

%o def A(m, n)

%o a = Array.new(2 * m + 1, 1)

%o ary = [1]

%o while ary.size < n + 1

%o i = (1..m).inject(1){|s, i| s + a[2 * i - 1]} * (1..m).inject(1){|s, i| s + a[2 * i]}

%o break if i % a[0] > 0

%o a = *a[1..-1], i / a[0]

%o ary << a[0]

%o end

%o ary

%o end

%o def A276284(n)

%o A(2, n)

%o end

%Y Cf. A276123.

%K nonn

%O 0,6

%A _Seiichi Manyama_, Aug 27 2016