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a(0) = a(1) = a(2) = a(3) = 1; for n > 3, a(n) = (a(n-1)^2+a(n-2)^2+a(n-3)^2+a(n-1)*a(n-2)*a(n-3))/a(n-4).
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%I #21 Oct 24 2018 08:08:52

%S 1,1,1,1,4,22,589,399253,41144206447,77387327118194895379,

%T 10169897514576967837097322386922878932,

%U 259050897146323086186965020577200627526185475088368701480903471601830

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

%H Seiichi Manyama, <a href="/A276124/b276124.txt">Table of n, a(n) for n = 0..15</a>

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

%t RecurrenceTable[{a[n] == (a[n - 1]^2 + a[n - 2]^2 + a[n - 3]^2 + a[n - 1] a[n - 2] a[n - 3])/a[n - 4], a[0] == a[1] == a[2] == a[3] == 1}, a, {n, 0, 11}] (* _Michael De Vlieger_, Aug 21 2016 *)

%o (Ruby)

%o def A(m, n)

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

%o ary = [1]

%o while ary.size < n + 1

%o i = a[1..-1].inject(0){|s, i| s + i * i} + a[1..-1].inject(:*)

%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 A276124(n)

%o A(4, n)

%o end # _Seiichi Manyama_, Aug 21 2016

%Y Cf. A165903, A072878.

%K nonn

%O 0,5

%A _Bruno Langlois_, Aug 21 2016