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A276097 A nonlinear recurrence of order 5: a(1)=a(2)=a(3)=a(4)=a(5)=1; a(n)=(a(n-1)+a(n-2)+a(n-3)+a(n-4))^2/a(n-5).

%I A276097 #23 Aug 25 2016 13:16:10
%S A276097 1,1,1,1,1,16,361,143641,20741472361,430214650013601071641,
%T A276097 11567790319010747187536221088708755344001,
%U A276097 370675271093071368960746074163948008803845834307486807769098691609909105887376
%N A276097 A nonlinear recurrence of order 5: a(1)=a(2)=a(3)=a(4)=a(5)=1; a(n)=(a(n-1)+a(n-2)+a(n-3)+a(n-4))^2/a(n-5).
%C A276097 All terms are perfect squares.
%H A276097 Seiichi Manyama, <a href="/A276097/b276097.txt">Table of n, a(n) for n = 1..15</a>
%F A276097 a(n) = A072879(n)^2.
%F A276097 a(n) = 25*a(n-1)*a(n-2)*a(n-3)*a(n-4) - 2a(n-1) - 2a(n-2) - 2a(n-3) - 2a(n-4) - a(n-5).
%F A276097 a(n)*a(n-1)*a(n-2)*a(n-3)*a(n-4) = ((a(n) + a(n-1) + a(n-2) + a(n-3) + a(n-4))/5)^2.
%t A276097 RecurrenceTable[{a[1] == a[2] == a[3] == a[4] == a[5] == 1, a[n] == (a[n-1] + a[n-2] + a[n-3] + a[n-4])^2 / a[n-5]}, a, {n, 15}] (* _Vincenzo Librandi_, Aug 21 2016 *)
%o A276097 (Ruby)
%o A276097 def A(m, n)
%o A276097   a = Array.new(m, 1)
%o A276097   ary = [1]
%o A276097   while ary.size < n
%o A276097     i = a[1..-1].inject(:+)
%o A276097     j = i * i
%o A276097     break if j % a[0] > 0
%o A276097     a = *a[1..-1], j / a[0]
%o A276097     ary << a[0]
%o A276097   end
%o A276097   ary
%o A276097 end
%o A276097 def A276097(n)
%o A276097   A(5, n)
%o A276097 end
%Y A276097 Cf. A072879, A072882, A276095.
%K A276097 nonn
%O A276097 1,6
%A A276097 _Seiichi Manyama_, Aug 18 2016