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A276095 A nonlinear recurrence of order 4: a(1)=a(2)=a(3)=a(4)=1; a(n)=(a(n-1)+a(n-2)+a(n-3))^2/a(n-4). 5

%I #35 Aug 26 2016 09:44:58

%S 1,1,1,1,9,121,17161,298978681,9933176210033041,

%T 815437979830770470704295274609,

%U 38747106750801481775941360512378545527545442200632960401

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

%C All terms are perfect squares.

%H Seiichi Manyama, <a href="/A276095/b276095.txt">Table of n, a(n) for n = 1..15</a>

%F a(n) = A072878(n)^2.

%F a(n) = 16*a(n-1)*a(n-2)*a(n-3) - 2a(n-1) - 2a(n-2) - 2a(n-3) - a(n-4).

%F a(n)*a(n-1)*a(n-2)*a(n-3) = ((a(n) + a(n-1) + a(n-2) + a(n-3))/4)^2.

%t RecurrenceTable[{a[n] == (a[n - 1] + a[n - 2] + a[n - 3])^2/a[n - 4], a[1] == a[2] == a[3] == a[4] == 1}, a, {n, 1, 12}] (* _Michael De Vlieger_, Aug 18 2016 *)

%o (Ruby)

%o def A(m, n)

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

%o ary = [1]

%o while ary.size < n

%o i = a[1..-1].inject(:+)

%o j = i * i

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

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

%o ary << a[0]

%o end

%o ary

%o end

%o def A276095(n)

%o A(4, n)

%o end

%Y Cf. A072878, A072882.

%K nonn

%O 1,5

%A _Seiichi Manyama_, Aug 18 2016

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Last modified April 25 03:15 EDT 2024. Contains 371964 sequences. (Running on oeis4.)