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a(n) = 4*a(n-1)*a(n-2) - a(n-3), with a(1) = a(2) = a(3) = 1.
3

%I #26 Sep 04 2016 23:26:46

%S 1,1,1,3,11,131,5761,3018753,69564144001,839987873581797251,

%T 233732149587751710483796746251,

%U 785328685279672432967483833110876164468741280003,734226246973363127354668827312570246092792043625372932024478449584047744277761

%N a(n) = 4*a(n-1)*a(n-2) - a(n-3), with a(1) = a(2) = a(3) = 1.

%H Seiichi Manyama, <a href="/A276258/b276258.txt">Table of n, a(n) for n = 1..18</a>

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

%F a(n) ~ 1/4 * c^(((1+sqrt(5))/2)^n), where c = 1.41452525081158447693692520473959... . - _Vaclav Kotesovec_, Aug 26 2016

%F a(n)*a(n+1)*a(n+2) = (a(n)^2+a(n+1)^2+a(n+2)^2+1)/4. - _Seiichi Manyama_, Sep 04 2016

%t RecurrenceTable[{a[n] == 4*a[n - 1]*a[n - 2] - a[n - 3], a[1] == 1,

%t a[2] == 1, a[3] == 1}, a, {n, 1, 10}] (* _G. C. Greubel_, Aug 25 2016 *)

%o (Ruby)

%o def A(m, n)

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

%o ary = [1]

%o while ary.size < n

%o a = *a[1..-1], *a[1..-1].inject(:*) * (m + 1) - a[0]

%o ary << a[0]

%o end

%o ary

%o end

%o def A276258(n)

%o A(3, n)

%o end

%Y Cf. A001519, A064098, A276256, A276259.

%K nonn

%O 1,4

%A _Seiichi Manyama_, Aug 25 2016