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a(n) = a(n-1)*(1 + a(n-1)/a(n-4)), with a(0) = a(1) = a(2) = a(3) = 1.
1

%I #17 Sep 04 2016 18:55:40

%S 1,1,1,1,2,6,42,1806,1632624,444245153520,4698898962968253924720,

%T 12225720633546031105793020748137513851120,

%U 91550929674875028299231929179221527919681972461210779957660001348767546720

%N a(n) = a(n-1)*(1 + a(n-1)/a(n-4)), with a(0) = a(1) = a(2) = a(3) = 1.

%H Seiichi Manyama, <a href="/A276416/b276416.txt">Table of n, a(n) for n = 0..16</a>

%F 0 = a(n)*(a(n+3) - a(n+4)) + a(n+3)*a(n+3) for all n>=0.

%F A133400(n) = a(n+1)/a(n).

%t RecurrenceTable[{a[n] == a[n - 1] (1 + a[n - 1]/a[n - 4]), a[0] == a[1] == a[2] == a[3] == 1}, a, {n, 0, 12}] (* _Michael De Vlieger_, Sep 04 2016 *)

%o (Ruby)

%o def A(m, n)

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

%o ary = [1]

%o while ary.size < n + 1

%o break if a[-1] % a[0] > 0

%o a = *a[1..-1], a[-1] * (1 + a[-1] / a[0])

%o ary << a[0]

%o end

%o ary

%o end

%o def A276416(n)

%o A(4, n)

%o end

%Y Cf. A133400, A250309.

%K nonn

%O 0,5

%A _Seiichi Manyama_, Sep 02 2016