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a(n) = (a(n-1) * a(n-6) + a(n-2) * a(n-3) * a(n-4) * a(n-5)) / a(n-7), with a(0) = a(1) = a(2) = a(3) = a(4) = a(5) = a(6) = 1.
3

%I #32 Nov 27 2016 06:59:30

%S 1,1,1,1,1,1,1,2,3,5,11,41,371,7507,429563,419408854,9811194604889,

%T 45615501062085527113,323645006689468299915979814409,

%U 217332607887523478570092794860281557159140687,8092345737591989154121803868154457767563221634145658745306515944569

%N a(n) = (a(n-1) * a(n-6) + a(n-2) * a(n-3) * a(n-4) * a(n-5)) / a(n-7), with a(0) = a(1) = a(2) = a(3) = a(4) = a(5) = a(6) = 1.

%C This sequence is one generalization of Dana Scott's sequence (A048736).

%C a(n) is integer for all n.

%C The recursion exhibits the Laurent phenomenon. See A278706 for the exponents of the denominator of the Laurent polynomial. - _Michael Somos_, Nov 26 2016

%H Seiichi Manyama, <a href="/A276532/b276532.txt">Table of n, a(n) for n = 0..26</a>

%F a(n) * a(n-7) = a(n-1) * a(n-6) + a(n-2) * a(n-3) * a(n-4) * a(n-5).

%F a(6-n) = a(n) for all n in Z.

%o (Ruby)

%o def A(k, n)

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

%o ary = [1]

%o while ary.size < n + 1

%o i = a[-1] * a[1] + a[2..-2].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 A276532(n)

%o A(7, n)

%o end

%Y Cf. A048736, A006721, A276531, A278706.

%K nonn

%O 0,8

%A _Seiichi Manyama_, Nov 16 2016