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A276175 a(n) = (a(n-1)+1)*(a(n-2)+1)*(a(n-3)+1)/a(n-4) with a(0) = a(1) = a(2) = a(3) = 1.

%I A276175 #65 Jan 12 2025 04:54:46
%S A276175 1,1,1,1,8,36,666,222111,685187756,2819713283228248,
%T A276175 644335913093223286486628176,
%U A276175 5604757351123068775966272886689217889936356651,14861563788248216173988661093334637018340529129342104300621091389266132702213641
%N A276175 a(n) = (a(n-1)+1)*(a(n-2)+1)*(a(n-3)+1)/a(n-4) with a(0) = a(1) = a(2) = a(3) = 1.
%C A276175 Conjecture: a(n) is an integer for all n >= 0. It has been checked by computer for n <= 40. A proof was proposed by 'mercio' as an answer to the MSE question, which however lacks details and heavily relies on computation. [Updated by _Max Alekseyev_, May 07 2023]
%H A276175 Seiichi Manyama, <a href="/A276175/b276175.txt">Table of n, a(n) for n = 0..16</a>
%H A276175 Math.StackExchange.com, <a href="http://math.stackexchange.com/q/1905063">Is A276175 integer-only?</a> Aug 27 2016.
%F A276175 Let b(n) = (b(n-1)*b(n-2)*b(n-3)+1)/b(n-4) with b(0) = 1/2, b(1) = 4, b(2) = b(3) = 1/2, then a(n) = b(n)*b(n+1)*b(n+2). - _Seiichi Manyama_, Sep 03 2016
%F A276175 The sequence 4*b(n) is given by A362884. Correspondingly, a(n) = A362884(n) * A362884(n+1) * A362884(n+2) / 64. - _Max Alekseyev_, May 07 2023
%F A276175 a(n) = a(3-n), 0 = a(n)*a(n+4)*(a(n+4)+1) - a(n+5)*a(n+1)*(a(n+1)+1) for all n in Z. - _Michael Somos_, Feb 23 2019
%F A276175 log(a(n)) ~ c * A289917^n, where c = 0.26774381278698... - _Vaclav Kotesovec_, Aug 27 2021
%t A276175 RecurrenceTable[{a[n] == (a[n - 1] + 1) (a[n - 2] + 1) (a[n - 3] + 1)/a[n - 4],
%t A276175 a[0] == a[1] == a[2] == a[3] == 1}, a, {n, 0, 12}] (* _Michael De Vlieger_, Aug 25 2016 *)
%t A276175 a[ n_] := With[{m = Max[3 - n, n]}, If[ m < 4, 1, (a[m - 1] + 1) (a[m - 2] + 1) (a[m - 3] + 1)/a[m - 4]]]; (* _Michael Somos_, Jun 02 2019 *)
%o A276175 (PARI) a(n) = if (n <=3, 1, (a(n-1)+1)*(a(n-2)+1)*(a(n-3)+1)/a(n-4)); \\ _Michel Marcus_, Aug 23 2016
%o A276175 (Ruby)
%o A276175 def A(m, n)
%o A276175   a = Array.new(m, 1)
%o A276175   ary = [1]
%o A276175   while ary.size < n + 1
%o A276175     i = a[1..-1].inject(1){|s, i| s * (i + 1)}
%o A276175     break if i % a[0] > 0
%o A276175     a = *a[1..-1], i / a[0]
%o A276175     ary << a[0]
%o A276175   end
%o A276175   ary
%o A276175 end
%o A276175 def A276175(n)
%o A276175   A(4, n)
%o A276175 end # _Seiichi Manyama_, Aug 23 2016
%Y A276175 Cf. A076839, A276123, A362884.
%K A276175 nonn
%O A276175 0,5
%A A276175 _Bruno Langlois_, Aug 23 2016