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A276531 a(n) = (a(n-1) * a(n-5) + a(n-2) * a(n-3) * a(n-4)) / a(n-6), with a(0) = a(1) = a(2) = a(3) = a(4) = a(5) = 1.

%I A276531 #41 Jan 12 2025 04:52:48
%S A276531 1,1,1,1,1,1,2,3,5,11,41,247,1498,39629,3121233,1344630757,
%T A276531 4527359876765,673384475958949877,12684198948982702826816701,
%U A276531 103442271685605704255863097581658042,12389248756108266360505757651017660004796444483503,657084395567781339286109602463271066924826185667810218784212689809097
%N A276531 a(n) = (a(n-1) * a(n-5) + a(n-2) * a(n-3) * a(n-4)) / a(n-6), with a(0) = a(1) = a(2) = a(3) = a(4) = a(5) = 1.
%C A276531 This sequence is the generalization of Dana Scott's sequence (A048736).
%C A276531 Conjecture: a(n) is an integer for all n. It has been checked by computer for 0 <= n <= 50.
%C A276531 The recursion has the Laurent property. If a(0), ..., a(5) are variables, then a(n) is a Laurent polynomial (a rational function with a monomial denominator). - _Michael Somos_, Nov 21 2016
%H A276531 Seiichi Manyama, <a href="/A276531/b276531.txt">Table of n, a(n) for n = 0..28</a>
%F A276531 a(n) * a(n-6) = a(n-1) * a(n-5) + a(n-2) * a(n-3) * a(n-4).
%F A276531 a(5-n) = a(n) for all n in Z.
%t A276531 RecurrenceTable[{a[n] == (a[n - 1] a[n - 5] + a[n - 2] a[n - 3] a[n - 4])/a[n - 6], a[0] == a[1] == a[2] == a[3] == a[4] == a[5] == 1}, a, {n, 0, 21}] (* _Michael De Vlieger_, Nov 21 2016 *)
%t A276531 nxt[{a_,b_,c_,d_,e_,f_}]:={b,c,d,e,f,(b*f+d*e*c)/a}; NestList[nxt,{1,1,1,1,1,1},30][[All,1]] (* _Harvey P. Dale_, Nov 21 2021 *)
%o A276531 (Ruby)
%o A276531 def A(k, n)
%o A276531   a = Array.new(k, 1)
%o A276531   ary = [1]
%o A276531   while ary.size < n + 1
%o A276531     i = a[-1] * a[1] + a[2..-2].inject(:*)
%o A276531     break if i % a[0] > 0
%o A276531     a = *a[1..-1], i / a[0]
%o A276531     ary << a[0]
%o A276531   end
%o A276531   ary
%o A276531 end
%o A276531 def A276531(n)
%o A276531   A(6, n)
%o A276531 end
%o A276531 (Magma) I:=[1,1,1,1,1,1]; [n le 6 select I[n] else (Self(n-1)*Self(n-5) + Self(n-2)*Self(n-3)*Self(n-4))/Self(n-6): n in [1..30]]; // _G. C. Greubel_, Jul 30 2018
%o A276531 (GAP) a:=[1,1,1,1,1,1];; for n in [7..25] do a[n]:=(a[n-1]*a[n-5]+a[n-2]*a[n-3]*a[n-4])/a[n-6]; od; a; # _Muniru A Asiru_, Jul 30 2018
%Y A276531 Cf. A048736, A006721.
%K A276531 nonn
%O A276531 0,7
%A A276531 _Seiichi Manyama_, Nov 16 2016