A112449 - OEIS

%I #23 Jun 26 2017 08:24:46

%S 1,1,2,10,505,12878813,4229958765311886322,

%T 5876687051603582015287706866081267480733704277890

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

%C A second-order recurrence with the Laurent property. This property is satisfied by any second-order recurrence of the form a(n+2) = f(a(n+1))/a(n) with f being a polynomial of the form f(x) = x*p(x) where p is a polynomial of degree d with integer coefficients such that p(0)=1 and p has the reciprocal property x^d*p(1/x) = p(x). Hence if a(0) = a(1) = 1 then a(n) is an integer for all n.

%C As n tends to infinity, log(log(a(n)))/n tends to log((3+sqrt(5))/2) or about 0.962 (A202543).

%H Seiichi Manyama, <a href="/A112449/b112449.txt">Table of n, a(n) for n = 0..10</a>

%H S. Fomin and A. Zelevinsky, <a href="http://dx.doi.org/10.1006/aama.2001.0770">The Laurent Phenomenon</a>, Advances in Applied Mathematics, 28 (2002), 119-144.

%F a(1-n) = a(n). - _Seiichi Manyama_, Nov 20 2016

%p a[0]:=1; a[1]:=1; f(x):=x^3+x;

%p for n from 0 to 8 do a[n+2]:=simplify(subs(x=a[n+1],f(x))/a[n]) od;

%p s[3]:=ln(10); s[4]:=ln(505);

%p for n from 3 to 10000 do s[n+2]:=evalf(3*s[n+1]+ln(1+exp(-2*s[n+1]))-s[n]): od: print(evalf(ln(s[10002])/(10002))): evalf(ln((3+sqrt(5))/2));

%p # s[n]=ln(a[n]); ln(s[n])/n converges slowly to 0.962...

%p f:=proc(n) option remember; local i,j,k,t1,t2,t3; if n <= 1 then RETURN(1); fi; (f(n-1)^3+f(n-1))/f(n-2); end;

%p # _N. J. A. Sloane_

%t nxt[{a_,b_}]:={b,(b^3+b)/a}; NestList[nxt,{1,1},10][[All,1]] (* _Harvey P. Dale_, Jun 26 2017 *)

%o (Ruby)

%o def A(l, m, n)

%o   a = Array.new(2 * m, 1)

%o   ary = [1]

%o   while ary.size < n + 1

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

%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 A112449(n)

%o   A(3, 1, n)

%o end # _Seiichi Manyama_, Nov 20 2016

%Y Cf. A101879, A112373, A202543.

%K nonn

%O 0,3

%A _Andrew Hone_, Dec 12 2005