A112449 a(n+2) = (a(n+1)^3 + a(n+1))/a(n) with a(0)=1, a(1)=1.

1, 1, 2, 10, 505, 12878813, 4229958765311886322, 5876687051603582015287706866081267480733704277890

Offset: 0

Author

Andrew Hone, Dec 12 2005

Keywords

• nonn

Comments

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.

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

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..10
• S. Fomin and A. Zelevinsky, The Laurent Phenomenon, Advances in Applied Mathematics, 28 (2002), 119-144.

Crossrefs

Cf. A101879, A112373, A202543.

Programs

Maple

a[0]:=1; a[1]:=1; f(x):=x^3+x;
for n from 0 to 8 do a[n+2]:=simplify(subs(x=a[n+1],f(x))/a[n]) od;
s[3]:=ln(10); s[4]:=ln(505);
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));
# s[n]=ln(a[n]); ln(s[n])/n converges slowly to 0.962...
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;
# N. J. A. Sloane

Mathematica

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

Ruby

def A(l, m, n)
  a = Array.new(2 * m, 1)
  ary = [1]
  while ary.size < n + 1
    i = a[1..-1].inject(:*) + a[m] ** l
    break if i % a[0] > 0
    a = *a[1..-1], i / a[0]
    ary << a[0]
  end
  ary
end
def A112449(n)
  A(3, 1, n)
end # Seiichi Manyama, Nov 20 2016

Formula

a(1-n) = a(n). - Seiichi Manyama, Nov 20 2016