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A112449
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a(n+2) = (a(n+1)^3 + a(n+1))/a(n) with a(0)=1, a(1)=1.
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1
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OFFSET
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0,3
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COMMENTS
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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).
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LINKS
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FORMULA
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MAPLE
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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;
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MATHEMATICA
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nxt[{a_, b_}]:={b, (b^3+b)/a}; NestList[nxt, {1, 1}, 10][[All, 1]] (* Harvey P. Dale, Jun 26 2017 *)
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PROG
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(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
A(3, 1, n)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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