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A112373
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a(n+2) =(a(n+1)^3+a(n+1)^2)/a(n) with a(0)=1, a(1)=1.
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5
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1, 1, 2, 12, 936, 68408496, 342022190843338960032, 584861200495456320274313200204390612579749188443599552, 584930341312971177408362523916802205808200279391306193256994497845597705097714126419509918543989061579501269006337310118967798603186717516416
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,3
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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) where f is a polynomial of the form f(x)=x^m*p(x) with m a positive integer and p arbitrary. So if p has integer coefficients and 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.
The Laurent property is satisfied by any second order recurrence of the form a(n+2)=f(a(n+1))/a(n) where f is a polynomial of the form f(x)=x^m*p(x) with m a positive integer greater than or equal to 2 and p arbitrary. In that case a(0)=a(1)=1 generates a sequence of integers and the ratios a(n+1)/a(n) and a(n+1)*a(n-1)/a(n)^2 are integers for all n. - Andrew Hone (anwh(AT)kent.ac.uk), Dec 12 2005
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REFERENCES
| S. Fomin and A. Zelevinsky, The Laurent phenomenon, Advances in Applied Math. 28 (2002), 119-144.
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MAPLE
| a[0]:=1; a[1]:=1; f(x):=x^3+x^2; for n from 0 to 8 do a[n+2]:=simplify(subs(x=a[n+1], f(x))/a[n]) od; s[3]:=ln(2^2*3); s[4]:=ln(2^3*3^2*13); for n from 3 to 10000 do s[n+2]:=evalf(3*s[n+1]+ln(1+exp(-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...
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CROSSREFS
| Sequence in context: A191555 A129933 A064320 * A058975 A057120 A112512
Adjacent sequences: A112370 A112371 A112372 * A112374 A112375 A112376
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KEYWORD
| nonn
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AUTHOR
| Andrew Hone (anwh(AT)kent.ac.uk), Dec 02 2005
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