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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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10
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OFFSET
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0,3
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COMMENTS
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As n tends to infinity, log(log(a(n)))/n tends to log((3+sqrt(5))/2) = A104457.
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 >= 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, Dec 12 2005
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LINKS
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Joshua Alman, Cesar Cuenca, and Jiaoyang Huang, Laurent phenomenon sequences, Journal of Algebraic Combinatorics, Vol. 43, No. 3 (2015), pp. 589-633.
Sergey Fomin and Andrei Zelevinsky, The Laurent Phenomenon, Advances in Applied Mathematics, Vol. 28, No. 2 (2002) pp. 119-144.
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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^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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MATHEMATICA
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RecurrenceTable[{a[n] == (a[n - 1]^3 + a[n - 1]^2)/a[n - 2], a[0] == a[1] == 1}, a, {n, 0, 7}] (* Michael De Vlieger, Jul 02 2015 *)
nxt[{a_, b_}]:={b, (b^3+b^2)/a}; NestList[nxt, {1, 1}, 8][[All, 1]] (* Harvey P. Dale, Apr 05 2019 *)
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PROG
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(Magma) I:=[1, 1]; [n le 2 select I[n] else (Self(n-1)^3+Self(n-1)^2)/Self(n-2): n in [1..10]]; // Vincenzo Librandi, Jul 02 2015
(Haskell)
a112373 n = a112373_list !! n
a112373_list = 1 : 1 : zipWith (\u v -> (u^3 + u^2) `div` v)
(tail a112373_list) a112373_list
(PARI) {a(n) = my(a=self()); if(n<0, a(1-n), n<2, 1, a(n-1)^2 * (1 + a(n-1)) / a(n-2))}; /* Michael Somos, Apr 19 2017 */
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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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