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A002814
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For n > 1: a(n) = a(n-1)^3 + 3a(n-1)^2 - 3; a(0) = 1, a(1) = 2.
(Formerly M2105 N0833)
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16
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OFFSET
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0,2
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COMMENTS
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An infinite coprime sequence defined by recursion. - Michael Somos, Mar 14 2004
The present sequence is the case x = 3 of the following general remarks. The recurrence equation a(n+1) = a(n)^3 + 3*a(n)^2 - 3 with the initial condition a(1) = x - 1 > 1 has the explicit solution a(n+1) = alpha^(3^n) + (1/alpha)^(3^n) - 1 for n >= 0, where alpha := {x + sqrt(x^2 - 4)}/2.
Two other recurrences satisfied by the sequence are a(n+1) = (a(1) + 3)*(Product_{k = 1..n} a(k)^2) - 3 and a(n+1) = 1 + (a(1) - 1)*Product_{k = 1..n} (a(k) + 2)^2, both with a(1) = x - 1.
The associated sequence b(n) := a(n) + 1 satisfies the recurrence equation b(n+1) = b(n)^3 - 3*b(n) with the initial condition b(1) = x. See A001999 for the case x = 3. The sequence c(n) := a(n) + 2 satisfies the recurrence equation c(n+1) = c(n)^3 - 3*c(n)^2 + 3 with the initial condition c(1) = x + 1.
The sequences a(n) and b(n) have been considered by Fine and Escott in connection with a product expansion for quadratic irrationals. We have the following identity, valid for x > 2: sqrt((x + 2)/(x - 2)) = (1 + 2/(x-1))*sqrt((y + 2)/(y - 2)), where y = x^3 - 3*x or, equivalently, y - 1 = (x - 1)^3 + 3*(x - 1)^2 - 3. Iterating the identity produces the product expansion sqrt((x+2)/(x-2)) = Product_{n >= 1} (1 + 2/a(n)), with a(1) = x - 1 and a(n+1) = a(n)^3 + 3*a(n)^2 - 3.
Conjecture: The sequence {a(n) - 2: n >= 1} is a strong divisibility sequence, that is, gcd(a(n) - 2, a(m) - 2) = a(gcd(n, m)) - 2 for n, m >= 1. - Peter Bala, Dec 08 2022
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REFERENCES
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L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 1, p. 397.
E. Lucas, Nouveaux théorèmes d'arithmétique supérieure, Comptes Rend., 83 (1876), 1286-1288.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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FORMULA
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a(n+1) = 5*(f(n))^2 - 3, where f(n) = Fibonacci(3^n) = product of first n entries. - Lekraj Beedassy, Jun 16 2003
a(n+2) = (G^(3^(n + 1)) - (1 - G)^(3^(n + 1)))/((G^(3^n)) - (1 - G)^(3^n)) where G = (1 + sqrt(5))/2;
a(n+1) = (1/2*(3 + sqrt(5)))^(3^n) + (1/2*(3 - sqrt(5)))^(3^n) - 1.
The sequence b(n):= a(n) + 2 is a solution to the recurrence b(n+1) = b(n)^3 - 3*b(n)^2 + 3 with b(1) = 4.
Other recurrence equations:
a(n+1) = -3 + 5*(Product_{k = 1..n} a(k)^2) with a(1) = 2.
a(n+1) = 1 + Product_{k = 1..n} (a(k) + 2)^2 with a(1) = 2.
Thus Y := Product_{k = 1..n} a(k) and X := Product_{k = 1..n} (a(k) + 2) give a solution to the Diophantine equation X^2 - 5*Y^2 = -4.
sqrt(5) = Product_{n >= 1} (1 + 2/a(n)). The rate of convergence is cubic. Fine remarks that twelve factors of the product would give well over 300,000 correct decimal digits for sqrt(5).
5 - {Product_{n = 1..N} (1 + 2/a(n))}^2 = 20/(a(N+1) + 3). (End)
a(n) = 2*T(3^(n-1),3/2) - 1 for n >= 1, where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. - Peter Bala, Dec 06 2022
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MATHEMATICA
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Join[{1}, NestList[#^3+3#^2-3&, 2, 5]] (* Harvey P. Dale, Apr 01 2011 *)
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PROG
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(PARI) a(n)=if(n<2, max(0, n+1), a(n-1)^3+3*a(n-1)^2-3)
(Maxima) a[0]:1$ a[1]:2$ a[n]:=a[n-1]^3 + 3*a[n-1]^2-3$ A002814(n):=a[n]$
(Haskell)
a002814 n = a002814_list !! n
a002814_list = 1 : zipWith div (tail xs) xs
where xs = map a000045 a000244_list
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CROSSREFS
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KEYWORD
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nonn,easy,nice
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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