A239900 -id:A239900 - cp's OEIS frontend

11, 111, 12211, 149096311, 22229709804712411, 494159998001727075769152612720511, 244194103625066907517263589918036880566782292998362610615987380611

Offset: 1

Artur Jasinski, Sep 21 2008

For the "exact" formula, compare the Aho-Sloane reference in A000058. - N. J. A. Sloane, Apr 07 2014

A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437.
A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437 (original plus references that F.Q. forgot to include - see last page!)
Mohammad K. Azarian, Sylvester's Sequence and the Infinite Egyptian Fraction Decomposition of 1, Problem 958, College Mathematics Journal, Vol. 42, No. 4, September 2011, p. 330.
Mohammad K. Azarian, Sylvester's Sequence and the Infinite Egyptian Fraction Decomposition of 1, Solution College Mathematics Journal, Vol. 43, No. 4, September 2012, pp. 340-342.

See A239900 for another version.

a = {}; r = 11; Do[AppendTo[a, r]; r = r^2 - r + 1, {n, 1, 10}]; a

a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 11.

a(n) ~ c^(2^n) where c = 3.242214... (see A144808).

2, 5, 9, 89, 8009, 64152089, 4115490587216009, 16937262773463574696951813104089

Offset: 0

N. J. A. Sloane, Apr 05 2014

			(9 + 1)*9 - 1 = 89, (89 + 1)*89 - 1 = 8009, (8009 + 1)*8009 - 1 = 64152089. - _Zak Seidov_, Apr 06 2014

I:=[2, 5, 9]; [n le 3 select I[n] else Self(n-1)*(Self(n-1)+1)-1: n in [1..10]]; // Vincenzo Librandi, May 22 2014

a(n) == 8 mod 9, for n > 2. - Ivan N. Ianakiev, Apr 06 2014

a(n) = (a(n-1)+1)*a(n-1)-1, for n > 2. - Zak Seidov, Apr 06 2014

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