A144784 Variant of Sylvester's sequence: a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 11.

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

Offset: 1

Author

Artur Jasinski, Sep 21 2008

Keywords

• nonn

Comments

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

Links

• 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.

Crossrefs

Cf. A000058, A082732, A144779, A144780, A144781, A144782, A144783, A144784, A144785, A144786, A144787, A144788, A144808.

See A239900 for another version.

Programs

Mathematica

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

Formula

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

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