cp's OEIS Frontend

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

%I A144784 #25 Jan 09 2025 09:53:20
%S A144784 11,111,12211,149096311,22229709804712411,
%T A144784 494159998001727075769152612720511,
%U A144784 244194103625066907517263589918036880566782292998362610615987380611
%N A144784 Variant of Sylvester's sequence: a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 11.
%C A144784 For the "exact" formula, compare the Aho-Sloane reference in A000058. - _N. J. A. Sloane_, Apr 07 2014
%H A144784 A. V. Aho and N. J. A. Sloane, <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Scanned/11-4/aho-a.pdf">Some doubly exponential sequences</a>, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437.
%H A144784 A. V. Aho and N. J. A. Sloane, <a href="/A000058/a000058.pdf">Some doubly exponential sequences</a>, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437 (original plus references that F.Q. forgot to include - see last page!)
%H A144784 Mohammad K. Azarian, <a href="http://www.jstor.org/stable/10.4169/college.math.j.42.4.329">Sylvester's Sequence and the Infinite Egyptian Fraction Decomposition of 1, Problem 958</a>, College Mathematics Journal, Vol. 42, No. 4, September 2011, p. 330.
%H A144784 Mohammad K. Azarian, <a href="http://www.jstor.org/stable/10.4169/college.math.j.43.4.337">Sylvester's Sequence and the Infinite Egyptian Fraction Decomposition of 1, Solution</a> College Mathematics Journal, Vol. 43, No. 4, September 2012, pp. 340-342.
%F A144784 a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 11.
%F A144784 a(n) ~ c^(2^n) where c = 3.242214... (see A144808).
%t A144784 a = {}; r = 11; Do[AppendTo[a, r]; r = r^2 - r + 1, {n, 1, 10}]; a
%Y A144784 Cf. A000058, A082732, A144779, A144780, A144781, A144782, A144783, A144784, A144785, A144786, A144787, A144788, A144808.
%Y A144784 See A239900 for another version.
%K A144784 nonn
%O A144784 1,1
%A A144784 _Artur Jasinski_, Sep 21 2008