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Variant of Sylvester's sequence: a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 8.
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%I #17 Sep 02 2020 14:42:58

%S 8,57,3193,10192057,103878015699193,10790642145601683494645152057,

%T 116437957914435303575899742229333045108448631998006179193,

%U 13557798043283806950297045269968250387897834581711367551819275131055206893868524458302302046950954641412419952057

%N Variant of Sylvester's sequence: a(n+1) = a(n)^2 - a(n) + 1, with a(1) = 8.

%H 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 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 a(n) ~ c^(2^n) where is c is 2.74167747444233776776... (A144805).

%t a = {}; k = 8; Do[AppendTo[a, k]; k = k^2 - k + 1, {n, 1, 10}]; a

%t NestList[#^2-#+1&,8,10] (* _Harvey P. Dale_, Jan 29 2017 *)

%Y Cf. A000058, A082732, A144805.

%K nonn

%O 1,1

%A _Artur Jasinski_, Sep 21 2008