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%I #14 Sep 08 2022 08:46:06
%S 4,7,18,123,5778,12752043,62113250390418,1473646213395791149646646123,
%T 829490056885282616312940022414182153153900944625970578,
%U 262813148121156922478324605390890951672774150584488451750823334086851733999224817160730017360019778038580843
%N a(n) = Lucas(2^n + 2).
%C Let x and b be positive real numbers. We define an Engel expansion of x to the base b to be a (possibly infinite) nondecreasing sequence of positive integers [a(1), a(2), a(3), ...] such that we have the series representation x = b/a(1) + b^2/(a(1)*a(2)) + b^3/(a(1)*a(2)*a(3)) + .... Depending on the values of x and b such an expansion may not exist, and if it does exist it may not be unique. When b = 1 we recover the ordinary Engel expansion of x.
%C Let Phi := 1/2*(sqrt(5) - 1) denote the reciprocal of the golden ratio. This sequence, apart from the initial term, gives an Engel expansion of x = Phi^6 to the base b = Phi^2. The associated Engel series expansion of Phi^6 to the base Phi^2 begins Phi^6 = Phi^2/7 + Phi^4/(7*18) + Phi^6/(7*18*123) + Phi^8/(7*18*123*5778) + ....
%C This result can be extended in two ways. Firstly, for k = 1,2,3,... , the sequence {Lucas(k*(2^n + 2))}n>=1 is an Engel expansion of Phi^(6*k) to the base Phi^(2*k). Secondly, for n = 1,2,3,..., the sequence [a(n),a(n+1),a(n+2),...] is an Engel expansion of Phi^(2^n + 4) to the base Phi^2. See below for some examples.
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Engel_expansion">Engel Expansion</a>
%F a(n) = A000032(2^n+2) = phi^(2^n+2) + (1/phi)^(2^n+2), where phi = 1/2*(1 + sqrt(5)) denotes the golden ratio A001622.
%F Recurrence equation: a(0) = 4 and a(n) = floor((2 - phi)*a(n-1)^2) for n >= 1.
%e Engel series expansion of Phi^(2^n + 4) to the base Phi^2 for n = 1 to 4.
%e n = 1
%e Phi^6 = Phi^2/7 + Phi^4/(7*18) + Phi^6/(7*18*123) + Phi^8/(7*18*123*5778) + ...
%e n = 2:
%e Phi^8 = Phi^2/18 + Phi^4/(18*123) + Phi^6/(18*123*5778) + ...
%e n = 3:
%e Phi^12 = Phi^2/123 + Phi^4/(123*5778) + Phi^6/(123*5778*12752043) + ...
%e n = 4:
%e Phi^20 = Phi^2/5778 + Phi^4/(5778*12752043) + ...
%t Table[LucasL[2^n + 2], {n, 0, 10}]
%o (PARI) for(n=0,10, print1(fibonacci(2^n+3) + fibonacci(2^n +1), ", ")) \\ _G. C. Greubel_, Dec 22 2017
%o (Magma) [Lucas(2^n +2): n in [0..10]]; // _G. C. Greubel_, Dec 22 2017
%Y Cf. A000032, A001622, A192223, A230600, A230602.
%K nonn,easy
%O 0,1
%A _Peter Bala_, Oct 28 2013
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