A230601 a(n) = Lucas(2^n + 2).

4, 7, 18, 123, 5778, 12752043, 62113250390418, 1473646213395791149646646123, 829490056885282616312940022414182153153900944625970578, 262813148121156922478324605390890951672774150584488451750823334086851733999224817160730017360019778038580843

Offset: 0

Author

Peter Bala, Oct 28 2013

Keywords

• nonn
• easy

Comments

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.

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) + ....

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.

Examples

			Engel series expansion of Phi^(2^n + 4) to the base Phi^2 for n = 1 to 4.
n = 1
Phi^6 = Phi^2/7 + Phi^4/(7*18) + Phi^6/(7*18*123) + Phi^8/(7*18*123*5778) + ...
n = 2:
Phi^8 = Phi^2/18 + Phi^4/(18*123) + Phi^6/(18*123*5778) + ...
n = 3:
Phi^12 = Phi^2/123 + Phi^4/(123*5778) + Phi^6/(123*5778*12752043) + ...
n = 4:
Phi^20 = Phi^2/5778 + Phi^4/(5778*12752043) + ...
		

Links

• Wikipedia, Engel Expansion

Crossrefs

Cf. A000032, A001622, A192223, A230600, A230602.

Programs

Magma

[Lucas(2^n +2): n in [0..10]]; // G. C. Greubel, Dec 22 2017

Mathematica

Table[LucasL[2^n + 2], {n, 0, 10}]

PARI

for(n=0,10, print1(fibonacci(2^n+3) + fibonacci(2^n +1), ", ")) \\ G. C. Greubel, Dec 22 2017

Formula

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.

Recurrence equation: a(0) = 4 and a(n) = floor((2 - phi)*a(n-1)^2) for n >= 1.