

A192223


a(n) = Lucas(2^n + 1).


7




OFFSET

0,1


COMMENTS

Product_{n>0} (1 + 1/a(n)) = 3  phi = A094874, where phi = (1+sqrt(5))/2 is the golden mean.
From Peter Bala, Oct 28 2013: (Start)
Compare with A230600(n) = Lucas(2^n  1).
Let x and b be positive real numbers. We define a Pierce expansion of x to the base b to be a (possibly infinite) increasing sequence of positive integers [a(1), a(2), a(3), ...] such that we have the alternating series representation x = b/a(1)  b^2/(a(1)*a(2)) + b^3/(a(1)*a(2)*a(3))  .... This definition generalizes the ordinary Pierce expansion of a real number 0 < x < 1, where the base b has the value 1. Depending on the values of x and b such a generalized Pierce expansion to the base b may not exist, and if it does exist it may not be unique.
Let Phi := 1/2*(sqrt(5)  1) denote the reciprocal of the golden ratio. This sequence, apart from the initial term, provides a Pierce expansion of Phi^4 to the base Phi. That is we have the identity Phi^4 = Phi/4  Phi^2/(4*11) + Phi^3/(4*11*76)  Phi^4/(4*11*76*3571) + ....
This result can be extended in two ways. Firstly, for k odd, the sequence {Lucas(k*(2^n + 1))}n>=1 gives a Pierce expansion of Phi^(4*k) to the base Phi^k. Secondly, for n = 1,2,3,..., the sequence [a(n),a(n+1),a(n+2),...] gives a Pierce expansion of Phi^(2^n + 2) to the base Phi. See below for some examples. (End)


LINKS

Table of n, a(n) for n=0..8.
J. Sondow, Evaluation of Tachiya's algebraic infinite products involving Fibonacci and Lucas numbers, Diophantine Analysis and Related Fields 2011  AIP Conference Proceedings, vol. 1385, pp. 97100.
Y. Tachiya, Transcendence of certain infinite products, J. Number Theory 125 (2007), 182200.
Eric Weisstein's World of Mathematics, Pierce Expansion


FORMULA

a(n) = A000032(2^n + 1).
From Peter Bala, Oct 28 2013: (Start)
a(n) = phi^(2^n + 1)  (1/phi)^(2^n + 1), where phi = 1/2*(1 + sqrt(5)) denotes the golden ratio A001622.
Recurrence equation: a(0) = 3, a(1) = 4 and a(n) = floor(1/phi*a(n1)^2) + 2 for n >= 2. (End)


EXAMPLE

Pierce series expansion of Phi^(2^n + 2) to the base Phi for n = 1 to 4:
n = 1:
Phi^4 = Phi/4  Phi^2/(4*11) + Phi^3/(4*11*76)  Phi^4/(4*11*76*3571) + ...
n = 2:
Phi^6 = Phi/11  Phi^2/(11*76) + Phi^3/(11*76*3571)  ...
n = 3:
Phi^10 = Phi/76  Phi^2/(76*3571) + Phi^3/(76*3571*7881196)  ...
n = 4:
Phi^18 = Phi/3571  Phi^2/(3571*7881196) + ...


MATHEMATICA

Table[LucasL[2^n + 1], {n, 0, 10}] (* T. D. Noe, Jan 11 2012 *)


CROSSREFS

Cf. A000032 (Lucas numbers L(n)), A094874 (decimal expansion of 3  phi), A192222 (Fibonacci(2^n + 1)). A001622, A058635, A230600, A230601, A230602.
Sequence in context: A327081 A201970 A102013 * A079940 A225205 A041299
Adjacent sequences: A192220 A192221 A192222 * A192224 A192225 A192226


KEYWORD

nonn,easy


AUTHOR

Jonathan Sondow, Jun 26 2011


STATUS

approved



