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A192223 a(n) = Lucas(2^n + 1). 7
3, 4, 11, 76, 3571, 7881196, 38388099893011, 910763447271179530132922476, 512653048485188394162163283930413917147479973138989971 (list; graph; refs; listen; history; text; internal format)
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. 97-100.

Y. Tachiya, Transcendence of certain infinite products, J. Number Theory 125 (2007), 182-200.

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(n-1)^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: A041947 A201970 A102013 * A079940 A225205 A041299

Adjacent sequences:  A192220 A192221 A192222 * A192224 A192225 A192226

KEYWORD

nonn,easy

AUTHOR

Jonathan Sondow, Jun 26 2011

STATUS

approved

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Last modified November 15 18:31 EST 2018. Contains 317240 sequences. (Running on oeis4.)