A230600 -id:A230600 - cp's OEIS frontend

A192223 a(n) = Lucas(2^n + 1).

3, 4, 11, 76, 3571, 7881196, 38388099893011, 910763447271179530132922476, 512653048485188394162163283930413917147479973138989971

Offset: 0

Jonathan Sondow, Jun 26 2011

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)

			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) + ...

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

Cf. A000032 (Lucas numbers L(n)), A094874 (decimal expansion of 3 - phi), A192222 (Fibonacci(2^n + 1)). A001622, A058635, A230600, A230601, A230602.

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

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)

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

Original entry on oeis.org

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

Offset: 0

Peter Bala, Oct 28 2013

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.

			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) + ...

Wikipedia, Engel Expansion

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

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

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

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

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.

A230602 a(n) = Lucas(2^n - 2).

Original entry on oeis.org

2, 3, 18, 843, 1860498, 9062201101803, 215002084978043708894524818, 121020968315000050139390193037122554865361969834971243, 38343921554607207587938114587587818441864732465057252794474861753545122655196096751375348482086938743684498

Offset: 1

Peter Bala, Oct 28 2013

Let phi := 1/2*(1 + sqrt(5)) denote the golden ratio. This sequence, apart from the initial term, gives an Engel expansion of 1 to the base phi^2 (see A230601 for a definition of this term). The associated Engel series expansion of 1 to the base phi^2 begins 1 = phi^2/3 + phi^4/(3*18) + phi^6/(3*18*843) + phi^8/(3*18*843*1860498) + .... This result can be extended in two ways. Firstly, the sequence Lucas(2*k*(2^n - 1)) for k = 1,2,3,... is an Engel expansion of 1 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^(4 - 2^n) to the base phi^2. Some example are given below.

			Engel series expansion of phi^(4 - 2^n) to the base phi^2 for n = 1 to 5.
n = 1:
phi^2 = phi^2/2 + phi^4/(2*3) + phi^6/(2*3*18) + phi^8/(2*3*18*843) + ...
n = 2:
1 = phi^2/3 + phi^4/(3*18) + phi^6/(3*18*843) + phi^8/(3*18*843*1860498) + ...
n = 3:
1/phi^4 = phi^2/18 + phi^4/(18*843) + phi^6/(18*843*1860498) + ...
n = 4:
1/phi^12 = phi^2/843 + phi^4/(843*1860498) + phi^6/(843*1860498*9062201101803) + ...
n = 5:
1/phi^28 = phi^2/1860498 + phi^4/(1860498*9062201101803) + ...

Wikipedia, Engel Expansion

Cf. A000032, A001622, A192223, A230600, A230601.

Mathematica
```
Table[LucasL[2^n - 2], {n, 1, 10}]
```

a(n) = A000032(2^n-2) = phi^(2^n-2) + (1/phi)^(2^n-2), where phi := 1/2*(1 + sqrt(5)).

Recurrence equation: a(1) = 2, a(2) = 3 and a(n) = floor(phi^2*a(n-1)^2) - 5 for n >= 3.

A232326 Pierce expansion of 1 to the base Pi.

Original entry on oeis.org

3, 69, 310, 1017, 36745, 214369, 966652, 11159821, 74039764, 550021544, 4481549430, 16543857917, 87205978613, 476981856953, 30989048525367, 203786458494160, 711639924282497, 3174772986229899, 29814569078896025, 100158574806804154

Offset: 0

Peter Bala, Nov 26 2013

Let r and b be positive real numbers. We define a Pierce expansion of r to the base b to be a (possibly infinite) increasing sequence of positive integers [a(0), a(1), a(2), ...] such that we have the alternating series representation r = b/a(0) - b^2/(a(0)*a(1)) + b^3/(a(0)*a(1)*a(2)) - .... Depending on the values of r and b such an expansion may not exist, and if it does exist it may not be unique. When b = 1 and 0 < r < 1 we recover the ordinary Pierce expansion of r.
See A058635, A192223 and A230600 for some predictable Pierce expansions to a base b other than 1.
In the particular case that the base b >= 1 and 0 < r < b then we can find a Pierce expansion of r to the base b as follows:
Define the map f(x) (which depends on the base b) by f(x) = x/b*ceiling(b/x) - 1 and let f^(n)(x) denote the n-th iterate of the map f(x), with the convention that f^(0)(x) = x.
For n = 0,1,2,... define a(n) = ceiling(b/f^(n)(-r)) until f^n(-r) = 0.
Then it can be shown that the sequence of positive integers |a(n)| is a Pierce expansion of r to the base b.
For the present sequence we apply this algorithm with r := 1 and with base b := Pi. See A232325 for an Engel expansion of 1 to the base Pi.

Eric Weisstein's World of Mathematics, Pierce Expansion

Cf. A014014, A006784, A058635, A061233, A192223, A230600, A232325, A232327, A232328.

# Define the n-th iterate of the map f(x) = x/b*ceiling(b/x) - 1
map_iterate := proc(n,b,x) option remember;
if n = 0 then
   x
else
   -1 + 1/b*thisproc(n-1,b,x)*ceil(b/thisproc(n-1,b,x))
end if
end proc:
# Define the (signed) terms of the expansion of x to the base b
a := n -> ceil(evalf(b/map_iterate(n,b,x))):
Digits:= 500:
# Choose values for x and b
x := -1: b:= Pi:
seq(abs(a(n)), n = 0..19);

a(n) = ceiling(Pi/f^(n)(-1)), where f^(n)(x) denotes the n-th iterate of the map f(x) = x/Pi*ceiling(Pi/x) - 1, with the convention that f^(0)(x) = x.
Pierce series expansion of 1 to the base Pi:
1 = Pi/3 - Pi^2/(3*69) + Pi^3/(3*69*310) - Pi^4/(3*69*310*1017) + ....
The associated power series F(z) := 1 - ( z/3 - z^2/(3*69) + z^3/(3*69*310) - z^4/(3*69*310*1017) + ...) has a zero at z = Pi. Truncating the series F(z) to n terms produces a polynomial F_n(z) with rational coefficients which has a real zero close to Pi.

cp's OEIS Frontend

A192223 a(n) = Lucas(2^n + 1).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A230602 a(n) = Lucas(2^n - 2).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A232326 Pierce expansion of 1 to the base Pi.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Formula