cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

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

Original entry on oeis.org

3, 4, 11, 76, 3571, 7881196, 38388099893011, 910763447271179530132922476, 512653048485188394162163283930413917147479973138989971
Offset: 0

Views

Author

Jonathan Sondow, Jun 26 2011

Keywords

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)

Examples

			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) + ...
		

Crossrefs

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

Programs

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

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)

A230600 a(n) = Lucas(2^n - 1).

Original entry on oeis.org

2, 1, 4, 29, 1364, 3010349, 14662949395604, 347880681146567910619198829, 195816040085094172011386545446645681141059001652009364, 62041768337314169100816125405238438263014895218124648720624536859920496229610874552659773465281966850403949
Offset: 0

Views

Author

Peter Bala, Oct 28 2013

Keywords

Comments

Compare with A192223(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 is taken equal to 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*(1 + sqrt(5)) denote the golden ratio A001622. This sequence, apart from the initial term, provides a Pierce expansion of 1 to the base phi. That is, we have the identity 1 = phi/1 - phi^2/(1*4) + phi^3/(1*4*29) - phi^4/(1*4*29*1364) + ....
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 1 to the base phi^k. Secondly, for n = 1,2,3,..., the sequence [a(n),a(n+1),a(n+2),...] provides a Pierce expansion of the quadratic irrational 1/phi^(2^n - 2) = Fibonacci(2^n - 1) - Fibonacci(2^n - 2)*phi to the base phi. Some examples are given below.

Examples

			A Pierce expansion of 1/phi^(2^n - 2) = Fibonacci(2^n - 1) - Fibonacci(2^n - 2)*phi to the base phi for n = 1 to 4.
n = 1: 1        = phi/1 - phi^2/(1*4) + phi^3/(1*4*29) - phi^4/(1*4*29*1364) + ...
n = 2: 1/phi^2  = phi/4 - phi^2/(4*29) + phi^3/(4*29*1364) - phi^4/(4*29*1364*3010349) + ...
n = 3: 1/phi^6  = phi/29 - phi^2/(29*1364) + phi^3/(29*1364*3010349) - ...
n = 4: 1/phi^14 = phi/1364 - phi^2/(1364*3010349) + ...
		

Crossrefs

Programs

  • Magma
    [Lucas(2^n -1): n in [0..10]]; // G. C. Greubel, Dec 22 2017
  • Mathematica
    Table[LucasL[2^n - 1], {n, 0, 10}]
  • PARI
    for(n=0,10, print1(fibonacci(2^n) + fibonacci(2^n -2), ", ")) \\ G. C. Greubel, Dec 22 2017
    

Formula

a(n) = Lucas(2^n - 1) = A000032(2^n-1) = phi^(2^n-1) + (-1/phi)^(2^n-1).
Recurrence equation: a(0) = 2, a(1) = 1 and a(n) = floor(phi*a(n-1)^2) + 4 for n >= 2.
Product {n = 2..k} (1 - 1/a(n)) = 1/2 + Fibonacci(2^k - 1)/(2*Lucas(2^k - 1)).
Product {n >= 2} (1 - 1/a(n)) = (5 + sqrt(5))/10.
a(n) = A000032(A000225(n)). - Omar E. Pol, Oct 28 2013

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

Original entry on oeis.org

4, 7, 18, 123, 5778, 12752043, 62113250390418, 1473646213395791149646646123, 829490056885282616312940022414182153153900944625970578, 262813148121156922478324605390890951672774150584488451750823334086851733999224817160730017360019778038580843
Offset: 0

Views

Author

Peter Bala, Oct 28 2013

Keywords

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) + ...
		

Crossrefs

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.
Showing 1-3 of 3 results.