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 1-4 of 4 results.

A232325 Engel expansion of 1 to the base Pi.

Original entry on oeis.org

4, 12, 72, 2111, 14265, 70424, 308832, 4371476, 320218450, 1101000257, 14020589841, 102772320834, 963205851651, 5997003656523, 50649135127796, 640772902021920, 2101002284323870, 35029677728070645, 176996397541889098, 1433436623499128186
Offset: 0

Views

Author

Peter Bala, Nov 25 2013

Keywords

Comments

Let r and b be positive real numbers. We define an Engel expansion of r to the base b to be a (possibly infinite) nondecreasing sequence of positive integers [a(0), a(1), a(2), ...] such that we have the 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 we recover the ordinary Engel expansion of r. See A181565 and A230601 for some predictable Engel 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 an Engel expansion of r to the base b using the following algorithm:
Choose values for r and b.
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 the integer a(n) = ceiling(b/f^(n)(r)) until f^n(r) = 0.
When b >= 1 and 0 < r < b the sequence a(n) produced by this algorithm provides an Engel expansion of r to the base b.
For the present sequence we apply this algorithm with r := 1 and with the base b := Pi.
We can also get an alternating series representation for r in powers of b (still assuming b >= 1 and 0 < r < b), called a Pierce series expansion of r to the base b, by running the above algorithm but now with input values -r and base b. See A232326.
In addition, we can obtain two further series expansions for r in powers of b by running the algorithm with either the input values r and base -b or with the input values -r and base -b. See examples below. See A232327 and A232328 for other examples of these types of expansions.

Examples

			Truncation F_5(z) = 1 - ( z/4 + z^2/(4*12) + z^3/(4*12*72) + z^4/(4*12*72*2111) + z^5/(4*12*72*2111*14265) ). The polynomial has a positive real zero at z = 3.14159 26535 (9...), which agrees with Pi to 10 decimal places.
Comparison of generalized Engel expansions of 1 to the base Pi.
A232325: Engel series expansion of 1 to the base Pi
1 = Pi/4 + Pi^2/(4*12) + Pi^3/(4*12*72) + Pi^4/(4*12*72*2111) + ....
A232326: 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) + - ....
Running the algorithm with the input values r = 1 and base -Pi produces the expansion
1 = Pi/3 - Pi^2/(3*70) - Pi^3/(3*70*740) + Pi^4/(3*70*740*6920) + - - + ....
Running the algorithm with the input values r = -1 and base -Pi produces the expansion
1 = Pi/4 + Pi^2/(4*11) - Pi^3/(4*11*73) - Pi^4/(4*11*73*560) + + - - ....
		

Crossrefs

Programs

  • Maple
    # 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 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(a(n), n = 0..19);

Formula

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.
Engel series expansion of 1 to the base Pi:
1 = Pi/4 + Pi^2/(4*12) + Pi^3/(4*12*72) + Pi^4/(4*12*72*2111) + ....
The associated power series F(z) := 1 - ( z/4 + z^2/(4*12) + z^3/(4*12*72) + z^4/(4*12*72*2111) + ...) 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. See below for an example.

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

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

Original entry on oeis.org

2, 3, 18, 843, 1860498, 9062201101803, 215002084978043708894524818, 121020968315000050139390193037122554865361969834971243, 38343921554607207587938114587587818441864732465057252794474861753545122655196096751375348482086938743684498
Offset: 1

Views

Author

Peter Bala, Oct 28 2013

Keywords

Comments

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.

Examples

			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) + ...
		

Crossrefs

Programs

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

Formula

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.
Showing 1-4 of 4 results.