A285494

The On-Line Encyclopedia of Integer Sequences gets regularly updated with new integer sequences. One of the recent updates was contributed by me, A285494.

A285494 is the list of all numbers k so that its digit sum equals its number of distinct prime factors.
A number’s digit sum is the sum of all of its decimal digits. The number 62831853, for example, has a digit sum of 6+2+8+3+1+8+5+3 = 36.
A number’s number of distinct prime factors is the number of different prime numbers that multiply together to result in the original number. As an example, 62831853 = 3^2 \cdot 7 \cdot 127 \cdot 7853, so it has five prime factors of which four are distinct.
Thereby one can conclude that 62831853 is not an entry in this sequence, as 36 \neq 4.

The sequence is certainly infinite, as the number k = 2 \cdot 10^n with n \in \mathbb{N}^* has a digit sum of 2 + (0 \cdot n) = 2 and — because k = 2^{n+1} \cdot 5^n — exactly two distinct prime factors.

In the encyclopedia entry, I provided a Mathematica one-liner to compute the first few entries of this sequence. Since then, I have also written a Python two-liner to achieve the same goal.


(* Mathematica *)
Select[Range[2,10000],Total[IntegerDigits[#]]==Length[FactorInteger[#]]&]
Out = {20, 30, 102, 120, 200, 300, 1002, 1200, 2000, 2001, 2002, 3000, 3010}
# Python 2.7
>>> def p(n):exec"i,f=2,set()\nwhile n>1:\n\tif n%i==0:f.add(i);n/=i;i=1\n\ti+=1";return len(f)
>>> print filter(lambda n:p(n)==sum(map(int,str(n))),range(2,10001))
[20, 30, 102, 120, 200, 300, 1002, 1200, 2000, 2001, 2002, 3000, 3010]
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Multibrot Set

The Mandelbrot Set is typically defined as the set of all numbers c \in \mathbb{C} for which — with z_0 = 0, z_{n+1} = f_c(z_n) and f_c(z) = z^2 + c — the limit \lim\limits_{n \to \infty} z_n converges. Visualizations of this standard Mandelbrot Set can be seen in three of my posts (Mandelbrot Set, Mandelbrot Set Miscalculations and Mandelbrot Set II).

f_c(z)=z^2+cHowever, one can extend the fractal’s definition beyond only having the exponent 2 in the function to be f_c(z)=z^\text{exp}+c with \text{exp} \in \mathbb{R}. The third post I mentioned actually has some generalization as it allows for \text{exp} \in \{2,3,4,5\}, although the approach used cannot be extended to real or even rational numbers.

f_c(z)=z^3+cThe method I used in the aforementioned post consists of manually expanding (a+b\cdot i)^n for each n. The polynomial (a+b\cdot i)^3, for example, would be expanded to (a^3 - 3 \cdot a \cdot b^2) + (3 \cdot a^2 \cdot b - b^3) \cdot i.
This method is not only tedious, error-prone and has to be done for every exponent (of which there are many), it also only works for whole-number exponents. To visualize real Multibrots, I had to come up with an algorithm for complex number exponentiation.

f_c(z)=z^4+cLuckily enough, there are two main ways to represent a complex number, Cartesian form z = a+b\cdot i and polar form z = k\cdot e^{\alpha\cdot i}. Converting from Cartesian to polar form is simply done by finding the number’s vector’s magnitude k = \sqrt{a^2+b^2} and its angle to the x-axis \alpha = \mbox{atan2}(\frac{a}{b}). (The function \mbox{atan2} is used in favor of \arctan to avoid having to divide by zero. View this Wikipedia article for more on the function and its definition.)
Once having converted the number to polar form, exponentiation becomes easy as z^\text{exp} = (k \cdot e^{\alpha\cdot i})^\text{exp} = k^\text{exp} \cdot e^{\alpha \cdot \text{exp} \cdot i}. With the exponentiated z^\text{exp} in polar form, it can be converted back in Cartesian form with z^\text{exp} = k^\text{exp} \cdot (\cos{(\alpha \cdot \text{exp})} + \sin{(\alpha \cdot \text{exp})} \cdot i \big).

f_c(z)=z^5+cUsing this method, converting the complex number to perform exponentiation, I wrote a Java program which visualizes the Multibrot for a given range of exponents and a number of frames.
Additionally, I added a new strategy for coloring the Multibrot Set, which consists of choosing a few anchor colors and then linearly interpolating the red, green and blue values. The resulting images have a reproducible (in contrast to randomly choosing colors) and more interesting (in contrast to only varying brightness) look.

f_c(z)=z^6+cThe family of Multibrot Sets can also be visualized as an animation, showing the fractal with an increasing exponent. The animated gif shown below was created using ImageMagick’s convert -delay <ms> *.png multibrot.gif command to stitch together the various .png files the Java application creates. To speed up the rendering, a separate thread is created for each frame, often resulting in 100% CPU-usage. (Be aware of this should you render your own Multibrot Sets!)

f_c(z)=z^10+cTo use the program on your own, either copy the source code listed below or download the .java file. The sections to change parameters or the color palette are clearly highlighted using block comments (simply search for ‘/*’).
To compile and execute the Java application, run (on Linux or MacOS) the command javac multibrot.java; java -Xmx4096m multibrot in the source code’s directory (-Xmx4096m tag optional, though for many frames at high quality it may be necessary as it allows Java to use more memory).
If you are a sole Windows user, I recommend installing the Windows 10 Bash Shell.

Multibrot animation (probably loading...)


// Java 1.8 Code
// Jonathan Frech, 11th of September 2016
//          edited 17th of April     2017
//          edited 18th of April     2017
//          edited 20th of April     2017
//          edited 21st of April     2017
//          edited 22nd of April     2017

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Collatz Conjecture

The Collatz conjecture states that every positive integer k>0 will — if you iteratively set k to f(k) — result in 1 (function shown beneath).
The graph beneath shows the path length of numbers from 1 to 10\,000. In this range 6170 is the number with the most steps, 261.
f(k)={\begin{cases}\frac{k}{2}&{\text{if }}k \mod 2 = 0\\3 \cdot k+1&{\text{if }}k \mod 2 = 1\end{cases}}

Path length in Collatz sequence [1 to 10**4]


# Python 2.7.7 Code
# Pygame 1.9.1 (for Python 2.7.7)
# Jonathan Frech 2nd of September, 2016

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