Last update: 27 December 2020
The immortal numbers were defined and researched by Daniel Marschall and are defined as integers, which contain themselves as suffix when raised to a power.
In this document we use following function that counts the amount of digits of an integer , represented in base :
Definition: A number is "immortal" (with power 2 at base 10), if it satisfies following equation:
Example: The number 625 is immortal at base 10, because 625 x 625 = 390625
Definition: A number is "immortal" with power at base , if it satisfies following equation:
Example: The number 2047 ( 3777(8) ) is immortal with power 3 at base 8, because 3777(8)3 = 77720013777(8)
A more general case with complex numbers and negative numbers can be found here.
The set contains all immortal numbers with power at base .
The "immortality" of a tuple is defined by the amount of immortal numbers divided by the maximum length of these numbers. The higher the maximum length is chosen, the more accurate is the immortality. The unit of measurement can be optionally written as ("Immortals per Digit"). Since the numbers are always immortal, they will be exluded from this calculation.
Example: For base 10, power 3, there are 1176 immortal numbers with length of 1..100 digits, exluding 0 and 1. The immortality of is therefore approximately
Here you can find a plot that compares the immortality of various tuples.
There are base/power-tuples where only the numbers are immortal. These base/power-tuples are called "pseudo-immortal". An example is .
Their immortality is zero because:
Here you can find a plot that shows the values of some pseudo-immortals.
Super immortal numbers are numbers which are immortal to every tuple. Only the numbers satisfy this property (see proof).
Immortal numbers can be written in graph with "branches".
In base 10, power 2, things are quite easy: There are only 4 branches 0, 1, 5 and 6. The branches 0 and 1 are trivial branches and have no children.
In branch 5, the immortal numbers are 5, 25, 625, 90625, 890625, 2890625, 12890625, 212890625, 8212890625, ...
In branch 6, the immortal numbers are 6, 76, 376, 9376, 109376, 7109376, 87109376, 787109376, 1787109376, ...
In this case, immortal numbers of base 10 power 2 can be generated using a tree node and adding a single digit in front of it. Additionally, the sum of the pairs of digits below branch 5 and 6 always add up to 9.
However, there are more complex graphs, for example base 10, power 3:
Credits: The graphs were created with mind-map-online.de
There has been a huge effort to find immortal numbers with several million digits. A special algorithm, optimized extensively using SSE/MMX technology, has been written to archive this task.
On July, 9th 2019, the immortal number of base 10, power 2 with 1.1 billion digits was found and on March, 30th 2020, the result was double-verified.
Download the giant immortal numbers:
Get the tool to calculate these numbers (Source codes written in C)
Credits: The search for giant immortal numbers at base-10-power-2 were only made possible by the users of the matheboard.de thread who contributed the algorithm for finding base-10-power-2 numbers as well as the contributor of this StackOverflow question who has showed me how to speed up the algorithm using SSE/MMX instructions.
A rather trivial search algorithm has been developed to find all immortal numbers between base 2 and 36 and power 2 till 10, with a maximum length of 100 digits.
Python source code
Get a list of immortal numbers with max length 100 digits, for powers 2 till 10, with following base:
02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
We want to find all
(1) Let be the number of unique prime factors of and the the prime decomposition of be:
being the valuation (how many times
can be divded by
(2) Let be the length of the number we want to find.
(3) The congruence equation for an immortal number with power 2, base and length can be written as:
(3b) and are coprime: .
(3c) Therefore, each , for , must divide into either just , giving that ; or , giving that . (see additional explanation)
(3d) So, the immortal number for a given length can be found using the following congruence equation:
(3e) Only the smallest canonical of the congruence equation is immortal (not: ).
(3f) The permutations of generate numbers of a specific branch.
Now can be found using the Chinese Remainder Theorem.
The permutations generate the different branches of the immortal numbers of :
Credits: John Omielan for the proof above and why this scheme cannot be easily applied to powers other than 2.
Gesucht sind die Zahlen , die zu der Basis unsterblich sind.
(1) Sei die Anzahl der verschiedenen Primfaktoren von und die kanonische Primfaktorzerlegung von :
wobei die -Bewertung von ist (wie oft durch teilbar ist)
(2) sei die Länge der unsterblichen Zahl, die wir suchen..
(3) Die Kongruenzgleichung für die unsterbliche Zahl mit Potenz 2, Basis und Länge kann geschrieben werden als:
(3b) und sind teilerfremd.
(3c) Deswegen muss für alle sich entweder durch teilen (ergibt ) oder durch (ergibt ). (Weitere Erklärungen auf Englisch)
(3d) Deswegen gilt die Kongruenzgleichung zur Findung aller mit jeweils Stellen:
(3e) Die unsterbliche Zahl der kleinste kanonische Wert der Kongruenzgleichung ist (nicht ).
(3f) Die Permutationen von erzeugen unsterbliche Zahlen eines bestimmten Stamms.
Es ergibt sich damit folgende Kongruenzgleichung:
kann mit dem Chinesischem Restsatz gefunden werden.
Die Permutationen erzeugen die verschiedenen Stämme der unsterblichen Zahlen von :
Python source code
Base 10 filtered:
Python source code
A red x denotes an immortality of zero. Dots denote an immortality.
The bigger the dots are, the greater is the immortality of
Green dots denote an immortality of .
Blue dots denote an immortality of .
Python source code
We want to prove that
Let's begin with:
With mathematical induction we can now prove that
Since all immortal numbers of power are also immortal in power , we have now proven that
Theorem: For every number there is a tuple so that . In other words, every number is immortal to some specific base and power.
Note: The theorem of complete immortality also leads to the conclusion:
(0) Let with an arbitary value.
(1) Let be a prime that is not a prime factor of ( ), then is not a prime factor of either ( ).
In other words: .
(2) Since we have , we can create following term using the Fermat-Euler theorem:
Transforming it a bit, we get:
(3) If we now define and insert the definition of from above, then we get:
which is equivalent to:
Credits: Many thanks to Finn from matheboard.de for showing me how to prove this theorem!
We want to prove that the only super-immortals are 0 and 1.
(1) First, we verify that 0 and 1 are immortal to every :
(2) To prove that the only super-immortals are 0 and 1, we want to find a tuple for every that will cause :
(3) The definition of an immortal number is
We can see that a number can only be immortal if is a number that has more digits (in the notation of base ) than .
(4) If we choose a high enough base , then and would have the same amounts of digits, and therefore could not be immortal.
So, we need to find a , so that .
If we choose , we get:
If , then and therefore:
, which is true, if .
(5) We continue by taking the general definition of an immortal number
which can also be written as:
Since and ( because if ) :
, which is false if .
If , then:
(7) Bringing together the results of step (1) and (6), we can conclude:
is prime, or the if the prime decomposition of
consists only of a single prime,
then there are no immortal numbers except .
Since the prime decomposition only consists of a single product
there is just one congruence equation (see Solution via Chinese Remainder Theorem above):
with and being the length of .
Therefore, the solutions of are
Definition: A complex number is "immortal" with power at base , if it satisfies following equation:
This general case also allows negative numbers, e.g. (-313+216i) is immortal with base 10 and power 2, because (-313+216i)2 = (51313-135216i).
The "complex immortality" of a tuple is defined by the amount of positive complex immortal numbers divided by the maximum length of the imaginary and real part of these complex numbers. The higher the maximum length is chosen, the more accurate is the immortality. The unit of measurement can be optionally written as ("Immortals per Digit"). Since the numbers are always immortal, they will be exluded from this calculation.
Example: For base 10, power 2, there are 27 immortal complex numbers (10 without imaginary part and 17 with imaginary part) with imaginary/real part with length of 1..5 digits, exluding numbers (0+0i) and (1+0i). The complex immortality of is therefore approximately
In comparison, the non-complex immortality is:
See a list of 1700 found immortal numbers with non-zero imaginary part here
Source code can be found here
Therefore, can only be 0.
Then, can only be 0, too, due to our previous assumption and additionally because .
So, the only possibility is
We can extend the definition of immortal numbers by searching the source number not only at the ending but at any position in the resulting power.
We define which is the amount of digits which is omitted at the right.
The definition of these immortal numbers is:
Example: because 146512 = 214651801
Here you can find a list of immortal numbers with .
Help and suggestions are welcome!
#3: Is there a base that does not have any non-trivial (not 0,1) immortal numbers?
#4: Is there a power that does not have any non-trivial (not 0,1) immortal numbers?
#5: The topic "Base B / Power P" is not researched very much. It is not known how to find "Base B / Power P" immortal numbers without brute-forcing them.
#6: some old notes from 2011 (in German):
Various notes and other stuff: runs, old_notes
Credits: The MathML equations are rendered by MathJax because Google Chrome does not care about MathML.