Blog
2015-08-29
A few weeks ago, I made OEISbot, a Reddit bot which posts information whenever an OEIS sequence is mentioned.
This post explains how OEISbot works. The full code can be found on GitHub.
Getting started
OEISbot is made in Python using PRAW (Python Reddit Api Wrapper). PRAW can be installed with:
bash
pip install prawBefore making a bot, you will need to make a Reddit account for your bot, create a Reddit app and obtain API keys. This python script can be used to obtain the necessary keys.
Once you have your API keys saved in your praw.ini file, you are ready to make a bot.
Writing the bot
First, the necessary imports are made, and test mode is activated if the script is run with test as an argument. We also define an exception that will be used later to kill the script once it makes a comment.
python
import prawimport re
import urllib
import json
from praw.objects import MoreComments
import sys
test = False
if len(sys.argv) > 1 and sys.argv[1] == "test":
test = True
print("TEST MODE")
class FoundOne(BaseException):
pass
To prevent OEISbot from posting multiple links to the same sequence in a thread, lists of sequences linked to in each thread can be loaded and saved using the following functions.
python
def save_list(seen, _id):print(seen)
with open("/home/pi/OEISbot/seen/"+_id, "w") as f:
return json.dump(seen, f)
def open_list(_id):
try:
with open("/home/pi/OEISbot/seen/" + _id) as f:
return json.load(f)
except:
return []
The following function will search a post for a mention of an OEIS sequence number.
python
def look_for_A(id_, text, url, comment):seen = open_list(id_)
re_s = re.findall("A([0-9]{6})", text)
re_s += re.findall("oeis\.org/A([0-9]{6})", url)
if test:
print(re_s)
post_me = []
for seq_n in re_s:
if seq_n not in seen:
post_me.append(markup(seq_n))
seen.append(seq_n)
if len(post_me) > 0:
post_me.append(me())
comment(joiner().join(post_me))
save_list(seen, id_)
raise FoundOne
The following function will search a post for a comma-separated list of numbers, then search for it on the OEIS. If there are 14 sequences or less found, it will reply. If it finds a list with no matches on the OEIS, it will message /u/PeteOK, as he likes hearing about possibly new sequences.
python
def look_for_ls(id_, text, comment, link, message):seen = open_list(id_)
if test:
print(text)
re_s = re.findall("([0-9]+\, *(?:[0-9]+\, *)+[0-9]+)", text)
if len(re_s) > 0:
for terms in ["".join(i.split(" ")) for i in re_s]:
if test:
print(terms)
if terms not in seen:
seen.append(terms)
first10, total = load_search(terms)
if test:
print(first10)
if len(first10)>0 and total <= 14:
if total == 1:
intro = "Your sequence (" + terms \
+ ") looks like the following OEIS sequence."
else:
intro = "Your sequence (" + terms + \
+ ") may be one of the following OEIS sequences."
if total > 4:
intro += " Or, it may be one of the " + str(total-4) \
+ " other sequences listed [here]" \
"(http://oeis.org/search?q=" + terms + ")."
post_me = [intro]
if test:
print(first10)
for seq_n in first10[:4]:
post_me.append(markup(seq_n))
seen.append(seq_n)
post_me.append(me())
comment(joiner().join(post_me))
save_list(seen, id_)
raise FoundOne
elif len(first10) == 0:
post_me = ["I couldn't find your sequence (" + terms \
+ ") in the [OEIS](http://oeis.org). "
"You should add it!"]
message("PeteOK",
"Sequence not in OEIS",
"Hi Peter, I've just found a new sequence (" \
+ terms + ") in [this thread](link). " \
"Please shout at /u/mscroggs to turn the " \
"feature off if its spamming you!")
post_me.append(me())
comment(joiner().join(post_me))
save_list(seen, id_)
raise FoundOne
def load_search(terms):
src = urllib.urlopen("http://oeis.org/search?fmt=data&q="+terms).read()
ls = re.findall("href=(?:'|\")/A([0-9]{6})(?:'|\")", src)
try:
tot = int(re.findall("of ([0-9]+) results found", src)[0])
except:
tot = 0
return ls, tot
The markup function loads the necessary information from OEIS and formats it. Each comment will end with the output of the me function. The ouput of joiner will be used between sequences which are mentioned.
python
def markup(seq_n):pattern = re.compile("%N (.*?)<", re.DOTALL|re.M)
desc = urllib.urlopen("http://oeis.org/A" + seq_n + "/internal").read()
desc = pattern.findall(desc)[0].strip("\n")
pattern = re.compile("%S (.*?)<", re.DOTALL|re.M)
seq = urllib.urlopen("http://oeis.org/A" + seq_n + "/internal").read()
seq = pattern.findall(seq)[0].strip("\n")
new_com = "[A" + seq_n + "](http://oeis.org/A" + seq_n + "/): "
new_com += desc + "\n\n"
new_com += seq + "..."
return new_com
def me():
return "I am OEISbot. I was programmed by /u/mscroggs. " \
"[How I work](http://mscroggs.co.uk/blog/20). " \
"You can test me and suggest new features at /r/TestingOEISbot/."
def joiner():
return "\n\n- - - -\n\n"
Next, OEISbot logs into Reddit.
python
r = praw.Reddit("OEIS link and description poster by /u/mscroggs.")access_i = r.refresh_access_information(refresh_token=r.refresh_token)
r.set_access_credentials(**access_i)
auth = r.get_me()
The subs which OEISbot will search through are listed. I have used all the math(s) subs which I know about, as these will be the ones mentioning sequences.
python
subs = ["TestingOEISbot","math","mathpuzzles","casualmath","theydidthemath","learnmath","mathbooks","cheatatmathhomework","matheducation",
"puremathematics","mathpics","mathriddles","askmath",
"recreationalmath","OEIS","mathclubs","maths"]
if test:
subs = ["TestingOEISbot"]
For each sub OEISbot is monitoring, the hottest 10 posts are searched through for mentions of sequences. If a mention is found, a reply is generated and posted, then the FoundOne exception will be raised to end the code.
python
try:for sub in subs:
print(sub)
subreddit = r.get_subreddit(sub)
for submission in subreddit.get_hot(limit = 10):
if test:
print(submission.title)
look_for_A(submission.id,
submission.title + "|" + submission.selftext,
submission.url,
submission.add_comment)
look_for_ls(submission.id,
submission.title + "|" + submission.selftext,
submission.add_comment,
submission.url,
r.send_message)
flat_comments = praw.helpers.flatten_tree(submission.comments)
for comment in flat_comments:
if ( not isinstance(comment, MoreComments)
and comment.author is not None
and comment.author.name != "OEISbot" ):
look_for_A(submission.id,
re.sub("\[[^\]]*\]\([^\)*]\)","",comment.body),
comment.body,
comment.reply)
look_for_ls(submission.id,
re.sub("\[[^\]]*\]\([^\)*]\)","",comment.body),
comment.reply,
submission.url,
r.send_message)
except FoundOne:
pass
Running the code
I put this script on a Raspberry Pi which runs it every 10 minutes (to prevent OEISbot from getting refusals for posting too often). This is achieved with a cron job.
bash
*/10 * * * * python /path/to/bot.pyMaking your own bot
The full OEISbot code is available on GitHub. Feel free to use it as a starting point to make your own bot! If your bot is successful, let me know about it in the comments below or on Twitter.
Edit: Updated to describe the latest version of OEISbot.
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2015-08-27
In 1961, Donald Michie built MENACE (Machine Educable Noughts And Crosses Engine), a machine capable of learning to be a better player of Noughts and Crosses (or Tic-Tac-Toe if you're American). As computers were less widely available at the time, MENACE was built from from 304 matchboxes.
![](javascript:showlimage("menace.jpg"))
Taken from Trial and error by Donald Michie [2]
The original MENACE.
To save you from the long task of building a copy of MENACE, I have written a JavaScript version of MENACE, which you can play against here.
How to play against MENACE
To reduce the number of matchboxes required to build it, MENACE always plays first. Each possible game position which MENACE could face is drawn on a matchbox. A range of coloured beads are placed in each box. Each colour corresponds to a possible move which MENACE could make from that position.
To make a move using MENACE, the box with the current board position must be found. The operator then shakes the box and opens it. MENACE plays in the position corresponding to the colour of the bead at the front of the box.
For example, in this game, the first matchbox is opened to reveal a red bead at its front. This means that MENACE (O) plays in the corner. The human player (X) then plays in the centre. To make its next move, MENACE's operator finds the matchbox with the current position on, then opens it. This time it gives a blue bead which means MENACE plays in the bottom middle.
![](javascript:showlimage("menace-example.png"))
The human player then plays bottom right. Again MENACE's operator finds the box for the current position, it gives an orange bead and MENACE plays in the left middle. Finally the human player wins by playing top right.
MENACE has been beaten, but all is not lost. MENACE can now learn from its mistakes to stop the happening again.
How MENACE learns
MENACE lost the game above, so the beads that were chosen are removed from the boxes. This means that MENACE will be less likely to pick the same colours again and has learned. If MENACE had won, three beads of the chosen colour would have been added to each box, encouraging MENACE to do the same again. If a game is a draw, one bead is added to each box.
Initially, MENACE begins with four beads of each colour in the first move box, three in the third move boxes, two in the fifth move boxes and one in the final move boxes. Removing one bead from each box on losing means that later moves are more heavily discouraged. This helps MENACE learn more quickly, as the later moves are more likely to have led to the loss.
After a few games have been played, it is possible that some boxes may end up empty. If one of these boxes is to be used, then MENACE resigns. When playing against skilled players, it is possible that the first move box runs out of beads. In this case, MENACE should be reset with more beads in the earlier boxes to give it more time to learn before it starts resigning.
How MENACE performs
In Donald Michie's original tournament against MENACE, which lasted 220 games and 16 hours, MENACE drew consistently after 20 games.
![](javascript:showlimage("menace-graph.jpg"))
Taken from Trial and error by Donald Michie [2]
![](javascript:showlimage("michie-copy-annot.png"))
Graph showing MENACE's performance in the original tournament. Edit: Added the redrawn graph on the left.
After a while, Michie tried playing some more unusual games. For a while he was able to defeat MENACE, but MENACE quickly learnt to stop losing. You can read more about the original MENACE in A matchbox game learning-machine by Martin Gardner [1] and Trial and error by Donald Michie [2].
You may like to experiment with different tactics against MENACE yourself.
Play against MENACE
I have written a JavaScript implemenation of MENACE for you to play against. The source code for this implementation is available on GitHub.
When playing this version of MENACE, the contents of the matchboxes are shown on the right hand side of the page. The numbers shown on the boxes show how many beads corresponding to that move remain in the box. The red numbers show which beads have been picked in the current game.
The initial numbers of beads in the boxes and the incentives can be adjusted by clicking Adjust MENACE's settings above the matchboxes. My version of MENACE starts with more beads in each box than the original MENACE to prevent the early boxes from running out of beads, causing MENACE to resign.
Additionally, next to the board, you can set MENACE to play against random, or a player 2 version of MENACE.
Edit: After hearing me do a lightning talk about MENACE at CCC, Oliver Child built a copy of MENACE. Here are some pictures he sent me:
![](javascript:showlimage("menace1s.jpg"))
![](javascript:showlimage("menace3s.jpg"))
![](javascript:showlimage("menace4s.jpg"))
Edit: Oliver has written about MENACE and the version he built in issue 03 of Chalkdust Magazine.
Edit: Inspired by Oliver, I have built my own MENACE. I took it to the MathsJam Conference 2016. It looks like this:
![](javascript:showlimage("my_menace.jpg"))
References
[2] Trial and error by Donald Michie. Penguin Science Survey, 1961.
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Comments in green were written by me. Comments in blue were not written by me.
This is very neat. I wonder how long it would take to use that many matches to get all those match boxes.×14 ×14 ×7 ×3 ×3
Duke Nukem
×14 ×14 ×7 ×3 ×3 "When playing against skilled players, it is possible that the first move box runs out of beads. In this case, MENACE should be reset with more beads in the earlier boxes to give it more time to learn before it starts resigning."
If someone were doing this, you could do this automatically to avoid the perception or temptation of the operator to help it along. Instead of "oh, it's dead, let's repopulate the boxes", you could just make it part of the inter-game cleanup, like a garbage collection routine. After all the bead deleting/adding whatever, but before the next game starts, look at all the boxes, make sure that each box contains at least one of each color. Now this weakens the learning algorithm moderately, but it guarantees that it will never get stuck.×3 ×7 ×3 ×4 ×4
If someone were doing this, you could do this automatically to avoid the perception or temptation of the operator to help it along. Instead of "oh, it's dead, let's repopulate the boxes", you could just make it part of the inter-game cleanup, like a garbage collection routine. After all the bead deleting/adding whatever, but before the next game starts, look at all the boxes, make sure that each box contains at least one of each color. Now this weakens the learning algorithm moderately, but it guarantees that it will never get stuck.
(anonymous)
×3 ×7 ×3 ×4 ×4 @Matthew: Thank you for such a quick response. Just to let you know that that link did not work after .../tree/master/output, but I managed to search around for the right files :). In these files MENACE plays the Nought right? and the user plays the Cross?
×2 ×2 ×2 ×2 ×2
(anonymous)
×2 ×2 ×2 ×2 ×2 @Finlay: You can find them at https://github.com/mscroggs/MENACE-pdf.... The files boxes0.pdf to boxes3.pdf are the boxes for a MENACE that plays first.×2 ×4 ×3 ×2 ×3
Matthew
×2 ×4 ×3 ×2 ×3 Add a Comment
2015-03-25
This is an article which I wrote for the
first issue of
Chalkdust. I highly recommend reading the rest of the magazine (and trying
to solve the crossnumber I wrote for the issue).
In the classic arcade game Pac-Man, the player moves the title character through a maze. The aim of
the game is to eat all of the pac-dots that are spread throughout the maze while avoiding the ghosts
that prowl it.
While playing Pac-Man recently, my concentration drifted from the pac-dots and I began to think
about the best route I could take to complete the level.
Seven bridges of Königsberg
In the 1700s, Swiss mathematician Leonhard Euler studied a related problem. The city of
Königsberg had seven bridges, which the residents would try to cross while walking around the
town. However, they were unable to find a route crossing every bridge without repeating one of them.
![](javascript:showlimage("7bridges.png"))
Diagram showing the bridges in Königsberg. If you have not
seen this puzzle before, you may like to try to find a route crossing them all exactly once before reading on.
In fact, the city dwellers could not find such a route because it is impossible to do so, as
Euler proved in 1735. He first simplified the map of the city, by making the islands into vertices (or nodes)
and the bridges into edges.
![](javascript:showlimage("graph.png"))
A graph of the seven bridges problem.
This type of diagram has (slightly confusingly) become known as a graph, the study of which is
called graph theory. Euler represented Königsberg in this way as he realised that the shape of the
islands is irrelevant to the problem: representing the problem as a graph gets rid of this useless information
while keeping the important details of how the islands are connected.
Euler next noticed that if a route crossing all the bridges exactly once was possible then whenever
the walker took a bridge onto an island, they must take another bridge off the island. In this way,
the ends of the bridges at each island can be paired off. The only bridge ends that do not need a
pair are those at the start and end of the circuit.
This means that all of the vertices of the graph except two (the first and last in the route) must have an even number of edges connected to them; otherwise there is no route around the graph
travelling along each edge exactly once. In Königsberg, each island is connected to an odd number of bridges. Therefore the route that
the residents were looking for did not exist (a route now exists due to two of the bridges being destroyed during World War II).
This same idea can be applied to Pac-Man. By ignoring the parts of the maze without pac-dots the pac-graph
can be created, with the paths and the junctions forming the edges and vertices respectively. Once this
is done there will be twenty-four vertices, twenty of which will be connected to an odd number of edges, and so it is impossible to eat
all of the pac-dots without repeating some edges or travelling along parts of the maze with no pac-dots.
![](javascript:showlimage("pac-graph.png"))
The Pac-graph. The odd nodes are shown in red.
This is a start, but it does not give us the shortest route we can take to eat all of the pac-dots: in
order to do this, we are going to have to look at the odd vertices in more detail.
The Chinese postman problem
The task of finding the shortest route covering all the edges of a graph has become known as the
Chinese postman problem as it is faced by postmen—they need to walk along each street to post
letters and want to minimise the time spent walking along roads twice—and it was first studied by
Chinese mathematician Kwan Mei-Ko.
As the seven bridges of Königsberg problem demonstrated, when trying to find a route, Pac-Man
will get stuck at the odd vertices. To prevent this from happening, all the vertices can be made into even
vertices by adding edges to the graph. Adding an edge to the graph corresponds to choosing an edge, or sequence
of edges, for Pac-Man to repeat or including a part of the maze without pac-dots. In order to complete the level
with the shortest distance travelled, Pac-Man wants to add the shortest total length of edges to the graph.
Therefore, in order to find the best route, Pac-Man must look at different ways to pair off the odd vertices and choose
the pairing which will add the least total distance to the graph.
The Chinese postman problem and the Pac-Man problem are slightly different: it is usually assumed
that the postman wants to finish where he started so he can return home. Pac-Man however can finish
the level wherever he likes but his starting point is fixed. Pac-Man may therefore leave one odd
node unpaired but must add an edge to make the starting node odd.
One way to find the required route is to look at all possible ways to pair up the odd vertices. With a low
number of odd vertices this method works fine, but as the number of odd vertices increases, the
method quickly becomes slower.
With four odd vertices, there are three possible pairings. For the Pac-Man problem there will be
over 13 billion (\(1.37\times 10^{10}\)) pairings to check. These pairings can be checked by a
laptop running overnight, but for not too many more vertices this method quickly becomes unfeasible.
With 46 odd nodes there will be more than one pairing per atom in the human body (\(2.53\times 10^{28}\)).
By 110 odd vertices there will be more pairings (\(3.47\times 10^{88}\)) than there are estimated to be atoms in the
universe. Even the greatest supercomputer will be unable to work its way through
all these combinations.
Better algorithms are known for this problem that reduce the amount of work on larger graphs. The number of
pairings to check in the method above increases like the factorial of the number of vertices. Algorithms are
known for which the amount of work to be done increases like a polynomial in the number of vertices. These algorithms
will become unfeasible at a much slower rate but will still be unable to deal with very large graphs.
Solution of the Pac-Man problem
For the Pac-Man problem, the shortest pairing of the odd vertices requires the edges marked in red to be
repeated. Any route which repeats these edges will be optimal. For example, the route in green will be
optimal.
![](javascript:showlimage("repeats.png"))
![](javascript:showlimage("best-route.png"))
One important element of the Pac-Man gameplay that I have neglected are the ghosts (Blinky, Pinky,
Inky and Clyde), which Pac-Man must avoid. There is a high chance that the ghosts will at some point
block the route shown above and ruin Pac-Man's optimality. However, any route repeating the red edges
will be optimal: at many junctions Pac-Man will have a choice of edges he could continue along. It may
be possible for a quick thinking player to utilise this freedom to avoid the ghosts and complete an optimal
game.
Additionally, the skilled player may choose when to take the edges that include the power pellets,
which allow Pac-Man to reverse the roles and eat the ghosts. Again cleverly timing these may allow
the player to complete an optimal route.
Unfortunately, as soon as the optimal route is completed, Pac-Man moves to the next level and the
player has to do it all over again ad infinitum.
A video
Since writing this piece, I have been playing Pac-Man using
MAME (Multiple Arcade Machine Emulator). Here is one game I played along
with the optimal edges to repeat for reference:
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@William: You're right. In a number of places I could've turned round a few pixels earlier.
There seems to be no world record for just one Pac-Man level (and I don't have time to get good enough to speed run all 255 levels before it crashes!)×2 ×3 ×3 ×3 ×3
There seems to be no world record for just one Pac-Man level (and I don't have time to get good enough to speed run all 255 levels before it crashes!)
Matthew
×2 ×3 ×3 ×3 ×3 This vid was billed as an "optimal" run but around 40 seconds in you eat one "pill" that you don't need to eat. Why don't you just speedrun the first level? This must have been done before. Can you beat the world record?×3 ×3 ×3 ×3 ×3
William
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2015-03-24
This is the fourth post in a series of posts about tube map folding.
A while ago, I made this (a stellated rhombicuboctahedron):
![](javascript:showlimage("rhombi.jpg"))
Here are some hastily typed instructions for
Matt Parker, who is making one
at this month's Maths Jam. Other people are
welcome to follow these instructions too.
You will need
- 48 tube maps
- glue
Making a module
First, take a tube map and fold the cover over. This will ensure that your
shape will have tube (map and not index) on the outside and you will have
pages to tuck your tabs between later.
![](javascript:showlimage("rh_mp1.jpg"))
Now fold one corner diagonally across to another corner. It does not matter
which diagonal you chose for the first piece but after this all following pieces
must be the same as the first.
![](javascript:showlimage("rh_mp2.jpg"))
Now fold the overlapping bit back over the top.
![](javascript:showlimage("rh_mp3.jpg"))
Turn it over and fold this overlap over too.
![](javascript:showlimage("rh_mp4.jpg"))
You have made one module.
![](javascript:showlimage("rh_mp5.jpg"))
You will need 48 of these and some glue.
Putting it together
By slotting three or four of these modules together, you can make a
pyramid with a triangle or square as its base.
![](javascript:showlimage("rh_mp6.jpg"))
![](javascript:showlimage("rh_mp7.jpg"))
A stellated rhombicuboctahedron is a rhombicuboctahedron with a pyramid, or
stellation on each face. In other words, you now need to build a
rhombicuboctahedron with the bases of pyramids like these. A rhombicuboctahedron
looks like this:
![](javascript:showlimage("rh_mp_wik.jpg"))
en.wiki User Cyp, CC BY-SA 3.0
More usefully, its net looks like this:
![](javascript:showlimage("rh_mp_net.png"))
To build a stellated rhombicuboctahedron, make this net, but with each shape
as the base of a pyramid. This is what it will look like 6/48 tube maps in:
![](javascript:showlimage("rh_mp8.jpg"))
If you make on of these, please tweet me a photo so I can see it!
![](javascript:showlimage("rh_mp_fin.jpg"))
Previous post in series
This is the fourth post in a series of posts about tube map folding.
Next post in series
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I wish you'd make the final stellation of the rhombicuboctahedron! And show us! I know the shapes of the faces but have been stuck two years on the assembly!×2 ×2 ×2 ×2 ×2
Roberts, David
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2015-03-15
A few months ago, I set
@mathslogicbot (and @logicbot@mathstodon.xyz and @logicbot.bsky.social) going on the
long task of tweeting all the tautologies (containing 140 characters or less)
in propositional calculus with the symbols \(\neg\) (not), \(\rightarrow\)
(implies), \(\leftrightarrow\) (if and only if), \(\wedge\) (and) and \(\vee\)
(or). My first post on logic bot contains a full
explanation of propositional calculus, formulae and tautologies.
An alternative method
Since writing the original post, I have written an alternative script to
generate all the tautologies.
In this new method, I run through all possible strings of length 1 made
with character in the logical language, then strings of length 2, 3 and so on.
The script then checks if they are valid formulae and, if so, if they are
tautologies.
In the new script, only formulae where the first appearances of variables
are in alphabetical order are considered. This means that duplicate tautologies
are removed. For example, \((b\rightarrow(b\wedge a))\) will now be counted as
it is the same as \((a\rightarrow(a\wedge b))\).
You can view or download this alternative code on
github.
All the terms of the sequence that I have calculated so far can be viewed
here and the tautologies for these terms are
here.
Sequence
One advantage of this method is that it generates the tautologies sorted by
the number of symbols they contain, meaning we can generate the sequence whose
\(n\)th term is the number of tautologies of length \(n\).
The first ten terms of this sequence are
$$0, 0, 0, 0, 2, 2, 12, 6, 57, 88$$
as there are no tautologies of length less than 5; and, for example two
tautologies of length 6 (\((\neg a\vee a)\) and \((a\vee \neg a)\)).
This sequence is listed as
A256120 on OEIS.
Properties
There are a few properties of this sequence that can easily be shown.
Throughout this section I will use \(a_n\) to represent the \(n\)th
term of the sequence.
Firstly, \(a_{n+2}\geq a_n\). This can be explained as follows: let \(A\)
be a tautology of length \(n\). \(\neg\neg A\) will be of length \(n+2\) and
is logically equivalent to \(A\).
Another property is \(a_{n+4}\geq 2a_n\): given a tautology \(A\) of length
\(n\), both \((a\vee A)\) and \((A\vee a)\) will be tautologies of length
\(n+4\). Similar properties could be shown for \(\rightarrow\),
\(\leftrightarrow\) and \(\wedge\).
Given properties like this, one might predict that the sequence will be
increasing (\(a_{n+1}\geq a_n\)). However this is not true as \(a_7\) is 12
and \(a_8\) is only 6. It would be interesting to know at how many points in
the sequence there is a term that is less than the previous one. Given the
properties above it is reasonable to conjecture that this is the only one.
Edit: The sequence has been published on OEIS!
Edit: Added Mastodon and Bluesky links
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Great project! Would be interesting to have a version of this for the sheffer stroke.×3 ×3 ×3 ×1 ×3
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