56
\$\begingroup\$

In this challenge, the goal is to recreate the On-Line Encyclopedia of Integer Sequences one sequence at a time. Similar to the Evolution of Hello World, each answer depends on a previous answer.

Over time, this challenge will create a "family tree" of the OEIS sequences. It is simple to add on to this tree.

  1. Find a previous answer, which can be at any depth N of the tree.
  2. Determine the first N numbers generated by that answer's sequence.
  3. Find a sequence in OEIS which starts with those same numbers and which hasn't been used before.
  4. Write a program to generate this new sequence you just found.
  5. Submit your answer as depth N+1

Since the level of your answer influences scoring, you should always add your answer onto the tree at the deepest level possible. If you cannot fit your answer anywhere on the tree, you can start a new branch of the tree and put your answer as depth 1.

Answer Requirements

There are a few ways to output a sequence.

The first option is to write a program or function that inputs a number (from STDIN or as an argument) and returns the Nth number in your chosen sequence. You can assume that the sequence will be defined for N and that N and S_N are "reasonably sized" (so it won't cause overflows). You can also use any reasonable indexing, such as 0 indexing, 1 indexing, or the indexing listed under "offset" on the sequence's OEIS page, that doesn't matter. The term produced by the first index must match the first term of the OEIS entry.

The second option is to write a program or function that inputs a number and returns the first N terms of the sequence. The first terms of the output must be the first terms of the OEIS entry (you can't leave off the first few terms). Consecutive terms must be delimited by arbitrary strings of non-digit characters, so 0,1 1.2/3,5;8,11 works but 011235811 does not count.

The third option is to create a program that outputs a continuous stream of numbers. Similarly to the second option, there must be delimiters between consecutive terms.

Your answer should contain a header like this to aid Stack Snippet parsing:

 # [language], [number] bytes, depth [number], A[new sequence] from A[old sequence] 

Your answer should contain the code to generate the sequence, along with the first few terms that any descendants will need to contain. These few terms should be preceded by the exact word terms: so that the controller can use them as part of the tree diagram. It is also recommended to write a description of the sequence you chose.

If your post is a depth 1 answer and thus has no ancestor, you should simply omit the from A[number] in your header.

Here is an example answer:

# Perl, 26 bytes, depth 3, A026305 from A084912

    various code here
    and here

The next answer should match the following terms:

    1, 4, 20

This sequence is .... and does ....

Chaining Requirements

In order to make this challenge more fair, there are restrictions on which answers you can chain yours to. These rules are mostly to prevent a single person from creating a whole branch of the tree by themselves or owning a lot of "root" nodes.

  • You cannot chain to yourself.
  • You cannot directly chain two of your answers to the same ancestor.
  • You cannot make more than one "Level 1" answer.

Also, if the ancestor was of depth N, your post must have depth N+1, even if more than the required number of terms agree.

Scoring

Your score as a user is the sum of the scores of all of your answers. The score of a single answer is determined by the following formula:

Answer Score = Sqrt(Depth) * 1024 / (Length + 256)

This scoring system should encourage users to submit a large number of deeper answers. Shorter answers are preferred over longer answers, but depth has a much larger influence.

Below is a stack snippet that generates a leaderboard as well as a tree diagram of all of the answers. I would like to thank Martin Büttner and d3noob as the sources for a lot of this code. You should click "Full screen" to see the complete results.

function answersUrl(t){return"https://api.stackexchange.com/2.2/questions/"+QUESTION_ID+"/answers?page="+t+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+ANSWER_FILTER}function getAnswers(){jQuery.ajax({url:answersUrl(page++),method:"get",dataType:"jsonp",crossDomain:!0,success:function(t){answers.push.apply(answers,t.items),t.has_more?getAnswers():process()}})}function shouldHaveHeading(t){var e=!1,r=t.body_markdown.split("\n");try{e|=/^#/.test(t.body_markdown),e|=["-","="].indexOf(r[1][0])>-1,e&=LANGUAGE_REG.test(t.body_markdown)}catch(a){}return e}function shouldHaveScore(t){var e=!1;try{e|=SIZE_REG.test(t.body_markdown.split("\n")[0])}catch(r){}return e}function getAuthorName(t){return t.owner.display_name}function decodeEntities(t){return $("<textarea>").html(t).text()}function process(){answers=answers.filter(shouldHaveScore).filter(shouldHaveHeading),answers.reverse();var t={},e=[],r=1,a=null,n=1,s=[];answers.forEach(function(t){var r=t.body_markdown.split("\n")[0],a=getAuthorName(t),n=r.match(SEQUENCE_REG)[0];n=n.trim();var o="from A000000";PARENT_REG.test(r)&&(o=r.match(PARENT_REG)[0]),o=o.substring(5).trim(),"A000000"==o&&(o="OEIS");var i="";SEQDATA_REG.test(t.body_markdown)&&(i=t.body_markdown.match(SEQDATA_REG)[1]);for(var u=!0,c=0;c<e.length;++c)u=u&&!(e[c]===n);for(var l=!0,c=0;c<e.length;++c)l=!(!l||e[c]===n||e[c]===n+a||e[c]===o+a);e.push(n),e.push(n+a),e.push(o+a),u&&data.push({name:n,parent:o,term:i+" : ",author:decodeEntities(a),URL:t.share_link}),l&&s.push(t)}),answers.sort(function(t,e){var r=t.body_markdown.split("\n")[0].match(SEQUENCE_REG),a=e.body_markdown.split("\n")[0].match(SEQUENCE_REG);return a>r?-1:r>a?1:void 0}),answers.forEach(function(e){var o=e.body_markdown.split("\n")[0],i=(o.match(NUMBER_REG)[0],(o.match(SIZE_REG)||[0])[0]),u=parseInt((o.match(DEPTH_REG)||[0])[0]).toString(),c=o.match(SEQUENCE_REG)[0],l="from A000000";PARENT_REG.test(o)&&(l=o.match(PARENT_REG)[0]),l=l.substring(5);var d=o.match(LANGUAGE_REG)[1];d.indexOf("]")>0&&(d=d.substring(1,d.indexOf("]")));for(var p=getAuthorName(e),E=!1,h=0;h<s.length;++h)E=E||s[h]===e;if(E){var f=jQuery("#answer-template").html();i!=a&&(n=r),a=i,++r;var m=1024*Math.pow(parseInt(u),.5)/(parseInt(i)+256);f=f.replace("{{SEQUENCE}}",c).replace("{{SEQUENCE}}",c).replace("{{NAME}}",p).replace("{{LANGUAGE}}",d).replace("{{SIZE}}",i).replace("{{DEPTH}}",u).replace("{{LINK}}",e.share_link),f=jQuery(f),jQuery("#answers").append(f),t[p]=t[p]||{lang:d,user:p,size:"0",numanswers:"0",link:e.share_link},t[p].size=(parseFloat(t[p].size)+m).toString(),t[p].numanswers=(parseInt(t[p].numanswers)+1).toString()}});var o=[];for(var i in t)t.hasOwnProperty(i)&&o.push(t[i]);o.sort(function(t,e){return parseFloat(t.size)>parseFloat(e.size)?-1:parseFloat(t.size)<parseFloat(e.size)?1:0});for(var u=0;u<o.length;++u){var c=jQuery("#language-template").html(),i=o[u];c=c.replace("{{RANK}}",u+1+".").replace("{{NAME}}",i.user).replace("{{NUMANSWERS}}",i.numanswers).replace("{{SIZE}}",i.size),c=jQuery(c),jQuery("#languages").append(c)}createTree()}function createTree(){function t(){var t=i.nodes(root).reverse(),e=i.links(t);t.forEach(function(t){t.y=180*t.depth});var r=c.selectAll("g.node").data(t,function(t){return t.id||(t.id=++o)}),a=r.enter().append("g").attr("class","node").attr("transform",function(t){return"translate("+t.y+","+t.x+")"});a.append("a").attr("xlink:href",function(t){return t.URL}).append("circle").attr("r",10).style("fill","#fff"),a.append("text").attr("x",function(){return 0}).attr("y",function(){return 20}).attr("dy",".35em").attr("text-anchor",function(){return"middle"}).text(function(t){return t.term+t.name}).style("fill-opacity",1),a.append("text").attr("x",function(){return 0}).attr("y",function(){return 35}).attr("dy",".35em").attr("text-anchor",function(){return"middle"}).text(function(t){return t.author}).style("fill-opacity",1);var n=c.selectAll("path.link").data(e,function(t){return t.target.id});n.enter().insert("path","g").attr("class","link").attr("d",u)}var e=data.reduce(function(t,e){return t[e.name]=e,t},{}),r=[];data.forEach(function(t){var a=e[t.parent];a?(a.children||(a.children=[])).push(t):r.push(t)});var a={top:20,right:120,bottom:20,left:120},n=3203-a.right-a.left,s=4003-a.top-a.bottom,o=0,i=d3.layout.tree().size([s,n]),u=d3.svg.diagonal().projection(function(t){return[t.y,t.x]}),c=d3.select("body").append("svg").attr("width",n+a.right+a.left).attr("height",s+a.top+a.bottom).append("g").attr("transform","translate("+a.left+","+a.top+")");root=r[0],t(root)}var QUESTION_ID=49223,ANSWER_FILTER="!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe",data=[{name:"OEIS",parent:"null",term:"",author:"",URL:"https://oeis.org/"}],answers=[],page=1;getAnswers();var SIZE_REG=/\d+(?=[^\d&]*(?:&lt;(?:s&gt;[^&]*&lt;\/s&gt;|[^&]+&gt;)[^\d&]*)*,)/,DEPTH_REG=/\d+, A/,NUMBER_REG=/\d+/,LANGUAGE_REG=/^#*\s*([^,]+)/,SEQUENCE_REG=/A\d+/,PARENT_REG=/from\s*A\d+/,SEQDATA_REG=/terms:\s*(?:(?:-)?\d+,\s*)*((?:-)?\d+)/;
body{text-align: left !important}#answer-list{padding: 10px; width: 550px; float: left;}#language-list{padding: 10px; width: 290px; float: left;}table thead{font-weight: bold;}table td{padding: 5px;}.node circle{fill: #fff; stroke: steelblue; stroke-width: 3px;}.node text{font: 12px sans-serif;}.link{fill: none; stroke: #ccc; stroke-width: 2px;}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script><script src="http://d3js.org/d3.v3.min.js"></script><link rel="stylesheet" type="text/css" href="//cdn.sstatic.net/codegolf/all.css?v=83c949450c8b"><div id="answer-list"> <h2>Sequence List</h2> <table class="answer-list"> <thead> <tr> <td>Sequence</td><td>Author</td><td>Language</td><td>Size</td><td>Depth</td></tr></thead> <tbody id="answers"></tbody> </table></div><div id="language-list"> <h2>Leaderboard</h2> <table class="language-list"> <thead> <tr> <td>Rank</td><td>User</td><td>Answers</td><td>Score</td></tr></thead> <tbody id="languages"></tbody> </table></div><table style="display: none"> <tbody id="answer-template"> <tr> <td><a href="https://oeis.org/{{SEQUENCE}}">{{SEQUENCE}}</a></td><td>{{NAME}}</td><td>{{LANGUAGE}}</td><td>{{SIZE}}</td><td>{{DEPTH}}</td><td><a href="{{LINK}}">Link</a> </td></tr></tbody></table><table style="display: none"> <tbody id="language-template"> <tr> <td>{{RANK}}</td><td>{{NAME}}</td><td>{{NUMANSWERS}}</td><td>{{SIZE}}</td></tr></tbody></table>

\$\endgroup\$
  • 5
    \$\begingroup\$ You know, I think this is might just be the single coolest codegolf.sx question I've ever seen asked. It's not just cool, but actually useful as an archive. \$\endgroup\$ – Todd Lehman Apr 26 '15 at 19:08
  • 3
    \$\begingroup\$ Given the OEIS is online, takes N terms of a sequence as a search term, and contains mathematica or maple code for many of the sequences, it would be possible to write a meta-entry which searched for the best scoring entry for which code exists in OEIS which is a descendent of any given entry here and posted it. \$\endgroup\$ – abligh Apr 26 '15 at 19:10
  • 2
    \$\begingroup\$ Can I recommend some way of marking on the graph the snippet generates that a node is terminal, ie there are no unused sequences of a greater depth available on the OEIS? \$\endgroup\$ – Claudiu Apr 27 '15 at 5:52
  • 1
    \$\begingroup\$ I think the only way to keep this challenge going would be to provide something where you give your username, and it lists the OEIS problems you could do, in order from highest depth to lowest. Otherwise it takes too long to find the next sequence to post. \$\endgroup\$ – Claudiu May 4 '15 at 22:38
  • 1
    \$\begingroup\$ The SVG is slightly too narrow. \$\endgroup\$ – CalculatorFeline Apr 8 '16 at 1:17

146 Answers 146

0
\$\begingroup\$

Python 2, 59 bytes, depth 4, A000123 from A000125

i=0;j=1;s=1,
while 1:print s[-1];s+=(s[-1]+s[i],);i+=j;j^=1

Try it online. Uses a modified version so you can actually read the output, since infinite output is hard to read.

The next answer should match the following terms:

1, 2, 4, 6
\$\endgroup\$
  • \$\begingroup\$ Haha. My score is 0... \$\endgroup\$ – mbomb007 Mar 9 '16 at 22:36
  • \$\begingroup\$ This is because it has a link in the answer title. This would be one aspect of the parser to improve if a similar challenge is ever posted in the future. \$\endgroup\$ – PhiNotPi Mar 19 '17 at 15:22
0
\$\begingroup\$

Haskell, 32 bytes, depth 7, A018819 from A016116

l=1:zipWith(+)l(tail$l>>=(:[0]))

This defines an infinite list containing all the values.

As more easily readable function that (could be golfed some more and) is much less efficient the sequence can be defined by

a 0=1; a m|odd m=a(m-1); a m=a(m-1)+a(m`div`2)

So it starts with 1, and then we always take the previous value and, in the even case, add the value from m/2. The list definition uses l>>=(:[0]) to create a list that alternates between values from l and zeroes ([l0,0,l1,0,l2,0,...]) to achieve the same effect.

The sequence is the binary partition function, giving the number of partitions of a number into powers of 2, for example for 5 it is 4, because 5 can be written in four ways: as 1+1+1+1+1, 1+1+1+2, 1+2+2 and 1+4.

The next sequence should match the following terms:

1, 1, 2, 2, 4, 4, 6
\$\endgroup\$
1
\$\begingroup\$

Python, 14 bytes, depth 3, A010872 from A000030

x=lambda x:x%3

The next sequence should start with the following terms:

0, 1, 2
\$\endgroup\$
0
\$\begingroup\$

Mathematica, 30 bytes, depth 9, A038668 from A057359

Floor[#/LucasL@a]~Sum~{a,#}-#&

An = ⌊n/L(2)⌋ + ⌊n/L(3)⌋ + ⌊n/L(4)⌋ + ⌊n/L(5)⌋ + ⋯, where L(n) denotes the nth term in the Lucas sequence. The next answer should start with the terms:

0, 0, 1, 2, 2, 3, 4, 5, 6
\$\endgroup\$
0
\$\begingroup\$

Mathematica, 48 bytes, depth 9, A025640 from A022336

SortBy[0~Range~#~Tuples~2,3^# 4^#2&@@#&][[#,1]]&

Exponent of 3 (value of i) in nth number of form 3i∙4j. The next answer should start with the terms:

0, 1, 0, 2, 1, 0, 3, 2, 1
\$\endgroup\$
0
\$\begingroup\$

Mathematica, 20 bytes, depth 10, A225545 from A098388

Floor@*Exp@*LambertW

Does what it says on the tin (takes ⌊eW(n)⌋). LambertW appears to be an alias for ProductLog. The next answer should start with the terms:

1, 1, 2, 2, 3, 3, 4, 4, 4, 5
\$\endgroup\$
2
\$\begingroup\$

Mathematica, 11 bytes, depth 15, A086388 from A157271

Round[#/2]&

A086388 is just a duplicate of A008619. However, the question didn't disallow dead sequences, so I assume that this is valid. The next answer should start with the terms:

1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8
\$\endgroup\$
3
\$\begingroup\$

TI-BASIC, 10 bytes, depth 5, A181857 from A001563

Input X
lcm(X²,X!

A181857 represents the least common multiple of the squares and factorials. The code length for TI-BASIC is always confusing, because it uses tokens, so most tokens are fewer bytes than their Unicode byte count. The next sequence should start with the terms:

0, 1, 4, 18, 48
\$\endgroup\$
  • \$\begingroup\$ Input X \n lcm( X ² , X \$\endgroup\$ – CalculatorFeline Apr 8 '16 at 1:11
  • \$\begingroup\$ @CatsAreFluffy Not understanding what you're saying... \$\endgroup\$ – NinjaBearMonkey Apr 8 '16 at 1:16
  • \$\begingroup\$ Aren't those the 9 bytes? Oh wait, missed the ! \$\endgroup\$ – CalculatorFeline Apr 8 '16 at 1:19
  • \$\begingroup\$ Oh wait you're right, lcm is 2 bytes. \$\endgroup\$ – NinjaBearMonkey Apr 8 '16 at 1:21
0
\$\begingroup\$

Mathematica, 98 bytes, depth 10, A175835 from A083912

Count[Normal@SeriesData[x,0,Reverse@RealDigits[EulerGamma,10,#][[1]],0,#,1]~NRoots~x,_Real,{1,2}]&

Pretty random sequence definition. Makes more sense if you read the sequence page. The next answer should start with the terms:

0, 1, 0, 1, 0, 1, 0, 1, 0, 1
\$\endgroup\$
0
\$\begingroup\$

Mathematica, 68 bytes, depth 11, A096088 from A096972

For[a=1,1>0,Print/@Union@Flatten[PowerModList[#,4,a]&/@0~Range~a++]]

Outputs terms, followed by newlines. The sequence definition is pretty complicated, so I won't go into it here. The next answer should start with the terms:

0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 3
\$\endgroup\$
0
\$\begingroup\$

Mathematica, 34 bytes, depth 10, A173923 from A045841

Floor[#~IntegerReverse~2/8]~Mod~2&

Uses a definition given with the sequence. Requires Mathematica 10.3 or higher to run. Starts at index 8. The next answer should start with the terms:

0, 1, 0, 1, 0, 1, 0, 1, 0, 0
\$\endgroup\$
0
\$\begingroup\$

Mathematica, 40 bytes, depth 10, A096125 from A026233

NestWhile[#+1&,a=#;1,!((a-#)!^2∣a!)&]&

Just increments k until the condition is met. See the sequence definition for more information. The next answer should start with the terms:

1, 1, 2, 2, 3, 3, 4, 4, 5, 4
\$\endgroup\$
1
\$\begingroup\$

Mathematica, 51 bytes, depth 11, A025765 from A226190

SeriesCoefficient[1/((1-x)(1-x^2)(1-x^9)),{x,0,#}]&

Just another infinite series. The next answer should start with the terms:

1, 1, 2, 2, 3, 3, 4, 4, 5, 6, 7
\$\endgroup\$
0
\$\begingroup\$

Mathematica, 58 bytes, depth 12, A029031 from A025766

SeriesCoefficient[1/Times@@(1-x^#&)/@{1,2,11,12},{x,0,#}]&

Just uses the definition of the series. The next answer should start with the terms:

1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 7
\$\endgroup\$
0
\$\begingroup\$

Mathematica, 11 bytes, depth 4, A004188 from A002275

(3#^3-#)/2&

The next answer should start with the terms:

0, 1, 11, 39
\$\endgroup\$
0
\$\begingroup\$

Mathematica, 14 bytes, depth 5, A129323 from A001563

BellB[#-1,2]#&

Derived from the Mathematica program given with the sequence. Ignore any messages. The next answer should start with the terms:

0, 1, 4, 18, 88
\$\endgroup\$
1
\$\begingroup\$

Ruby, 30 bytes, depth 6, A010051 from A209229

->n{(2..n).one?{|x|n%x<1}?1:0}

The next answer should start with the terms:

0,1,1,0,1,0
\$\endgroup\$
0
\$\begingroup\$

𝔼𝕊𝕄𝕚𝕟, noncompeting, depth 4, A001614 from A000125

17 chars / 25 bytes:

2*ï-⌊ ½+½*√ 8*ï-7

Try it here (Firefox only).

The next answer should match:

1, 2, 4, 5
\$\endgroup\$
  • \$\begingroup\$ Language made after this challenge. \$\endgroup\$ – Mama Fun Roll Mar 9 '16 at 21:34
  • 1
    \$\begingroup\$ I think that non-competing answers shouldn't be posted for this. At the very least, it's messing up the snippet/tree. Also, you'd need to move "noncompeting" somewhere else. Your answer has to follow the format. \$\endgroup\$ – mbomb007 Mar 9 '16 at 22:29
0
\$\begingroup\$

Seriously, 14 bytes, depth 3, A000125 from A000027

,;3@^@5*+6+6@\

Try it online!

The next answer should match the following terms:

1, 2, 4
\$\endgroup\$
  • \$\begingroup\$ @ՊՓԼՃՐՊՃՈԲՍԼ It's not required to chain off of the most recent answer. \$\endgroup\$ – Mego Feb 1 '16 at 4:59
6
\$\begingroup\$

Clip, 0 bytes, depth 2, A000027 from A000012

Given a number n, prints the nth number in the sequence 1, 2, 3, 4...

The next answer should start with the terms:

1, 2
\$\endgroup\$
1
\$\begingroup\$

Ruby, 26 bytes, depth 4, A006882 from A000142

a=->n{n>1?n*a.call(n-2):1}

The next answer should start with the terms:

1,1,2,3

The Double factorial sequence.

\$\endgroup\$
1
\$\begingroup\$

Ruby, 35 bytes, depth 6, A008998 from A014137

a=->n{n<0?0:n>0?2*a(n-1)+a(n-3):1}

The next sequence should contain the following terms:

1, 2, 4, 9, 20, 44

Defined recursively as:a(n) = 2*a(n-1) + a(n-3).

\$\endgroup\$
1
\$\begingroup\$

Ruby, 75 bytes, depth 2, A125710 from A000211

def r n;n==2??0:n%2==0??0+r(n/2):?1+r(n*3+1);end;p (r 2*gets.to_i+1).to_i 2

The next sequence should contain the following terms:

4, 80

This is a complicated sequence so I will just quote OEIS:

In the "3x+1" problem, let 0 denote a halving step and 1 denote an x->3x+1 step. Then a(n) is obtained by writing the sequence of steps needed to reach 1 from 2n+1 and reading it as a decimal number.

My code is not short; but It was fun to make.

\$\endgroup\$
1
\$\begingroup\$

Java, 30 bytes, depth 6, A016116 from A000010

int f(int n){return 1<<(n/2);}

This sequence is the powers of two, twice. This is a zero-indexed function. The next answer should match the following terms:

1, 1, 2, 2, 4, 4
\$\endgroup\$
1
\$\begingroup\$

Java, 29 bytes, depth 5, A000325 from A000108

int f(int n){return(1<<n)-n;}

This sequence is 2n-n for all n >= 0. This is a zero-indexed function. The next answer should start with the following terms:

1, 1, 2, 5, 12
\$\endgroup\$
1
\$\begingroup\$

Java, 186 bytes, depth 10, A073719 from A072490

int f(int n){return g(1<<n,0)/g(1<<n,1);}int g(int n,int o){int k=2,c=1;while(c<n)if(h(++k)>0&&o<1||h(k)<1&&o>0)c++;return k;}int h(int n){for(int k=2;k<n;)if(n%k++<1)return 0;return 1;}

Long version:

int f(int n) {
    return g(1<<n,0) / g(1<<n,1);
}

int g(int n, int o) {
    int k=2, c=1;
    while(c<n)
        if (h(++k)>0 && o<1 || h(k)<1 && o>0)
            c++;
    return k;
}

int h(int n) {
    for (int k=2; k<n;)
        if (n % k++ < 1)
            return 0;
    return 1;
}

This is the sequence of the 2nth prime numbers divided by the 2nth composite numbers for all natural numbers greater than 1, rounded down. Method g finds the 2nth prime or composite number while method h checks if numbers are prime or composite. Due to the inefficiency of the algorithm, lag becomes apparent at n = 12.

This is a one-indexed function (n starts at 1). The next answer should start with the following terms:

0, 0, 1, 2, 2, 3, 4, 5, 5, 6
\$\endgroup\$
  • \$\begingroup\$ Oh that's what prime(2^n) meant... really had trouble figuring that out for some reason. \$\endgroup\$ – Claudiu Apr 29 '15 at 2:40
  • \$\begingroup\$ I was also confused by that at first. What really hung me up was the formula for composite numbers; the denominator of the first one made no sense. :P \$\endgroup\$ – TNT Apr 29 '15 at 3:13
1
\$\begingroup\$

Piet, 96 bytes, depth 5, A209229 from A010060

Piet program

with size 10 codels for legibility. Tested on PietDev. The next sequence should start with the terms:

0, 1, 1, 0, 1

This is the characteristic function of powers of 2 (i. e. it outputs 1 if n is a power of 2 and 0 if not). It starts on the red codel in the first row: it takes input as stdin and pushes it to the stack n, it then duplicates it twice n n nand pushes 1 to the stack n n n 1 [the loop starts on the red codel at the end of the second row] that it also duplicates twice n n n 1 1 1 it then rolls the first one near the bottom n 1 n n 1 1 and again n 1 1 n n 1 then check if the first two are equal (if yes than it is a power of 2 so it change direction and outputs 1), then check if the first number of the stack (on first iteration, 1) is smaller than the second. If yes it stops and output 0 and if no it continues on the loop (at that point the stack only contains n 1 as the first two pairs of n 1 were used for the comparisons). It pushes 2 on the stack and multiply n 1 2->n 2 then rolls it at the bottom, 2 n, duplicates twice 2 n n n rolls the bottom number on top of the stack n n n 2 and reaches back the beginning of the loop.

\$\endgroup\$
1
\$\begingroup\$

R, 79 bytes, depth 3, A000945 from A000043

f=function(n)if(n-1){m=prod(sapply(1:(n-1),f))+1;p=2;while(m%%p)p=p+1;p}else{2}

The next sequence should contain the following terms:

2, 3, 7

The sequence is such that a(n+1) is the smallest prime factor of prod(a(k) for k in 1:n)+1. f=function(n)if(n-1){m=prod(sapply(1:(n-1),f))+1;p=(2:m)[!m%%2:m];p[1]}else{2} would be one character shorter but can't perform f(9) because it can't create of vector of length 38 709 183 810 570.

\$\endgroup\$
0
\$\begingroup\$

Python 3, 66 bytes, depth 2, A000043 from A000040

n=2
while n:k=2**n-1;any(k%d<1for d in range(2,k))or print(n);n+=1

The next answer should match the following terms:

2, 3

These are the positive integers n such that 2^n - 1 is prime, i.e. n is the exponent of a Mersenne prime. It's easy to see that n itself must be prime, since if d divides n then 2^d - 1 is a divisor of 2^n - 1.

Equivalently, by the Euclid-Euler theorem, these are also the integers n such that 2^(n-1) (2^n - 1) is an even perfect number (sum of divisors equals itself).

This Python program uses trial division, so it only manages to get up to 19 before it starts taking a while. For comparison, simply replacing the primality check with the probabilistic Miller-Rabin manages up to 607 in a reasonable amount of time.

\$\endgroup\$
3
\$\begingroup\$

Piet, 56 bytes, depth 4, A000124 from A000079

Piet program

with size 10 codels for legibility. The next sequence should start with the terms:

1, 2, 4, 7

This sequence is the "maximal number of pieces formed when slicing a pancake with n cuts" (i. e. sum(1:n) + 1 ). The sequence was so simple so i thought i could increase the difficulty a little by doing it in Piet. I tested it on PietDev.

It pushes 1 on the stack 1, then the number provided with stdin 1 n, duplicate the latter 1 n n [here the loop begins = bright green codel], pushes 1 1 n n 1, substracts it 1 n n-1, duplicates the result twice 1 n n-1 n-1 n-1 (the deeper one will be then one we ll add to n, the second one the one we ll compare to zero and the third one the loop counter), pushes 4 and then 1 1 n n-1 n-1 n-1 4 1 uses the last two to roll the "loop counter" to the bottom 1 n-1 n n-1 n-1, check if n-1 is equal to 0: if no the stack becomes 1 n-1 n n-1 0, which we negates 1 n-1 n n-1 1 and uses the 1 to change the pointer direction 1 n-1 n n-1, adds n-1 to n 1 n-1 (n+n-1) which we roll to the bottom with a 2 and 1 that we just pushed on the stack thus leaving the stack in this state 1 (n+n-1) n-1 when it starts the loop over. When the counter indeed reaches 0, it exits the loop with a stack looking like this 1 0 Sum(1:n) 0, it then pops the first number, adds the rest and outputs as a numeric.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.