One OEIS after another

Question

As of 13/03/2018 16:45 UTC, the winner is answer #345, by Khuldraeseth na'Barya. This means the contest is officially over, but feel free to continue posting answers, just so long as they follow the rules.

As well, just a quick shout out to the top three answerers in terms of numbers of answers:

1. NieDzejkob - 41 answers

2. KSmarts - 30 answers

3. Hyper Neutrino - 26 answers

---

This is an answer chaining question that uses sequences from OEIS, and the length of the previous submission.

This answer chaining question will work in the following way:

• I will post the first answer. All other solutions must stem from that.
• The next user (let's call them userA) will find the OEIS sequence in which its index number (see below) is the same as the length of my code.
• Using the sequence, they must then code, in an unused language, a program that takes an integer as input, n, and outputs the nth number in that sequence.
• Next, they post their solution after mine, and a new user (userB) must repeat the same thing.

The nth term of a sequence is the term n times after the first, working with the first value being the first value given on its OEIS page. In this question, we will use 0-indexing for these sequences. For example, with A000242 and n = 3, the correct result would be 25.

However!

This is not a code-golf, so shortest code doesn't matter. But the length of your code does still have an impact. To prevent the duplication of sequences, your bytecount must be unique. This means that no other program submitted here can be the same length in bytes as yours.

If there isn't a sequence for then length of the last post, then the sequence for your post is the lowest unused sequence. This means that the sequences used also have to be unique, and that the sequence cannot be the same as your bytecount.

After an answer has been posted and no new answers have been posted for more than a week, the answer before the last posted (the one who didn't break the chain) will win.

Input and Output

Generic input and output rules apply. Input must be an integer or a string representation of an integer and output must be the correct value in the sequence.

Formatting

As with most answer-chaining questions, please format your answer like this

# N. language, length, [sequence](link)

`code`

[next sequence](link)

*anything else*

Rules

• You must wait for at least 1 hour before posting an answer, after having posted.
• You may not post twice (or more) in a row.
• The index number of a sequence is the number after the A part, and with leading zeros removed (e.g. for A000040 the index number is 40)
• You can assume that neither the input nor the required output will be outside your languages numerical range, but please don't abuse this by choosing a language that can only use the number 1, for example.
• If the length of your submission is greater than 65536 characters long, please provide a link to a way to access the code (pastebin for example).
• n will never be larger than 1000, or be out of bounds for the sequence, simply to prevent accuracy discrepancies from stopping a language from competing.
• Every 150 (valid) answers, the number of times a language may be used increases. So after 150 solutions have been posted, every language may be used twice (with all previous answers counting towards this). For instance, when 150 answers have been posted, Python 3 may be used twice, but due to the fact that it has already been used once, this means it can only be used once more until 300 answers have been posted.
• Please be helpful and post a link to the next sequence to be used. This isn't required, but is a recommendation.
• Different versions of languages, e.g. Python 2 and Python 3 are different languages. As a general rule, if the different versions are both available on Try It Online, they are different languages, but keep in mind that this is a general rule and not a rigid answer.
• It is not banned, but please try not to copy the code from the OEIS page, and actually try to solve it.
• Hardcoding is only allowed if the sequence is finite. Please note that the answer that prompted this (#40) is the exception to the rule. A few answers early in the chain hardcode, but these can be ignored, as there is no good in deleting the chain up to, say, #100.

Answer chain snippet

var QUESTION_ID=133754,OVERRIDE_USER=66833;function shareUrl(i){return"https://codegolf.stackexchange.com/a/"+i}function answersUrl(e){return"https://api.stackexchange.com/2.2/questions/"+QUESTION_ID+"/answers?page="+e+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+ANSWER_FILTER}function commentUrl(e,s){return"https://api.stackexchange.com/2.2/answers/"+s.join(";")+"/comments?page="+e+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+COMMENT_FILTER}function getTemplate(s){return jQuery(jQuery("#answer-template").html().replace("{{PLACE}}",s.index+".").replace("{{NAME}}",s.user).replace("{{LANGUAGE}}",s.language).replace("{{SEQUENCE}}",s.sequence).replace("{{SIZE}}",s.size).replace("{{LINK}}",s.link))}function search(l,q){m=jQuery("<tbody id='answers'></tbody>");e.forEach(function(s){if(!q||(l==0&&RegExp('^'+q,'i').exec(s.lang_name))||(l==1&&q===''+s.size)){m.append(jQuery(getTemplate(s)))}});jQuery("#answers").remove();jQuery(".answer-list").append(m)}function sortby(ix){t=document.querySelector('#answers');_els=t.querySelectorAll('tr');els=[];for(var i=0;i<_els.length;i++){els.push(_els[i]);}els.sortBy(function(a){a=a.cells[ix].innerText;return ix==0||ix==4?Number(a):a.toLowerCase()});for(var i=0;i<els.length;i++)t.appendChild(els[i]);}function checkSize(x){if(!x)return jQuery("#size-used").text("");var i=b.indexOf(+x);if(i<0)return jQuery("#size-used").text("Available!");var low=+x,high=+x;while(~b.indexOf(low))low--;while(~b.indexOf(high))high++;jQuery("#size-used").text(("Not available. The nearest are "+low+" and "+high).replace("are 0 and","is"))}function checkLang(x){}function getAnswers(){jQuery.ajax({url:answersUrl(answer_page++),method:"get",dataType:"jsonp",crossDomain:!0,success:function(e){answers.push.apply(answers,e.items),answers_hash=[],answer_ids=[],e.items.forEach(function(e){e.comments=[];var s=+e.answer_id;answer_ids.push(s),answers_hash[s]=e}),e.has_more||(more_answers=!1),comment_page=1,getComments()}})}function getComments(){jQuery.ajax({url:commentUrl(comment_page++,answer_ids),method:"get",dataType:"jsonp",crossDomain:!0,success:function(e){e.items.forEach(function(e){e.owner.user_id===OVERRIDE_USER&&answers_hash[e.post_id].comments.push(e)}),e.has_more?getComments():more_answers?getAnswers():process()}})}function getAuthorName(e){return (e.owner.user_id==OVERRIDE_USER?"<span id='question-author'>"+e.owner.display_name+"</span>":e.owner.display_name)}function process(){b=[];c=[];answers.forEach(function(s){var r=s.body;s.comments.forEach(function(e){OVERRIDE_REG.test(e.body)&&(r="<h1>"+e.body.replace(OVERRIDE_REG,"")+"</h1>")});var a=r.match(SCORE_REG);if(a){e.push({user:getAuthorName(s),size:+a[4],language:a[2],lang_name:a[3],index:+a[1],sequence:a[5],link:shareUrl(s.answer_id)});if(b.indexOf(+a[4])>=0&&c.indexOf(+a[4])<0){c.push(+a[4])};b.push(+a[4])}else{jQuery('#weird-answers').append('<a href="'+shareUrl(s.answer_id)+'">This answer</a> is not formatted correctly. <b>Do not trust the information provided by this snippet until this message disappears.</b><br />')}}),e.sortBy(function(e){return e.index});e.forEach(function(e){jQuery("#answers").append(getTemplate(e))});var q="A"+("000000"+e.slice(-1)[0].size).slice(-6);jQuery("#next").html("<a href='http://oeis.org/"+q+"'>"+q+"</a>");c.forEach(function(n){jQuery('#weird-answers').append('The bytecount '+n+' was used more than once!<br />')})}Array.prototype.sortBy=function(f){return this.sort(function(a,b){if(f)a=f(a),b=f(b);return(a>b)-(a<b)})};var ANSWER_FILTER="!*RB.h_b*K(IAWbmRBLe",COMMENT_FILTER="!owfmI7e3fd9oB",answers=[],answers_hash,answer_ids,answer_page=1,more_answers=!0,comment_page,e=[];getAnswers();var SCORE_REG=/<h\d>\s*(\d+)\.\s*((?:<a [^>]+>\s*)?((?:[^\n,](?!<\/a>))*[^\s,])(?:<\/a>)?),.*?(\d+)(?=[^\n\d<>]*(?:<(?:s>[^\n<>]*<\/s>|[^\n<>]+>)[^\n\d<>]*)*, ((?:<a[^>]+>\s*)?A\d+(?:\s*<\/a>)?)\s*<\/h\d>)/,OVERRIDE_REG=/^Override\s*header:\s*/i;

body{text-align:left!important;font-family:Roboto,sans-serif}#answer-list,#language-list{padding:10px;/*width:290px*/;float:left;display:flex;flex-wrap:wrap;list-style:none;}table thead{font-weight:700}table td{padding:5px}ul{margin:0px}#board{display:flex;flex-direction:column;}#language-list li{padding:2px 5px;}#langs-tit{margin-bottom:5px}#byte-counts{display:block;margin-left:15px;}#question-author{color:purple;text-shadow: 0 0 15px rgba(128,0,128,0.1);}#label-info{font-weight: normal;font-size: 14px;font-style: italic;color: dimgray;padding-left: 10px;vertical-align: middle; }

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script><link rel="stylesheet" type="text/css" href="//cdn.sstatic.net/codegolf/all.css?v=83c949450c8b"><p id="weird-answers"></p><p>Currently waiting on <span id="next"></span></p><span>Search by Byte Count: <input id="search" type="number" min=1 oninput="checkSize(this.value);search(1,this.value)" onclick="document.getElementById('search2').value='';!this.value&&search(0,'')"/> <span id="size-used"></span></span><br><span>Search by Language: <input id="search2" oninput="checkLang(this.value);search(0,this.value)" onclick="document.getElementById('search').value='';!this.value&&search(0,'')"/> <span id="language-used"></span></span><h2>Answer chain <span id="label-info">click a label to sort by column</span></h2><table class="answer-list"><thead><tr><td  onclick="sortby(0)">#</td><td onclick="sortby(1)">Author</td><td onclick="sortby(2)">Language</td><td onclick="sortby(3)">Sequence</td><td onclick="sortby(4)">Size</td></tr></thead><tbody id="answers"></tbody></table><table style="display: none"><tbody id="answer-template"><tr><td>{{PLACE}}</td><td>{{NAME}}</td><td>{{LANGUAGE}}</td><td>{{SEQUENCE}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr></tbody></table><table style="display: none"><tbody id="language-template"><tr><td>{{LANGUAGE}}</td><td>{{NAME}}</td><td>{{SEQUENCE}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr></tbody></table>

Answer 1

46. Jellyfish, 21 bytes, A000190

p
d^0
1|&*&*r
 E    i

Try it online!

Next sequence

Answer 2

47. awk, 194 bytes, A000021

{
 n=0;
 for (i=1; i<=2**$1; i++) {
  stop = 0;
  for (x=0; x<=i && !stop; x++) {
   for (y=0; y<=i && !stop; y++) {
    if (x**2 + 12*y**2 == i){
     n++; stop=1;
    }
   }
  }
 }
 print n;
}

Try it online!

Next sequence

Answer 3

51. Burlesque, 8 bytes, A000026

fCf:FLpd

Try it online!

Next sequence

Answer 4

52. Ceylon, 484 bytes, A000008

Integer f(Integer n) {
	variable Integer coin1 = 0;
	variable Integer coin2 = 0;
	variable Integer coin5 = 0;
	variable Integer coinT = 0;
	variable Integer count = 0;
	while (coin1 <= n) {
		while (coin2 <= n / 2) {
			while (coin5 <= n / 5) {
				while (coinT <= n / 10) {
					if (n == coin1 + 2 * coin2 + 5 * coin5 + 10 * coinT) {
						count++;
					}
					coinT++;
				}
				coin5++;
				coinT = 0;
			}
			coin2++;
			coin5 = 0;
		}
		coin1++;
		coin2 = 0;
	}
	return count; 
}

Try it online!

huehue I'm a Ceylon noob

Next Sequence (c'mon this is so easy)

Answer 5

55. ><>, 41 bytes, A000196

:1>::*$@(?vv
  ^ +1@@:$ <
          \1-n;

Try it online!

Next sequence

Answer 6

21. M, 24 bytes, A000015

Ṫ*Ḣ
l2Rp²ÆR$µÇ€<³$ÐḟṂ
‘Ç

Try it online!

Next Sequence

Sorry for changing the next sequence; I had to fix my indexing.

Explanation

Ṫ*Ḣ                !!!!!!! Helper Link
Ṫ                  Last element
 *                 To the power of
  Ḣ                First element
l2Rp²ÆR$µÇ€<³$ÐḟṂ  !!!!!!! Main Link
l2R                Range up to log_2(input); this is the largest power we need to handle because 2**k will be at least `input` this way
   p               Cartesian Product with
    ²ÆR$           All primes up to `input squared`
        µ          New Monadic Link
         ǀ        Call the last link on each element (all prime powers)
              Ðḟ   Filter to discard
           <³$                       elements smaller than the input
                Ṃ  Minimum Value
‘Ç                 !!!!!!! Called Link
‘                  Decrement
 Ç                 Call the Main Link

Answer 7

58. 2sable, 11 bytes, A000119

<DÅFÙïæOrQO

Try it online!

<DÅFÙïæOrQO  Program
<            Decrement for 0-indexing
 D           Duplicate top of stack (for later use)
  ÅF         List of fibonacci numbers up to the current number
    Ù        Remove duplicates
     ï       Convert each element to an integer (they become strings for whatever reason ;_;)
      æ      Powerset
       O     Sum of each subset in the powerset
        r    Reverse stack (I don't think this is necessary, but 10 bytes isn't allowed anyway)
         Q   Equality check; vectorizes
          O  Gives number of occurrences; done

Next Sequence

Answer 8

62. S.I.L.O.S, 57 bytes, A000005

readIO
n=i
lblb
r=n%i
r/r
r=1-r
s+r
i-1
if i b
printInt s

Try it online!

Next Sequence

---

^{*LeakyNun asked me to post this, again.}

Answer 9

68. Pascal (FPC), 761 bytes, A000837

function Mobius(n: integer): integer;
type
	iarray = array of integer;
var
	result: iarray;
	i, j, sum: integer;
begin
	setlength(result,n+1);
	result[1] := 1;
	for i := 2 to n do
	begin
		sum := 0;
		for j := 1 to i-1 do
			if (i mod j = 0) then
				sum := sum + result[j];
		result[i] := -sum;
	end;
	Mobius := result[n];
end;

function A000837(n: integer): integer;
type
	iarray = array of integer;
var
	partitions: iarray;
	i, j, sum: integer;
begin
	setlength(partitions,n+1);
	partitions[0] := 1;
	for i := 1 to n do
		for j := i to n do
			partitions[j] := partitions[j] + partitions[j-i];
	sum := 0;
	for i := 1 to n do
		if (n mod i = 0) then
			sum := sum + partitions[i] * Mobius(n div i);
	if (n = 0) then
		A000837 := 1
	else
		A000837 := sum;
end;

Try it online!

Next sequence

How it works

It generates the partitions using the trivial gf:

prod(q=[1,infty], 1/(1-x^q))

and then uses the Möbius Transform.

Answer 10

74. Racket, 77 bytes, A000363

(define(a n)(/(-(expt 5 n)(*(sub1(* 2 n))(expt 3 n))(* 2 n)2(-(* 2 n n)))16))

Try it online!

Next sequence

Answer 11

75. GolfScript, 36 bytes, A000077

~2\?):P.*,{.P/.*\P%.*6*+}%.&{P<},0-,

Online demo

Next sequence

Answer 12

82. Cubix, 14 bytes, A000290

....I:*....u@O

Try it online!

Next sequence

Answer 13

95. Taxi, 478 bytes, A000027

Go to Post Office:w 1 l 1 r 1 l.
Pickup a passenger going to The Babelfishery.
Go to The Babelfishery:s 1 l 1 r.
Pickup a passenger going to Addition Alley.
1 is waiting at Starchild Numerology.
Go to Starchild Numerology:n 1 l 1 l 1 l 2 l.
Pickup a passenger going to Addition Alley.
Go to Addition Alley:w 1 r 1 l 1 r 3 r 1 r.
Pickup a passenger going to The Babelfishery.
Go to The Babelfishery:n 1 r 1 r.
Pickup a passenger going to Post Office.
Go to Post Office:n 1 l 1 r.

Adds 1 to the input, since this has to be 0-based.

Try it online!

Next Sequence

Answer 14

102. Perl 5, 158 bytes, A000335

@b[0]=1;
while($n++<=$ARGV[0]){
  for($d=1;$d<=$n;$d++){
    for($k=$d;$k<=$n;$k+=$d){@b[$n]+=$d*$d*($d+1)*($d+2)/6*@b[$n-$k]}
  }
  @b[$n]/=$n
}
print @b[-1]

Online demo. This is a full program which takes input as a command-line parameter.

Next Sequence

Answer 15

104. Vala, 267 bytes, A000713

static int main(string[] args) {
	int n = int.parse(stdin.read_line());
	int[] a = new int[n+1];
	a[0] = 1;
	for(int i=1;i<=n;i++){
		int iter = i==1?3:2;
		for(int j=0;j<iter;j++)
			for(int k=i;k<=n;k++) a[k] += a[k-i];
	}
	stdout.printf("%d\n", a[n]);
	return 0;
}

Try it online!

Next sequence.

Explanation

Uses the g.f. 1/[(1-x)^3 (1-x^2)^2 (1-x^3)^2 (1-x^4)^2 ...].

Answer 16

114. Crystal, 1560 bytes, A000228

Dirs = [0,1,-1].permutations.to_a

def addhp(a,b)
  (0..2).map{ |i| a[i]+b[i] }
end

def hexh(n,use,seen,possible)
  if n==0
    testhex(use) ? 1 : 0
  else
    if possible.empty?
      0
    else
      elem = possible[0]
      remain = possible[1..-1]
      newseen = [elem]+seen
      c = hexh(n,use,newseen,remain)
      newpossible = Dirs.map{ |d| addhp(elem,d) }
      newpossible = newpossible.select \
         { |p| ! ((seen.includes? p) || (remain.includes? p)) }
      c + hexh(n-1,use+[elem],newseen,remain+newpossible)
    end
  end
end

def hex(n)
  empty = [] of Array(Int32)
  hexh(n,empty,empty,[[0,0,0]])
end

def rot(hl)
  hl.map{ |h| h.rotate }
end

def neg(hl)
  hl.map{ |h| h.map{ |v| -v } }
end

def mirr(hl)
  hl.map{ |h| [0,2,1].map{ |i| h[i] } }
end

def normhex(hl)
  x = hl.map{ |h| h[0] }.min 
  hl = hl.map{ |h| addhp(h,[-x,0,x]) }
  x = hl.map{ |h| h[1] }.min
  hl.map{ |h| addhp(h,[0,-x,x]) }
end

def check(shl,chl)
  ( shl <=> chl.sort ) == 1
end

def testonesided(hl,ohl)
  return false if check(hl,ohl)
  nhl = normhex(rot(ohl))
  return false if check(hl,nhl)
  nhl = normhex(rot(nhl))
  return false if check(hl,nhl)
  nhl = normhex(neg(ohl))
  return false if check(hl,nhl)
  nhl = normhex(rot(nhl))
  return false if check(hl,nhl)
  nhl = normhex(rot(nhl))
  return false if check(hl,nhl)
  true
end

def testhex(hl)
  hl = normhex(hl)
  return false if hl[0] != hl.min
  hl.sort!
  mhl = normhex(mirr(hl))
  testonesided(hl,hl) && testonesided(hl,mhl)
end

n = ARGV.size==1 ? ARGV[0].to_i : 7
puts "f(#{n}) = #{hex(n)}"

Next sequence

Try it online!

This is a full program that takes input from the command line and computes the value for n=7 if no argument was given, because that has a nice value. I don't really know this language, it may show. And I don't know how to give command line arguments to TIO, you may just replace the assignment to n in the line before the last one instead.

Strangely, it compiles fine (on old debian jessie using crystal's external debian repository) with crystal build hex.cr, but the optimizing call crystal build hex.cr --release fails with what seems to be an internal error.

Answer 17

168. Python 2 (PyPy), 261 bytes, A000101

import sympy

def A000101(n):
  i = 0
  p = 2
  max_difference = 0
  while i <= n:
    next_p = sympy.nextprime(p)
    if next_p-p > max_difference:
      max_difference = next_p - p
      i = i + 1
    p = next_p    
  return p

print A000101(int(raw_input()))

Next Sequence

Try it online!

Others want to starve the challenge of the use of Python, so I'll help.

Answer 18

171. TrumpScript, 541 bytes, A000204

Putin hears all your life
Putin is safe because its Putin
Trump is, 1000001 minus 1000000; great
Our beautiful America is, 1000003 minus 1000000; the greatest
Of course everything is, Trump plus Trump;
Every million is for Trump;
If you want, Putin is Trump?;:
Say Putin!
Otherwise: if for you, Putin is Everything?;:
Say America!
Otherwise:
Money is Everything
As long as, our Money less than Putin;:
Russia is America
America is, Trump plus America;
Trump is Russia
Money is, Money plus Million;!
Say America with force!!
America is great.

Try it online!

Next sequence

Pseudocode for explanation:

var Putin = int(input())
var Trump = 1
var America = 3
var Everything = Trump + Trump
var Million = Trump
if(Putin == Trump)
    print(Putin)
else if(Putin == Everything)
    print(America)
else
    var Money = Everything
    while(Money < Putin)
        var Russia = America
        America = Trump + America
        Trump = Russia
        Money = Money + Million
    print(America)
America is great.

Answer 19

174. C++ (gcc), 54 bytes, A000257

int f(int n){return n?~-~-~-~-(n<<3)*f(~-n)/-~-~n:01;}

• Try it online!
• Next sequence: A000054 :P
• Alternative C solution (as C (clang) and C (gcc) are not considered to be different languages, even if Python interpreters are, I switched to C++)

Answer 20

185. Octave, 17 bytes, A000142

@(n)factorial(n);

Try it online!

Next sequence

Answer 21

199. Kotlin, 92 bytes, A000466

fun main(args: Array<String>){
    val n = Integer.valueOf(args[0]);
    println(4*n*n-1);
}

Try it online!

Next Sequence

Answer 22

207. ARBLE, 389 bytes, A003275

coprime = eq(max((#table.unpack)(where(range(1,max(a,b)),eq(d%first(a,b,c,d,e),0)*eq(e%a,0),a,b))),1)
phi = len(where(range(1,n),coprime(a,first(n,b,c)),n))
A003275 = list()
local A003275_n = 0
return function(n)
	while(#A003275 < n)do
		A003275_n = A003275_n + 1
		if(phi(A003275_n) == phi(A003275_n+1))then
			A003275[#A003275+1] = phi(A003275_n)
		end
	end
	return A003275[#A003275]
end

Try it online!

Next Sequence

Answer 23

225. VBA, 80 bytes, A001232

Sub s(n)
While k<n+1
i=i+1
If i*9=StrReverse(i) Then k=k+1
Wend
MsgBox i
End Sub

Next sequence

Output is a popup message. A not-very-exciting sequence deserves a not-very-exciting solution, I think.

Answer 24

235. Pascal (FPC), 3295 bytes, A000512

4 days passed without anyone answered.

Type
	int = LongInt; // Can be changed to Integer

Var
	n, n_column, matrices_len, i, j, k: int;
	row_values, matrices: Array of array of int;
	row_id, column_sum: Array of int;
	valid: Boolean;

Function next_permutation (var A : Array of int): Boolean;
var x, y, z, temp: int;
begin
	if Length(A) <> n then writeln('Error!!');
	x := Length(A) - 1; // or High(A)
	z := x;
	while (x > 0) and (A[x-1] > A[x]) do Dec(x);
	y := x;
	while y < z do
	begin
		temp := A[z]; A[z] := A[y]; A[y] := temp;
		Inc(y); Dec(z);
	end;
	if (x = 0) then
		Exit(false);
	y := x; temp := A[x-1];
	while A[y] < temp do Inc(y);
	A[x-1] := A[y]; A[y] := temp;
	Exit(true);
end;

Function equivalent (mat1, mat2: int): Boolean;
{ Use highly sub-optimal solution, O(n!^2 x n^2).
  O( n! x n^2 x log(n) ) solution is possible by trying all row permutation
  and sort the columns, and check if the matrices are equal.
  Sort the rows and the columns alternatively does not work, though.
}
var
	xP, yP: Array of int;
	x, y: int;
	different: Boolean;
begin
	SetLength(xP, n);
	SetLength(yP, n);
	for x := 0 to n-1 do
	begin
		xP[x] := x;
		yP[x] := x;
	end;
	repeat
		repeat
			different := false;
			for x := 0 to n-1 do
			begin
				for y := 0 to n-1 do
				begin
					if row_values[matrices[mat1, x], y] <>
					row_values[matrices[mat2, xP[x]], yP[y]] then
					begin
						different := true;
						break;
					end;
				end;
				if different then break;
			end;
			if not different then Exit(true);
		until not next_permutation(xP);
	until not next_permutation(yP);
	Exit(false);
end;

Begin
	Read(n); Inc(n);
	If n < 3 then
	begin
		Writeln(0);
		Halt;
	end;
	SetLength(row_values, (n * (n - 1) * (n - 2)) div 6);
	n_column := 0;
	for i := 0 to n-1 do
		for j := i+1 to n-1 do
			for k := j+1 to n-1 do
			begin
				SetLength(row_values[n_column], n);
				row_values[n_column, i] := 1;
				row_values[n_column, j] := 1;
				row_values[n_column, k] := 1;
				Inc(n_column);
			end;
	// Dynamic array in FP is 0-based
	if n * (n - 1) * (n - 2) <> 6 * n_column then
	begin
		WriteLn('Error!');
		exit();
	end;

	matrices_len := 0;
	SetLength(matrices, 4);

	SetLength(row_id, n);
	SetLength(column_sum, n);
	while true do
	begin
		for i := 0 to n-1 do column_sum[i] := 0;

		for i := 0 to n-1 do
		begin
			for j := 0 to n-1 do
				Inc(column_sum[j], row_values[row_id[i], j]);
		end;

		valid := true;
		for i := 0 to n-1 do
		begin
			if column_sum[i] <> 3 then // column_sum[i] = sum of all values on column i
			begin
				valid := false;
				break;
			end;
		end;
		if valid then
		begin
			if matrices_len = length(matrices) then
				SetLength(matrices, 2 * length(matrices));
			matrices[matrices_len] := copy(row_id, 0, n);
			Inc(matrices_len);
		end;

		i := n - 1;
		while (i <> -1) and (row_id[i] = n_column - 1) do
			Dec(i);
		if i = -1 then
			Break;
		Inc(row_id[i]); Inc(i);
		while i < n do
		begin
			row_id[i] := row_id[i-1];
			Inc(i);
		end;
	end;

	row_id := nil; column_sum := nil;

	k := 0;
	for i := 0 to matrices_len-1 do
	begin
		valid := true;
		for j := 0 to i-1 do
		begin
			if matrices[j, 0] <> -1 then
			begin
				if equivalent(i, j) then
				begin
					valid := false;
					break;
				end;
			end;
		end;
		if valid then
			inc(k)
		else
			matrices[i, 0] := -1;
	end;
	Writeln(k);
End.

Try it online!

Next sequence.

I can easily delete a byte to get this sequence instead (which seems to be much easier to understand), but I didn't do that.

Answer 25

236. Axiom, 341 bytes, A003295

qpc(n,k) ==  -- inspired by 3rd PARI code for A010815
  qr := divide(n,k)
  if qr.remainder>0 then return 0
  sq := perfectSqrt(24*qr.quotient+1)
  sq case "failed" => 0
  jacobi(12,sq)

qp(k) == series(n+->qpc(n,k),'q=0)

q := series q

F := q*qp(3)*(qp(33)/(qp(1)*qp(11)))

s := q*(-11+(1+3*F)^2*(1/F+1+3*F))

f(n) == coefficients(s).(n+1)

Next sequence.

Put it in a file called a003295.input and, at the axiom prompt, type )read a003295. Now you can for example call f 25. The first time it will give a warning before still giving the correct result. Here it is computing the first values:

(9) -> [f n for n in 0..20]
   Cannot compile map: f 
   We will attempt to interpret the code.

   (9)
   [1, - 5, 17, 46, 116, 252, 533, 1034, 1961, 3540, 6253, 10654, 17897, 29284,
    47265, 74868, 117158, 180608, 275562, 415300, 620210]
                                              Type: List(Expression(Integer))

qp(k) computes the series of the q-Pochhammer symbol for q^k. The definition of F and s follows the Mathematica code.

Answer 26

262. Cubically, 91 bytes, A000115

$:7*1+4/1**1+55/1RFD'/2%:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::

Try it online!

Next sequence.

Use the formula

a(n) = round((n+4)^2/20).

listed on the OEIS page.

Originally I intended to format the code as a cube, something like

  $:7*1
 +    4
/1**1 +
5   5 /
1   RF
D'/2%

but it does not work out too well, because Cubically does not have comments and it is not allowed to put spaces between command and parameters, unfortunately.

More explanation later.

Answer 27

261. Alice, 115 bytes, A000504

/
 i                            >'Qa*:\ o
  /.!.h*?2Ea*?a5+*+2+*?2*3+.!w=
                              >?*?2-.!K @

Try it online!

Next sequence.

Uses the formula used by the Mathematica snippet on OEIS:

a[n_] := n (n+1) (10n^2+15n+2) (2n+3)!! / 810

In detail:

/      Switch to Ordinal mode.
i      Read all input as a string.
/      Switch back to Cardinal mode.
.!     Implicitly convert the input to an integer N and store a copy on the tape.
.h*    Multiply N by (N+1).
?2E    Retrieve a copy of N and square it.
a*     Multiply by 10.
?a5+*  Retrieve another copy of multiply it by 15.
+      Add the last two results.
2+     Add 2.
*      Multiply the initial value by this, giving N*(N+1)*(10N^2+15N+2)
?2*3+  Retrieve another copy of N and compute 2N+3, which will be our
       iterator for the double factorial.
.!     Store a copy of this on the tape.
w      Push the current IP address to the return address stack. This starts
       a loop.
  =      Sign junction. Pops the current value of the double factorial iterator.
         If it's positive, turn right (continuing the loop). If it's negative,
         turn left (exiting the loop).
  >      Redirect the IP eastward again.
  ?*     Multiply our result by the current value of the iterator.
  ?2-    Retrieve another copy of the iterator and subtract 2.
  .!     Store a copy on the tape to update the iterator.
K      Jump back to the w to get to the next iteration.

       Once we exit the loop, we continue here:
>      Redirect the IP eastward again.
'Qa*:  Divide the result by 810 (computed as 81*10).
\      Switch to Ordinal mode.
o      Implicitly convert the result to a string and print it.
@      Terminate the program.

Answer 28

260. Brain-Flak, 504 bytes, A001333

<>       # switch to another stack
(())(()) # push two 1s. Assuming top = a n-1, second top = a n-2 apart from the input
<>({}<>) # transfer the input to this stack
{        # repeat input times
    ({}[()]<     # subtract 1, ignore 
      (
        ({}<>)   # move top a n-1 to another stack
        ({})     # + top a n-1
        <>{}<>   # add to a n-2 on the other stack
      )          # and push 2 * a n-1 + a n-2
    >)           # and push back again
}
{}{}     # remove redundant values on top 

Try it online!

Next sequence.

The actual code is only 52 bytes: Try it online!

Use the recurrence relation

a(n) = 2a(n-1) + a(n-2)

described on the OEIS page.

---

The mathematical information about that sequence is pretty simple and elementary, and not worth the effort I write the explanation, so I decide to not write explanation for this sequence.

Answer 29

194. ALGOL 68 (Genie), 589 bytes, A000160

BEGIN
  LONG LONG INT n   := read int + 4;
  LONG LONG INT r   := 0;
  LONG LONG REAL ma := ( n / 4.0 ) ** 1.5;
  LONG LONG REAL mb := ( n / 3.0 ) ** 1.5;
  LONG LONG REAL mc := ( n / 2.0 ) ** 1.5;
  LONG LONG REAL x  := 2.0 / 3;
  INT a             := 1;
  WHILE a <= ma DO
    INT b := a;
    WHILE b <= mb DO
      INT c := b;
      WHILE c <= mc DO
        INT d := c;
        WHILE a ** x + b ** x + c ** x + d ** x <= n DO
          r := r + 1;
          d := d + 1
        OD;
        c := c + 1
      OD;
      b := b + 1
    OD;
    a := a + 1
  OD;
  print ( whole ( r , 0 ))
END

Try it online!

Next Sequence.

Based on the comments of the sequence, counts the solutions to the inequality

a^(2/3)+b^(2/3)+c^(2/3)+d^(2/3) ≤ n

for any four integers 1 ≤ a ≤ b ≤ c ≤ d.

I also gave a fairly straightforward next sequence, to make up for last time.

Edit: This was originally written for Java 8, but I missed that there were already two Java 8 answers when I wrote this. Probably because one was labeled "Java (OpenJDK 8)" and the other was just "Java 8". Whatever the reason, this answer was invalid. I re-wrote it in a different language, with the same byte count, to fix this.

Answer 30

284. Hexagony, 211 bytes, A000384

     ? } 2 \ . .
    . . . " . . .
   . @ ! * = < . .
  . . . > ( { / . .
 . . . . . . . . . .
. . . . . . . . . . .
 . . . . . . . . . .
  . . . . . . . . .
   . . . . . . . .
    . . . . . . .
     . . . . . .

Try it online!

Next sequence!

What better language is there to compute hexagonal numbers?

Explanation

I decided to make this code use each character (except ., space and newline) only once, because I had too much time on my hands. Let's label the edge the MP starts on A, the one to the left of it L and the one to the right - R. The following linear code gets executed:

?}2"*({=*!@
?           A = input()
 }2         R = 2
   "*       L = A * R = 2 * n
     (      L = L - 1 = 2 * n - 1
      {=*   R = A * L = n(2n-1) = a(n)
         !  Print the result
          @ End execution