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

169. Chez Scheme, 84 bytes, A000261

(define (a n)
(if (< n 1)
 n 
 (+ (* (+ n 1) (a (- n 1))) (* (- n 2) (a (- n 2))))))

Try it online!

Next sequence

Answer 2

170. Python 3 (PyPy), 204 bytes, A000084

n=int(input())+1
a=[0,1,1,2]
p=[0,1,3,7]
for m in range(4,n+1):s=sum([d*a[d]for d in range(1,m)if m%d==0]);u=p[m-1]+2*sum([p[k]*a[m-k]for k in range(1,m-1)])+s;a+=[u//m];p+=[s+u]
print(2*a[n]if n>1else 1)

Try it online!

Next sequence!

4 Pythons down, 2 to go. Wait, there's Python 0 but no one used it in the first 150 answers. 3 to go.

Ported the second Maple program for A000669. Next sequence is Lucas numbers - as easy as Fibonacci, but nicer.

Answer 3

172. Python 1, 55 bytes, A000541

S=0
for s in range(-~input()):S=S+s*s*s*s*s*s*s
print S

Try it online!

Next Sequence: A000055

Answer 4

173. Python 2 (IronPython), 257 bytes, A000055

c={0:0,1:1}
def b(n):
 if n not in c:c[n]=sum([sum([d*b(d)for d in range(1,j+1)if j%d==0])*b(n-j)for j in range(1,n)])//(n-1)
 return c[n]
def a(n):return 1if n==0 else b(n)-(sum([b(k)*b(n-k)for k in range(n)])-(b(n//2)if n%2==0 else 0))//2
print a(input())

Try it online!

Next sequence!

2 versions of Python on the wall, 2 versions of Python. Take one down and pass it around, 1 version of Python on the wall.

Caches the results of b because it seemed like it would take forever to compute for big n without it. Didn't test whether it does anything useful, but it probably does.

Answer 5

176. JavaScript (ES6), 184 bytes, A000122

function A000122(n) {
 if (n === 0) { return 1; }
 squares = [];
 for (i = 0; i <= 32; i++) {
  squares.push(i*i);
 }
 if (~squares.indexOf(n)) {
  return 2;
 } else {
  return 0;
 }
}

Try it online!

Next Sequence

Answer 6

188. Groovy, 105 bytes, A000321

def a
a = { n -> 
	if (n == 0G) { 1 }
	else if (n == 1G) { -1 }
	else {
		-a(n-1G)-2G*(n-1G)*a(n-2G)
	}
}

Try it online!

Next Sequence

Answer 7

189. C# (Visual C# Compiler), 1182 bytes, A000105

Three parts - the function (824 bytes) + comparer class used in function (273 bytes) + imports (85 bytes)

n=>{var t=new Func<P,P>[]{p=>new P(p.Y,-p.X),p=>new P(-p.X,-p.Y),p=>new P(-p.Y,p.X),p=>new P(-p.X,p.Y),p=>new P(-p.Y,-p.X),p=>new P(p.X,-p.Y),p=>new P(p.Y,p.X)};Func<P[],P[][]>r=l=>new[]{l}.Concat(new int[7].Select((x,i)=>l.Select(p=>t[i](p)).ToArray())).ToArray();Func<int,P[][]>k=null;k=a=>a<2?new[]{new[]{new P(0,0)}}:k(a-1).SelectMany(l=>l.SelectMany(p=>new[]{new P(p.X-1,p.Y),new P(p.X+1,p.Y),new P(p.X,p.Y-1),new P(p.X,p.Y+1)}).Where(p=>!l.Contains(p)).Distinct().Select(p=>r(l.Concat(new[]{p}).ToArray()).Select(o=>o.Select(c=>new P(c.X-o.Select(q=>q.X).Min(),c.Y-o.Select(q=>q.Y).Min())).OrderBy(q=>q.X).ThenBy(q=>q.Y).ToArray()).Aggregate((m,o)=>(m==null||string.Concat(o.Select(x=>$"({x.X},{x.Y})")).CompareTo(string.Concat(m.Select(x=>$"({x.X},{x.Y})")))<0?o:m)))).Distinct(new C()).ToArray();return k(n).Length;}

class C:IEqualityComparer<P[]>{public bool Equals(P[] f,P[] s)=>f==s?1>0:(f==null|s==null)?1<0:f.OrderBy(p=>p.X).ThenBy(p=>p.Y).Zip(s.OrderBy(p=>p.X).ThenBy(p=>p.Y),(a,b)=>a==b).All(x=>x);public int GetHashCode(P[]l)=>l.Aggregate(0,(a,b)=>a.GetHashCode()^b.GetHashCode());}

using System.Collections.Generic;using P=System.Windows.Point;namespace System.Linq{}

Try it online!

Next sequence

I accidentaly deleted the extended version after I golfed it.... Also the code has 2 unneeded spaces added because A001180 was not fun.

Answer 8

193. Hodor, 160 bytes, A000326

HODOR! = HoDoRHoDoR( HODOR? ) {
	HOdor!!!(HODOR? == 0)
	HODOR:: 0;
	HODOR = 1;
	Hodor! = 2;
	hodor = 3;
	HODOR:: HODOR? * (hodor * HODOR? - HODOR ) / Hodor! ;
}

Try it online!
Next sequence

It's a little inefficient, but please don't hold it against Hodor. He is a little slow, but he's a beautiful gentle giant.

Answer 9

202. Oasis, 50 bytes, A000213

bc+d+$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$111

$s are filler.

Try it online!

Next Sequence

Answer 10

209. APL (Dyalog), 100 bytes, A000075

{
    dims ← 1 + 2 * ⍵
    table ← ∘.{+/2 3×⍺ ⍵*2}⍨ ⍳dims
    uniq ← ∪ (dims*2)⍴ table
    +/ (××(2*⍵)≥⊢) uniq
}

using ⎕IO←0.

Try it online!

Next sequence

Answer 11

213. Befunge-98 (PyFunge), 117 bytes, A000079

v   v2<
    * \
    \ -  did you know that comments in befunge start and end with nothing?
      1
>&1\>:|
      >$.@

Try it online!

Next sequence!

Answer 12

214. Julia 0.4, 383 bytes, A000117

Because of sequence relationships, I've implemented A000013 four times now.

function EulerPhi(n)
  x = 0
  for i = 1:n
    if gcd(i,n) == 1
      x = x + 1
    end
  end
  return x
end

function A000013(n)
  if n <= 1
    return 1
  end
  x = 0
  for d = 1:n
    if n/d == n÷d
      x = x + (EulerPhi(2*d) * 2^(n/d))/(2*n)
    end
  end
  return x
end

function A000011(n)
  return (A000013(n) + 2^(n÷2))/2
end

function A000117(n)
  return A000011(2*n)
end

Next Sequence

Try it online!

Answer 13

212. JavaScript (SpiderMonkey), 79 bytes, A000078

a = n => n < 3 ? 0 : n == 3 ? 1 : a(--n) + a(--n) + a(--n) + a(--n)

// balloon

Try it online!

Next sequence.

Answer 14

219. Maxima, 127 bytes, A000702

load("ratpow")$
p(n):=num_partitions(n);
q(n):=ratp_dense_coeffs( product(1+x^(2*k-1),k,1,n) ,x)[n];
a(n):=(p(n+2)+3*q(n+3))/2;

Next Sequence

Try it online!

This uses the first formula on the OEIS page, with a built-in for p and the generating function for q. It's basically the same as the Mathematica code.

Answer 15

220. Visual Basic .NET (.NET Core), 107 bytes, A000127

Function A000172(n As Integer) As Integer
n += 1
Return(n^4 - 6*n^3 + 23*n^2 - 18*n + 24) / 24
End Function

Try it online!

Next Sequence

Note that the implementation of VB.NET on TIO does not have BigIntegers (I tried it and it wants a .dll) so I can use 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.

Answer 16

234. Rust, 512 bytes, A000141

fn a(n: i64) -> i64 {
    let mut ret = 0;

    for a in !n..-!n {
        for b in !n..-!n {
            for c in !n..-!n {
                for d in !n..-!n {
                    for e in !n..-!n {
                        for f in !n..-!n {
                            if a * a + b * b + c * c + d * d + e * e + f * f == n {
                                ret += 1;
                            }
                        }
                    }
                }
            }
        }
    }

    return ret;
}

Try it online!

Next sequence.

Nice byte count, eh?

Answer 17

238. LOLCODE, 429 bytes, A000089

HAI 1.2

I HAS A INPUT
GIMMEH INPUT
INPUT IS NOW A NUMBR
INPUT R SUM OF INPUT AN 1

I HAS A COUNT ITZ 0
I HAS A INDEX ITZ 1
I HAS A TEMP ITZ 0

IM IN YR LOOP
TEMP R PRODUKT OF INDEX AN INDEX
TEMP R SUM OF TEMP AN 1
TEMP R MOD OF TEMP AN INPUT

BOTH SAEM TEMP AN 0
O RLY?
 YA RLY
  COUNT R SUM OF COUNT AN 1
OIC

BOTH SAEM INDEX AN INPUT
O RLY?
 YA RLY
  GTFO
OIC

INDEX R SUM OF INDEX AN 1
IM OUTTA YR LOOP

VISIBLE COUNT
KTHXBYE

Next Sequence

Try it online!

This just directly counts the solutions. None of that Legendre symbol nonsense.

Answer 18

240. CHICKEN Scheme, 256 bytes, A000346

(define (factorial x)
(cond
((< x 2) 1)
(else (* x (factorial (- x 1))))))
(define (choose n k)
(cond
((= n 0) 1)(else
(/ (factorial n) (* (factorial k) (factorial (- n k)))))))
(define (a000346 n)
(- (expt 2 (+ 1 (* 2 n))) (choose (+ 1 (* n 2)) (+ n 1))))

Try it online!

Next sequence

Answer 19

153. Emojicode, 1344 bytes, A000136

🐋🍨🍇
 👴 Find the index of e in the list by comparing the elements with c
 🐖🔎e Element c🍇Element Element➡👌🍉➡🍬🚂🍇
  🔂i⏩0🐔🐕🍇 🍊🍭c e🍺🐽🐕i🍇🍎i🍉🍉
  🍎⚡
 🍉
🍉

🐇☯🍇
 👴 Checks whether a permutation is a possible way to fold stamps
 🐇🐖🅱p🍨🐚🚂➡👌🍇
  🔂k⏩0➖🐔p 1🍇
   🔂r⏭🚮k 2➖🐔p 1 2🍇
    🍦c🍇a🚂b🚂➡👌🍎😛a b🍉
    🍦A🍺🔎p k c
    🍦B🍺🔎p r c
    🍦C🍺🔎p➕k 1 c
    🍦D🍺🔎p➕r 1 c
    🍊🎉🎉🎉🎊🎊◀A B◀B C◀C D🎊🎊◀B C◀C D◀D A🎊🎊◀C D◀D A◀A B🎊🎊◀D A◀A B◀B C🍇🍎👎🍉
   🍉
  🍉
  🍎👍
 🍉

 👴 Iterates over the permutations
 🐇🐖🅰n🚂➡🚂🍇
  🍦a🔷🍨🐚🚂🐧n
  🍦c🔷🍨🐚🚂🐧n
  🔂i⏩0 n🍇
   🐷a i i
   🐷c i 0
  🍉

  🍮i 0
  🍮r 1
  🔁◀i n🍇
   🍊◀🍺🐽c i i🍇
    🍮j 0
    🍊😛🚮i 2 1🍇🍮j🍺🐽c i🍉
    🍦t🍺🐽a j
    🐷a j🍺🐽a i
    🐷a i t
    🍊🍩🅱☯a🍇🍮➕r 1🍉
    🐷c i➕1🍺🐽c i
    🍮i 0
   🍉🍓🍇
    🐷c i 0
    🍮➕i 1
   🍉
  🍉
  🍎r
 🍉
🍉

🏁🍇
 🍊🍦i🚂🔷🔡😯🔤🔤10🍇
  😀🔡🍩🅰☯➕1 i 10
 🍉
🍉

Try it online!

Next sequence!

I was going to use something esoteric, but then I changed my mind. Pretty self explanatory, I don't think an explanation is necessary.

Note: for inputs larger than 7, you have to specify a heap size larger than 512 MB during the compilation of the Real-Time Engine. For example, use cmake .. -DheapSize=1073741824 for 1 GB. If you run out of heap space, the VM segfaults, probably because someone does not check for NULL after malloc.

Edit: make this faster and less memory-hungry. Now I wonder whether the garbage collector is working properly

Edit almost 2 months later: this code had to work around a bug in the implementation of the ⏩ type. Now that this bug is fixed and TIO updated, I had to offset the range parameters again. I also managed to add some comments, all while keeping the bytecount the same.

Answer 20

249. PicoLisp, 171 bytes, A000904

(de a000904 (n)
 (cond
  ((= n 0) 0)
  ((= n 1) 3)
  ((= n 2) 13)
  (T 
  (+ (* (+ n 2) (a000904 (- n 1)))
     (* (+ n 3) (a000904 (- n 2)))
	 (a000904 (- n 3)))
  )
 )
)

Try it online!

next sequence

Woo! eleventh page!

Answer 21

157. C (gcc), 376 bytes, A002466

int f(int n)
{
    int i = 0, s[100] = {0};
    s[0] = 1;
    s[1] = 1;
    s[2] = 2;

    while (i < 97) {
        if ((i % 5) == 3) {
            s[3+i] = s[i] + s[2+i];
        }
        else if ((i % 5) == 4) {     
            s[3+i] = s[1+i] + s[2+i];
        }
        else {
            s[3+i] = s[i] + s[1+i] + s[2+i];
        }
        i++;
    }

    return s[n];
}

It's like an obnoxious Fibonacci! 0-indexed.

Next sequence

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The function for this sequence said this:

a(1) = a(2) = 1, a(3) = 2, a(5*k+2) = a(5*k+1) + a(5*k-1), a(5*k+3) = a(5*k+2) + a(5*k+1), a(5*k+b) = a(5*k+b-1) + a(5*k+b-2) + a(5*k+b-3) for b=-1,0,1

It's completely unhelpful. However, I noticed that the integers in the sequence were usually the previous three terms added together, but more obnoxious. Here's my work:

1.  1,    1,    2    -> 4    (ok)
2.  1,    2,    4    -> 7    (ok)
3.  2,    4,    7    -> 13   (ok)
4.  4,    7,    13   -> 17   (no) sum does not include second value
5.  7,    13,   17   -> 30   (no) sum does not include first value
6.  13,   17,   30   -> 60   (ok)
7.  17,   30,   60   -> 107  (ok)
8.  30,   60,   107  -> 197  (ok)
9.  60,   107,  197  -> 257  (no) sum does not include second value
10. 107,  197,  257  -> 454  (no) sum does not include first value
11. 197,  257,  454  -> 908  (ok)
12. 257,  454,  908  -> 1619 (ok)
13. 454,  908,  1619 -> 2981 (ok)
14. 908,  1619, 2981 -> 3889 (no) sum does not include second value
15. 1619, 2981, 3889 -> 6870 (no) sum does not include first value

So for each five terms, the last two would be calculated differently. The modulo's and if-statements in the while loop of the code handle that.

Answer 22

255. Perl 5, 454 bytes, A000098

sub part {
 my $S;
 if ($_[1]==0) { $S = 1 } elsif ($_[1] < $_[0]) { $S = 0 } else { $S = part($_[0],$_[1]-$_[0])+part($_[0]+1,$_[1]) }
 $S;
}

sub partsum {
 my @a = (0..$_[0]);
 my $S = 0;
 for my $i (@a) {
  $S += part(1,$i);
 }
 $S;
}

sub A000097 {
 my @a = (0..$_[0]//2);
 my $S = 0;
 for my $i (@a) {
  $S += partsum($_[0]-2*$i);
 }
 $S;
}

sub A000098 {
 my @a = (0..$_[0]//3);
 my $S = 0;
 for my $i (@a) {
  $S += A000097($_[0]-3*$i);
 }
 $S;
}

Next Sequence

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There's a lot of redundant subroutine calls, so memoization would speed this up a lot.

Answer 23

257. TI-NSpire CX Basic, 144 bytes, A000562

Define a562(n)=
Func
:If n=4 Then
:  Return 9
:Else
:  Return ((27)/(8))*n^(4)-((135)/(4))*n^(3)+((921)/(8))*n^(2)-((539)/(4))*n
:EndIf
:EndFunc

Next sequence

Uses Plouffe's conjecture mentioned in the OEIS entry, note that this is different than TI-Nspire CAS Basic as they are two different models of calculator. Next sequence hopefully is easy.

This allows the language to be used again before the 300th.

Answer 24

263. Visual Basic .NET (.NET Core), 493 bytes, A000091

It is a multiplicative function. Using a simple prime checking function.

Now VB .NET was used twice.

Function IsPrime(p As Integer) As Boolean
 For i = 2 To p - 1
  If p Mod i = 0 Then Return False
 Next
 Return True
End Function

Function A000091(n As Integer) As Integer
 n += 1
 If n = 1 Then Return 1
 If n Mod 9 = 0 Then Return 0

 Dim ans As Integer = 1
 If n Mod 2 = 0 Then ans *= 2
 If n Mod 3 = 0 Then ans *= 2
 While n Mod 2 = 0
  n /= 2
 End While
 For i = 5 To n
 If IsPrime(i) And n Mod i = 0 Then
  If i Mod 3 = 1 Then ans *= 2 Else Return 0
 End If
 Next
 Return ans
End Function

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Next sequence

I chose a simple sequence. The smallest unused sequence until now is A000099.

Answer 25

215. MATL, 95 bytes, A000383

'this is just a random string for padding aagghhhh need at least 80 chars!'xT6Y"i:"t-5:0)sh]GQ)

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Next sequence

'...'x                 string and delete from stack
T6Y"                   push 6 ones
i:"                    iterate user input number of times
   t-5:0)sh            dup, index last 6, sum, horzcat
           ]           end loop
            GQ)        copy user input, increment, index (return)

Answer 26

264. PowerShell, 102 bytes, A000493

function A000493($n) {return [math]::floor([math]::sin($n))}

# Please, let's get this to 102. Please?

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Next sequence.

Answer 27

270. Husk, 103 bytes, A000118

←→-1+1-1+1-1+1-1+1-1+1-1+1-1+1-1+1-1+1-1+1-1+1-1+1-1+1-1+1-1+1-1+1+*1=⁰0*±⁰≠*8Σu†ΠṖp⁰*<1%4⁰*32Σu†ΠṖp/4⁰

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Next Sequence.

Ignore the madness that precedes the actual code, it’s just fluff to make this 103 bytes.

How it works

Husk does not have a “divisors” built-in, so (Ḋ has been implemented) this was actually fun to solve.

• First off, it computes the divisors of N using prime factorization (prime factors, powerset, product of each, deduplicate) and sum them up. Then, it multiplies the result by 8.

• After that, it computes the sum of N / 4’s divisors, and multiplies it by 32 if N is divisible by 4, or evaluates the whole thing to 0 otherwise.

• Takes the absolute difference between the above terms.

• For handling the special case of 0, this just multiplies our result by the sign of N, and adds 1 iff N is 0.

To put it simply, it calculates (N==0 ? 1 : 0)+sgn(N)(8σ(N)+32σ(N/4)(N%4==0 ? 1:0)).

Answer 28

135. Husk, 29 bytes, A000277

*1*1*1*1*1*1*1+5-*2⌊√+5*4⁰*3⁰

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Next Sequence.

Answer 29

144. Chez Scheme, 68 bytes, A000120

(define(e n)(cond((< n 1)0)((+(e(fx/ n 2))(remainder n 2)))))       

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Next sequence

Answer 30

273. Ohm v2, 123 bytes, A000129

2¬1+³ⁿ1 2¬-³ⁿ-2¬d/¦                                                                                                        

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Next Sequence.

I didn't want to nearly-kill the challenge again, so I chose A000123, which isn't that hard.

Implements a(n) = ((1+√2)ⁿ - (1-√2)ⁿ) / (2√2), rounded to the nearest integer (because there are small floating-point inaccuracies).