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

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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

110. Nim, 442 bytes, A000081

from strutils import parseint
import math

proc A000081(index: int): int =
 if index <= 0: return 0
 elif index <= 2: return 1
 elif index == 3: return 2
 elif index == 4: return 4
 else:
  var n: int = index - 1
  result = 0
  for k in 1..n:
   var t: int = 0
   for d in 1..k:
    if gcd(d,k)==d:
     t = t + (d * A000081(d))
   result = result + t*A000081(n-k+1)
  result = int(result / n)
 return

echo A000081(parseint(readline(stdin)))

Try it online!

Next sequence

Answer 2

113. Arcyóu, 228 bytes, A000049

(: grp (F(d) d))
(: q (# (q)))
(: t (? q (] 
 (r * (* (_ 2 3) q))) 2))
(: a 0)
(f x (_ 1 t) (grp
 (: b 0)
 (f y (_ t) (grp
  (f z (_ t) (grp
   (? (= (+ (* 3 (* y y)) (* 4 (* z z))) x) (: b 1) 1)
  ))
 ))
 (: a (+ a b))
))
(p a)

This is slow for terms past about the fifth, but Arcyóu is implemented in Python, which does support arbitrary precision numbers. Try it online!

Next sequence (let's try some geometry ...)

Answer 3

122. MIT/GNU Scheme, 2057 bytes, A000525

So I'm implementing A000081 again...

(define (add f g)
  (stream-map + f g) )

(define (scale c f)
  (stream-map (lambda (v) (* c v)) f) )

(define (adddiag sf)
  (let ((f0 (stream-first sf)))
    (cons-stream (stream-first f0)
         (add (stream-rest f0) (adddiag (stream-rest sf))) ) ) )

(define (mult f g)
  (adddiag (stream-map (lambda (v) (scale v g)) f)) )

(define (prependzeros n f)
  (if (= 0 n) f (cons-stream 0 (prependzeros (-1+ n) f))) )

(define (intersperse f n)
  (cons-stream (stream-first f)
           (prependzeros n (intersperse (stream-rest f) n)) ) )

(define (iterate fun init)
  (cons-stream init (iterate fun (fun init))) )

(define (from n)
  (cons-stream n (from (1+ n))) )

(define (scanl fun init s)
  (cons-stream init (scanl fun (fun init (stream-first s)) (stream-rest s))) )

(define (expx f)
  (cons-stream 1
           (adddiag (stream-map (lambda (s g) (scale (/ 1 s) g))
                    (scanl * 1 (from 2))
                    (iterate (lambda (g) (mult f g)) f) )) ) )

(define (eu f)
  (expx (adddiag (stream-map (lambda (n) (scale (/ 1 n)
                                                (intersperse f (-1+ n)) ))
                             (from 1) ))) )

(define gf000081s (cons-stream 1 (stream-rest (eu gf000081s))))

(define gf000081 (cons-stream 0 gf000081s))

(define (recip1 f)
  (let ((msf (scale -1 (stream-rest f))))
    (cons-stream 1 (adddiag (iterate (lambda (g) (mult msf g)) msf))) ) )

(define (div1 f g)
  (mult f (recip1 g)) )

(define (pow7 f)
  (let* ((p2 (mult f f))
         (p4 (mult p2 p2))
         (p6 (mult p4 p2)) )
    (mult p6 f) ) )

(define (addconst c f)
  (cons-stream (+ c (stream-first f))
               (stream-rest f) ) )

(define gf
  (let* ((p2 (mult gf000081 gf000081))
         (p3 (mult p2 gf000081))
         (p4 (mult p3 gf000081)) )
    (div1 (mult p4 (addconst 64
                             (add (add (scale -79 gf000081)
                                       (scale  36 p2) )
                                  (scale -6 p3) ) ))
          (pow7 (addconst 1 (scale -1 gf000081))) ) ) )

(define (f n) (stream-ref gf (+ 4 n)))

In debian, MIT/GNU Scheme is in the package mit-scheme.

Here's how to use it when the program is in a file ps.scm:

$ scheme --load ps.scm
MIT/GNU Scheme running under GNU/Linux
[...]
1 ]=> (f 20)

;Value: 90039381031273

Next sequence

Answer 4

130. TI-Nspire CAS Basic, 90 bytes, A000482

Define f(n)=
Prgm
:Disp polyCoeffs(taylor((−ln(1-x))^(5),x,n),x)[1]*((n!)/(5!))
:EndPrgm

Next Sequence

Code in Action

Fun fact: I tried to implement 3 other sequences, but was too late as other answers had been made

Answer 5

148. Python 2 (IronPython), 173 bytes, A000695

def to_b_4(k):
 r=""
 while k>0:
  r=`k%4`+r
  k //=4
 return r
n=input()
i=j=b=0
while i<=n:
 b=set(to_b_4(j))
 if not("2"in b or"3"in b or"4"in b):
  i+=1
 j+=1
print(j-1)

Try it online!

Next Sequence

Answer 6

150. Julia 0.6, 125 bytes, A001993

function A001993(n)
 m=zeros(15,15)
 for i=1:14
 m[i,i+1]=1
 end
 m[:,1]=[1;2;0;-2;-4;1;3;3;1;-4;-2;0;2;1;-1]
 (m^n)[1,1]
end

Next sequence

Try it online!

The next one is easy, and all languages are back on the table, so somebody may have to re-work the stack snippet to allow for multiple uses (I don't know enough javascript to tell if that's already there or not).

Implements the matrix product formula given in the OEIS page. It was pretty fun learning a bit of Julia; it's quite similar to Matlab.

Answer 7

154. Verilog, 262 bytes, A001344

module OEIS1344 (input c, input [99:0] n, output reg [99:0] r=0, output reg v=1'b1);
reg [99:0] i,f;
always @ (posedge c) if(v)begin f=n-((n>0)?1:0);v=(f==0);i=f-1;r=(n*(n+3))+((n==0)?2:1);end else begin v=(i==0)||(i==-1);f=v?f:f*i;i=i-1;r=v?r*f:r;end 
endmodule

Next sequence!

The Verilog module above calculates OEIS A001344.

The code below is the same code but prettified, and a testbench module added to allow you to run the code in a Verilog simulator such as modelsim.

I have verified inputs up to the 22nd element in the sequence, beyond which OEIS runs out of examples, and shortly afterwards the numbers start overflowing 100bit arithmetic. I believe that is allowed as per the rules about assuming input and output numbers will never overflow the language. Originally I had used 32bit arithmetic which caused overflow at 13, but without changing the byte count (just 3 numbers from 31 to 99) we can get up to 22-ish.

When simulating, the input number is clocked in on the rising edge. v may go low if it is going to take more than one clock cycle to calculate. Once v is high after the clock edge that loaded the number, the output r can be considered a valid number of the OEIS A001344 sequence.

The code is actually making use of the related OEIS sequence, A028387 which calculates n + (n+1)^2. By multiplying the n+1'th element of that sequence by n! we arrive at the A001344 sequence. All that is required is a little fiddling around with the value of n to make it 0-indexed as per the requirements and we have our result.

To do this in Verilog is a little bit of a pain as it requires a factorial. The simplest approach is to use an iterative approach over n clock cycles to calculate the factorial and output the final result.

I should probably point out that both the OEIS1344 module and its testbench are very naïve approaches to get the job done. But since joining PPCG I've wanted to be able to use Verilog in an answer, and I found a way to shoehorn it into this one.

---

The following is an example output from ModelSim:

The following is the testbench code:

module OEIS1344_test;

reg c = 0;
always begin
    #10 c = !c;
end

reg  [99:0] n;
wire [99:0] r;
wire v;

OEIS1344 dut (
    .c(c),
    .n(n),
    .r(r),
    .v(v)
);

reg  [99:0] i;
initial begin
    for (i = 0; i < 23; i=i+1) begin
        n = i;
        @(posedge c);
        while (!v) @(posedge c);
    end
end

endmodule

module OEIS1344 (
    input c,
    input [99:0] n,
    output reg [99:0] r = 0,
    output reg v = 1'b1
);

reg [99:0] i,f;
always @ (posedge c) begin
    if (v) begin
        f = n - ((n > 0) ? 1 : 0);
        v = (f == 0);
        i = f - 1;
        r = (n * (n + 3)) + ((n == 0) ? 2 : 1);
    end else begin
        v = (i == 0) || (i == -1);
        f = v ? f : f * i;
        i = i - 1;
        r = v ? r * f : r;
    end
end
endmodule

Answer 8

160. Bash, 315 bytes, A000778

#!/bin/bash

array=()
array[0]=1
i=1
read n

while [ $i -le $(expr 2 * $n + 2) ]
do
    let array[$i]=array[$(expr $i - 1)]*$i
    let i=i+1
done

echo $(expr ${array[$(expr 2 * $n)]} / ${array[$n]} / ${array[$(expr $n + 1)]} + ${array[$(expr 2 * $n + 2)]} / ${array[$(expr $n + 1)]} / ${array[$(expr $n + 2)]} - 1)

Try it online!

A000315

Answer 9

166. Scratch 2, 638 bytes, A000114

Or, when written as text in the ScratchBlocks2 format:

when green flag clicked
ask [] and wait
set [a v] to ((answer) + (2))
if <(a) = [2]> then 
  say [3]
else 
  set [b v] to (((a) * (a)) / (2))
  set [d v] to [2]
  repeat until <(d) = (a)> 
    if <((a) mod (d)) = [0]> then 
      set [c v] to [2]
      set [prime v] to [1]
      repeat until <<(prime) = [0]> or <(c) > ([sqrt v] of (d))>> 
        if <((d) mod (c)) = [0]> then 
          set [prime v] to [0]
        end
        set [c v] to ((c) + (1))
      end
      if <(prime) = [1]> then 
        set [b v] to ((b) * ((1) - ((1) / ((d) * (d)))))
      end
    end
    set [d v] to ((d) + (1))
  end
  say ((b) - ((b) mod (1)))
end

Next sequence

Answer 10

167. Magma, 101 bytes, A000638

a000638:=function(n);if n eq 0 then return 1;else return#SubgroupClasses(Sym(n));end if;end function;

Next sequence

You're probably thinking "This is taken from the OEIS site!", but hold on. While checking Magma documentation, I found something interesting here. One of OEIS descriptions of this sequence is number of conjugacy classes of subgroups of symmetric group S_n, and the documentation speaks of a build-in! And it's a different one than the one used on OEIS SubgroupLattice(), this one is SubgroupClasses(). So it happens that both functions create the same amount of subgroups (possibly how the math of this works - I'm not sure), but SubgroupClasses() is actually THE correct way. Thus I felt that this answer is eligible.

You can try it here

Code usage:

Just paste the function and then call it by a000638(5); or any other number instead of 5 (note, anything higher than 11 will time out). You can also use this to output 11 first integers in the sequence:

for i := 0 to 10 by 1 do a000638(i); end for;

Code explanation:

a000638 := function(n);           // Create a function taking one argument
    if n eq 0 then                // If argument is 0
        return 1;                 //     Then return 1
    else                          // Otherwise
        return #                  //     Count and return the number of
               SubgroupClasses(   //         Conjugacy classes of subgroups of
                   Sym(n)         //             Symmetric group of order n
               );
    end if;
end function;

Answer 11

190. COBOL (GNU), 725 bytes, A001182

       IDENTIFICATION DIVISION.
       PROGRAM-ID.  A001182.
       AUTHOR.  KSmarts.

       DATA DIVISION.

       WORKING-STORAGE SECTION.
       01  Num1                                PIC 999   VALUE ZEROS.
       01  Num2                                PIC 999   VALUE ZEROS.
       01  Iterator                            PIC 999   VALUE ZEROS.
       01  Result                              PIC 9(6)  VALUE ZEROS.

       PROCEDURE DIVISION.

       ACCEPT Num1.
       MOVE Num1 TO Iterator
       ADD 1 TO Num1
       MOVE 1 TO Num2
       PERFORM Summing Iterator TIMES
       DISPLAY Result.
       STOP RUN.

       Summing.
           COMPUTE Result = Result + (Num1**2 - Num2**2)**.5
           ADD 1 TO Num2.

Next Sequence

Try it online!

COBOL was originally designed with ease-of-use as the goal. We've come a long way since then.

Answer 12

195. ><>, 210 bytes, A000589

5+::6+}2*}5 -{2[\]b*$,n;
//?={:{:/?= 0:  /
:\      >    ~~1v
\{:}-12[v   /~]{ 22[v/~]*$32[v
v       <           <        >
            /:2=    ?/\
            ?
>:?\~11>*l2\
-1:/ v$\?- /
     >  :1) ^   v,$]~ /

Try it here - input given as initial stack
Next sequence

Fixed previous version. Byte count changed so I decided to delete and repost.
For some reason, doesn't work on TIO.

The code has 3 parts:

Red - computing main sequence;
Orange - computing binomial using factorial;
Blue - computing factorial;

Answer 13

196. Java (OpenJDK 9), 56 bytes, A000210

int f(int n) {return (int)Math.floor((n+1)*(Math.E-1));}

Try it online!

Next Sequence

Answer 14

205. Commata, 109 bytes, A000053

Due to the fact that the proper name breaks the snippet, it has been changed. The proper name is ,,,

14 18 23 28 34 42 50 59 66 72 79 86 96 103 110 116 125 137 145 157 168 181 191 207 215 225 231 238 242 ↺•

Try it online!

Next sequence. (wtf)

Hello darkness,,, my old friend.

Answer 15

204. Positron, 53 bytes, A000096

f = function {
      return ( $1 * ( $1 + 3 ) ) / 2
}

Try it online!

Next Sequence

shout out to all the New Yorkers here! xD

N.B. The next one can be hardcoded because it is a finite sequence

Answer 16

211. 05AB1E, 78 bytes, A000076

2sm©ÝDâεDDн2m4*sP4*+sθ2m5*+}Ùʒ®›1s-}gqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq

Try it online!

Explanation

2sm©ÝDâεDDн2m4*sP4*+sθ2m5*+}Ùʒ®›1s-}gqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq  5
2                                                                               5, 2
 s                                                                              2, 5
  m                                                                             32
   ©                                                                            32 [Copy to register]
    Ý                                                                           [0, ..., 32]
     D                                                                          [0, ..., 32], [0, ..., 32]
      â                                                                         [[0, 0], [0, 1], ..., [32, 31], [32, 32]]
       ε                   }                                                    For each pair [x, y]
        D                                                                       [x, y], [x, y]
         D                                                                      [x, y], [x, y], [x, y] # yes I realize I can triplicate
          н                                                                     [x, y], [x, y], x
           2                                                                    [x, y], [x, y], x, 2
            m                                                                   [x, y], [x, y], x^2
             4                                                                  [x, y], [x, y], x^2, 4
              *                                                                 [x, y], [x, y], 4x^2
               s                                                                [x, y], 4x^2, [x, y]
                P                                                               [x, y], 4x^2, xy
                 4                                                              [x, y], 4x^2, xy, 4
                  *                                                             [x, y], 4x^2, 4xy
                   +                                                            [x, y], 4x^2+4xy
                    s                                                           4x^2+4xy, [x, y]
                     θ                                                          4x^2+4xy, y
                      2                                                         4x^2+4xy, y, 2
                       m                                                        4x^2+4xy, y^2
                        5                                                       4x^2+4xy, y^2, 5
                         *                                                      4x^2+4xy, 5y^2
                          +                                                     4x^2+4xy+5y^2
                            Ù                                                   Now we have a list of all numbers formed by this; uniquify
                             ʒ     }                                            Keep numbers `n` where the result of the codeblock is 1
                              ®                                                 n, 32
                               ›                                                n > 32
                                1                                               n > 32, 1
                                 s                                              1, n > 32
                                  -                                             1 - (n > 32)
                                    g                                           Length; final number of numbers formed
                                     qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq  `q` terminates the program

Next Sequence

Answer 17

210. Pyth, 76 bytes, A000100

 "This is dummy text. Can be replaced with whatever you want!"l{s.pMfq3eST./

See the first few terms.

Next Sequence.

Explanation

• (leading space) : Makes the interpreter ignore the next command.

• "This is dummy text.... !" : Pushes a dummy String, which is ignored.

• l{s.pMfq3eST./ : The thing which does the actual job.

  ./     @ Integer partitions.
  fq3eST @ Filter the partitions with maximal integer `3`.
  .pM    @ Get all the permutations of each.
  s      @ Flattens.
  {      @ Deduplicate.
  l      @ Length.
  

Answer 18

217. Mathematica, 174 bytes, A000411

L[a_, s_] := Sum[JacobiSymbol[-a, 2k + 1] / (2k + 1)^s, {k, 0, Infinity}]
d[a_, n_] := L[-a, 2n] * (2a / Pi)^(2n) / Sqrt[a] * (2n - 1)!
A000411[n_] := Round @ N @ d[6, n + 1]

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Next sequence.

Note that this takes 30 seconds for n = 2 on TIO. Paper that was both useful and useless at the same time.

Answer 19

218. ExtraC, 702 bytes, A000174

long N be readint
long R be 0
for(long A be 0 AND not(A times A greater N) AND increment A)do
  for(long B be A AND not(A times A plus B times B greater N) AND increment B)do
    for(long C be B AND not(A times A plus B times B plus C times C greater N) AND increment C)do
      for(long D be C AND not(A times A plus B times B plus C times C plus D times D greater N) AND increment D)do
        for(long E be D AND not(A times A plus B times B plus C times C plus D times D plus E times E greater N) AND increment E)do
          if(A times A plus B times B plus C times C plus D times D plus E times E equals N)do
            increment R
          end
        end
      end
    end
  end
end

print(R)

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Next sequence!

Answer 20

208. Befunge 93, 75 bytes, A000389

&:5-v
vp00<v`g00<
    v_:v
    >1 >/v:
>:1-:v v*<^_$5432***/.@
^    _$> \:^

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Next sequence

Since it has been pointed out that MATLAB might be is overkill for such a simple sequence I decided to give Befunge a try, and free MATLAB for future use.

Hoping it's not against the rules I'm changing the language of my answer maintaining the same bytecount.

Answer 21

221. Pony, 670 bytes, A000107

fun b(n: U128): U128 =>
  if n == 0 then
    return 0
  elseif n == 1 then
    return 1
  else
    var result: U128 = 0
    var j: U128 = 1
    
    while j < n do
      var d: U128 = 1
      var t: U128 = 0
      while d <= j do
        if (j % d) == 0 then
          t = t + (d * b(d))
        end
        d = d+1
      end
      result = result + (t * b(n-j))
      j = j+1
    end
    return result / (n-1)
  end

fun a(n: U128): U128 =>
  if n == 0 then
    return 0
  elseif n == 1 then
    return 1
  else
    var result: U128 = 0
    var i: U128 = 1
    while i < n do
      result = result + (a(n-i) * b(i))
      i = i+1
    end
    return result + b(n)
  end

Next Sequence

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Answer 22

223. GolfScript, 88 bytes, A000110

{
 # Recurrence a(n+1) = sum a(k) * binomial(n, k)
 [1]{..,({[.(;\);]zip{~+}%}*+}@*-1=
}

Online demo

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Answer 23

224. Racket, 1232 bytes, A000088

#lang racket

(define (partsm n m)
  (if (= n 0)
      '(())
      (for*/list ([k (range 1 (add1 (min n m)))]
                  [r (partsm (- n k) k)] )
        (cons k r) ) ) )

(define (parts n)
  (partsm n n) )

(define (fact n)
  (for/product ([i (range 1 (add1 n))]) i) )

(define (c2 n) ; ways to choose 2 elements from n
  (* n (sub1 n) 1/2) )

(define (cycHash cl)
  (let ([h (make-hash)])
     (for ([i cl])
       (hash-update! h i add1 0) )
   h ) )

(define (halfes ch)
  (for/hash ([(l m) ch]
             #:when (even? l) )
    (values (/ l 2) m) ) )

(define (centralizerSize ch)
  (for/product ([(l m) ch])
    (* (expt l m) (fact m)) ) )

(define (fixp ch n)
  (for/sum ([(l m) ch]
            #:when (= (remainder n l) 0) )
    (* l m) ) )

(define (s2orb hch n)
  (for/sum ([(hl m) hch]
            #:when (let-values ([(q r) (quotient/remainder n hl)])
                     (and (= 0 r) (odd? q)) ) )
    (* hl m) ) )

(define (edgeOrbs ch)
  (let* ([hch (halfes ch)]
         [order (apply lcm (hash-keys ch))])
    (/ (for/sum ([k order])
         (+ (c2 (fixp ch k)) (s2orb hch k)) )
       order ) ) )

(define (f n)
  (for/sum ([ch (map cycHash (parts n))])
    (/ (expt 2 (edgeOrbs ch)) (centralizerSize ch)) ) )

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Next sequence

Yes, it was Burnside's lemma time again! Symmetric group acts on n nodes, inducing an action on the possible edges and on the possible graphs. Count the orbits of the last action, yada yada. Run through the elements by conjugacy classes represented by partitions of n. Number of graphs fixed by an element g is 2 to the number of orbits of edges under the action of the cyclic group generated by g. So use Burnside's lemma again and find the number of edges fixed by each g^k. An edge is fixed if its endpoints are on fixed points or on points that are interchanged by the group element. Feel free to ask for more details!

Answer 24

228. Hodor, 361 bytes, A000861

HoDoRHoDoR HODOR 000861(HODOR? ) {
  $HODOR: Hodor? = 0;
  $HODOR: HODOR?! = 0;
  hodor............ (Hodor? < HODOR? ) {
    HODOR?! = HODOR?! +1;
    $HODOR: d = HODOR?! % 10;
    $HODOR: h = HODOR?! % 100;
    HOdor!!! ( h==12 || ((h>20 || h<10) && ((d==1) || (d==2) || (d==3) || (d==5) || (d==9)))) {
      Hodor? = Hodor? +1;
    }
  }
  HODOR:: HODOR?! ;
}

Next Sequence

Hodor!

Answer 25

227. C++ (clang), 861 bytes, A004903

Thanks to user202729 for C++ help :)

# include <iostream>
# include <algorithm>
# include <vector>
# include <math.h>

std::vector<int> getsums(std::vector<int> values, int times) {
	if (times == 1) {
		return values;
	} else {
		std::vector<int> results;
		for (int i = 0; i < values.size(); i++) {
			std::vector<int> subs = getsums(values, times - 1);
			for (int j = 0; j < subs.size(); j++) {
				results.push_back(subs[j] + values[i]);
			}
		} return results;
	}
}

// This method is O(log_8(n)) I think.

int main() {
	int input; std::cin >> input;
	int size = (int) pow(input / 8, 0.1) + 1;
	std::vector<int> values;
	for (int i = 0; i <= size; i++) values.push_back(pow(i, 10));
	std::vector<int> sums = getsums(values, 8);
	std:sort(sums.begin(), sums.end());
	sums.erase(std::unique(sums.begin(), sums.end()), sums.end());
	for (int i = 0; i < input; i++) std::cout << sums[i] << " ";
}

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My first C++ program that actually does something interesting \o/

Next Sequence

Answer 26

237. M, 89 bytes, A000341

‘ḤŒ!s€2Ṣ€Ṣ$€QS€ÆPa/$$ÐfL                                                                 

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Explanation

‘ḤŒ!s€2Ṣ€Ṣ$€QS€ÆPa/$$ÐfL  Main Link
‘                         Increment (for 0-indexing)
 Ḥ                        Double
  Œ!                      All permutations of {1 .. 2n} (implicit range)
    s€2                   Slice each permutation into subslices of length 2
          $€              For each subconfiguration
       Ṣ€                 Sort each pair
         Ṣ                Then sort the entire thing
            Q             Then remove duplicates
                   $$Ðf   Filter pairings using the last three links as a condition
                 a/       All (reduce over logical AND because M doesn't have the ALL builtin)
             S€           Sums
               ÆP         Are Prime
                       L  How many pairings are there?

There are a bunch of spaces at the end

Next Sequence

Answer 27

242. FriCAS, 111 bytes, A000230

a: (Integer) -> Integer
a(n) ==
 i:=2
 repeat
  j:=nextPrime(i)
  if j-i = 2*n or n = 0 then
   return i
  i:=j

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Pretty straightforward.

Answer 28

243. Ly, 112 bytes, A000111

1<<<<111<n2-0G*
[
 >&s0>l>lr>1>&s>lp
 [
  sp<l+sp<l>>
 ]
 <p<1&s>l<
 [
  sp<l*sp<l*sp<l+>>>
 ]
 <p<p<<,
]
p>u&p;

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Next sequence

This uses the recurrence a(n) = Sum_{i=0..n-2} binomial(n-2,i)*a(i)*a(n-1-i).

Using the recurrence for one value needs all the binomials from one row in Pascal's tringle, and then the next value needs the next row, so we compute one row from the previous one by addition, avoiding factorials and divisions, so Ly will stay in the realm of integers and not switch to floats.

Ly uses an infinite strip of stacks, of which we're using seven. For reference, let's call them, from left to right: count, vals, oldvals, oldvalsR, bins, oldbins, and oldbinsT. The stack that we are at in the beginning will be used as oldbins, just because this is the rightmost that needs initialisation. We can move between the stacks with < and >.

We initialise oldbins with 1, the first row of Pascal's triangle. Then we initialise vals with the first three values, which are all 1. Next, we read the input into count and subtract 2, then 0G* will turn that into 0 if it was negative.

Now count tells us how many more values we need. Next we'll have a loop with this loop invariant: vals contains some initial values of the series, we are at count which holds one nonnegative number that tells how many more values are needed, oldbins contains the row of Pascal's triangle that was used for the last value, and the other stacks are empty.

(We didn't really use the first row of the triangle, but we would have needed it if we had computed a(2) from the recurrence. We don't start with only two values, because there is no 0th row from which we can compute the first without complicating the algorithm. We could also store the row we need next, but then we would needlessly compute one row more than needed.)

The looping construct [...] loops while the stack is not empty and its top element isn't zero. The outer loop will be terminated when count is zero, all other loops will be terminated by empty stacks.

To prepare for the work in the loop, we copy the vals stack to the backup cell using &s, then add a zero to this stack as initial value for the sum. Then we store the value of the backup cell to oldvals (with l) and also to oldvalsR, where we reverse it with r. Then we push the first value 1 of the new row of Pascal's triangle to bins, copy oldbins to oldbinsT and drop its first element there (with p).

We can think of the computation of bins as equivalent to this Haskell computation:

[1] ++ zipWith(+) oldbins oldbinsT ++ [1]   where oldbinsT = tail oldbins

The inner loop then copies (s) the top element from oldbinsT to the backup cell and drops it from the stack (p), and puts it on top of the oldbins stack, where it is added to the former top element. This sum is then moved (again with sp<l) to the bins stack. The loop body is completed by going back to the oldbinsT stack. (Note that at the first iteration of the outer loop, this loop will terminate immediately.)

After this loop, the remaining value from oldbins is dropped, and the final 1 value appended to bins. Then bins is copied to oldbins.

The second inner loop is like this Haskell code:

sum ( zipWith3 (\a b c->a*b*c) oldvals oldvalsR bins )

Note that bins has one value less than the two other stacks, so we use this to control the loop. In it, we move the top value to oldvalsR and multiply, then move the result to oldvals and multiply again. This product is moved to vals and added to the accumulated sum on its top. (Remember that we already pushed a zero here.)

When the loop is done, we drop the remaining values from oldvalsR and oldvals and decrement count (with ,).

When the outer loop is done, we needlessly drop the zero from count and output (and drop) the top element from vals. Finally, we delete this whole stack with &p and explicitely end the program with ;. We don't need both, but if we didn't do either, implicit output would print all values from vals.

Answer 29

245. Java 7, 1272 bytes, A001524

import java.util.*;

public class Main {
	public static List<List<Long>> partitions(long val) {
		List<List<Long>> list = new ArrayList<>();
		if (val == 0) {
			List<Long> sublist;
			list.add(sublist = new ArrayList<>());
		} else {
			for (int i = 1; i <= val; i++) {
				for (List<Long> ext : partitions(val - i)) {
					List<Long> sublist;
					list.add(sublist = new ArrayList<>());
					sublist.add((long) i);
					sublist.addAll(ext);
				}
			}
		} return list;
	}

	public static List<List<Long>> filter(List<List<Long>> partitions) {
		List<List<Long>> list = new ArrayList<>();
		outerloop: for (List<Long> sublist : partitions) {
			for (int i = 0; i < sublist.size() - 1; i++) {
				if (sublist.get(i) <= sublist.get(i + 1)) continue outerloop;
			} list.add(sublist);
		} return list;
	}

	public static long variants(List<Long> stack) {
		int variants = 1;
		for (int i = 0; i < stack.size() - 1; i++) {
			variants *= stack.get(i) - stack.get(i + 1);
		} return variants;
	}

	public static long A001524(long val) {
		long out = 0;
		for (List<Long> stack : filter(partitions(val))) out += variants(stack);
		return out;
	}

	public static void main(String[] args) {
		for (long i = 0; i < 10; i++) {
			System.out.println(i + ": " + A001524(i));
		}
	}
}

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This isn't too slow though it probably could be improved. This sequence was fairly simple but 1272 bytes because Java.

Next Sequence

Answer 30

246. Coconut, 240 bytes, A001272

import math

def p(a): return not (a < 2 or any(a % x == 0 for x in range(2, int(a**0.5) + 1)))

def A(n):
 m=n+1
 i=k=0
 while k<m:
  i= i + 1
  x=0
  for j in range(1,i+1):
   x = x+(-1)**(i-j)*math.factorial(j)
  if p(x): k=k+1
 return i

Next Sequence

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