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; }

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Answer 1

256. Pygmy Forth, 562 bytes, A000454

CODE PICK  dpush(_stack[~dpop()]) END-CODE
: COEFFICIENT-A(N-1) 2 * 5 - 2 * ;
: COEFFICIENT-A(N-2) DUP 6 - * 6 * 55 + ;
: COEFFICIENT-A(N-3) DUP DUP 7 - * 2 * 25 + SWAP 2 * 7 - * ;
: COEFFICIENT-A(N-4) 4 - DUP * DUP * ;
: ADVANCE DUP COEFFICIENT-A(N-1) 2 PICK * OVER COEFFICIENT-A(N-2) 4 PICK * - OVER COEFFICIENT-A(N-3) 5 PICK * + OVER COEFFICIENT-A(N-4) 6 PICK * - SWAP 1 + ;
CODE ADVANCE-LOOP  for i in range(dpop()): ADVANCE() END-CODE
CODE CLEAN-STACK  global _stack; _stack = [_stack[-1]] END-CODE
: A000454 0 0 0 1 5 5 PICK ADVANCE-LOOP DROP CLEAN-STACK ;

Next sequence!

Nice, round number of answers! Computes up to 16 A454 on 64-bit Forths, and you can Try It Online! on gforth with some modifications. However, Pygmy Forth is arbitrary precision and it can compute n=1000 in a blink of an eye.

Explanation

Forth is a stack-based language, like 05AB1E or Actually, but it was not designed that way to be esoteric. Forth was designed in 1970, and being stack based allowed the interpreter to fit in extremely small space - even now, an x86 interpreter could fit in the MBR and VBR (two 512 byte sectors) of a hard disk, with the code loaded from a FAT32 filesystem (first-hand experience). The most important consequences of the stack-oriented design are the postfix notation in which all calculations are described, and having to keep the words ("functions") small so you don't get lost. Breaking up a word into smaller chunks is called factoring.

ADVANCE expects the last 4 numbers of the sequence on the stack, with a cherry the sequence index n on top, and calculates the next number in the sequence as well as increments n. The sequence can start as 0, 0, 0, 1, and the next number will be assigned the index of 5, so you can calculate a specific element of the sequence like this:

0 0 0 1 5 ADVANCE ADVANCE ADVANCE ADVANCE ADVANCE
DROP ( the index )
. ( print )

Now we only need to loop ADVANCE a given number of times, and this is exactly what A454 does. The 0 0 0 1 5 at the beginning initializes the stack, then 5 PICK brings the argument this word was given to the top. Normally in Forth, one would do limit start ?DO stuff LOOP, which is a construct equivalent to Python's for i in range(start, limit): stuff. However, Pygmy Forth is missing a lot of words, so I have to make up for it with CODE, which is used to define words in Python. At the end of A454, DROP is used to remove the sequence index, leaving the result at the top of the stack.

Answer 2

124. ArnoldC, 796 bytes, A000278

IT'S SHOWTIME

HEY CHRISTMAS TREE input
YOU SET US UP @I LIED
HEY CHRISTMAS TREE result
YOU SET US UP @I LIED
HEY CHRISTMAS TREE x
YOU SET US UP @I LIED
HEY CHRISTMAS TREE y
YOU SET US UP @NO PROBLEMO
HEY CHRISTMAS TREE temp1
YOU SET US UP @I LIED

GET YOUR ASS TO MARS input
DO IT NOW
I WANT TO ASK YOU A BUNCH OF QUESTIONS AND I WANT TO HAVE THEM ANSWERED IMMEDIATELY

STICK AROUND input

GET TO THE CHOPPER temp1
HERE IS MY INVITATION x
ENOUGH TALK

GET TO THE CHOPPER x
HERE IS MY INVITATION x
GET UP y
ENOUGH TALK

GET TO THE CHOPPER y
HERE IS MY INVITATION temp1
YOU'RE FIRED temp1
ENOUGH TALK

GET TO THE CHOPPER input
HERE IS MY INVITATION input
GET DOWN 1
ENOUGH TALK

CHILL

GET TO THE CHOPPER result
HERE IS MY INVITATION x
ENOUGH TALK

TALK TO THE HAND result
YOU HAVE BEEN TERMINATED

Try it Online!

Next Sequence

Answer 3

309. Haskell, 699 bytes, A003025

import Control.Monad
import Data.List

-- (n, es) has n nodes; es is a list of (i,j) where 1 <= i < j <= n.
type Dag = (Int, [(Int, Int)])

both f (a,b) = (f a, f b)
powerset = filterM (const [True, False])

-- All dags with n nodes.
dags :: Int -> [Dag]
dags n = map ((,) n) $ powerset [(i,j) | i <- [1..n], j <- [i+1..n]]

-- Is there only one node with no exits?
ok :: Dag -> Bool
ok (n, es) = not $ any (\i -> not $ any ((==i) . fst) es) [1..n-1]

-- All ways to relabel the given dag.
rewrites :: Dag -> [Dag]
rewrites (n, es) = [(n, sort $ map (both ((p!!) . pred)) es) | p <- permutations [1..n]]

-- A003025
f :: Int -> Integer
f = genericLength . nub . concatMap rewrites . filter ok . dags

Try it online!

Enumerates the n-node labeled acyclic digraphs with 1 out-point.

print (f 5) produces the correct output (16885) after 20 seconds when compiled with -O3.

Next sequence

Answer 4

311. ALGOL 68 (Genie), 334 bytes, A000148

BEGIN
  LONG LONG INT n   := read int + 2;
  LONG LONG INT r   := 0;
  LONG LONG REAL ma := ( n / 2.0 ) ** 1.5;
  LONG LONG REAL x  := 2.0 / 3;
  INT a             := 1;
  WHILE a <= ma DO
    INT b := a;
    WHILE a ** x + b ** x <= n DO
      r := r + 1;
      b := b + 1
    OD;
    a := a + 1
  OD;
  print ( whole ( r , 0 ) )
END

Next Sequence

Try it online!

Answer 5

312. Python 2 (PyPy), 1811 bytes, A000334

partition_result_set = []
location_result_set = []
big_max = 1

def p_value(coord, value):
 global big_max
 i = value
 for d in coord:
  i = i * big_max + d
 return i

def partition(number, max, row = []):
 global partition_result_set
 if number == 0:
  partition_result_set[len(row) - 1] += [row]
  return
 
 for i in range(max, 0, -1):
  if number >= i:
   partition(number - i, i, row + [i])

def init_partitions(number):
 global partition_result_set
 for i in range(number):
  partition_result_set += [[]]

 partition(number, number)

def enumerate_locations(number_of_blocks, dimension, have_locations = [], possible_locations = []):
 global g, result_set, y, big_max, partition_result_set, location_result_set
 
 if possible_locations == []: 
  possible_locations = [[0] * dimension]
  y = [0] * number_of_blocks
  temp = number_of_blocks
  while temp > 0:
   big_max *= 10
   temp //= 10
 
 for location in possible_locations:
  new_have_locations = have_locations + [location]
  new_possible_locations = [l for l in possible_locations if l != location]
  for i in range(dimension):
   m = [location[d] + (1 if i == d else 0) for d in range(dimension)]
   is_ok = True
   for j in range(dimension):
    n = [m[d] - (1 if j == d else 0) for d in range(dimension)]
    if [d for d in n if d < 0] == [] and n not in new_have_locations: is_ok = False
   if is_ok: new_possible_locations += [m]

  len1 = len(new_have_locations)
  for x in partition_result_set[len1 - 1]:
   location_result_set += [repr(sorted([p_value(new_have_locations[k], x[k]) for k in range(len1)]))]
  if number_of_blocks > 1:
   enumerate_locations(number_of_blocks - 1, dimension, new_have_locations, new_possible_locations)

def A000334(n):
 init_partitions(n + 1)
 enumerate_locations(n + 1, 4)
 print len(set(location_result_set))

Try it online!

Next sequence: A001811, Laguerre Polynomial. At least there is a closed-form formula

Sorry for Python again, but when the question comes to enumeration I think Python is the easiest way to deal the question with.

Answer 6

314. M, 4166 bytes, A000146

I hope I didn’t accidentally add any bytes outside of the M code page in the code.

The actual source code is the markdown source of this post. To prevent infinite recursion, the source can be found here.

Try it online! (with anti-infinite-recursion)

Next sequence!

Explanation

® ÆD ‘ ÆPÐf İ           ḷ“   First link. Calculate Sum_{(p-1)|2n} 1/p, with
                             the value of 2n stored in the register.
® ÆD                         All divisors of 2n (possible values of (p-1).
     ‘                  ḷ“   Increment. Get all possible values of (p).
       ÆPÐf                  Filter, keep only prime values.
            İ                Inverse.       ”
-* ×c@ × *®¥            ḷ“   Second (dyadic) link. Given v and k, calculate
                             (-1)^v * (k choose v) * v^m / (k+1). (m = 2*n),
                             a term from the closed-form formula.
-                            Literal -1.
 *                           Power. Current value = (-1)^v.
   ×c@                       Multiply (×) by the combination (c) with swapped (@)
                             arguments applied on k and v.
       × *®¥                 Multiply by the value of dyad applied on v...
         *                     Raise to power...
          ®                    register value (currently have value m = 2n)   ”
‘ Ḥ© 0r µ0r ç€ ÷‘µ€ ;-£ FS ḷ“ Main link.
‘                          ḷ“ Increment.
  Ḥ©                          Unhalve (double) the value, and store to register.
     0r                       Generate range from 0.
        µ........µ€           For each value (k),
         0r                     generate range from 0 (values (v)),
            ç€                  apply previous range for each element,
               ÷‘          ḷ“   and divide by (k+1).
                    ;-£       Concatenate the resulting list with the value of
                                the first link. x£ is "call link index x as a
                                nilad, excluding the main link, and wrap around".
                       FS     Flatten, and sum.         ”

Answer 7

315. Befunge-98 (PyFunge), 147 bytes, A004166

very simple :)
>"HTRF"4(1&>:!#v_1v <-- this loop here
  @.$<--out^\*3 \-< <-- calculates 3^n
 >;#_^#:/a\+%aO<\;#z<-- and here the digits are summed

Try it online!

Next sequence!

Answer 8

332. tinylisp, 852 bytes, A000766

(d in (q ((x xs) (i (e x (h xs)) 1 (i xs (in x (t xs)) 0

(d vec (q ((x y z) (c x (c y (c z (
(d + (q ((x y) (i x (c (a (h x) (h y)) (+ (t x) (t y))) (

(d map (q ((f xs) (i xs (c (f (h xs)) (map f (t xs))) (

(d sub (q ((exp pat rep) (i (e exp pat) rep (i (e (type exp) (q List)) (i exp (c (sub (h exp) pat rep) (sub (t exp) pat rep)) ()) exp
(d qt (q ((x) (c (q q) (c x (

(d - (s 0 1
(d adj (c (vec 1 1 0) (c (vec - 1 0) (c (vec 1 - 0) (c (vec - - 0) (c (vec 1 0 1) (c (vec - 0 1) (c (vec 1 0 -) (c (vec - 0 -) (c (vec 0 1 1) (c (vec 0 - 1) (c (vec 0 1 -) (c (vec 0 - -) (

(d sum (q ((ls) (i ls (a (h ls) (sum (t ls))) 0

(d sol (q ((ve pre n) (i (in ve pre) 0 (i n (sum (map (sub (sub (sub (q ((aj) (sol (+ VEC aj) PRE N))) (q VEC) (qt ve)) (q PRE) (qt (c ve pre))) (q N) (qt (s n 1))) adj)) (e (h ve) 1

(d main (q ((x) (sol (vec 0 0 0) () (a 1 x

Try it online!

Next sequence! Intentionally chosen to be easy. (somewhat weird, too)

... It was (not) fun learning this language.

Defines a function main which calculates the required function. tinylisp doesn't support reading from standard input.

Brute force approach. That's why it's terribly slow.

---

... There is library to load? In an esoteric language? Surprise!

Answer 9

335. Python 2 (Cython), 620 bytes, A000157

from itertools import*
from math import*
from functools import*
n=int(input())+1#offset 1
a=0#answer
for p in permutations(range(n)):
 for i in range(2**n):#inversion
  v=[0]*(2**n)#visited
  c=0#number of cycles
  e=1#all is even
  for x in range(2**n):
   if v[x]:continue
   w=1#1 if this cycle is even
   while 1>v[x]:
    v[x]=1;w^=1
    x=reduce(lambda x,y:x+x+y,[1&(x>>s)for s in p])^i
   e&=w;c+=1
  a+=2**c*(1+e)
print(a//(2**n*factorial(n)*4))

#Come on... it's not that hard.
#Time complexity: 4**n*factorial(n)
#Memory complexity: 2**n
#(which is actually order of magnitude
#faster than the naive algorithm)

Try it online!

Next sequence

---

Originally intended to work in Python3, but Python 3 (Cython) raises an error "deallocating None".

Sorry for using Python, but programming on mobile is not easy. Posted 2 days ago in chat while waiting for someone else to post, probably with another language.

The next sequence has some chemistry-related things, but the recurrence relation is easy to implement. Ideally I hope someone will explain what the sequence is about. (I don't even explain most of my own answers...)

(Side note: Syntax highlighting fails for some comments)

Answer 10

364. Wenyan (文言), 2534 bytes, A001669

吾有一術。名之曰「開方本源圖」。欲行是術。必先得一數。曰「甲」。乃行是術曰。
　批曰。「「西人謂之曰「巴斯噶三角」者是也。」」
　吾有一列。名之曰「層」。充「層」以一。
　為是「甲」遍。
　　吾有一列。名之曰「下層」。
　　吾有一數。曰零。名之曰「上廉」。
　　凡「層」中之「廉」。
　　　加「廉」以「上廉」。名之曰「下廉」。充「下層」以「下廉」。
　　　昔之「上廉」者。今「廉」是矣。
　　云云。
　　充「下層」以一。
　　昔之「層」者。今「下層」是矣。
　云云。
　乃得「層」。
是謂「開方本源圖」之術也。

吾有一術。名之曰「指數」。欲行是術。必先得一列。曰「天」。乃行是術曰。
　批曰。「「形式冪級數之指數者也。不計其常數項。」」
　夫「天」之長。名之曰「階」。
　吾有一列。名之曰「地」。
　吾有一數。曰零。名之曰「甲」。
　為是「階」遍。
　　若「甲」等於零者。
　　　充「地」以一。
　　若非。
　　　減「甲」以一。施「開方本源圖」於其。名之曰「開方本源」。
　　　吾有一數。曰零。名之曰「乙」。
　　　吾有一數。曰零。名之曰「和」。
　　　凡「開方本源」中之「廉」。
　　　　減「甲」以「乙」。加其以一。名之曰「丙」。
　　　　加一於「乙」。昔之「乙」者。今其是矣。
　　　　夫「地」之「乙」。名之曰「項」。
　　　　夫「天」之「丙」。乘其以「項」。乘其以「廉」。
　　　　加其於「和」。昔之「和」者。今其是矣。
　　　云云。
　　　充「地」以「和」。
　　云云。
　　加一於「甲」。昔之「甲」者。今其是矣。
　云云。
　乃得「地」。
是謂「指數」之術也。

吾有一術。名之曰「序列零零壹陸陸玖」。欲行是術。必先得一數。曰「甲」。乃行是術曰。
　批曰。「「OEIS之序列A001669者是也。」」
　加一於「甲」。昔之「甲」者。今其是矣。
　吾有一列。名之曰「天」。為是「甲」遍。充「天」以一也。
　為是六遍。
　　施「指數」於「天」。昔之「天」者。今其是矣。
　云云。
　夫「天」之「甲」。名之曰「乙」。
　乃得「乙」。
是謂「序列零零壹陸陸玖」之術也。

Next sequence!

Time to write some Classical Chinese.

You can try it on this Online IDE. Add the following snippets to the end to print the first 20 terms. Uncheck "Print Hanzi" in the settings to show the results in Arabic numerals.

吾有一數。曰零。名之曰「甲」。
為是二十遍。
　施「序列零零壹陸陸玖」於「甲」。書之。
　加一於「甲」。昔之「甲」者。今其是矣。
云云。

Answer 11

48. SOGL, 3 bytes, A000194

.√Ρ

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Answer 12

53. Coffeescript, 34 bytes, A000484

alert Math.round Math.cos prompt()

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Answer 13

54. C# (.NET Core), 196 bytes, A000034

using System;
namespace A000008
{
    class Program
    {
        static void Main(string[] args)
        {
           Console.WriteLine(1 + Int32.Parse(Console.ReadLine()) % 2);
        }
    }
}

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Answer 14

60. Oasis, 5 + 2 = 7 bytes, A000059

+2 bytes for the N and o flags.

Code:

x4m>p

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Explanation:

x      # Double the number
 4m    # Raise to the power of 4
   >   # Add one
    p  # Check for primality

Used with the following flags:

• N: Find the Nth number that satisfies each line of code
• o: Shift the sequence by 1 (N = N + 1).

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Answer 15

61. √ å ı ¥ ® Ï Ø ¿, 5 bytes, A000007

I0^o 

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Answer 16

64. LiveScript, 152 bytes, A000108

f = (n) ->
  ns = [1]
  for i from 0 to n-1
    ns.push 0
    s = 0
    for k from 0 to ns.length-1
      s += ns[k]
      ns[k] =  s
  ns[ns.length-1]

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Answer 17

65. Fortran (GFortran), 371 bytes, A000152

function A000152(n)
	integer :: n, i, a, sum, index, A000152
	integer, dimension(n+1) :: gf
	gf(1) = 1
	do i=2,n+1
		gf(i) = 0
	end do
	do a=1,16
		do i=n+1,2,-1
			sum = 0
			index = 1
			do while (index * index < i)
				sum = sum + 2 * gf(i - index * index)
				index = index + 1
			end do
			gf(i) = gf(i) + sum
		end do
	end do
	A000152 = gf(n+1)
end function A000152

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How it works

It uses the following gf:

(sum(m=[-infty,infty], x^(m^2)))^16

Answer 18

66. Yacas, 43 bytes, A000371

f(n):=Sum(k,0,n,(-1)^(n-k)*Bin(n,k)*2^2^k);

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Answer 19

56. Pony, 138 bytes, A000041

fun f(n: U64, q: U64 = 1): U64 =>
  if n>0 then
    if n>=q then
      f(n-q, q)+f(n,q+1)
    else
      0
    end
  else
    1
  end //a

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Answer 20

70. Elm, 1134 bytes, A000333

import Html exposing (text)

f n s = let l = takeWhile (\x -> (s + sqrt (toFloat x)) <= n) (List.range 1 (floor (n*n)))
        in List.map (\x -> [x]) l 
        ++ (List.concatMap (\x -> List.map (\y -> x::y) ( f n (s+sqrt (toFloat x)) )) l)

t n = List.length(List.map List.head(group(List.sort(List.map List.sort(f(n+1)0)))))

main = text (toString (List.length (t 5)))


takeWhile : (a -> Bool) -> List a -> List a
takeWhile predicate list =
  case list of
    []      -> []
    x::xs   -> if (predicate x) then x :: takeWhile predicate xs
              else []

dropWhile : (a -> Bool) -> List a -> List a
dropWhile predicate list =
  case list of
    []      -> []
    x::xs   -> if (predicate x) then dropWhile predicate xs
              else list

span : (a -> Bool) -> List a -> (List a, List a)
span p xs = (takeWhile p xs, dropWhile p xs)

groupBy : (a -> a -> Bool) -> List a -> List (List a)
groupBy eq zs =
  case zs of
    [] -> []
    (x::xs) -> let (ys,zs) = span (eq x) xs
              in (x::ys)::groupBy eq zs

group : List a -> List (List a)
group = groupBy (==)

This is much prettier in Haskell, but it has been used.


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Answer 21

71. F# (Mono), 401 bytes, A001134

That was quite an experience, using a language I've never tried before which is based on a bunch of programming paradigms I've never experienced before...

let isprime p =
  let mutable q = 2
  while q * q < p && p % q > 0 do
    q <- q + 1
  p >= 2 && q * q > p

let isA001134 p =
  let mutable i = 1
  let mutable g = 2
  while g <> 1 do
    g <- g * 2 % p
    i <- i + 1
  i * 4 = p - 1 && isprime p

let A001134 n =
  let mutable q = n
  let mutable result = 113
  while q > 0 do
    result <- result + 4
    if isA001134 result then q <- q - 1
  result

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Answer 22

77. Braingolf, 51 bytes, A004147

m32&gM9784&gM7571840&gM11140566368&gMvvcv1+[R<v]v_;

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Answer 23

78. anyfix, 12 bytes, A000051

        2«*‘

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Note: TIO says 15 bytes but that's because I haven't asked Dennis to configure Anyfix to use the Jelly codepage yet.

Answer 24

79. Commentator, 13 bytes, A000012

 */          

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A space increments the accumulator and */ outputs it as a number. As the program has outputted, the spaces at the end are essentially no-ops.

Answer 25

87. Neim, 16 bytes, A000040

>𝐋1𝕗><><><><><><

Output is wrapped in braces []. Yes, those are random filler bytes (that increment and decrement continuously), but I like to think of them as fish having fun with code.

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Answer 26

90. Open Shading Language (OSL), 66 bytes, A000201

shader abc(int i=1, output float o=1){o=floor(i*((1+sqrt(5))/2));}

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Answer 27

40. Rust, 3416 bytes, A000001

The rules seem to imply hardcoding the first 1001 cases is fine.

EDIT: I’m editing this post to give some commenters the chance to adjust their vote on it.

static COUNT: [i32; 1001] =
[0, 1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2,
1, 15, 2, 2, 5, 4, 1, 4, 1, 51, 1, 2, 1, 14, 1, 2, 2, 14, 1, 6, 1, 4,
2, 2, 1, 52, 2, 5, 1, 5, 1, 15, 2, 13, 2, 2, 1, 13, 1, 2, 4, 267, 1,
4, 1, 5, 1, 4, 1, 50, 1, 2, 3, 4, 1, 6, 1, 52, 15, 2, 1, 15, 1, 2, 1,
12, 1, 10, 1, 4, 2, 2, 1, 231, 1, 5, 2, 16, 1, 4, 1, 14, 2, 2, 1, 45,
1, 6, 2, 43, 1, 6, 1, 5, 4, 2, 1, 47, 2, 2, 1, 4, 5, 16, 1, 2328, 2,
4, 1, 10, 1, 2, 5, 15, 1, 4, 1, 11, 1, 2, 1, 197, 1, 2, 6, 5, 1, 13,
1, 12, 2, 4, 2, 18, 1, 2, 1, 238, 1, 55, 1, 5, 2, 2, 1, 57, 2, 4, 5,
4, 1, 4, 2, 42, 1, 2, 1, 37, 1, 4, 2, 12, 1, 6, 1, 4, 13, 4, 1, 1543,
1, 2, 2, 12, 1, 10, 1, 52, 2, 2, 2, 12, 2, 2, 2, 51, 1, 12, 1, 5, 1,
2, 1, 177, 1, 2, 2, 15, 1, 6, 1, 197, 6, 2, 1, 15, 1, 4, 2, 14, 1,
16, 1, 4, 2, 4, 1, 208, 1, 5, 67, 5, 2, 4, 1, 12, 1, 15, 1, 46, 2, 2,
1, 56092, 1, 6, 1, 15, 2, 2, 1, 39, 1, 4, 1, 4, 1, 30, 1, 54, 5, 2,
4, 10, 1, 2, 4, 40, 1, 4, 1, 4, 2, 4, 1, 1045, 2, 4, 2, 5, 1, 23, 1,
14, 5, 2, 1, 49, 2, 2, 1, 42, 2, 10, 1, 9, 2, 6, 1, 61, 1, 2, 4, 4,
1, 4, 1, 1640, 1, 4, 1, 176, 2, 2, 2, 15, 1, 12, 1, 4, 5, 2, 1, 228,
1, 5, 1, 15, 1, 18, 5, 12, 1, 2, 1, 12, 1, 10, 14, 195, 1, 4, 2, 5,
2, 2, 1, 162, 2, 2, 3, 11, 1, 6, 1, 42, 2, 4, 1, 15, 1, 4, 7, 12, 1,
60, 1, 11, 2, 2, 1, 20169, 2, 2, 4, 5, 1, 12, 1, 44, 1, 2, 1, 30, 1,
2, 5, 221, 1, 6, 1, 5, 16, 6, 1, 46, 1, 6, 1, 4, 1, 10, 1, 235, 2, 4,
1, 41, 1, 2, 2, 14, 2, 4, 1, 4, 2, 4, 1, 775, 1, 4, 1, 5, 1, 6, 1,
51, 13, 4, 1, 18, 1, 2, 1, 1396, 1, 34, 1, 5, 2, 2, 1, 54, 1, 2, 5,
11, 1, 12, 1, 51, 4, 2, 1, 55, 1, 4, 2, 12, 1, 6, 2, 11, 2, 2, 1,
1213, 1, 2, 2, 12, 1, 261, 1, 14, 2, 10, 1, 12, 1, 4, 4, 42, 2, 4, 1,
56, 1, 2, 1, 202, 2, 6, 6, 4, 1, 8, 1, 10494213, 15, 2, 1, 15, 1, 4,
1, 49, 1, 10, 1, 4, 6, 2, 1, 170, 2, 4, 2, 9, 1, 4, 1, 12, 1, 2, 2,
119, 1, 2, 2, 246, 1, 24, 1, 5, 4, 16, 1, 39, 1, 2, 2, 4, 1, 16, 1,
180, 1, 2, 1, 10, 1, 2, 49, 12, 1, 12, 1, 11, 1, 4, 2, 8681, 1, 5, 2,
15, 1, 6, 1, 15, 4, 2, 1, 66, 1, 4, 1, 51, 1, 30, 1, 5, 2, 4, 1, 205,
1, 6, 4, 4, 7, 4, 1, 195, 3, 6, 1, 36, 1, 2, 2, 35, 1, 6, 1, 15, 5,
2, 1, 260, 15, 2, 2, 5, 1, 32, 1, 12, 2, 2, 1, 12, 2, 4, 2, 21541, 1,
4, 1, 9, 2, 4, 1, 757, 1, 10, 5, 4, 1, 6, 2, 53, 5, 4, 1, 40, 1, 2,
2, 12, 1, 18, 1, 4, 2, 4, 1, 1280, 1, 2, 17, 16, 1, 4, 1, 53, 1, 4,
1, 51, 1, 15, 2, 42, 2, 8, 1, 5, 4, 2, 1, 44, 1, 2, 1, 36, 1, 62, 1,
1387, 1, 2, 1, 10, 1, 6, 4, 15, 1, 12, 2, 4, 1, 2, 1, 840, 1, 5, 2,
5, 2, 13, 1, 40, 504, 4, 1, 18, 1, 2, 6, 195, 2, 10, 1, 15, 5, 4, 1,
54, 1, 2, 2, 11, 1, 39, 1, 42, 1, 4, 2, 189, 1, 2, 2, 39, 1, 6, 1, 4,
2, 2, 1, 1090235, 1, 12, 1, 5, 1, 16, 4, 15, 5, 2, 1, 53, 1, 4, 5,
172, 1, 4, 1, 5, 1, 4, 2, 137, 1, 2, 1, 4, 1, 24, 1, 1211, 2, 2, 1,
15, 1, 4, 1, 14, 1, 113, 1, 16, 2, 4, 1, 205, 1, 2, 11, 20, 1, 4, 1,
12, 5, 4, 1, 30, 1, 4, 2, 1630, 2, 6, 1, 9, 13, 2, 1, 186, 2, 2, 1,
4, 2, 10, 2, 51, 2, 10, 1, 10, 1, 4, 5, 12, 1, 12, 1, 11, 2, 2, 1,
4725, 1, 2, 3, 9, 1, 8, 1, 14, 4, 4, 5, 18, 1, 2, 1, 221, 1, 68, 1,
15, 1, 2, 1, 61, 2, 4, 15, 4, 1, 4, 1, 19349, 2, 2, 1, 150, 1, 4, 7,
15, 2, 6, 1, 4, 2, 8, 1, 222, 1, 2, 4, 5, 1, 30, 1, 39, 2, 2, 1, 34,
2, 2, 4, 235, 1, 18, 2, 5, 1, 2, 2, 222, 1, 4, 2, 11, 1, 6, 1, 42,
13, 4, 1, 15, 1, 10, 1, 42, 1, 10, 2, 4, 1, 2, 1, 11394, 2, 4, 2, 5,
1, 12, 1, 42, 2, 4, 1, 900, 1, 2, 6, 51, 1, 6, 2, 34, 5, 2, 1, 46, 1,
4, 2, 11, 1, 30, 1, 196, 2, 6, 1,10,1,2,15,199];

fn count(i: usize) -> i32 { COUNT[i] }

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Answer 28

93. Clojure, 35 bytes, A000093

#(Math/floor (Math/sqrt (* % % %)))

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Answer 29

98. Befunge-98 (PyFunge), 153 bytes, A001772

&v >1 >1+:00g\%kvv>1\>1-\2*\:kvv
 #   vw\-1g00: < >$v ^       < $
>0p^ >$01g1-:01v> #> > 1+vv-1*b<
^>1+01p         ^ ^:     <>0
               >p  k^.@zz

Try it online! The two zs after the @ are nop's that never get executed but I added in the make the bytecount correspond to an easier sequence. Becuase of what I think is a quirk of the interpreter and the fact that TIO doesn't provide a trailing newline, input needs to be given with a trailing space.

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Answer 30

100. Pyke, 20 bytes, A000030

Yaay, the 100th answer! 0-indexed.

                  `h

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