This challenge is about recursion (Cops' thread)

Question

Cops' thread

In this thread, your task is to make a recursion-based program/function to generate any integer series. Robbers will try and find a shorter non-recursive solution over in the Robbers' thread.

Challenge synopsis

In many languages, recursive functions can significantly simplify a programming task. However, the syntax overhead for a proper recursion may limits its usability in code-golf.

The cops will create a program or function taking a single integer n, which will generate the first n entries of an integer series, using only recursion¹. They should also make sure there is a shorter nonrecursive way to generate the sequence in order to mark their entry as safe.

The robbers will try to find a shorter program or function in the same language, generating the same integer series, using no recursion².

If the cops' submission is not cracked within ten days (240 hours), the cop will prove it was in fact possible to have a shorter non-recursive approach by revealing their own solution. They may then mark their submission as safe.

The winner of the cops challenge will be the shortest (according to code-golf) recursion-based submission marked safe.

The winner of the robbers challenge will be the robber who cracked the most solutions.

^{1: It only needs to be recursive in syntax; you don't need to worry about for example tail call optimization.}

^{2: Again, non-recursive in syntax; so you can't post a recursive solution and claim its compiled to a loop thanks to tail call optimization.}

Submission requirements

Each submission will take a single integer n (zero- or one-based). The submission will then output or return the first n entries of an integer series of choice. (note that this integer series must not depend on n). The input and output method may differ between the recursive and non-recursive approach. The integer series may be any deterministic series with a length of at least 5. The series should be explained properly.

Your submission does not have to work for arbitrary large n, but should work for at least n=5. The non-recursive approach must be able to work up to at least the same n as the recursive approach, or up to n=2^15-1, whichever is smaller.

Recursion

For the sake of this challenge, recursion is defined as creating the desired sequence using a function (or function-like construct) that calls itself (or calls a sequence of functions that ends up calling itself; this includes constructs like the Y combinator). The recursion depth should go to infinity as n goes to infinity. The non-recursive approach is anything that is not recursive.

Answer 1

Python 3, 65 bytes (Safe)

f=lambda n,a=3,b=0,c=6,d=6:n*[1]and[a+b]+f(n-1,c,d,2*c+d,2*a+3*b)

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Another try in Python.

The sequence is "the number of ways to fill a 2-by-n board with dominoes in three colors, so that no two same-colored dominoes touch each other". Not on OEIS.

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Let's say n=6. The board looks like:

######
######

and these are valid domino tilings in three colors (1-3 represent a color each):

123123 122331 212332 212121 113311
123123 133221 212112 212121 331133

but these are not (two same-colored dominoes are touching each other):

112323 332333 211113
112323 112311 233223

The sequence counts all possible domino tilings that satisfy the rules for each n.

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Intended solution, 58 bytes

n=int(input());a=3;b=12
for _ in[0]*n:print(a);a,b=b,a*4+b

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Unfortunately it looks like no one bothered to simplify the recurrence relation, which was clearly shown in the recursive code. Making a program with the given double recurrence as-is doesn't work since it's Python 3.

Answer 2

Octave, 47 bytes, cracked by l4m2

@(n)(f=@(r,m){@()[r(r,m-1),m],[]}{~m+1}())(f,n)

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As an example, here is an Octave entry which generates the first n positive integers, https://oeis.org/A000027.

Answer 3

Forth (gforth), 39 bytes, cracked by NieDzejkob

: | dup 1 > if dup 1 - recurse then . ;

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Answer 4

Python 3, 75 bytes, cracked by xnor

f=lambda n,a=[1]:a*n and[a[0]]+f(n-1,sorted({*a[1:],a[0]*2,a[0]*3,a[0]*5}))

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The famous Hamming numbers, a.k.a. 5-smooth numbers (A051037).

Cracked solution, 51 bytes

lambda n:[k for k in range(1,2**n)if 60**k%k<1][:n]

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Intended solution, 74 bytes

lambda n:sorted(2**(i%n)*3**(i//n%n)*5**(i//n**2)for i in range(n**3))[:n]

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Answer 5

Röda, 40 bytes

f x,a=1,b=2{[a];f x-1,a=b,b=a+b if[x>1]}

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This function gives the following finite sequence (the 90 first Fibonacci numbers):

1
2
3
5
8
13
21
34
55
89
144
233
377
610
987
1597
2584
4181
6765
10946
17711
28657
46368
75025
121393
196418
317811
514229
832040
1346269
2178309
3524578
5702887
9227465
14930352
24157817
39088169
63245986
102334155
165580141
267914296
433494437
701408733
1134903170
1836311903
2971215073
4807526976
7778742049
12586269025
20365011074
32951280099
53316291173
86267571272
139583862445
225851433717
365435296162
591286729879
956722026041
1548008755920
2504730781961
4052739537881
6557470319842
10610209857723
17167680177565
27777890035288
44945570212853
72723460248141
117669030460994
190392490709135
308061521170129
498454011879264
806515533049393
1304969544928657
2111485077978050
3416454622906707
5527939700884757
8944394323791464
14472334024676221
23416728348467685
37889062373143906
61305790721611591
99194853094755497
160500643816367088
259695496911122585
420196140727489673
679891637638612258
1100087778366101931
1779979416004714189
2880067194370816120
4660046610375530309

I know it can generate more Fibonacci numbers, but for purposes of this challenge it is enough to produce these numbers.

Answer 6

JavaScript (Node.js), 91 bytes, Cracked by l4m2

f=x=>[w=~-x&&(g=(n,y=2)=>~-n&&(n<y?1:n%y?g(n,y+1):1+g(n/y,y)))(x)+f(x-1),console.log(w)][0]

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Prints the first n terms of the OEIS sequence A022559 (starting from i=1).

l4m2 fit 3 for loops into 74 72 bytes and cracked my cop post:

n=>{for(i=s=0;j=i++<n;console.log(s))for(x=i;j++<i;)for(;x%j<1;x/=j)s++}

However, my intended answer actually has only 2 for loops:

n=>{for(i=c=0;i++<n;console.log(c))for(p=2,z=i;p<=z;z%p?p++:(z/=p,c++));}

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Answer 7

x86 .COM function, 12 bytes, Cracked by NieDzejkob

0000 52                     push dx
0001 4A                     dec dx
0002 7403                   je 0007
0004 E8F9FF                 call 0000
0007 58                     pop ax
0008 F7E0                   mul ax
000A AB                     stosw


000B C3                     ret

Input DX, Output [DI] ~ [DI+2*DX-1]

Cracker's solution:

0: 31 C0    xor ax, ax
2: BF 01 00 mov di, 1
5: 01 F8    add ax, di
7: AB       stosw
8: E2 FB    loop 5
A: C3       ret

Intended solution:

  xor bx,bx
c:inc bx
  mov ax,bx
  mul ax
  stosw
  loop c
  ret

Answer 8

Gol><>, 15 bytes, cracked by mbomb007

I1AZZ;
M:K:?ZNB

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The series is 0,1,2,3,4,5 but each element is followed by that many 0s.

For example, the first few values are:

 1: 0  First element, followed by 0 zeroes
 2: 1  Followed by 1 zero
 3: 0
 4: 2  Followed by 2 zeroes
 5: 0
 6: 0
 7: 3  Followed by 3 zeroes
 8: 0
 9: 0
10: 0
    etc.

Answer 9

Python 3, 62 bytes, Cracked by mwchase

def f(x):
 if(x<1):return[1]
 return f(x-1)+[sum(f(x-1)[-2:])]

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I feel like this one is going to be too easy...

The sequence is the Fibonacci sequence f(n) = f(n-1) + f(n-2) with f(0) = f(1) = 1

Answer 10

JavaScript, 63 bytes, Cracked

f=x=>(x?[f(x-1)[0]&&1?3*f(x-1)[0]+1:f(x-1)[0]/2,...f(x-1)]:[7])

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Returns the first n elements in a reversed array

Answer 11

Windows .BAT, 80 Bytes

@set /a n=%1-1
@echo 8%3
@if 0 neq %n% @call %0 %n% 2%3 6%2%3

Usage:

CD <PATH>
<FILENAME> <N_1>
<FILENAME> <N_2>
<FILENAME> <N_3>

The loop version can assume in the current dictionary, but have to init or reset

Answer 12

Python, 82 bytes; cracked

This is a recursive Python implementation of OEIS sequence A004001 in 82 bytes. More background on this series can be found on Wolfram's Mathworld.

def A(n):
 if n in[1,2]:return[1]*n
 S=A(n-1);return S+[S[S[n-2]-1]+S[n-S[n-2]-1]]

The first 30 numbers in this sequence are:

1, 1, 2, 2, 3, 4, 4, 4, 5, 6, 7, 7, 8, 8, 8, 8, 9, 10, 11, 12, 12, 13, 14, 14, 15, 15, 15, 16, 16, 16